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Table of Contents, , 0 Introduction ............................................................................................................................. 1, Mechanics, 1 Measurement and Mathematics ............................................................................................. 7, 2 Motion in One Dimension ..................................................................................................... 24, 3 Vectors.................................................................................................................................. 52, 4 Motion in Two and Three Dimensions .................................................................................. 65, 5 Force and Newton's Laws .................................................................................................... 88, 6 Work, Energy, and Power................................................................................................... 121, 7 Momentum.......................................................................................................................... 145, 8 Uniform Circular Motion...................................................................................................... 166, 9 Rotational Kinematics......................................................................................................... 175, 10 Rotational Dynamics........................................................................................................... 198, 11 Static Equilibrium and Elasticity.......................................................................................... 205, 12 Gravity and Orbits............................................................................................................... 220, 13 Fluid Mechanics.................................................................................................................. 247, Mechanical Waves, 14 Oscillations and Harmonic Motion ...................................................................................... 275, 15 Wave Motion....................................................................................................................... 294, 16 Sound ................................................................................................................................. 308, 17 Wave Superposition and Interference ................................................................................ 322, Thermodynamics, 18 Temperature and Heat ....................................................................................................... 335, 19 Kinetic Theory of Gases ..................................................................................................... 360, 20 First Law of Thermodynamics, Gases, and Engines.......................................................... 373, 21 Second Law of Thermodynamics, Efficiency, and Entropy ................................................ 383, , Electricity and Magnetism, 22 Electric Charge and Coulomb's Law .................................................................................. 400, 23 Electric Fields ..................................................................................................................... 419, 24 Electric Potential ................................................................................................................. 436, 25 Electric Current and Resistance ......................................................................................... 457, 26 Capacitors........................................................................................................................... 476, 27 Direct Current Circuits ........................................................................................................ 489, 28 Magnetic Fields................................................................................................................... 505, 29 Electromagnetic Induction .................................................................................................. 539, 30 Electromagnetic Radiation.................................................................................................. 557, Light and Optics, 31 Reflection............................................................................................................................ 574, 32 Refraction ........................................................................................................................... 596, 33 Lenses ................................................................................................................................ 607, 34 Interference......................................................................................................................... 627, Modern Physics, 35 Special Relativity ................................................................................................................ 638, 36 Quantum Physics Part One ................................................................................................ 659, 37 Quantum Physics Part Two ................................................................................................ 688, 38 Nuclear Physics .................................................................................................................. 698, , Answers to selected problems............................................................................................. 721
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Copyright 2000-2007 © Kinetic Books Company, All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or, mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the, publisher., This printed book is sold only in combination with its digital counterpart obtained via CD or subscription over the internet., For additional information on the product(s) and purchasing, please contact Kinetic Books Company., www.kineticbooks.com, 1-877-4kbooks (877) 452-6657
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0.0 - Welcome to an electronic physics textbook!, Textbooks, like this one, contain words and illustrations. In an ordinary textbook, the, words are printed and the illustrations are static, but in this book, many of the, illustrations are animations and many words are spoken. Altogether, this textbook, contains more than 600,000 words, 150 simulations, 1000 animations, 5000, illustrations, 15 hours of audio narration, and 35,000 lines of Java and JavaScript, code., All this is designed so that you will experience more physics. You will race cars, around curves, see the forces between charged particles, dock a space craft,, generate electricity by moving a wire through a magnetic field, control waves in a, string to “make music”, measure the force exerted by an electric field, and much, more. These simulations and animations are designed to allow you to “see” more, physics and make it easier for you to assess your learning, since many of them, pose problems for you to solve., The foundation of this textbook is the same as a traditional textbook: text like this, and illustrations. Concepts like “velocity” or “Newton’s second law” are explained as, they are in traditional textbooks. From there we go a step further, taking advantage, of the computer and giving you additional ways to learn about physics. The textbook, has many features: simulations; problems where the computer checks your answers, and then works with you step by step; animations that are narrated; search, capability; and much more. We will start with some simulations. In subsequent, sections, we will show you how we use animations and narration to teach physics,, and how a computer will help you solve problems., At the right are three examples of how we take advantage of an interactive, simulation engine. Click on any of the illustrations to start an interactive simulation; it, will open a separate window. When you are done, you can close the window. The, window that contains this text will remain open., In the first simulation, you aim the monkey’s banana bazooka so that the banana, will reach the professor. The instant the banana is fired, the professor lets go of the, tree and falls toward the ground. You aim the banana bazooka by dragging the, head of the arrow shown on the right. Aim the bazooka and then press GO. Press, RESET to try again. (And do not worry: We, too, value physics professors, so the, professor will emerge unscathed.), This is an animated version of a classic physics problem and appears about halfway, through a chapter of the textbook. The majority of our simulations require the, calculation of precise answers, but like this one, they are all great ways to see a, concept at work., In the second simulation, you can extend a simple circuit. The initial circuit shown, on the right contains a battery and a light bulb. You can add light bulbs or more wire, segments by dragging them near the desired location. Once there, they will snap, into place. You can also use an ammeter to measure the amount of current flowing, through a section of a wire, and a voltmeter to measure the potential difference, across a light bulb or the battery., There are many experiments you can conduct with these simple tools. For instance,, place a light bulb in the horizontal segment above the one which already contains, the light bulb and connect it to the circuit with two additional vertical wire segments., Does this alter the power flowing through the first light bulb? The brightness of each, light bulb is roughly proportional to the power the circuit supplies to it., How do the potential differences across the light bulbs compare to one another? To the potential difference across the battery? Measure the, current flowing through a piece of wire immediately adjacent to the battery, and through each of the wire segments that contains a light bulb., Do you see a mathematical relationship between these three values?, You will be asked to make observations in many simulations like this, and as you learn physics, to apply what you have learned to answer, problems posed by the simulations. You will use your knowledge to do everything from juggle to dock a spaceship!, In the third simulation, you experiment with a simple electric generator. When the crank is turned, the rectangular wire loop shown in the, illustration turns in a magnetic field. The straight lines you see are called magnetic field lines. Turning the handle of the generator creates an, electric current and what is called an emf. The emf is measured in volts., After you launch the simulation, you can change your point-of-view with a slider. The illustration you see to the right provides a conceptual, overview of what a generator is. If you change the viewing angle, you can better see the angle between the wire and the field, and how that, affects the current. A device called an oscilloscope is used to measure the emf created by the generator., , Copyright 2000-2007 Kinetic Books Co. Chapter 00, , 1
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The electric generator is an advanced topic, and if you are just beginning your study of physics, it presents you with many unfamiliar concepts., However, the simulation shows how we can take advantage of software to allow you to change the viewing angle and to view processes that, change over time., If you want to see more simulations we enjoyed creating: “dragging” a ball to match a graph, sliding a block up a plane, electromagnetic, induction, electric potential, space docking mission and wave interference. You can click on any of these topics and the link will take you to that, section. There are many simulations; to see even more of them, you can click on the table of contents, pick a chapter, and then click on any, section whose name starts with “interactive problem.”, To move to the next section, click on the right arrow in the black bar above or below, the arrow to the right of 0.0., , 0.1 - Whiteboards, Right now, you are reading the text of this textbook. Its design is similar to that found in, traditional textbooks. By “text,” we mean the words you are reading and the illustrations, and writing you see to the right., As you read the words, study the illustrations and work the problems, you may feel as, though you are using a traditional textbook. (We like to think it is well-conceived and, well-written, but that is for you to judge.) You can print out this textbook and use it as, you would a traditional print textbook., When you use the electronic version of this book, however, you have access to an, entirely different way of learning the material. It starts with what we call the whiteboards., You launch the whiteboards by clicking on the illustrations to the right. They present the, same material discussed in the text, but do so using a sequence of narrated animations., The text and the whiteboards cover essentially the same material. You could learn, physics exclusively through the whiteboards, or you could learn it all via the words and, pictures you now see. The text sometimes contains additional material: the history of, the topic, an application of a principle and so on. Everything found in the whiteboards is, always found in the text, so you do not have to click through them unless you find them, a useful way to learn. The point is: You have a choice. You may also find a combination, of the two particularly useful, especially for topics you find challenging., , Whiteboards, Illustrate physics concepts, Explain with narration, , If you are reading this on a computer, try clicking on the illustration titled “Concept 1” to, the right. This will open the whiteboard in a separate window. Each whiteboard is, equipped with animations, audio and its own set of controls. Both this textbook and the, whiteboards can be used simultaneously. If you do not have headphones or, speakers, click on the “show text” button after you open the whiteboard.This will, allow you to read the whiteboard narration., The electronic format provides a visually compelling way for you to learn what can be, complex concepts and formulas. For instance, instead of a static diagram that, represents a car rounding a curve, our format allows us to actually show the car moving, and turning. We can also show you a greater amount of information í for example, how, the horizontal and vertical velocities of the car change over time., , Whiteboard components, Concept slides explain idea, , Typical sections throughout the book feature three graphic elements on the right side of, the page, corresponding to three parts of the whiteboard. The first introduces the, concept: For instance, what does the term “displacement” mean? The second contains, the equation: How is displacement calculated? The third, located at the bottom right,, then works an example problem to test your understanding of the concept and the, equation., The textbook contains hundreds of these whiteboards. If you would like to view some, more to get a sense of how animation and audio play together, you can browse any, chapter. You can explore topics like displacement, graphing simple harmonic motion,, hitting a baseball, electric field diagrams, determining the type of image produced by a, mirror and the force of a magnetic field on a moving charged particle., To move to the next section, click on the right-arrow in the black bar above or below,, the arrow to the right of 0.1., , 2, , Equations provide formula(s), , Copyright 2000-2007 Kinetic Books Co. Chapter 00
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Examples work basic problems, , 0.2 - Interactive problems, In the first section of this chapter, we encouraged you to try various simulations. We, call simulations where you set values and watch the results interactive problems. In, this section, we explain in more detail how they work., A sample interactive problem can be launched by clicking on the graphic on the, right. When the correct x (horizontal) and y (vertical) velocities are supplied, the, juggler will juggle the three balls. Before you proceed, you may wish to read the, instructions below for using these interactive simulations., As mentioned, you launch the simulations by clicking on the graphic. Typically, you, will be asked to enter a value in the simulation. Sometimes you fill in a value in a, text entry box, and other times you select a value using spin dials that have up and, down arrow buttons., Then, you typically push the GO button and the simulation begins í things begin to, move. Most simulations have a RESET button that allows you to start again. Many, have a PAUSE button that makes things go three times faster (just kidding í they, pause the simulation so you can record data)., Many simulations, especially those at the beginning of the chapters, just ask you to observe how entering different values changes the results., Simulations often come with gauges that display variables as they change, such as a speedometer to keep track of a car’s speed as it goes, around a track. You may observe the relationship between mass and the amount of gravitational force, for instance. Other simulations provide, direct feedback if you succeed: the juggler juggles, you beat another racecar, and so forth., Simulations later in the chapter often ask you to perform calculations in order to achieve a particular goal. These simulations are designed to, make trial-and-error an ineffective tactic since they require a great amount of precision in the answer., Enough preamble: Try the juggling simulation to the right. Enter any values you like for the initial y and x velocities, using the spin dials. Then, press GO and watch as the juggler begins to juggle. Press RESET to enter a different set of values. A hint: One pair of values that will enable, you to juggle is 6.0 m/s (meters per second) for the y velocity and 0.6 m/s for the x velocity., , 0.3 - Sample problems and derivations, The mouse goes 11.8 meters in 3.14, seconds at a constant acceleration of, 1.21 m/s2., What is its velocity at the beginning, and end of the 11.8 meters?, , In addition to text and interactive problem sections, this textbook contains sections with sample problems and derivations of equations., Sample problems often demonstrate a useful problem-solving technique. You see a typical sample problem above. Derivations show how, an equation new to you can be created from equations you have already learned., We follow the same sequence of steps in sample problems and derivations. (You will also follow this same sequence when you work through, problems called interactive checkpoints; more on this type of problem in the next section.) Sample problems, derivations and interactive, checkpoints all have some or all of the following: a diagram, a table of variables, a statement of the problem-solving strategy, the principles and, equations used, and a step-by-step solution., To show how these are organized, we work through a sample problem from the study of linear motion. The problem is stated above., Draw a diagram, , Copyright 2000-2007 Kinetic Books Co. Chapter 00, , 3
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It is often helpful to draw a diagram of the problem, with important values labeled. Although almost every problem is stated using an illustration,, we sometimes find it useful to draw an additional diagram., Variables, We summarize the variables relating to the problem in a table. Some of these have values given in the problem statement or illustration. If we, do not know the value of a variable, we enter the variable symbol. A variable table for the problem stated above is shown., displacement, , ǻx = 11.8 m, , acceleration, , a = 1.21 m/s2, , elapsed time, , t = 3.14 s, , initial velocity, , vi, , final velocity, , vf, , There are two reasons we write the variables. One is so that if you see a variable with which you are unfamiliar, you can quickly see what it, represents. The other is that it is another useful problem-solving technique: Write down everything you know. Sometimes you know more than, you think you know! Some variables may also prompt you to think of ways to solve the problem., After these two steps, we move to strategy., What is the strategy?, The strategy is a summary of the sequence of steps we will follow in solving the problem. Some students who used this book early in its, development called the strategy section “the hints,” which is another way to think of the strategy. There are typically many ways to solve a, problem; our strategy is the one we chose to employ. (As we point out in the text when we actually solve this problem, there is another efficient, manner in which to solve it.), For the problem above, our strategy was:, 1., , There are two unknowns, the initial and final velocities, so choose two equations that include these two unknowns and the values you, do know., , 2., , Substitute known values and use algebra to reduce the two equations to one equation with a single unknown value., , Principles and equations, Principles and equations from physics and mathematics are often used to solve a problem. For the problem above, for example, these two, linear motion equations that apply when acceleration is constant are useful:, , vf = vi + at, ǻx = ½(vi + vf)t, The physics principles are the crucial points that the problems are attempting to reinforce. If they look quite familiar to you at some point: Great!, Step-by-step solution, We solve the problem (or work through the derivation) in a series of steps. We provide a reason for each step. If you want a more detailed, explanation, you can click on a step, which causes a more detailed text explanation to appear on the right. Some students find the additional, information quite helpful; others prefer the very brief explanation. It also varies depending on the difficulty of the problem í everyone can use a, little help sometimes., Here are the first three steps that we used to solve the problem above., , Step, , Reason, , 1., , vf = vi + at, , 2., , vf = vi + (1.21 m/s2) (3.14 s) substitute values, , 3., , vf = vi + 3.80 m/s, , first motion equation, , multiply, , 0.4 - Interactive checkpoints, The great pyramid of Cheops has a, square base with edges that are, almost exactly 230 m long. The side, faces of the pyramid make an angle, of 51.8° with the ground., The apex of the pyramid is directly, above the center of the base. Find its, height., , 4, , Copyright 2000-2007 Kinetic Books Co. Chapter 00
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This section shows you an example of an interactive checkpoint. We chose a problem that uses mathematics you may be familiar with in case, you would like to solve the problem yourself., In interactive checkpoints, all of the problem-solving elements are initially hidden. You can open any element by clicking [Show] below., You can check your answer at any time by entering it at the top and pressing [Check] to see if you are right. You will find this is far more, efficient than keying all the information into the computer, which provides a good motivation for you to solve the problem yourself. However, if, you are stuck, you can always have the computer help you., In the parts called Variables, Strategy, and Physics principles and equations, the computer will show you the information you need when you, ask. In the Physics principles and equations section, we show you the principles and equations you need to solve the problem, as well as some, that do not apply directly to the problem., In the Step-by-step solution, you choose from the equations by clicking on the one you think you need to use. You must enter the correct, values in each Step-by-step part of the solution to proceed to the next step., , Answer:, , h=, , m, , 0.5 - Quizboards, Each chapter has a quizboard containing several multiple-choice conceptual and, quantitative problems. There are over 250 quizboard problems throughout the textbook., Quizboards allow you to test your understanding of a chapter. You see a quizboard on, the right., Quizboards appear between the summary section and the problems section in every, chapter. The quizboard for a chapter can be launched from the quizboard section by, clicking on the image on the right side of the page. The quizboards are designed to, enable you to review many of the crucial ideas in a chapter., Each problem in a quizboard consists of four parts: the question, the answer choices,, the hints, and the solution. If you think you know the answer to the problem, choose it, and click the “Check answer” button. A message will appear telling you whether you are, correct. You can keep trying until you get the problem right. If you are having trouble,, click “Give me a hint”. Every time you click this button, a new hint appears until there, are no more hints available. You can always click “Show solution” if you find yourself, completely stuck., Use the “next” and “previous” buttons on the gray bar at the bottom of the window to navigate between problems. You do not have to answer a, problem correctly before moving on, so you can skip problems and come back to them later. If you use the “previous” button to go back to a, problem, it will appear unanswered (even if you answered it before) so that you can try the problem again., Click on the image to the right to use a sample quizboard. You do not need to know any physics to answer these questions. Good luck!, , Copyright 2000-2007 Kinetic Books Co. Chapter 00, , 5
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0.6 - Highlighting and notes, You can add notes or highlight text on most sections of the textbook. Notes always appear at the top of the section. You can use a note to write, short messages about key elements of a section, or to remind yourself not to forget the extra soccer practice or to pick up the groceries., As the note above says, you insert notes by pressing Add Note at the bottom of the page. You remove a note by clicking on the Delete button, located next to the note. Modify a note by pressing the Edit button., The text you are reading now is highlighted. To highlight text, click the Highlight button at the bottom of the page to switch it from “Off” to “On”., When it is “On”, any text you select (by clicking on your mouse and dragging) will be highlighted. You can remove all highlighting from a section, by pressing the “Clear” button., If the text above is not highlighted, your operating system or browser does not enable us to offer this feature. For instance, the feature is not, available on the Macintosh operating system 0S X 10.2., If you use a shared computer, the highlighting and notes features may be turned off. The preferences page allows you to enable or disable, either of these features. You will find the preferences page by clicking on the Preferences button at the bottom of the page., Notes and highlighting are not supported on our Web Access option or the trial version of the product on our web site., , 0.7 - Online Homework, This textbook was designed to support online assessment of homework. Instructors can assign specific problems online, and you submit your, responses over the Internet to a central computer. You can work offline and submit the answers when you are ready., The computer checks the answers, and sends a report about your efforts, and the efforts of your peers, to your instructor. Your instructor can, configure this service in a variety of fashions. For instance, they can set deadlines for homework assignments or decide if you are allowed to, try answering a question a few times., Online Homework is an optional feature; not all instructors will use it. If your instructor has supplied you with a login ID or told you to sign up for, Online Homework, please log in now. If you are unsure, please check with your instructor., If you want to learn more about on-line assessment in general, click here., , 0.8 - Finding what you need in this book, You can navigate through the book using the Table of Contents button. When you roll your mouse over it, you will see three links., The Chapter TOC link takes you to the table of contents for the chapter you are currently in. You will see more sections than you might see in a, typical physics textbook table of contents. We chose to make it very easy to navigate to each element of the textbook by listing sample, problems, derivations and other elements discretely., The Main TOC link takes you to the list of all the chapters in the textbook. Clicking on the third link, Physics Factbook, opens a reference tool, containing useful information including mathematics review topics and formulas, unit conversion factors, fundamental physical constants,, properties of the elements, astronomical data, and physics equations. The Factbook also has a built-in search feature to help you find, information quickly., This textbook has no index, but likely you will find that entering text in the “search box” is more useful. Search is located at the bottom of each, Web page. Search performs its task by looking at the name of each section, at the first (or essential) time any term is defined, and at some, other types of text. Typing in a phrase like “kinetic energy” will produce a number of useful results., When you use search, you do not have to worry about sequence: You do not have to guess whether we listed something under, say, “average, velocity” or “velocity average.” Search looks for the terms and presents them to you along with some of their context., That is it for logistics. The people who worked on this textbook í about 50 of us í hope you enjoy it. We have a passion for physics, and we, hope some of that carries on to you., To explore the rest of the book, move your mouse over a Table of Contents button at the top or bottom of this page, and select the Main TOC., , 6, , Copyright 2000-2007 Kinetic Books Co. Chapter 00
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1.0 - Introduction, Heavyweight, lightweight, overweight, slender. Small,, tall, vertically impaired, “how’s the weather up there?”, Gifted, average, 700 math/600 verbal, rocket, scientist. Gorgeous, handsome, hunk, babe., Humans like to measure things. Whether it is our, body size, height, IQ or looks, everything seems to be, fair game., Physics will teach you to measure even more things., For example, quantities such as displacement,, velocity and acceleration are crucial to understanding, motion. Other topics have yet more things to quantify:, Mass and period are concepts required to understand the movement of planets; resistance and current are used for analyzing electric circuits., Just as you have developed a vocabulary for the things you measure, so have physicists., There are many different units for measuring different properties. It is possible to go all the way from A through Z in units: amperes, bars,, centimeters, dynes, ergs, farads, grams, hertz, inches, joules, kilograms, liters, meters, newtons, ohms, pascals, quintals, rydbergs, slugs,, teslas, unit magnetic poles, volts, webers, x units, years, and zettabars. (OK, we had to stretch for X, but it is a real unit.), Physicists have so many units of measure at their disposal because they have plenty to measure. Physicists use amperes to tell how much, electric current flows through a wire, “pascals” quantify pressure, and “teslas” are used to measure the strength of a magnetic field. If you so, desired, you could become a units expert and impress (or worry) your classmates by casually noting that the U.S. tablespoon equals 1.04, Canadian tablespoons, or deftly differentiating between the barrel, U.K. Wine, versus the barrel, U.S. federal spirits, or the barrel, U.S. federal,, all of which define slightly different volumes. Or you could become an international sophisticate, telling friends that one German doppelzentner, equals about 77,162 U.K. scruples, which of course equals approximately 101.97 metric glugs, which comes out to3120 ukies, a Libyan unit, used for the sole purpose of measuring ostrich feathers and wool., Fortunately, you do not need to learn units such as the ones mentioned immediately above, and you will learn the others over time. Textbooks, like this one provide tables that specify the relationships between commonly used units and you will use these tables to convert between units., , 1.1 - The metric system and the Système International d’Unités, , Metric system: The dominant system of, measurement in science and the world., Historically, people chose units of measure related to everyday life (the “foot” is one, example). Scientists continued this tradition, developing units such as “horsepower” to, measure power., The French challenged this philosophy of measurement during their Revolution, when, they decided to give measurement a more scientific foundation. Instead of basing their, system on things that change í the length of a person’s foot changes during her, lifetime, for example í the French based their system on what they viewed as constant., To accomplish this, they created units such as the meter, which they defined as a, certain fraction of the Earth’s circumference. (To be specific: one ten-millionth of the, meridian passing through Paris from the equator to the North Pole. It turns out that the, distance from the equator to the North Pole does vary, but the metric system’s intent of, consistency and measurability was exactly on target.), , Metric system and the Système, International, System defines fundamental units, Larger/smaller units based on powers of, 10, , The metric system is also based on another inspired idea: units of measurement should, be based on powers of 10. This differs from the British system, which provides more variety: 12 inches in a foot, 5280 feet to a mile and so, forth., The metric system makes conversions much simpler to perform. For example, in order to calculate the number of inches in a mile, you would, typically multiply by 5280 (for feet in a mile) and then by 12 (for inches in a foot). However, in the metric system, to convert between units, you, typically multiply by a power of 10. For instance, to convert from kilometers to meters, you multiply by 1000. The prefix “kilo” means 1000., The revolutionaries were a little extreme (as revolutionaries tend to be) and they held onto their position of power for only a decade or so. While, some of their legacy (including their political art, rather mediocre as is much political art) has been forgotten, their clever and sensible metric, system endures. Most scientists, and most countries, use the metric system today., Scientists continue to update and refine the metric system. This expanded and updated system of measurement used today is called the, Système International d’Unités, or SI. We typically use SI units in this textbook; several times, though, we refer to different units that may be, better known to you or are commonly used in the sciences. We will discuss some of the SI units further in this chapter., Over the years, scientists have refined measurement systems, making the definition of units ever more precise. For example, instead of being, , Copyright 2000-2007 Kinetic Books Co. 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based on the Earth’s circumference, the meter is now defined as the distance light travels in a vacuum during the time interval of, 1/299,792,458 of a second. Although perhaps not as memorable as the initial standard, this definition is important because it is constant,, precise, indestructible, and can be reproduced in laboratories around the world., In addition to using meters for length, the Système International uses seconds (time), kilograms (mass), amperes (electric current), kelvins, (temperature), moles (amount of substance) and candelas (luminous intensity). Many other derived units are based on these fundamental, units. For instance, a newton measures force and is equal to kilograms times meters per second squared. On Earth, the force of gravity on a, small apple is about one newton., At the risk of drowning you in terminology, we should point out that you might also encounter references to the MKS (meter/kilogram/second), and CGS (centimeter/gram/second) systems. These systems are named for the units they use for length, mass and time., , 1.2 - Prefixes, Metric units often have prefixes. Kilometers and centimeters both have prefixes before, the word “meter.” The prefixes instruct you to multiply or divide by a power of 10: kilo, means multiply by 1000, so a kilometer equals 1000 meters. Centi means divide by, 100, so a centimeter is one one-hundredth of a meter. In other words, there are, 100 centimeters in a meter. The table in Equation 1 on the right lists the values for the, most common prefixes., Prefixes allow you to describe the unimaginably vast and small and everything in, between. To illustrate, every day the City of New York produces 10 gigagrams of, garbage. The distance between transistors in a microprocessor is less than a, micrometer. The power of the Sun is 400 yottawatts (a yotta corresponds to the factor of, 1024). It takes 3.34 nanoseconds for light to travel one meter. The electric potential, difference across a nerve cell is about 70 millivolts., These prefixes can apply to any unit. You can use gigameters to conveniently quantify a, vast distance, gigagrams to measure the mass of a huge object, or gigavolts to describe, a large electrical potential difference., , Prefixes, Create larger, smaller units, , Some of the most common prefixes í kilo, mega, and giga í are commonly used to, describe the specifications of computers. The speed of a computer microprocessor is, measured by how many computational cycles per second it can perform., Microprocessor speeds used to be specified in megahertz (one million cycles per, second) but are now specified in gigahertz (one billion cycles per second). Modem, speeds have increased from kilobits to megabits per second. (Although bits are not part, of the metric system, computer scientists use the same prefixes.), The units of measurement you use are a matter of both convenience and convention., For example, snow skis are typically measured in centimeters; a ski labeled “170” is 170, centimeters long. However, it could also be called a 1.7-meter ski or a 1700-millimeter, ski. The ski industry has decided that centimeters are reasonable units and has settled, on their use as a convention., , Common prefixes for powers of 10, , In this textbook, you are most likely to encounter kilo, mega and giga on the large side, of things and centi, milli, micro and nano on the small. Some other prefixes are not as, common because they just do not seem that useful. Is it easier to say “a decameter”, than the more straightforward 10 meters? And, for the extremely large and small,, scientists often use another technique called scientific notation rather than prefixes., , What is the distance between the, towns in kilometers?, 1000 m = 1 km, , 8, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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1.3 - Scientific notation, , Scientific notation: A system, based on powers, of 10, most useful for expressing very large and, very small numbers., Physicists like to measure the very big, the very small and everything in between. To, express the results efficiently and clearly, they use scientific notation., Scientific notation expresses a quantity as a number times a power of 10. Why is this, useful? Here’s an example: the Earth is about 149,000,000,000 meters from the Sun., You could express that distance as we just did, with a long string of zeros, or you could, use scientific notation to write it as 1.49×1011 meters. The latter method has proven, itself to be clearer and less prone to error., The value on the left (1.49) is called the leading value. The power of 10 is typically, chosen so the leading value is between one and 10. In the example immediately above,, we multiplied by 1011 so that we could use 1.49. We also could have written 14.9×1010, or 0.149×1012 since all three values are equal, but a useful convention is to use a, number between one and 10., , Scientific notation, Number between 1 and 10 (leading, value), Multiplied by power of 10, , In case you have forgotten how to use exponents, here’s a quick review. Ten is the, base number. Ten to the first power is 10; 102 is ten to the second (ten squared) or 100;, ten to the third is 10 times 10 times 10, or 1000. A positive exponent tells you how many, zeros to add after the one. When the exponent is zero, the value is one: 100 equals one., As mentioned, scientists also measure the very small. For example, a particle known as, a muon has a mean lifespan of about 2.2 millionths of a second. Scientific notation, provides a graceful way to express this number: 2.2×10í6 (2.2 times 10 to the negative, sixth). To review the mathematics: ten to the minus one is 1/10; ten to the minus two is, 1/100; ten to the minus three is 1/1000, and so forth., You can also write 1.49×1011 as 1.49e11. The two are equivalent. You may have seen, this notation in computer spreadsheet programs such as Microsoft® Excel. We do not, use this "e" notation in the text of the book, but if you submit answers to homework, problems or interactive checkpoints, you will use it there., , How do you write the numbers, above in scientific notation?, 45 = 4.5 × 10 = 4.5×101, 0.012 = 1.2 × 1/100 = 1.2×10í2, , 1.4 - Standards and constants, , Standard: A framework for establishing, measurement units., Physical constant: An empirically based value., Physicists establish standards so they can measure things consistently; how they define, a standard can change over time. For example, the length of a meter is now based on, how far light travels in a precise interval of time. This replaces a standard based on the, wavelength of light emitted by krypton-86. Prior to that, the meter was defined as the, distance between enscribed marks on platinum-iridium bars. Advances in technology,, and the requirement for increased precision, cause scientists to change the method, used to define the standard. Scientists strive for precise standards that can be, reproduced as needed and which will not change., , Standards, Establish benchmarks for measurement, , By choosing standards, scientists can achieve consistent results around the globe and compare the results of their experiments. Well-equipped, labs can measure time using atomic clocks like the one shown in Concept 1 on the right. These clocks are based on a characteristic frequency, of cesium atoms. You can access the official time, as maintained by an atomic clock, by clicking here., You will encounter two types of constants in this textbook. First, there are mathematical constants like ʌ or the number 2. Second, there are, physical constants, such as the gravitational constant, which is represented with a capital G in equations. We show its value in Concept 2 on, the right. Devices such as the torsion balance shown are used to gather data to determine the value of G. This is an active area of research, as, G is the least precisely known of the major physical constants., Constants such as G are used in many equations. G is used in Sir Isaac Newton’s law of gravitation, an equation that relates the attractive, force between two bodies to their masses and the square of the distance between them. You see this equation on the right., , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 9
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Physical constants, Empirically determined values, G = 6.674 2×10í11 N·m2/kg2, , 1.5 - Length, If you live in a country that uses the metric system, you already have an intuitive sense, of how long a meter is. You are likely taller than one meter, and probably shorter than, two. If you are a basketball fan, you know that male professional basketball centers, tend to be taller than two meters while female professional centers average about two, meters., If you live in a country, such as the United States, that still uses the British system, you, may not be as familiar with the meter. A meter equals about 3.28 feet, or 39.4 inches,, which is to say a meter is slightly longer than a yard. The kilometer is another unit of, length commonly used in metric countries. You may have noticed that cars often have, speedometers that show both miles per hour and kilometers per hour. A kilometer, (1000 meters) equals about 0.621 miles. In track events, a metric mile is, 1.50 kilometers, which is about 93% of a British mile., Centimeters (one one-hundredth of a meter) are also frequently used metric units. One, inch equals 2.54 centimeters, so a centimeter is about four tenths of an inch. One foot, equals 30.48 centimeters. You see some common abbreviations in Equation 1 to the, right: “m” for meters, “km” for kilometers and “cm” for centimeters., , Length, Measured in meters (m), Distance light travels in, 3.335 640 95×10í9 seconds, , 1 meter (m) = 3.28 feet, 1 kilometer (km) = 0.621 miles, 1 centimeter (cm) = 0.394 inches, , 10, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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1.6 - Time, Despite the appeal of measuring a lifetime in daylights, sunsets, midnights, cups of, coffee, inches, miles, laughter, or in strife, physicists still choose “seconds.” How, refreshingly simple!, However, as you might expect, physicists have developed a precise way to define a, second. Atomic clocks, such as the one shown in Concept 1 to the right, rely on the fact, that cesium-133 atoms undergo a transition when exposed to microwave radiation at a, frequency of 9,192,631,770 cycles per second. These clocks are extremely accurate., Thousands of years would pass before two such clocks would differ even by a second., If you are an exceedingly precise person, you might want to consider buying a, wristwatch that calibrates itself via radio signals from an atomic clock. For now, though,, you can visit a web site that displays the current time as measured by an atomic clock., In addition to being used to measure a second, atomic clocks are used to keep, time. The length of a day on Earth, measured by the time to complete one rotation, is, not constant. Why? The frictional force of tides causes the Earth to spin more slowly., This means that the day is getting longer (does it not just feel that way sometimes?)., Every fifteen months or so since 1978, a leap second has been added to official timekeeping clocks worldwide to compensate for increased time it takes the Earth to, complete a revolution., , Time, Measured in seconds (s), 1 second = 9,192,631,770 cycles, , 1 hour = 3600 seconds, 1 day = 86,400 seconds, , 1.7 - Mass, , Mass, Measured in kilograms (kg), Resistance to change in motion, Not weight!, , The standard unit of mass is the kilogram. (The British system equivalent is the slug,, which is perhaps another reason to go metric.), Physicists define mass as the property of an object that measures its resistance to a, change in motion. A car has more mass than a bicycle. The three people shown, straining at the car above will cause it to accelerate slowly; if they were pushing a, bicycle instead, they could increase its speed much more quickly. Once they do set the, car in motion, if they are not careful, its mass might prevent them from stopping it., The official kilogram, the International Prototype Kilogram, is a cylinder of platinumiridium alloy that resides at France’s International Bureau of Weights and Measures., Copies of this kilogram reside in other secure facilities in different countries and are, occasionally brought back for comparison to the original., A liter of water has a mass of about one kilogram. A typical can of soda contains about, 354 milliliters and has a mass of 0.354 kilograms., , One kilogram = one liter of water, One gram = about 25 raindrops, , It is tempting to write that one kilogram equals about 2.2 pounds, but this is wrong. The, pound is a unit of weight; kilograms and slugs are units for mass. Weight measures the force of gravity that a planet exerts on an object, while, mass reflects that object’s resistance to change in motion. A classic example illustrates the difference: Your mass is the same on the Earth and, the Moon, but you weigh less on the Moon because it exerts less gravitational force on you. On Earth, the force of gravity on one kilogram is, 2.2 pounds but the force of gravity on a kilogram is only 0.36 pounds on the Moon., Kilogram is abbreviated as kg. We typically use kilograms in this book, not grams (which are abbreviated as g)., The three units you need in order to understand motion, force and energy, the topics that start a physics textbook, are meters, kilograms and, seconds. Other units used in studying these topics are derived from these fundamental units., , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 11
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1.8 - Converting units, At times, you will need to convert units. Some, conversion factors you know, such as 60 seconds in a, minute, 12 inches in a foot, etc. Others, such as the, number of seconds in a year, require a bit of, calculation., Keep in mind that if you do not use consistent units,, troubles will arise. NASA dramatically illustrated the, cost of such errors when it lost a spacecraft in 1999., A company supplied data to NASA based on British, units (pounds) when NASA engineers expected, metric units (newtons). Oops. That, alas, was the end, of that space probe (and about $125 million and, one, suspects, some engineer’s NASA career)., , Speedometers often show speeds in both mi/h and km/h, , As NASA’s misfortune indicates, you need to make sure you use the correct units when, solving physics problems. If a problem presents information about a quantity like time in, different units, you need to convert that information to the same units., You convert units by:, 1., , Knowing the conversion factor (say, 12 inches to a foot; 2.54 centimeters to an, inch; $125 million to a spacecraft)., , 2., , Multiplying by the conversion factor (such as 3.28 feet/1.00 meter) so that you, cancel units in both the numerator and denominator. For example, to convert, meters to feet, you multiply by 3.28 ft/m so that the meter units cancel. This may, be easier to understand by viewing the example on the right., , In conversions, it is easy to make mistakes so it is good to check your work. To make, sure you are applying conversions correctly, make sure the appropriate units cancel. To, do this, you note the units associated with each value and each conversion factor., As is shown on the right, a unit that is in both a denominator and a numerator cancels., You should look to see that the units that remain “uncancelled” are the ones that you, desired. For instance, in the example problem, fluid ounces cancel out, and the desired, units, milliliters, remain., , Converting units, Choose appropriate conversion factor, Multiply by conversion factor as a, fraction, Make sure units cancel!, , How many milliliters of orange, juice in the bottle?, 1 mL = 0.0338 fl. oz., , 12, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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1.9 - Sample problem: conversions, Express the car's acceleration in, m/s2, using scientific notation. For an, extra challenge, state the result in, terameters per second squared., , Variables, We will use a to represent the acceleration we are converting., What is the strategy?, 1., , Express the car's acceleration in scientific notation., , 2., , Using conversion factors in scientific notation, convert miles to yards and then to meters., , 3., , Convert hours squared to minutes squared and then to seconds squared., , 4., , Go for the challenge! Convert to terameters., , Conversion factors and prefixes, 1 mi = 1760 yd, 1 yd = 0.914 m, 1 h = 60 min, 1 min = 60 s, tera = 1012, Step-by-step solution, We first write the acceleration in scientific notation and convert miles to yards to meters., , Step, , 1., , Reason, , a = 9430 mi/h2 = 9.43×103 mi/h2, , scientific notation, , 2., , multiply by factor converting miles to yards, , 3., , multiply by factor converting yards to meters, , Now we convert hours squared to minutes squared to seconds squared. Because the units we are converting are squared, we square the, conversion factors. As requested, we state the final result in scientific notation., , Step, , Reason, , 4., , multiply by square of factor converting hours to minutes, , 5., , multiply by square of factor converting minutes to seconds, , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 13
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Finally, we convert to terameters per second squared, or Tm/s2. "Tera" means 1012., , Step, , Reason, , 6., , multiply by conversion factor for terameters, , 1.10 - Interactive checkpoint: cheeseburgers or gasoline?, A double cheeseburger contains 591, kilocalories of energy. A gallon of, gasoline contains 159 MJ of energy., A compact car makes 39.0, miles/gallon on the highway., Assuming that it could run as, efficiently on fast food as on gasoline,, find the number of cheeseburgers, needed to propel the car from New, York to Chicago (719 miles)., A kilocalorie is a unit used to measure food energy, and is often called a Calorie, spelled with a capital C. One calorie (small c) is equal to 4.19, J (joules)., , Answer:, , N=, , cheeseburgers, , 1.11 - Pythagorean theorem, As you proceed through your physics studies, you will find it necessary to understand, the Pythagorean theorem, which is reviewed in this section. At the right, you see a right, triangle (a triangle with a 90° angle). The Pythagorean theorem states that the square, of the hypotenuse (the side opposite the right angle) equals the sum of the squares of, the two legs. This equation is shown in Equation 1 to the right., There are two specific right triangles that occur frequently in physics homework, problems. In an isosceles right triangle, the two legs are the same length and the, hypotenuse is the length of either leg times the square root of two. The angles of an, isosceles right triangle are 45, 45 and 90 degrees, so it is also called a 45-45-90, triangle., When the angles of the triangle measure 30, 60 and 90 degrees, it is called a 30-60-90, triangle. The shorter leg, the one opposite the 30° angle, is one half the length of the, hypotenuse. This relationship makes for an easy mathematical calculation and makes, this triangle a favorite in homework problems., , Pythagorean theorem, Relates hypotenuse to legs of right, triangle, , c 2 = a2 + b2, c = length of hypotenuse, a, b = lengths of legs, 14, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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What is the length of the, hypotenuse?, c 2 = a2 + b2 = 52 + 122 = 25 + 144, c 2 = 169, , 1.12 - Trigonometric functions, , Trigonometric functions, sin ș = opposite / hypotenuse, cos ș = adjacent / hypotenuse, tan ș = opposite / adjacent, , You will often encounter trigonometric functions in physics. You need to understand the, basics of the sine, cosine and tangent, and their inverses: the arcsine, arccosine and, arctangent., The illustration above depicts an angle and three sides of a triangle. The sine (sin) of, the angle ș (the Greek letter theta, pronounced “thay-tuh”) equals the ratio of the side, opposite the angle divided by the triangle's hypotenuse. “Opposite” means the leg, across from the angle, as the diagram reflects., The cosine (cos) of șequals the ratio of the side of the triangle adjacent to the angle,, divided by the hypotenuse. “Adjacent” means the leg that forms one side of the angle., Finally, the tangent (tan) of șequals the ratio of the opposite side divided by the, adjacent side., These three ratios are constant for a given angle in a right triangle, no matter what the, size of the triangle. They are useful because you are often given information such as, the length of the hypotenuse and the size of an angle, and then asked to calculate one, of the legs of the triangle. For instance, if asked to calculate the opposite leg, you would, multiply the sine of the angle by the hypotenuse., , What is sin ș? cos ș? tan ș?, sin ș = opposite / hypotenuse = 3/5, cos ș = adjacent / hypotenuse = 4/5, tan ș = opposite / adjacent = 3/4, , You may also be asked to use the arcsine, the arccosine or the arctangent. These are, often written as siní1, cosí1 and taní1. These are not the reciprocals of the sine, cosine, and tangent! Rather, they supply the size of the angle when the value of the, trigonometric function is known. For example, since sin 30° equals 0.5, the arcsine, of 0.5 (or siní1 0.5) equals 30°. This is also often written as arcsin(0.5); arccos and, arctan are the abbreviations for arccosine and arctangent., In the old days, scientists consulted tables for these trigonometric values. Today,, calculators and spreadsheets can calculate them for you., , Inverse trigonometric functions, ș = arcsin (opposite / hypotenuse), , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 15
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ș = arccos (adjacent / hypotenuse), ș = arctan (opposite / adjacent), Often written: siní1, cosí1, taní1, , If the tangent of ș is 1, what is ș?, ș = arctangent(1) = 45°, 1.13 - Radians, , Radian measure: A measurement of angles, based on a ratio of lengths., Angles are often measured or specified in degrees, but another unit, the radian, is, useful in many computations. The radian measure of an angle is the ratio of two lengths, on a circle. The angle and lengths are perhaps most easily understood by looking at the, diagram in Equation 1 on the right. The arc length is the length of the arc on the, circumference cut off by the angle when it is placed at the circle’s center. The other, length is the radius of the circle. The radian measure of the angle equals the arc length, divided by the radius., A 360° angle equals 2ʌ radians. Why is this so? The angle 360° describes an entire, circle. The arc length in this case equals the circumference (2ʌr) of a circle divided by, the radius r of the circle. The radius factor cancels out, leaving 2ʌ as the result., Radians are dimensionless numbers. Why? Since a radian is a ratio of two lengths, the, length units cancel out. However, we follow a radian measure with "rad" so it is clear, what is meant., , Radian measure, Angle = arc length / radius = s/r, 360° = 2ʌ rad, ·Radians are dimensionless, ·Units: radians (rad), , What is the angle's measure in, radians?, Angle = arc length / radius, ș = (ʌ/2 m)/(2 m), ș = ʌ/4 rad, , 16, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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1.14 - Sample problem: trigonometry, What is the length of side b of this, triangle?, What is ș in degrees? in radians?, , The lengths of two sides of a right triangle are shown., Variables, short leg, , a = 8.16 m, , long leg, , b, , hypotenuse, , c = 17.2 m, , angle, , ș, , What is the strategy?, 1., , Use the Pythagorean theorem to calculate the length of the third side., , 2., , Calculate the cosine of ș and then use the arccosine function on a calculator (or consult a table) to determine, , 3., , Convert ș to radians., , ș in degrees., , Mathematics principles, , c2 = a2 + b2, cos ș = adjacent/hypotenuse, 360° = 2ʌ rad, Step-by-step solution, First we use the Pythagorean theorem to find the length of the longer leg., , Step, , Reason, , 1., , c2 = a2 + b2, , 2., , (17.2 m)2 = (8.16 m)2 + b2 enter values, , 3., , b2 = 17.22 – 8.162 = 229, , solve for b2, , 4., , b = 15.1 m, , take square root, , Pythagorean theorem, , Now, we use the lengths of the sides to find the value of the cosine of ș, and then look up the arccosine of this ratio., , Step, , Reason, , 5., , cos ș = adjacent/hypotenuse definition of cosine, , 6., , cos ș = (8.16 m)/(17.2 m), , enter values, , 7., , cos ș = 0.474, , divide, , 8., , ș = arccos(0.474), , definition of arccosine, , 9., , ș = 61.7°, , use calculator, , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 17
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Finally, we convert the angle from degrees to radians., , Step, , Reason, , 10. ș = (61.7°)(2ʌ rad/360°) multiply by conversion factor, 11. ș = 1.08 rad, , multiplication and division, , There are other ways to solve this problem. For example, you could first find, find the length of the third side., , ș using the arccosine function, and then use the tangent ratio to, , 1.15 - Interactive checkpoint: trigonometry, The great pyramid of Cheops has a, square base with edges that are, almost exactly 230 m long. The side, faces of the pyramid make an angle, of 51.8° with the ground., The apex of the pyramid is directly, above the center of the base. Find its, height., , Answer:, , h=, , m, , 1.16 - Gotchas, The goal of “gotchas” is to help you avoid common errors. (Not that your teacher’s tests would ever try to make you commit any of these, errors!), Confusing weight and mass. You do not weigh 70 kilograms, or 80, or 60. However, those values could very well be your mass, which is an, unchanging value that reflects your resistance to a change in motion., Converting units with factors incorrectly oriented. There could probably be an essay written on this topic. Make sure the units cancel! is, probably the best advice we can give. For example, if you are converting meters per second to miles per hour, begin by multiplying by a, conversion fraction of 3600 seconds over one hour. This will cause the seconds to cancel and hours to be in the right place. (If this is not clear,, write it down and strike out units. If it is still unclear, do some practice problems.) In any physics calculation, checking that the units on each, side are consistent is a good technique., , 18, , Copyright 2000-2007 Kinetic Books Co. Chapter 01
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1.17 - Summary, Scientists use the Système International d’Unités, also known as the metric system, of measurement. Examples of metric units are meters, kilograms, and seconds., , Prefixes, In these systems, units that measure the same property, for example units for mass,, are related to each other by powers of ten. Unit prefixes tell you how many powers, of ten. For example, a kilogram is 1000 grams and a kilometer is 1000 meters, while, a milligram is one one-thousandth of a gram, and a millimeter is one-thousandth of, a meter., , giga (G) = 109, mega (M) = 106, kilo (k) = 103, , Numbers may be expressed in scientific notation. Any number can be written as a, number between 1 and 10, multiplied by a power of ten. For example,, 875.6 = 8.756×102., , centi (c) = 10–2, , A standard is an agreed-on basis for establishing measurement units, like defining, the kilogram as the mass of a certain platinum-iridium cylinder that is kept at the, International Bureau of Weights and Measures, near Paris. A physical constant is, an empirically measured value that does not change, such as the speed of light., , micro (ȝ) = 10–6, , In the metric system, the basic unit of length is the meter; time is measured in, seconds; and mass is measured in kilograms., , Pythagorean Theorem, , Sometimes a problem will require you to do unit conversion. Work in fractions so, that you can cancel like units, and make sure that the units are of the same type (all, are units of length, for instance)., The Pythagorean theorem states that the square of the hypotenuse of a triangle is, equal to the sum of the squares of the two legs., , c2 = a2 + b2, Trigonometric functions, such as sine, cosine and tangent, relate the angles of a, right triangle to the lengths of its sides., , milli (m) = 10–3, nano (n) = 10–9, , c2 = a2 + b2, Trigonometric functions, , sin ș = opposite / hypotenuse, cos ș = adjacent / hypotenuse, tan ș = opposite / adjacent, Radian measure, , Radians (rad) measure angles. The radian measure of an angle located at the, center of a circle equals the arc length it cuts off on the circle, divided by the radius, of the circle., , Angle = arc length / radius = s/ r, 360° = 2ʌ rad, , Copyright 2000-2007 Kinetic Books Co. Chapter 01, , 19
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Chapter 1 Problems, , Conceptual Problems, C.1, , Why do scientists not define the standard for the second to be a fraction of a day?, , C.2, , What are the pitfalls in attempting to convert pounds into kilograms?, , C.3, , When multiplying by a conversion factor, how do you determine which unit belongs in the numerator and which in the, denominator?, , C.4, , Can a leg of a right triangle be longer than the hypotenuse? Why or why not?, , C.5, , Consider the sine and cosine functions. (a) As an angle increases from 0° to 90°, does the sine of the angle increase or, decrease? (b) How about the cosine?, , Yes, , C.6, , No, , (a), , Increases, , Decreases, , (b), , Increases, , Decreases, , You walk along the edge of a large circular lawn. You walk clockwise from your starting location until you have moved an, angle of ʌ radians. What geometric shape is defined by the following three points: your starting position, your final position,, and the center point of the lawn?, i. Isoceles triangle, ii. Right triangle, iii. Straight line, , Section Problems, Section 2 - Prefixes, 2.1, , How many centimeters are there in a kilometer?, cm, , 2.2, , A carbon-carbon triple bond has a length of 120 picometers. What is its length in nanometers?, nm, , 2.3, , The 2000 Gross Domestic Product of the United States was $9,966 billion. If you were to regard the dollar as a scientific unit,, how would you write this using the standard prefixes?, i. 9966, ii. 9.966, iii. 99.66, , i., ii., iii., iv., , pico, milli, kilo, tera, , dollars, , 2.4, , A gremlin is a tiny mythical creature often blamed for mechanical failures. Suppose for a moment that a shrewd physicist, catches one such gremlin and makes it work for her instead of against her. She determines that the gremlin can produce 24, milliwatts of power. How many gremlins are required to produce 24,000 megawatts of power?, , 2.5, , A drug company has just manufactured 50.0 kg of acetylsalicylic acid for use in aspirin tablets. If a single tablet contains 500, mg of the drug, how many tablets can the company make out of this batch?, , gremlins, , tablets, 2.6, , The diameter of an aluminum atom is about 0.24 nm, and the nuclear diameter is about 7.2 fm (femtometer = 10í15 meter)., (a) If the atom's diameter were expanded to the length of an American football field (91.44 m) and the nuclear diameter, expanded proportionally, what would be the nuclear diameter in meters? (b) Is the saying that "the atom is mostly empty, space" confirmed by these figures?, (a), (b), , m, Yes, , No, , Section 3 - Scientific notation, 3.1, , 20, , An electron can tunnel through an energy barrier with probability 0.0000000000375. (This is a concept used in quantum, mechanics.) Express this probability in scientific notation., , Copyright 2007 Kinetic Books Co. Chapter 1 Problems
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3.2, , An atom of uranium-235 has a mass 235.043924 amu (atomic mass units). Using scientific notation, state the mass in amu of, ten million such atoms., amu, , 3.3, , Sara has lived 18.0 years. How many seconds has she lived? Express the answer in scientific notation. Use 365.24 days per, year for your calculations., , 3.4, , Assume a typical hummingbird has a lifespan of 4.0 years and an average heart rate of 1300 beats per minute. (a) Calculate, the number of times a hummingbird's heart beats in its life, and express it in scientific notation. Use 365.24 days per year. (b), A long-lived elephant lives for 61 years, and has an average heart rate of 25 beats per minute. Calculate the number of, heartbeats in that elephant's lifetime in scientific notation., , s, , 3.5, , (a), , beats, , (b), , beats, , A PC microprocessor runs at 2.40 GHz. A hertz (Hz) is a unit meaning "one cycle per second." A movie projector displays, images at a rate of 24.0 Hz. What is the ratio of the microprocessor's rate in cycles per second to the update rate of a movie, projector? Answer the question using scientific notation., , Section 5 - Length, 5.1, , Which of the following coins has a diameter of 17.91 mm: a nickel, a dime or a quarter?, i. Nickel, ii. Dime, iii. Quarter, , Section 8 - Converting units, 8.1, , 11.2 meters per second is how many miles per hour?, mi/h, , 8.2, , Freefall acceleration g is the acceleration due to gravity. It equals 9.80 meters per second squared near the Earth's surface., (a) What does it equal in feet per second squared? (b) In miles per second squared? (c) In miles per hour squared?, (a), , ft/s2, , (b), , mi/s2, , (c), , mi/h2, , 8.3, , A historian tells you that a cubit is an ancient unit of length equal to 0.457 meters. If you traveled 585 kilometers from Venice,, Italy to Frankfurt, Germany, how many cubits did you cover?, , 8.4, , You are on the phone with a friend in Greece, who tells you that he has just caught a fish 65 cm long in the Mediterranean, Sea. Assuming he is telling the truth, what is the length of the fish in inches?, , 8.5, , In an attempt to rid yourself of your little brother's unwanted attention, you tell him to count to a million and then come find, you. If he were to start counting at a rate of one number per second, would he be finished in a year?, , cubits, , in, , Yes, 8.6, , No, , Light travels at 3.0×108 m/s in a vacuum. Find its speed in furlongs per fortnight. There are roughly 201 meters in a furlong,, and a fortnight is equal to 14 days., furlongs/fortnight, , 8.7, , Mercury orbits the Sun at a mean distance of 57,900,000 kilometers. (a) What is this distance in meters? Use scientific, notation to express your answer and state it with three significant figures. (b) Pluto orbits at a mean distance of 5.91×1012, meters from the Sun. What is this distance in kilometers?, m, , (a), (b), , 8.8, , i., ii., iii., iv., v., , 5 910 000 km, 5 910 000 000 km, 591 000 000 000 km, 5 910 000 000 000 km, 591 000 000 000 000 km, , In 2003, Bill Gates was worth 40.7 billion dollars. (a) Express this figure in dollars in scientific notation. (b) Assume a dollar is, , Copyright 2007 Kinetic Books Co. Chapter 1 Problems, , 21
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worth 2,060 Italian lire. State Bill Gates's net worth in lire in scientific notation., , 8.9, , (a), , dollars, , (b), , lire, , The world's tallest man was Robert Pershing Wadlow, who was 8 feet, 11.1 inches tall. There are 2.54 centimeters in an inch, and 12 inches in a foot. How tall was Robert in meters?, m, , 8.10 Romans and Greeks used their stadium as a unit of measure. An estimate of the ancient Roman unit stadium is 185 m. A, heroic epic describes a battle in which the Roman army marched 201 stadia to defend Rome from invading barbarians. In, kilometers, how far did they march?, km, 8.11 You are carrying a 1.3 gallon jug full of cold water. Given that the water has a density of 1.0 grams per milliliter, and one, gallon is equal to 3.8 liters, state the mass of the water in kilograms., kg, , Section 11 - Pythagorean theorem, 11.1 A crew of piano movers uses a 6.5-foot ramp to move a 990 lb Steinway concert grand up onto a stage that is 1.8 feet higher, than the floor. How far away from the base of the stage should they set the end of the ramp so that the other end exactly, reaches the stage?, ft, 11.2 Suzy is holding her kite on a string 25.0 m long when the kite hits the top of a flagpole, which is 15.3 m higher than her, hands. Assuming that the string is taut and forms a straight line, what is the horizontal distance from her hands to the, flagpole?, m, 11.3 A right triangle has a hypotenuse of length 17 cm and a leg of length 15 cm. What is the length of the other leg?, cm, 11.4 A Pythagorean triple is a set of three integers (a,b,c) that could form three sides of a right triangle. (3,4,5) and (5,12,13) are, two examples. There exists a Pythagorean triple of the form (7,n,n+1). Find n., , 11.5 A hapless motorist is trying to find his friend's, house, which is located 2.00 miles west of the, intersection of State Street and First Avenue., He sets out from the intersection, and drives, the same number of kilometers due south, instead. How far in kilometers is he from his, intended destination? The sketch is not to, scale., km, , Section 12 - Trigonometric functions, 12.1 You want to estimate the height of the Empire State Building. You start at its base and walk 15 m away. Then you, approximate the angle from the ground at that point to the top to be 88 degrees. How tall do you estimate the Empire State, Bulding to be?, m, 12.2 For an angle ș, write tan ș in terms of sin ș and cos ș., tan ș = cos ș / sin ș, tan ș = sin ș / cos ș, tan ș = (sin ș)(cos ș), 12.3 A window washer is climbing up a ladder to wash a window. The end of the 10.5-foot ladder exactly touches the windowsill, and the ladder makes a 70.0° angle with the ground. How far off the ground is the windowsill?, ft, , 22, , Copyright 2007 Kinetic Books Co. Chapter 1 Problems
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12.4 The 28 ft flagpole outside Martin Luther King,, Jr. Elementary School casts a 42 ft shadow. In, degrees, at what angle above the horizon is, the Sun? The diagram may not be to scale., °, , Section 13 - Radians, 13.1 Find the radian measure of one of the angles of a regular pentagon. (The angles are all the same in a regular pentagon.) In, degrees, the formula for the sum of the interior angles of an n-sided polygon is (n í 2) ⋅ 180°., radians, 13.2 A right triangle has sides of length 1, 2 and the square root of 3. In radians, what is its smallest angle?, , ʌ/3 radians, ʌ/6 radians, ʌ/12 radians, , Copyright 2007 Kinetic Books Co. Chapter 1 Problems, , 23
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2.0 - Introduction, Objects move: Balls bounce, cars speed, and spaceships accelerate. We are so, familiar with the concept of motion that we use sophisticated physics terms in, everyday language. For example, we might say that a project has reached “escape, velocity” or, if it is going less well, that it is in “free fall.”, In this chapter, you will learn more about motion, a field of study called kinematics., You will become familiar with concepts such as velocity, acceleration and, displacement. For now, the focus is on how things move, not what causes them to, move. Later, you will study dynamics, which centers on forces and how they affect, motion. Dynamics and kinematics make up mechanics, the study of force and, motion., Two key concepts in this chapter are velocity and acceleration. Velocity is how fast, something is moving (its speed) and in what direction it is moving. Acceleration is, the rate of change in velocity. In this chapter, you will have many opportunities to, learn about velocity and acceleration and how they relate. To get a feel for these, concepts, you can experiment by using the two simulations on the right. These, simulations are versions of the tortoise and hare race. In this classic parable, the, steady tortoise always wins the race. With your help, though, the hare stands a, chance. (After all, this is your physics course, not your literature course.), In the first simulation, the tortoise has a head start and moves at a constant velocity, of three meters per second to the right. The hare is initially stationary; it has zero, velocity. You set its acceleration í in other words, how much its velocity changes, each second. The acceleration you set is constant throughout the race. Can you set, the acceleration so that the hare crosses the finish line first and wins the race? To, try, click on Interactive 1, enter an acceleration value in the entry box in the, simulation, and press GO to see what happens. Press RESET if you want to try, again. Try acceleration values up to 10 meters per second squared. (At this, acceleration, the velocity increases by 10 meters per second every second. Values, larger than this will cause the action to occur so rapidly that the hare may quickly, disappear off the screen.), It does not really matter if you can cause the hare to beat this rather fast-moving, tortoise. However, we do want you to try a few different rates of acceleration and, see how they affect the hare’s velocity. Nothing particularly tricky is occurring here;, you are simply observing two basic properties of motion: velocity and acceleration., In the second simulation, the race is a round trip. To win the race, a contestant needs to go around the post on the right and then return to the, starting line. The tortoise has been given a head start in this race. When you start the simulation, the tortoise has already rounded the post and, is moving at a constant velocity on the homestretch back to the finish line., In this simulation, when you press GO the hare starts off moving quickly to the right. Again, you supply a value for its acceleration. The, challenge is to supply a value for the hare's acceleration so that it turns around at the post and races back to beat the tortoise. (Hint: Think, negative! Acceleration can be either positive or negative.), Again, it does not matter if you win; we want you to notice how acceleration affects velocity. Does the hare's velocity ever become zero?, Negative? To answer these questions, click on Interactive 2, enter the acceleration value for the hare in the gauge, press GO to see what, happens, and RESET to try again. You can also use PAUSE to stop the action and see the velocity at any instant. Press PAUSE again to, restart the race., We have given you a fair number of concepts in this introduction. These fundamentals are the foundation of the study of motion, and you will, learn much more about them shortly., , 24, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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2.1 - Position, , Position: The location of an object; in physics,, typically specified with graph coordinates., Position tells you location., There are many ways to describe location: Beverly Hills 90210; “...a galaxy far, far, away”; “as far away from you as possible.” Each works in its own context., Physicists often use numbers and graphs instead of words and phrases. Numbers and, graphs enable them (and you) to analyze motion with precision and consistency. In this, chapter, we will analyze objects that move in one dimension along a line, like a train, moving along a flat, straight section of track., To begin, we measure position along a number line. Two toy figures and a number line, are shown in the illustrations to the right. As you can see, the zero point is called the, origin. Positive numbers are on the right and negative numbers are on the left., , Position, Location of an object, Relative to origin, , By convention, we draw number lines from left to right. The number line could reflect an, object's position in east and west directions, or north and south, or up and down; the, important idea is that we can specify positions by referring to points on a line., When an object moves in one dimension, you can specify its position by its location on, the number line. The variable x specifies that position. For example, as shown in the, illustration for Equation 1, the hiker stands at position x = 3.0 meters and the toddler is, at position x = í2.0 meters., Later, you will study objects that move in multiple dimensions. For example, a, basketball free throw will initially travel both up and forward. For now, though, we will, consider objects that move in one dimension., , x represents position, Units: meters (m), , What are the positions of the, figures?, Hiker: x = 2.0 m, Toddler: x = í3.0 m, 2.2 - Displacement, , Displacement: The direction and distance of the, shortest path between an initial and final, position., You use the concept of distance every day. For example, you are told a home run, travels 400 ft (122 meters) or you run the metric mile (1.5 km) in track (or happily watch, others run a metric mile)., Displacement adds the concept of direction to distance. For example, you go, approximately 954 mi (1540 km) south when you travel from Seattle to Los Angeles;, the summit of Mount Everest is 29,035 ft (8849.9 meters) above sea level. (You may, have noticed we are using both metric and English units. We will do this only for the first, , Displacement, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 25
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part of this chapter, with the thought that this may prove helpful if you are familiarizing, yourself with the metric system.), Sometimes just distance matters. If you want to be a million miles away from your, younger brother, it does not matter whether that’s east, north, west or south. The, distance is called the magnitude í the amount í of the displacement., , Distance and direction, Measures net change in position, , Direction, however, can matter. If you walk 10 blocks north of your home, you are at a, different location than if you walk 10 blocks south. In physics, direction often matters., For example, to get a ball to the ground from the top of a tall building, you can simply, drop the ball. Throwing the ball back up requires a very strong arm. Both the direction, and distance of the ball’s movement matters., The definition of displacement is precise: the direction and length of the shortest path, from the initial to the final position of an object’s motion. As you may recall from your, mathematics courses, the shortest path between two points is a straight line. Physicists, use arrows to indicate the direction of displacement. In the illustrations to the right, the, arrow points in the direction of the mouse’s displacement., Physicists use the Greek letter ǻ (delta) to indicate a change or difference. A change in, position is displacement, and since x represents position, we write ǻx to indicate, displacement. You see this notation, and the equation for calculating displacement, to, the right. In the equation, xf represents the final position (the subscript f stands for final), and xi represents the initial position (the subscript i stands for initial)., Displacement is a vector. A vector is a quantity that must be stated in terms of its, direction and its magnitude. Magnitude means the size or amount. “Move five meters to, the right” is a description of a vector. Scalars, on the other hand, are quantities that are, stated solely in terms of magnitude, like “a dozen eggs.” There is no direction for a, quantity of eggs, just an amount., , ǻx = xf í xi, ǻx = displacement, xf = final position, xi = initial position, Units: meters (m), , In one dimension, a positive or negative sign is enough to specify a direction. As, mentioned, numbers to the right of the origin are positive, and those to the left are, negative. This means displacement to the right is positive, and to the left it is negative., For instance, you can see in Example 1 that the mouse’s car starts at the position +3.0, meters and moves to the left to the position í1.0 meters. (We measure the position at, the middle of the car.) Since it moves to the left 4.0 meters, its displacement is í4.0, meters., Displacement measures the distance solely between the beginning and end of motion., We can use dance to illustrate this point. Let’s say you are dancing and you take three, steps forward and two steps back. Although you moved a total of five steps, your, displacement after this maneuver is one step forward., It would be better to use signs to describe the dance directions, so we could describe, forward as “positive” and backwards as “negative.” Three steps forward and two steps, back yield a displacement of positive one step., Since displacement is in part a measure of distance, it is measured with units of length., Meters are the SI unit for displacement., , What is the mouse car’s, displacement?, ǻx = xf íxi, ǻx = í1.0 m í 3.0 m, ǻx = í4.0 m, , 2.3 - Velocity, , Velocity: Speed and direction., You are familiar with the concept of speed. It tells you how fast something is going:, 55 miles per hour (mi/h) is an example of speed. The speedometer in a car measures, speed but does not indicate direction., When you need to know both speed and direction, you use velocity. Velocity is a vector., It is the measure of how fastandin which direction the motion is occurring. It is, represented by v. In this section, we focus on average velocity, which is represented by, v with a bar over it, as shown in Equation 1., A police officer uses the concepts of both speed and velocity in her work. She might, issue a ticket to a motorist for driving 36 mi/h (58 km/h) in a school zone; in this case,, speed matters but direction is irrelevant. In another situation, she might be told that a, suspect is fleeing north on I-405 at 90 mi/h (149 km/h); now velocity is important, because it tells her both how fast and in what direction., , Velocity, Speed and direction, , To calculate an object’s average velocity, divide its displacement by the time it takes to move that displacement. This time is called the elapsed, time, and is represented by ǻt. The direction for velocity is the same as for the displacement., For instance, let’s say a car moves positive 50 mi (80 km) between the hours of 1 P.M. and 3 P.M. Its displacement is positive 50 mi, and, , 26, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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two hours elapse as it moves that distance. The car’s average velocity equals +50 miles, divided by two hours, or +25 mi/h (+40 km/h). Note that the direction is positive because, the displacement was positive. If the displacement were negative, then the velocity, would also be negative., At this point in the discussion, we are intentionally ignoring any variations in the car’s, velocity. Perhaps the car moves at constant speed, or maybe it moves faster at certain, times and then slower at others. All we can conclude from the information above is that, the car’s average velocity is +25 mi/h., Velocity has the dimensions of length divided by time; the units are meters per second, (m/s)., , ǻx = displacement, ǻt = elapsed time, Units: meters/second (m/s), , What is the mouse’s velocity?, , 2.4 - Average velocity, , Average velocity: Displacement divided by, elapsed time., Average velocity equals displacement divided by the time it takes for the displacement, to occur., For example, if it takes you two hours to move positive 100 miles (160 kilometers), your, average velocity is +50 mi/h (80 km/h). Perhaps you drive a car at a constant velocity., Perhaps you drive really fast, slow down for rush-hour traffic, drive fast again, get pulled, over for a ticket, and then drive at a moderate speed. In either case, because your, displacement is 100 mi and the elapsed time is two hours, your average velocity is, +50 mi/h., , Average velocity, Displacement divided by elapsed time, , Since the average velocity of an object is calculated from its displacement, you need to, be able to state its initial and final positions. In Example 1 on the right, you are shown, the positions of three towns and asked to calculate the average velocity of a trip. You must calculate the displacement from the initial to final, position to determine the average velocity., A classic physics problem tempts you to err in calculating average velocity. The problem runs like this: “A hiker walks one mile at two miles per, hour, and the next mile at four miles per hour. What is the hiker's average velocity?” If you average two and four and answer that the average, velocity is three mi/h, you will have erred. To answer the problem, you must first calculate the elapsed time. You cannot simply average the two, velocities. It takes the hiker 1/2 an hour to cover the first mile, but only 1/4 an hour to walk the second mile, for a total elapsed time of 3/4 of an, hour. The average velocity equals two miles divided by 3/4 of an hour, which is a little less than three miles per hour., , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 27
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ǻx = displacement, ǻt = elapsed time, , A plane flies Acme to Bend in 2.0, hrs, then straight back to Cote in, 1.0 hr. What is its average, velocity for the trip in km/h?, , 2.5 - Instantaneous velocity, , Instantaneous velocity: Velocity at a specific, moment., Objects can speed up or slow down, or they can change direction. In other words, their, velocity can change. For example, if you drop an egg off a 40-story building, the egg’s, velocity will change: It will move faster as it falls. Someone on the building’s 39th floor, would see it pass by with a different velocity than would someone on the 30th., When we use the word “instantaneous,” we describe an object’s velocity at a particular, instant. In Concept 1, you see a snapshot of a toy mouse car at an instant when it has a, velocity of positive six meters per second., , Instantaneous velocity, , The fable of the tortoise and the hare provides a classic example of instantaneous, Velocity, versus average velocity. As you may recall, the hare seemed faster because it could, achieve a greater instantaneous velocity than could the tortoise. But the hare’s long, naps meant that its average velocity was less than that of the tortoise, so the tortoise won the race., , at a specific moment, , When the average velocity of an object is measured over a very short elapsed time, the result is close to the instantaneous velocity. The, shorter the elapsed time, the closer the average and instantaneous velocities. Imagine the egg falling past the 39th floor window in the example, , 28, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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we mentioned earlier, and let’s say you wanted to determine its instantaneous velocity, at the midpoint of the window., You could use a stopwatch to time how long it takes the egg to travel from the top to the, bottom of the window. If you then divided the height of the window by the elapsed time,, the result would be close to the instantaneous velocity. However, if you measured the, time for the egg to fall from 10 centimeters above the window’s midpoint to 10, centimeters below, and used that displacement and elapsed time, the result would be, even closer to the instantaneous velocity at the window’s midpoint. As you repeated this, process "to the limit" í measuring shorter and shorter distances and elapsed times, (perhaps using motion sensors to provide precise values) í you would get values closer, and closer to the instantaneous velocity., , v = instantaneous velocity, ǻx = displacement, ǻt = elapsed time, , To describe instantaneous velocity mathematically, we use the terminology shown in, Equation 1. The arrow and the word “lim” mean the limit as ǻt approaches zero. The, limit is the value approached by the calculation as it is performed for smaller and smaller intervals of time., , To give you a sense of velocity and how it changes, let’s again use the example of the egg. We calculate the velocity at various times using an, equation you may have not yet encountered, so we will just tell you the results. Let’s assume each floor of the building is four meters (13 ft), high and that the egg is being dropped in a vacuum, so we do not have to worry about air resistance slowing it down., One second after being dropped, the egg will be traveling at 9.8 meters per second; at three seconds, it will be traveling at 29 m/s; at, five seconds, 49 m/s (or 32 ft/s, 96 ft/s and 160 ft/s, respectively.), After seven seconds, the egg has an instantaneous velocity of 0 m/s. Why? The egg hit the ground at about 5.7 seconds and therefore is not, moving. (We assume the egg does not rebound, which is a reasonable assumption with an egg.), Physicists usually mean “instantaneous velocity” when they say “velocity” because instantaneous velocity is often more useful than average, velocity. Typically, this is expressed in statements like “the velocity when the elapsed time equals three seconds.”, , 2.6 - Position-time graph and velocity, , Position-time graph, Shows position of object over time, Steeper graph = greater speed, , A graph of an object's position over time is a useful tool for analyzing motion. You see, such a position-time graph above. Values on the vertical axis represent the mouse car's, position, and time is plotted on the horizontal axis. You can see from the graph that the, mouse car started at position x = í4 m, then moved to the position x = +4 m at about, t = 4.5 s, stayed there for a couple of seconds, and then reached the position, x = í2 m again after a total of 12 seconds of motion., Where the graph is horizontal, as at point B, it indicates the mouse’s position is not, changing, which is to say the mouse is not moving. Where the graph is steep, position, is changing rapidly with respect to time and the mouse is moving quickly., Displacement and velocity are mathematically related, and a position-time graph can be, used to find the average or instantaneous velocity of an object. The slope of a straight, line between any two points of the graph is the object’s average velocity between them., Why is the average velocity the same as this slope? The slope of a line is calculated by, dividing the change in the vertical direction by the change in the horizontal direction,, “the rise over the run.” In a position-time graph, the vertical values are the x positions, and the horizontal values tell the time. The slope of the line is the change in position,, which is displacement, divided by the change in time, which is the elapsed time. This is, the definition of average velocity: displacement divided by elapsed time., , Average velocity, Slope of line between two points, , You see this relationship stated and illustrated in Equation 1. Since the slope of the line shown in this illustration is positive, the average, velocity between the two points on the line is positive. Since the mouse moves to the right between these points, its displacement is positive,, which confirms that its average velocity is positive as well., The slope of the tangent line for any point on a straight-line segment of a position-time graph is constant. When the slope is constant, the, velocity is constant. An example of constant velocity is the horizontal section of the graph that includes the point B in the illustration above., The slope of a tangent line at different points on a curve is not constant. The slope at a single point on a curve is determined by the slope of, the tangent line to the curve at that point. You see a tangent line illustrated in Equation 2. The slope as measured by the tangent line equals, the instantaneous velocity at the point. The slope of the tangent line in Equation 2 is negative, so the velocity there is negative. At that point,, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 29
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the mouse is moving from right to left. The negative displacement over a short time, interval confirms that its velocity is negative., , Instantaneous velocity, Slope of tangent line at point, , What is the average velocity, between points A and B?, , Consider the points A, B and C., Where is the instantaneous, velocity zero? Where is it, positive? Where is it negative?, Zero at B, Positive at A, Negative at C, , 30, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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2.7 - Interactive problem: draw a position-time graph, In this section, you are challenged to match a pre-drawn position-time graph by, moving a ball along a number line. As you drag the ball, its position at each instant, will be graphed. Your challenge is to get as close as you can to the target graph., When you open the interactive simulation on the right, you will see a graph and a, coordinate system with x positions on the vertical axis and time on the horizontal, axis. Below the graph is a ball on a number line. Examine the graph and decide how, you will move the ball over the 10 seconds to best match the target graph. You may, find it helpful to think about the velocity described by the target graph. Where is it, increasing? decreasing? zero? If you are not sure, review the section on positiontime graphs and velocity., You can choose to display a graph of the velocity of the motion of the ball as, described by the target graph by clicking a checkbox. We encourage you to think, first about what the velocity will be and use this checkbox to confirm your, hypothesis., Create your graph by dragging the ball and watching the graph of its motion. You, can press RESET and try again as often as you like., , 2.8 - Acceleration, , Acceleration: Change in, velocity., When an object’s velocity changes, it accelerates., Acceleration measures the rate at which an object, speeds up, slows down or changes direction. Any of, these variations constitutes a change in velocity. The, letter a represents acceleration., Acceleration is a popular topic in sports car, commercials. In the commercials, acceleration is, A racing car accelerates., often expressed as how fast a car can go from zero to, 60 miles per hour (97 km/h, or 27 m/s). For example, a current model Corvette®, automobile can reach 60 mi/h in 4.9 seconds. There are even hotter cars than this in, production., To calculate average acceleration, divide the change in instantaneous velocity by the, elapsed time, as shown in Equation 1. To calculate the acceleration of the Corvette,, divide its change in velocity, from 0 to 27 m/s, by the elapsed time of 4.9 seconds. The, car accelerates at an average rate of 5.5 m/s per second. We typically express this as, 5.5 meters per second squared, or 5.5 m/s2. (This equals 18 ft/s2, and with this, observation we will cease stating values in both measurement systems, in order to, simplify the expression of numbers.) Acceleration is measured in units of length divided, by time squared. Meters per second squared (m/s2) express acceleration in SI units., Let’s assume the car accelerates at a constant rate; this means that each second the, Corvette moves 5.5 m/s faster. At one second, it is moving at 5.5 m/s; at two seconds,, 11 m/s; at three seconds, 16.5 m/s; and so forth. The car’s velocity increases by 5.5 m/s, every second., , Acceleration, Change in velocity, , Since acceleration measures the change in velocity, an object can accelerate even, while it is moving at a constant speed. For instance, consider a car moving around a, curve. Even if the car’s speed remains constant, it accelerates because the change in, the car’s direction means its velocity (speed plus direction) is changing., Acceleration can be positive or negative. If the Corvette uses its brakes to go from +60, to 0 mi/h in 4.9 seconds, its velocity is decreasing just as fast as it was increasing, before. This is an example of negative acceleration., You may want to think of negative acceleration as “slowing down,” but be careful! Let’s, say a train has an initial velocity of negative 25 m/s and that changes to negative, 50 m/s. The train is moving at a faster rate (speeding up) but it has negative, acceleration. To be precise, its negative acceleration causes an increasingly negative, velocity., Velocity and acceleration are related but distinct values for an object. For example, an, object can have positive velocity and negative acceleration. In this case, it is slowing, down. An object can have zero velocity, yet be accelerating. For example, when a ball, bounces off the ground, it experiences a moment of zero velocity as its velocity changes, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 31
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from negative to positive, yet it is accelerating at this moment since its velocity is, changing., , ǻv = change in instantaneous velocity, ǻt = elapsed time, Units: meters per second squared, (m/s2), , What is the average acceleration, of the mouse between 2.5 and, 4.5 seconds?, , 2.9 - Average acceleration, , Average acceleration: The change in, instantaneous velocity divided by the elapsed, time., Average acceleration is the change in instantaneous velocity over a period of elapsed, time. Its definition is shown in Equation 1 to the right., We will illustrate average acceleration with an example. Let's say you are initially driving, a car at 12 meters per second and 8 seconds later you are moving at 16 m/s. The, change in velocity is 4 m/s during that time; the elapsed time is eight seconds. Dividing, the change in velocity by the elapsed time determines that the car accelerates at an, average rate of 0.5 m/s2., Perhaps the car's acceleration was greater during the first four seconds and less during, the last four seconds, or perhaps it was constant the entire eight seconds. Whatever the, case, the average acceleration is the same, since it is defined using the initial and final, instantaneous velocities., , 32, , Average acceleration, Change in instantaneous velocity, divided by elapsed time, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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ǻv = change in instantaneous velocity, ǻt = elapsed time, , What is the mouse’s average, acceleration?, , 2.10 - Instantaneous acceleration, , Instantaneous acceleration: Acceleration at a, particular moment., You have learned that velocity can be either average or instantaneous. Similarly, you, can determine the average acceleration or the instantaneous acceleration of an object., We use the mouse in Concept 1 on the right to show the distinction between the two., The mouse moves toward the trap and then wisely turns around to retreat in a hurry., The illustration shows the mouse as it moves toward and then hurries away from the, trap. It starts from a rest position and moves to the right with increasingly positive, velocity, which means it has a positive acceleration for an interval of time. Then it slows, to a stop when it sees the trap, and its positive velocity decreases to zero (this is, negative acceleration). It then moves back to the left with increasingly negative velocity, (negative acceleration again). If you would like to see this action occur again in the, Concept 1 graphic, press the refresh button in your browser., , Instantaneous acceleration, Acceleration at a particular moment, , We could calculate an average acceleration, but describing the mouse's motion with, instantaneous acceleration is a more informative description of that motion. At some, instants in time, it has positive acceleration and at other instants, negative acceleration., By knowing its acceleration and its velocity at an instant in time, we can determine, whether it is moving toward the trap with increasingly positive velocity, slowing its rate, of approach, or moving away with increasingly negative velocity., Instantaneous acceleration is defined as the change in velocity divided by the elapsed, time as the elapsed time approaches zero. This concept is stated mathematically in, Equation 1 on the right., , a = instantaneous acceleration, ǻv = change in velocity, , Earlier, we discussed how the slope of the tangent at any point on a position-time graph, ǻt = elapsed time (approaches 0), equals the instantaneous velocity at that point. We can apply similar reasoning here to, conclude that the instantaneous acceleration at any point on a velocity-time graph, equals the slope of the tangent, as shown in Equation 2. Why? Because slope equals the rate of change, and acceleration is the rate of change, of velocity., In Example 1, we show a graph of the velocity of the mouse as it approaches the trap and then flees. You are asked to determine the sign of, the instantaneous acceleration at four points; you can do so by considering the slope of the tangent to the velocity graph at each point., , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 33
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Instantaneous acceleration, Slope of line tangent to point on, velocity-time graph, , The graph shows the mouse's, velocity versus time. Describe, the instantaneous acceleration at, A, B, C and D as positive,, negative or zero., a positive at A, a negative at B, a zero at C, a negative at D, 2.11 - Sample problem: velocity and acceleration, The mouse car goes 10.3 meters in, 4.15 seconds at a constant velocity,, then accelerates at 1.22 m/s2 for 5.34, more seconds., What is its final velocity?, , Solving this problem requires two calculations. The mouse car's velocity during the first part of its journey must be calculated. Using that value, as the initial velocity of the second part of the journey, and the rate of acceleration during that part, you can calculate the final velocity., Draw a diagram, , 34, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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Variables, Part 1: Constant velocity, displacement, , ǻx = 10.3 m, , elapsed time, , ǻt = 4.15 s, , velocity, , v, , Part 2: Constant acceleration, initial velocity, , vi = v (calculated above), , acceleration, , a = 1.22 m/s2, , elapsed time, , ǻt = 5.34 s, , final velocity, , vf, , What is the strategy?, 1., , Use the definition of velocity to find the velocity of the mouse car before it accelerates. The velocity is constant during the first part of the, journey., , 2., , Use the definition of acceleration and solve for the final velocity., , Physics principles and equations, The definitions of velocity and acceleration will prove useful. The velocity and acceleration are constant in this problem. In this and later, problems, we use the definitions for average velocity and acceleration without the bars over the variables., , v = ǻx/ǻt, a = ǻv/ǻt = (vf í vi)/ǻt, Step-by-step solution, We start by finding the velocity before the engine fires., , Step, , Reason, , 1., , v = ǻx/ǻt, , 2., , v = (10.3 m)/(4.15 s) enter values, , 3., , v = 2.48 m/s, , definition of velocity, , divide, , Next we find the final velocity using the definition of acceleration. The initial velocity is the same as the velocity we just calculated., , Step, , 4., , a = (vf – vi)/ǻt, , Reason, , definition of acceleration, enter given values, and velocity from step 3, , 5., 6., , 6.51 m/s = vf – 2.48 m/s, , multiply by 5.34 s, , 7., , vf = 8.99 m/s, , solve for vf, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 35
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2.12 - Interactive checkpoint: subway train, A subway train accelerates along a, straight track at a constant 1.90 m/s2., How long does it take the train to, increase its speed from 4.47 m/s to, 13.4 m/s?, , Answer:, , ǻt =, , s, , 2.13 - Interactive problem: what’s wrong with the rabbits?, You just bought five rabbits. They were supposed to be constant acceleration, rabbits, but you worry that some are the less expensive, non-constant acceleration, rabbits. In fact, you think two might be the cheaper critters., You take them home. When you press GO, they will run or jump for five seconds, (well, one just sits still) and then the simulation stops. You can press GO as many, times as you like and use the PAUSE button as well., Your mission: Determine if you were ripped off, and drag the “½ off” sale tags to the, cheaper rabbits. The simulation will let you know if you are correct. You may decide, to keep the cuddly creatures, but you want to be fairly charged., Each rabbit has a velocity gauge that you can use to monitor its motion in the, simulation. The simplest way to solve this problem is to consider the rabbits one at a, time: look at a rabbit’s velocity gauge and determine if the velocity is changing at a, constant rate. No detailed mathematical calculations are required to solve this, problem., If you find this simulation challenging, focus on the relationship between, acceleration and velocity. With a constant rate of acceleration, the velocity must change at a constant rate: no jumps or sudden changes. Hint:, No change in velocity is zero acceleration, a constant rate., , 2.14 - Derivation: creating new equations, Other sections in this chapter introduced some of the fundamental equations of motion., These equations defined fundamental concepts; for example, average velocity equals, the change in position divided by elapsed time., Several other helpful equations can be derived from these basic equations. These, equations enable you to predict an object’s motion without knowing all the details. In, this section, we derive the formula shown in Equation 1, which is used to calculate an, object’s final velocity when its initial velocity, acceleration and displacement are known,, but not the elapsed time. If the elapsed time were known, then the final velocity could, be calculated using the definition of velocity, but it is not., , Deriving a motion equation, vf2 = vi2 + 2aǻx, vi = initial velocity, vf = final velocity, , This equation is valid when the acceleration is constant, an assumption that is used in, many problems you will be posed., , a = constant acceleration, , Variables, , ǻx = displacement, , We use t instead of ǻt to indicate the elapsed time. This is simpler notation, and we will, use it often., , 36, , acceleration (constant), , a, , initial velocity, , vi, , final velocity, , vf, , elapsed time, , t, , displacement, , ǻx, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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Strategy, First, we will discuss our strategy for this derivation. That is, we will describe our overall plan of attack. These strategy points outline the major, steps of the derivation., 1., , We start with the definition of acceleration and rearrange it. It includes the terms for initial and final velocity, as well as elapsed time., , 2., , We derive another equation involving time that can be used to eliminate the time variable from the acceleration equation. The condition, of constant acceleration will be crucial here., , 3., , We eliminate the time variable from the acceleration equation and simplify. This results in an equation that depends on other variables,, but not time., , Physics principles and equations, Since the acceleration is constant, the velocity increases at a constant rate. This means the average velocity is the sum of the initial and final, velocities divided by two., , We will use the definition of acceleration,, , a = (vf í vi)/t, We will also use the definition of average velocity,, , Step-by-step derivation, We start the derivation with the definition of average acceleration, solve it for the final velocity and do some algebra. This creates an equation, with the square of the final velocity on the left side, where it appears in the equation we want to derive., , Step, , Reason, , 1., , a = (vf – vi)/t, , definition of average acceleration, , 2., , vf = vi + at, , solve for final velocity, , 3., , vf2 = (vi + at)2, , square both sides, , 4., , vf2 = vi2 + 2viat + a2t2, , expand right side, , 5., , vf2 = vi2 + at(2vi + at), , factor out at, , 6., , vf2 = vi2 + at(vi + vi + at) rewrite 2vi as a sum, , 7., , vf2 = vi2 + at(vi + vf), , substitution from equation 2, , The equation we just found is the basic equation from which we will derive the desired motion equation. But it still involves the time variable t í, multiplied by a sum of velocities. In the next stage of the derivation, we use two different ways of expressing the average velocity to develop a, second equation involving time multiplied by velocities. We will subsequently use that second equation to eliminate time from the equation, above., , Step, , Reason, , 8., , average velocity is average of initial and final velocities, , 9., , definition of average velocity, , 10., , set right sides of 8 and 9 equal, , 11. t(vi + vf) = 2ǻx rearrange equation, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 37
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We have now developed two equations that involve time multiplied by a sum of velocities. The left side of the equation in step 11 matches an, expression appearing in equation 7, at the end of the first stage. By substituting from this equation into equation 7, we eliminate the time, variable t and derive the desired equation., , Step, , Reason, 2, , 12. vf =, , vi2, , + a(2ǻx) substitute right side of 11 into 7, , 13. vf2 = vi2 + 2aǻx, , rearrange factors, , We have now accomplished our goal. We can calculate the final velocity of an object when we know its initial velocity, its acceleration and its, displacement, but do not know the elapsed time. The derivation is finished., , 2.15 - Motion equations for constant acceleration, The equations above can be derived from the, fundamental definitions of motion (equations such as, a = ǻv/ǻt). To understand the equations, you need, to remember the notation: ǻx for displacement, v for, velocity and a for acceleration. The subscripts i and f, represent initial and final values. We follow a common, convention here by using t for elapsed time instead of, ǻt. We show the equations above and below on the, right., Note that to hold true these equations all require a, constant rate of acceleration. Analyzing motion with a, ǻx = displacement, v = velocity, a = acceleration, t = elapsed time, varying rate of acceleration is a more challenging, task. When we refer to acceleration in problems, we, mean a constant rate of acceleration unless we explicitly state otherwise., To solve problems using motion equations like these, you look for an equation that, includes the values you know, and the one you are solving for. This means you can, solve for the unknown variable., In the example problem to the right, you are asked to determine the acceleration, required to stop a car that is moving at 12 meters per second in a distance of 36, meters. In this problem, you know the initial velocity, the final velocity (stopped =, 0.0 m/s) and the displacement. You do not know the elapsed time. The third motion, equation includes the two velocities, the acceleration, and the displacement, but does, not include the time. Since this equation includes only one value you do not know, it is, the appropriate equation to choose., , Applying motion equations, Determine the “knowns” and the, “unknown(s)”, ·Find other knowns from situation, Choose an equation with those, variables, , Motion equations, , 38, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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What acceleration will stop the, car exactly at the stop sign?, vf2 = vi2 + 2aǻx, a = (vf2 ívi2)/2ǻx, , a = í144/72 m/s2, a = í2.0 m/s2, 2.16 - Sample problem: a sprinter, What is the runner's velocity at the, end of a 100-meter dash?, , You are asked to calculate the final velocity of a sprinter running a 100-meter dash. List the variables that you know and the one you are asked, for, and then consider which equation you might use to solve the problem. You want an equation with just one unknown variable, which in this, problem is the final velocity., The sprinter’s initial velocity is not explicitly stated, but he starts motionless, so it is zero m/s., Draw a diagram, , Variables, displacement, , ǻx = 100 m, , acceleration, , a = 0.528 m/s2, , initial velocity, , vi = 0.00 m/s, , final velocity, , vf, , What is the strategy?, 1., , Choose an appropriate equation based on the values you know and the one you want to find., , 2., , Enter the known values and solve for the final velocity., , Physics principles and equations, Based on the known and unknown values, the equation below is appropriate. We know all the variables in the equation except the one we are, asked to find, so we can solve for it., , vf2 = vi2 + 2aǻx, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 39
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Step-by-step solution, , Step, , 1., , Reason, 2, , vf =, , vi2, , + 2aǻx, , motion equation, enter known values, , 2., 3., , vf2 = 106 m2/s2, , multiplication and addition, , 4., , vf = 10.3 m/s, , take square root, , In step 4, we take the square root of 106 to find the final velocity. We chose the positive square root, since the runner is moving in the positive, direction. When there are multiple roots, you look at the problem to determine the solution that makes sense given the circumstances. If the, runner were running to the left, then a negative velocity would be the appropriate choice., , 2.17 - Interactive checkpoint: passenger jet, A passenger jet lands on a runway, with a velocity of 71.5 m/s. Once it, touches down, it accelerates at a, constant rate of í3.17 m/s2., How far does the plane travel down, the runway before its velocity is, decreased to 2.00 m/s, its taxi speed, to the landing gate?, , Answer:, , ǻx =, , m, , 2.18 - Free-fall acceleration, , Free-fall acceleration: Rate of acceleration due, to the force of Earth's gravity., Galileo Galilei is reputed to have conducted an interesting experiment several hundred, years ago. According to legend, he dropped two balls with different masses off the, Leaning Tower of Pisa and found that both landed at the same time. Their differing, masses did not change the time it took them to fall. (We say he was “reputed to have”, because there is little evidence that he in fact conducted this experiment. He was more, of a “roll balls down a plane” experimenter.), Today this experiment is used to demonstrate that free-fall acceleration is constant: that, the acceleration of a falling object due solely to the force of gravity is constant,, regardless of the object’s mass or density. The two balls landed at the same time, because they started with the same initial velocity, traveled the same distance and, accelerated at the same rate. In 1971, the commander of Apollo 15 conducted a version, of the experiment on the Moon, and demonstrated that in the absence of air resistance,, a hammer and a feather accelerated at the same rate and reached the surface at the, same moment., , Free-fall acceleration, Acceleration due to gravity, , In Concept 1, you see a photograph that illustrates free-fall acceleration. Pictures of a, freely falling egg were taken every 2/15 of a second. Since the egg’s speed constantly, increases, the distance between the images increases over time. Greater displacement, over the same interval of time means its velocity is increasing in magnitude; it is, accelerating., Free-fall acceleration is the acceleration caused by the force of the Earth’s gravity,, ignoring other factors like air resistance. It is sometimes stated as the rate of, acceleration in a vacuum, where there is no air resistance. Near the Earth’s surface, its, magnitude is 9.80 meters per second squared. The letter g represents this value. The, value of g varies slightly based on location. It is less at the Earth's poles than at the, equator, and is also less atop a tall mountain than at sea level., , 40, , Galileo's famous experiment, Confirmed by Apollo 15 on the Moon, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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The acceleration of 9.80 m/s2 occurs in a vacuum. In the Earth’s atmosphere, a feather, and a small lead ball dropped from the same height will not land at the same time, because the feather, with its greater surface area, experiences more air resistance., Since it has less mass than the ball, gravity exerts less force on it to overcome the, larger air resistance. The acceleration will also be different with two objects of the same, mass but different surface areas: A flat sheet of paper will take longer to reach the, ground than the same sheet crumpled up into a ball., By convention, “up” is positive, and “down” is negative, like the values on the y axis of a, graph. This means when using g in problems, we state free-fall acceleration as, negative 9.80 m/s2. To make this distinction, we typically use a or ay when we are, using the negative sign to indicate the direction of free-fall acceleration., Free-fall acceleration occurs regardless of the direction in which an object is moving., For example, if you throw a ball straight up in the air, it will slow down, accelerating at, í9.80 m/s2 until it reaches zero velocity. At that point, it will then begin to fall back, toward the ground and continue to accelerate toward the ground at the same rate. This, means its velocity will become increasingly negative as it moves back toward the, ground., , Free-fall acceleration on Earth, g = 9.80 m/s2, g = magnitude of free-fall acceleration, , The two example problems in this section stress these points. For instance, Example 2, on the right asks you to calculate how long it will take a ball thrown up into the air to, reach its zero velocity point (the peak of its motion) and its acceleration at that point., , What is the egg's velocity after, falling from rest for 0.10, seconds?, vf = vi + at, vf = (0 m/s) + (í9.80 m/s2)(0.10 s), vf = í0.98 m/s, , How long will it take the ball to, reach its peak? What is its, acceleration at that point?, vf = vi + at, t = (vf ívi)/a, t = (0 m/s í 4.9 m/s)/(í9.80 m/s2), t = 4.9/9.80 s = 0.50 s, acceleration = í9.80 m/s2, , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 41
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2.19 - Interactive checkpoint: penny drop, You drop a penny off Taiwan’s Taipei, 101 tower, which is 509 meters tall., How long does it take to hit the, ground? Ignore air resistance,, consider down as negative, and the, ground as having zero height., , Answer:, , t=, , s, , 2.20 - Gotchas, Some errors you might make, or that tests or teachers might try to tempt you to make:, Switching the order in calculating displacement. Remember: It is the final position minus the initial position. If you start at a position of three, meters and move to one meter, your displacement is negative two meters. Be sure to subtract three from one, not vice versa., Confusing distance traveled with displacement. Displacement is the shortest path between the beginning point and final point. It does not, matter how the object got there, whether in a straight line or wandering all over through a considerable net distance., Forgetting the sign. Remember: displacement, velocity and acceleration all have direction. For one-dimensional motion, they require signs, indicating the direction. If a problem says that an object moves to the left or down, its displacement is typically negative. Be sure to note the, signs of displacement, velocity or acceleration if they are given to you in a problem. Make sure you are consistent with signs. If up is positive,, then upward displacement is positive, and the acceleration due to gravity is negative., Confusing velocity and acceleration. Can an object with zero acceleration have velocity? Yes! A train barreling down the tracks at 150 km/h has, velocity. If that velocity is not changing, the train’s acceleration is zero., Can an object with zero velocity have acceleration? Yes again: a ball thrown straight up has zero velocity at the top of its path, but its, acceleration at that instant is í9.80 m/s2., Confusing constant acceleration with constant velocity. If an object has constant acceleration, it has a constant velocity, right? Quite wrong, (unless the constant acceleration is zero). With a constant acceleration other than zero, the velocity is constantly changing., Misunderstanding negative acceleration. Can something that is “speeding up” also have a negative acceleration? Yes. If something is moving, in the negative direction and moving increasingly quickly, it will have a negative velocity and a negative acceleration. When an object has, negative velocity and experiences negative acceleration, it will have increasing speed. In other words, negative acceleration is not just “slowing, something down.” It can also mean an object with negative velocity moving increasingly fast., , 42, , Copyright 2000-2007 Kinetic Books Co. Chapter 02
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2.21 - Summary, Position is the location of an object relative to a reference point called the origin,, and is specified by the use of a coordinate system., Displacement is a measure of the change in the position of an object. It includes, both the distance between the object’s starting and ending points, and the direction, from the starting point to the ending point. An example of displacement would be, “three meters west” or “negative two meters”., Similarly, velocity expresses an object’s speed and direction, as in “three meters, per second west.” Velocity has a direction. In one dimension, motion in one direction, is represented by positive numbers, and motion in the other direction is negative., , vf = vi + at, , An object’s velocity may change while it is moving. Its average velocity is its, displacement divided by the elapsed time. In contrast, its instantaneous velocity is, its velocity at a particular moment. This equals the displacement divided by the, elapsed time for a very small interval of time, as the time interval gets smaller and, smaller., , ǻx = vit + ½ at2, , Acceleration is a change in velocity. Like velocity, it has a direction and in one, dimension, it can be positive or negative. Average acceleration is the change in, velocity divided by the elapsed time, and instantaneous acceleration is the, acceleration of an object at a specific moment., , vf2 = vi2 +2aǻx, ǻx = ½ (vi + vf)t, , There are four very useful motion equations for situations where the acceleration is constant. They are the last four equations shown on the, right., Free-fall acceleration, represented by g, is the magnitude of the acceleration due to the force of Earth’s gravity. Near the surface of the Earth,, falling objects have a downward acceleration due to gravity of 9.80 m/s2., , Copyright 2000-2007 Kinetic Books Co. Chapter 02, , 43
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Chapter 2 Problems, , Conceptual Problems, C.1, , A toddler has become lost in the forest and her father is trying to retrieve her. He is currently located to the north of a large, tree and he hears her shouts coming from the south. Do we know from this information whether the toddler is north or south, of the tree?, Yes, , No, , C.2, , Assume Waterville is precisely 100 miles due east of Seattle. (a) With Seattle at the origin, draw a number line that shows, only the east-west positions of both cities, with east as positive. (b) Draw a different number line that shows the east-west, positions of the cities with Waterville at the origin, and east still positive., , C.3, , Julie is a citizen of Country A and she is describing the nearby geography: "Our capital is at the origin. Our famous beaches, are 200 km directly west of our capital. Country B's capital is 200 km directly east of our capital. Country B's largest mountain, is 700 km to the east of Country A's capital." (a) Draw the four geographical features on a number line from Julie's, perspective. (b) A citizen of Country B begs to differ. He believes that his capital is the origin. Draw the features on a number, line from his perspective., , C.4, , A long straight highway has markers along the side that represent the distance north (in miles) from the start of the highway., In terms of displacement, is there any difference in moving (a) from the "3" marker to the "7" marker, or from "4" to "8"? (b), From "1" to "3", or from "3" to "1"?, , C.5, , C.6, , (a), , Yes, , No, , (b), , Yes, , No, , A diver is standing on a diving board that is 15 meters above the surface of the water. He jumps 0.50 meters straight up into, the air, dives into the water and goes 3.0 meters underwater before returning to the surface. (a) Assume that "up" is the, positive direction. What is the vertical displacement of the diver from when he is standing on the diving board to when he, emerges from underwater? (b) Now assume that "down" is the positive direction. Determine the displacement, as before., (a), , m, , (b), , m, , Can a car with negative velocity move faster than a car with positive velocity? Explain., Yes, , C.7, , Two people drive their cars from Piscataway, New Jersey, to Perkasie, Pennsylvania, about 35 miles east, leaving at the, same time and using the same route. The first driver travels at a constant speed the whole trip. The second driver stops for a, while for doughnuts and coffee in Lambertville. They arrive in Perkasie at the same time. Do the cars have the same average, velocity? Explain., Yes, , C.8, , No, , No, , Elaine wants to return a video she rented at the video store, which is 5.0 kilometers away in the positive direction. It takes her, 10 minutes to drive to the store, 1.0 minute to deposit the videotape, and 9.0 more minutes to drive home. What is her, average velocity for the entire trip?, m/s, , C.9, , An airplane starts out at the Tokyo Narita Airport (x = 0) destined for Bangkok, Thailand (x = 4603 km). After departing from, the gate, the plane has to wait ten minutes on the runway before taking off. When the plane is halfway to Bangkok, and, cruising at its top speed, which of these is greater: instantaneous velocity, or average velocity since leaving the gate?, i. Instantaneous velocity, ii. Average velocity, iii. They're the same, , C.10 The position versus time graph for a man, trapped on an island is shown. Is he traveling, at a constant velocity? Explain., Yes, , 44, , No, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems
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C.11 A truck full of corn is parked at x = 0 and is, pointed in the negative direction. If the driver, puts it into reverse and holds down on the, accelerator, will the following quantities be, positive, zero, or negative after one second?, (a) position, (b) velocity, (c) acceleration., (a), , (b), , (c), , i., ii., iii., i., ii., iii., i., ii., iii., , Positive, Negative, Zero, Positive, Negative, Zero, Positive, Negative, Zero, , C.12 David is driving a minivan to work, and he is stopped at a red light. The light turns green and David drives to the next red, light, where he stops again. Is David's average acceleration from light to light positive, negative, or zero?, i. Positive, ii. Negative, iii. Zero, C.13 If you know only the initial position, the final position, and the constant acceleration of an object, can you calculate the final, velocity? Explain., Yes, , No, , C.14 A football is thrown vertically up in the air at 10 m/s. If it is later caught at the same spot it was thrown from, will the speed be, greater than, less than, or the same as when it was thrown? Ignore air resistance., i. Greater than, ii. Less than, iii. Same, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following question. What kind of acceleration will, cause the hare to change directions?, i. Positive, ii. Negative, iii. Positive or negative, , Section 1 - Position, 1.1, , Anita and Nick are playing tug-of-war near a mud puddle. They are each holding on to an end of a taut rope that has a knot, exactly in the middle. Anita's position is 6.2 meters east of the center of the puddle and Nick's position is 3.0 meters west of, the center of the puddle. What is the location of the knot relative to the center of the puddle? Treat east as positive and west, as negative., meters, , Section 2 - Displacement, 2.1, , A photographer wants to take a picture of a particularly interesting flower, but he is not sure how far away to place his, camera. He takes three steps forward, four back, seven forward, then five back. Finally, he takes the photo. As measured in, steps, what was his displacement? Assume the forward direction is positive., step(s), , Copyright 2007 Kinetic Books Co. Chapter 2 Problems, , 45
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2.2, , The school bus picks up Brian in front of his house and takes him on a straight-line 2.1 km bus ride to school in the positive, direction. He walks home after school. If the front of Brian's house is the origin, (a) what is the position of the school, (b) what, is his displacement on the walk home, and (c) what is his displacement due to the combination of the bus journey and his, walk home?, (a), , 2.3, , 2.4, , km, , (b), , km, , (c), , km, , A strange number line is measured in meters to the left of the origin, and in kilometers to the right of the origin. An object, moves from í2200 m to 3.1 km. Find its displacement (a) in kilometers, and (b) in meters., (a), , km, , (b), , m, , The Psychic Squishy Sounds band is on tour in Texas and now are resting in Houston. They traveled 60.0 miles due south, from their first concert to Huntsville. If Huntsville is 80.0 miles due north of Houston, what is their displacement from the first, concert to Houston? Use the convention that north is positive, and south is negative., miles, , Section 3 - Velocity, 3.1, , To estimate the distance you are from a lightning strike, you can count the number of seconds between seeing the flash and, hearing the associated thunderclap. For this purpose, you can consider the speed of light to be infinite (it arrives instantly)., Sound travels at about 343 m/s in air at typical surface conditions. How many kilometers away is a lightning strike for every, second you count between the flash and the thunder?, km, , 3.2, , A jogger is moving at a constant velocity of +3.0 m/s directly towards a traffic light that is 100 meters away. If the traffic light is, at the origin, x = 0 m, what is her position after running 20 seconds?, m, , 3.3, , A slug has just started to move straight across a busy street in Littletown that is 8.0 meters wide, at a constant speed of 3.3, millimeters per second. The concerned drivers on the street halt until the slug has reached the opposite side. How many, seconds elapse until the traffic can start moving again?, s, , 3.4, , Light travels at a constant speed of 3.0×108 m/s in a vacuum. (a) It takes light about 1.3 seconds to travel from the Earth to, the Moon. Estimate the distance of the Moon from the Earth's surface, in meters. (b) The astronomical unit (abbreviated AU), is equal to the distance between the Earth and the Sun. One AU is about 1.5×1011 m. If the Sun suddenly ceased to emit, light, how many minutes would elapse until the Earth went dark?, (a), , m, , (b), , min, , Section 4 - Average velocity, 4.1, , 4.2, , In 1271, Marco Polo departed Venice and traveled to Kublai Khan's court near Beijing, approximately 7900 km away in a, direction we will call positive. Assume that the Earth is flat (as some did at the time) and that the trip took him 4.0 years, with, 365 days in a year. (a) What was his average velocity for the trip, in meters per second? (b) A 767 could make the same trip, in about 9.0 hours. What is the average velocity of the plane in meters per second?, (a), , m/s, , (b), , m/s, , An airport shuttle driver is assigned to drive back and forth between a parking lot (located at 0.0 km on a number line) and the, airport main terminal (located at +4.0 km). The driver starts out at the terminal at noon, arrives at the lot at 12:15 P.M., returns, to the terminal at 12:30 P.M., and arrives back at the lot at 12:45 P.M. What is the average velocity of the shuttle between (a), noon and 12:15, (b) noon and 12:30, and (c) noon and 12:45? Express all answers in meters per second., (a), , 46, , m/s, , (b), , m/s, , (c), , m/s, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems
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4.3, , 4.4, , You are driving in one direction on a long straight road. You drive in the positive direction at 126 km/h for 30.0 minutes, at, which time you see a police car with someone pulled over, presumably for speeding. You then drive in the same direction at, 100 km/h for 45.0 minutes. (a) How far did you drive? (b) What was your average velocity in kilometers per hour?, (a), , km, , (b), , km/h, , You made a journey, and your displacement was +95.0 km. Your initial velocity was +167 km/h and your final velocity was, í26.0 km/h. The journey took 43.0 minutes. What was your average velocity in kilometers per hour?, km/h, , 4.5, , The first controlled, sustained flight in a heavier-than-air craft was made by Orville Wright on December 17, 1903. The plane, took off at the end of a rail that was 60 feet long, and landed 12 seconds later, 180 feet away from the beginning of the rail., Assume the rail was essentially at the same height as the ground. (a) Calculate the average velocity of the plane in feet per, second while it was in the air. (b) What is the average velocity in kilometers per hour?, (a), , ft/s, , (b), , km/h, , 4.6, , A horse is capable of moving at four different speeds: walk (1.9 m/s), trot (5.0 m/s), canter (7.0 m/s), and gallop (12 m/s). Ann, is learning how to ride a horse. She spends 15 minutes riding at a walk and 2.4 minutes at each other speed. If she traveled, the whole way in the positive direction, what was her average velocity over the trip?, , 4.7, , A vehicle is speeding at 115 km/h on a straight highway when a police car moving at 145 km/h enters the highway from an, onramp and starts chasing it. The speeder is 175 m ahead of the police car when the chase starts, and both cars maintain, their speeds. How much time, in seconds, elapses until the police car overtakes the speeder?, , m/s, , s, , Section 5 - Instantaneous velocity, 5.1, , 5.2, , The tortoise and the hare start a race from the same starting line, at the same time. The tortoise moves at a constant, 0.200 m/s, and the hare at 5.00 m/s. (a) How far ahead is the hare after five minutes? (b) How long can the hare then snooze, until the tortoise catches up?, (a), , m, , (b), , s, , You own a yacht which is 14.5 meters long. It is motoring down a canal at 10.6 m/s. Its bow (the front of the boat) is just, about to begin passing underneath a bridge that is 30.0 m across. How much time is required until its stern (the end of the, boat) is no longer under the bridge?, s, , 5.3, , The velocity versus time graph of a unicycle is, shown. What is the instantaneous velocity of, the unicycle at (a) t = 1.0 s, (b) t = 3.0 s, and, (c) t = 5.0 s?, (a), , 5.4, , m/s, , (b), , m/s, , (c), , m/s, , Two boats are initially separated by distance d and head directly toward one another. The skippers of the boats want to arrive, at the same time at the point that is halfway between their starting points. Boat 1 moves at a speed v and boat 2 moves at, twice the speed of boat 1. Because it moves faster, boat 2 starts at time t later than boat 1. The skippers want to know how, much later boat 2 should start than boat 1. Provide them with an equation for t in terms of d and v., t = d/4v, t = 3d/2v, t = 3d/4v, t = d/2v, , 5.5, , This problem requires you to apply some trigonometry. A friend of yours is 51.0 m directly to your left. You and she start, running at the same time, and both run in straight lines at constant speeds. You run directly forward at 5.00 m/s for 125 m., She runs to the same final point as you, and wants to arrive at the same moment you do. How fast must she run?, m/s, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems, , 47
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Section 6 - Position-time graph and velocity, 6.1, , A fish swims north at 0.25 m/s for 3.0 seconds, stops for 2.0 seconds, and then swims south at 0.50 m/s for 4.0 seconds., Draw a position-time graph of the fish's motion, using north as the positive direction., , 6.2, , A renegade watermelon starts at x = 0 m and rolls at a constant velocity to x = í3.0 m at time t = 5.0 seconds. Then, it bumps, into a wall and stops. Draw its position versus time graph from t = 0 s to t = 10 s., , Section 7 - Interactive problem: draw a position-time graph, 7.1, , Use the information given in the interactive problem in this section to answer the following questions. Assume that the, positive x direction is to the right. (a) Is the ball moving to the left or right from 0 s to 3.0 s? (b) Is the velocity from 3.0 s to, 6.0 s positive, negative or zero? (c) What is a good strategy to match the graph from 7.0 s to 10.0 s? Test your answers using, the simulation., (a), , (b), , (c), , i., ii., iii., i., ii., iii., i., ii., iii., iv., , Left, Right, It is not moving, Positive, Negative, Zero, Start the ball moving slowly to the left and then increase its speed., Start the ball quickly moving to the left remaining at a constant speed., Start the ball slowly moving to the right and then increase its speed., Start the ball moving quickly to the right and then decrease its speed., , Section 8 - Acceleration, 8.1, , The speed limit on a particular freeway is 28.0 m/s (about 101 km/hour). A car that is merging onto the freeway is capable of, accelerating at 2.25 m/s2. If the car is currently traveling forward at 13.0 m/s, what is the shortest amount of time it could take, the vehicle to reach the speed limit?, s, , 8.2, , A sailboat is moving across the water at 3.0 m/s. A gust of wind fills its sails and it accelerates at a constant 2.0 m/s2. At the, same instant, a motorboat at rest starts its engines and accelerates at 4.0 m/s2. After 3.0 seconds have elapsed, find the, velocity of (a) the sailboat, and (b) the motorboat., (a), , m/s, , (b), , m/s, , Section 9 - Average acceleration, 9.1, , A rail gun uses electromagnetic energy to accelerate objects quickly over a short distance. In an experiment, a 2.00 kg, projectile remains on the rails of the gun for only 2.10eí2 s, but in that time it goes from rest to a velocity of 4.00×103 m/s., What is the average acceleration of the projectile?, m/s2, , 9.2, , A baseball is moving at a speed of 40.0 m/s toward a baseball player, who swings his bat at it. The ball stays in contact with, the bat for 5.00×10í4 seconds, then moves in essentially the opposite direction at a speed of 45.0 m/s. What is the magnitude, of the ball's average acceleration over the time of contact? (These figures are good estimates for a professional baseball, pitcher and batter.), m/s2, , 9.3, , 9.4, , A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion, as straight-line motion. What is the average acceleration of the shuttle (a) during the first 2.0 minutes, (b) during the 8.0, minutes the shuttle moves, and (c) during the entire 10 minute period?, (a), , m/s2, , (b), , m/s2, , (c), , m/s2, , A particle's initial velocity is í24.0 m/s. Its final velocity, 3.12 seconds later, is í14.0 m/s. What was its average acceleration?, m/s2, , 48, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems
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9.5, , 9.6, , The velocity versus time graph of an ant is, shown. What is the ant's acceleration at (a), t = 1.0 s, (b) t = 3.0 s, and (c) t = 5.0 s?, (a), , cm/s2, , (b), , cm/s2, , (c), , cm/s2, , The driver of a jet-propelled car is hoping to break the world record for ground-based travel of 341 m/s. The car goes from, rest to its top speed in 26.5 seconds, undergoing an average acceleration of 13.0 m/s2. Upon reaching the top speed, the, driver immediately puts on the brakes, causing an average acceleration of í5.00 m/s2, until the car comes to a stop. (a) What, is the car's top speed? (b) How long will the car be in motion?, , 9.7, , (a), , m/s, , (b), , s, , A particle's initial velocity is v. Every second, its velocity doubles. Which of the following expressions would you use to, calculate its average acceleration after five seconds of motion?, 32v/5, , 2v, , 31v, , 31v/5, , Section 10 - Instantaneous acceleration, 10.1 The velocity versus time graph for a pizza, delivery driver who is frantically trying to, deliver a pizza is shown. (a) During what time, interval is he traveling at a constant velocity?, (b) During what time interval is his, acceleration 5.0 m/s2? (c) During what time is, his acceleration negative?, (a), , (b), , (c), , i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., , 0 to 1.0 s, 1.0 s to 2.0 s, 2.0 to 4.0 s, During no time interval, 0 to 1.0 s, 2.0 to 4.0 s, 4.0 to 5.0 s, During no time interval, 0 to 1.0 s, 2.0 to 4.0 s, 4.0 to 5.0 s, During no time interval, , Section 13 - Interactive problem: what’s wrong with the rabbits?, 13.1 Use the information given in the interactive problem in this section to find the rabbits that do not have a constant acceleration., Which are the faulty rabbits? Test your answer using the simulation., The brown rabbit, The tan rabbit, The white rabbit, The grey rabbit, The black rabbit, , Section 15 - Motion equations for constant acceleration, 15.1 The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the, ball, giving it an initial velocity of 3.6 m/s. The ball rolls a distance of 5.0 m and is then intercepted by a South Korean player., If the ball accelerates at í0.50 m/s2 while rolling along the grass, find its velocity at the time of interception., m/s, 15.2 Which one of the following equations could be used to calculate the time of a journey when the initial velocity is zero, and the, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems, , 49
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constant acceleration and displacement are known?, A., B., C., , D., A, , B, , C, , D, , 15.3 The city is trying to figure out how long the traffic light should stay yellow at an intersection. The speed limit on the road is, 45.0 km/h and the intersection is 23.0 m wide. A car is traveling at the speed limit in the positive direction and can brake with, an acceleration of í5.20 m/s2. (a) If the car is to stop on the white line, before entering the intersection, what is the minimum, distance from the line at which the driver must apply the brakes? (b) How long should the traffic light stay yellow so that if the, car is just closer than that minimum distance when the light turns yellow, it can safely cross the intersection without having to, speed up?, (a), , m, , (b), , s, , 15.4 The brochure advertising a sports car states that the car can be moving at 100.0 km/h, and stop in 37.19 meters. What is its, average acceleration during a stop from that velocity? Express your answer in m/s2. Consider the car's initial velocity to be a, positive quantity., m/s2, 15.5 Engineers are designing a rescue vehicle to catch a runaway train. When the rescue vehicle is launched from a stationary, position, the train will be at a distance d meters away, moving at constant velocity v meters per second. The rescue vehicle, needs to reach the train in t seconds. Write an equation for the constant acceleration needed for the rescue vehicle in terms, of d, v, and t., a = 2(vt + d)/t2, a = (vt í d)/t, a = 2(dv + t)/vt2, a = t(v í d), 15.6 Two spacecraft are 13,500 m apart and moving directly toward each other. The first spacecraft has velocity 525 m/s and, accelerates at a constant í15.5 m/s2. They want to dock, which means they have to arrive at the same position at the same, time with zero velocity. (a) What should the initial velocity of the second spacecraft be? (b) What should be its constant, acceleration?, (a), , m/s, , (b), , m/s2, , Section 18 - Free-fall acceleration, 18.1 An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an, unknown velocity when the cable snaps. The elevator falls 1.10 meters before hitting the bottom of the shaft. The elevator, was in free fall for 0.900 seconds. Determine its velocity when the cable snapped. As usual, up is the positive direction., m/s, 18.2 A watermelon cannon fires a watermelon vertically up into the air at a velocity of + 9.50 m/s, starting from an initial position, 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is (a) its velocity, (b) its acceleration,, (c) the elapsed time, and (d) its height above the ground?, (a), , m/s, , (b), , m/s2, , (c), , s, , (d), , m, , 18.3 A croissant is dropped from the top of the Eiffel Tower. The height of the tower is 300.5 meters (ignoring the antenna, and, this figure changes slightly with temperature). Ignoring air resistance, at what speed will the croissant be traveling when it hits, the ground?, m/s, , 50, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems
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18.4 On a planet that has no atmosphere, a rocket 14.2 m tall is resting on its launch pad. Freefall acceleration on the planet is, 4.45 m/s2. A ball is dropped from the top of the rocket with zero initial velocity. (a) How long does it take to reach the launch, pad? (b) What is the speed of the ball just before it reaches the ground?, (a), , s, , (b), , m/s, , 18.5 To determine freefall acceleration on a moon with no atmosphere, you drop your handkerchief off the roof of a baseball, stadium there. The roof is 113 meters tall. The handkerchief reaches the ground in 18.2 seconds. What is freefall acceleration, on this moon? (State the result as a positive quantity.), m/s2, 18.6 You are a bungee jumping fanatic and want to be the first bungee jumper on Jupiter. The length of your bungee cord is, 45.0 m. Freefall acceleration on Jupiter is 23.1 m/s2. What is the ratio of your speed on Jupiter to your speed on Earth when, you have dropped 45.0 m? Ignore the effects of air resistance and assume that you start at rest., , 18.7 On the Apollo 15 space mission, Commander David R. Scott verified Galileo's assertion that objects of different masses, accelerate at the same rate. He did so on the Moon, where the acceleration due to gravity is 1.62 m/s2 and there is no air, resistance, by dropping a hammer and a feather at the same time. Assume they were 1.25 meters above the surface of the, Moon when he released them. How long did they take to land?, s, 18.8 You stand near the edge of Half Dome in Yosemite, reach your arm over the railing, and (thoughtlessly, since what goes up, does come down and there are people below) throw a rock upward at 8.00 m/s. Half Dome is 1460 meters high. How long, does it take for the rock to reach the ground? Ignore air resistance., s, 18.9 To get a check to bounce 0.010 cm in a vacuum, it must reach the ground moving at a speed of 6.7 m/s. At what velocity, toward the ground must you throw it from a height of 1.4 meters in order for it to it have a speed of 6.7 m/s when it reaches, the ground? Treat downward as the negative direction, and watch the sign of your answer., m/s, 18.10 Two rocks are thrown off the edge of a cliff that is 15.0 m above the ground. The first rock is thrown upward, at a velocity of, +12.0 m/s. The second is thrown downward, at a velocity of í12.0 m/s. Ignore air resistance. Determine (a) how long it takes, the first rock to hit the ground and (b) at what velocity it hits. Determine (c) how long it takes the second rock to hit the ground, and (b) at what velocity it hits., (a), , s, , (b), , m/s, , (c), , s, , (d), , m/s, , 18.11 A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of, 12.0 m/s. Ignore the effects of air resistance. (a) How long until the ball reaches its highest point? (b) How high above the, ground does the ball go?, (a), , s, , (b), , m, , 18.12 To determine how high a cliff is, a llama farmer drops a rock, and then 0.800 s later, throws another rock straight down at a, velocity of í10.0 m/s. Both rocks land at the same time. How high is the cliff?, m, , Copyright 2007 Kinetic Books Co. Chapter 2 Problems, , 51
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3.0 - Introduction, Knowing “how far” or “how fast” can often be useful, but “which way” sometimes, proves even more valuable. If you have ever been lost, you understand that, direction can be the most important thing to know., Vectors describe “how much” and “which way,” or, in the terminology of physics,, magnitude and direction. You use vectors frequently, even if you are not familiar, with the term. “Go three miles northeast” or “walk two blocks north, one block east”, are both vector descriptions. Vectors prove crucial in much of physics. For example,, if you throw a ball up into the air, you need to understand that the initial velocity of, the ball points “up” while the acceleration due to the force of gravity points “down.”, In this chapter, you will learn the fundamentals of vectors: how to write them and, how to combine them using operations such as addition and subtraction., On the right, a simulation lets you explore vectors, in this case displacement, vectors. In the simulation, you are the pilot of a small spaceship. There are three, locations nearby that you want to visit: a refueling station, a diner, and the local, gym. To reach any of these locations, you describe the displacement vector of the, spaceship by setting its x (horizontal) and y (vertical) components. In other words, you set how far horizontally you want to travel, and how far, vertically. This is a common way to express a two-dimensional vector., There is a grid on the drawing to help you determine these values. You, and each of the places you want to visit, are at the intersection of two, grid lines. Each square on the grid is one kilometer across in each direction. Enter the values, press GO, and the simulation will show you, traveling in a straight line í along the displacement vector í according to the values you set. See if you can reach all three places. You can do, this by entering displacement values to the nearest kilometer, like (3, 4) km. To start over at any time, press RESET., , 3.1 - Scalars, , Scalar: A quantity that states only an amount., Scalar quantities state an amount: “how much” or “how many.” At the right is a picture of, a dozen eggs. The quantity, a dozen, is a scalar. Unlike vectors, there is no direction, associated with a scalar í no up or down, no left or right í just one quantity, the, amount. A scalar is described by a single number, together with the appropriate units., Temperature provides another example of a scalar quantity; it gets warmer and colder,, but at any particular time and place there is no “direction” to temperature, only a value., Time is another commonly used scalar., Speed and distance are yet other scalars. A speed like 60 kilometers per hour says how, fast but not which way. Distance is a scalar since it tells you how far away something is,, but not the direction., , Scalars, Amount, Only one value, , Examples of scalars, 12 eggs, Temperature is í5º C, , 52, , Copyright 2000-2007 Kinetic Books Co. Chapter 03
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3.2 - Vectors, , Vector: A quantity specified, by both magnitude and, direction., Vectors have both magnitude (how much) and, direction. For example, vectors can be used to supply, traveling instructions. If a pilot is told “Fly 20, kilometers due south,” she is being given a, displacement vector to follow. Its magnitude is 20, kilometers and its direction is south. Vector, magnitudes are positive or zero; it would be confusing, to tell somebody to drive negative 20 kilometers south., , A spelunker (cave explorer) uses both distance and direction to navigate., , Many of the fundamental quantities in physics are vectors. For instance, displacement,, velocity and acceleration are all vector quantities. Physicists depict vectors with arrows., The length of the arrow is proportional to the vector’s magnitude, and the arrow points, in the direction of the vector. The horizontal vector in Concept 1 on the right represents, the displacement of a car driving from Acme to Dunsville., You see two displacement vectors in Concept 1. The displacement vector of a drive, from Acme to Dunsville is twice as long as the displacement vector from Chester to, Dunsville. This is because the distance from Acme to Dunsville is twice that of Chester, to Dunsville., Even if they do not begin at the same point, two vectors are equal if they have the same, magnitude and direction. For instance, the vector from Chester to Dunsville in Concept, 1 represents a displacement of 100 km southeast. That vector could be moved without, changing its meaning. Perhaps it is 100 km southeast from Edwards to Frankville, as, well. A vector’s meaning is defined by its length and direction, not by its starting point., , Vectors, Magnitude and direction, Represented by arrows, Length proportional to magnitude, , Now that we have introduced the concept of vectors formally, we will express vector, quantities in boldface. For instance, F represents force, v stands for velocity, and so, on. You will often see F and v, as well, representing the magnitudes of the vectors,, without boldface. Why? Because it is frequently useful to discuss the magnitude of the, force or the velocity without concerning ourselves with its direction. For instance, there, may be several equations that determine the magnitude of a vector quantity like force,, but not its direction., , It is half as far from Baker to, Chester as from Acme to, Dunsville. Describe the, displacement vector from Baker, to Chester., Displacement: 100 km, east, , Copyright 2000-2007 Kinetic Books Co. Chapter 03, , 53
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3.3 - Polar notation, , Polar notation: Defining a vector by its angle, and magnitude., Polar notation is a way to specify a vector. With polar notation, the magnitude and, direction of the vector are stated separately. Three kilometers due north is an example, of polar notation. “Three kilometers” is the magnitude and “north” is the direction. The, magnitude is always stated as a positive value. Instead of using “compass” or map, directions, physicists use angles. Rather than saying “three kilometers north,” a, physicist would likely say “three kilometers directed at 90 degrees.”, The angle is most conveniently measured by placing the vector’s starting point at the, origin. The angle is then typically measured from the positive side of the x axis to the, vector. This is shown in Concept 1 to the right., , Polar notation, Magnitude and angle, , Angles can be positive or negative. A positive angle indicates a counterclockwise, direction, a negative angle a clockwise direction. For example, 90° represents a quarter, turn counterclockwise from the positive x axis. In other words, a vector with a 90°, angle points straight up. We could also specify this angle as í270°., The radian is another unit of measurement for angles that you may have seen before., We will use degrees to specify angles unless we specifically note that we are using, radians. (Radians do prove essential at times.), , Polar notation, v is magnitude, ș is angle, Written v = (v, ș), , Write the velocity vector of the, car in polar notation., v = (v, ș), v = (5 m/s, 135º), , 54, , Copyright 2000-2007 Kinetic Books Co. Chapter 03
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3.4 - Vector components and rectangular notation, , Rectangular notation: Defining a vector by its, components., Often what we know, or want to know, about a particular vector is not its overall, magnitude and direction, but how far it extends horizontally and vertically. On a graph,, we represent the horizontal direction as x and the vertical direction as y. These are, called Cartesian coordinates. The xcomponent of a vector indicates its extent in the, horizontal dimension and the ycomponent its extent in the vertical dimension., Rectangular notation is a way to describe a vector using the components that make up, the vector. In rectangular notation, the x and y components of a vector are written, inside parentheses. A vector that extends a units along the x axis and b units along the, y axis is written as (a, b). For instance (3, 4) is a vector that extends positive three in, the x direction and positive four in the y direction from its starting point., , Vector components and, rectangular notation, x component and y component, , The components of vectors are scalars with the direction indicated by their sign: x, components point right (positive) or left (negative), and y components point up (positive), or down (negative). You see the x and y components of a car’s velocity vector in, Concept 1 at the right, shown as “hollow” vectors. The x and y values define the vector,, as they provide direction and magnitude., For a vector A, the x and y components are sometimes written as Ax and Ay. You see, this notation used for a velocity vector v in Equation 1 and Example 1 on the right., Consider the car shown in Example 1 on the right. Its velocity has an x component vx of, 17 m/s and a y component vy of í13 m/s. We can write the car’s velocity vector as, (17, í13) m/s., A vector can extend in more than two dimensions: z represents the third dimension., Sometimes z is used to represent distance toward or away from you. For instance, your, computer monitor’s width is measured in the x dimension, its height with y and your, distance from the monitor with z. If you are reading this on a computer monitor and, punch your computer screen, your fist would be moving in the z dimension. (We hope, we’re not the cause of any such aggressive feelings.) Three-dimensional vectors are, written as (x, y, z)., , Rectangular notation, vx is horizontal component, vy is vertical component, Written v = (vx, vy), , The z component can also represent altitude. A Tour de France bike racer might believe, the z dimension to be the most important as he ascends one of the competition’s, famous climbs of a mountain pass., , What is the car’s velocity vector, in rectangular notation?, v = (vx, vy) m/s, v = (17, í13) m/s, , Copyright 2000-2007 Kinetic Books Co. Chapter 03, , 55
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3.5 - Adding and subtracting vectors graphically, Vectors can be added and subtracted. In this section, we show how to do these, operations graphically. For instance, consider the vectors A and B shown in Concept 1, to the right. The vector labeled A + B is the sum of these two vectors., It may be helpful to imagine that these two vectors represent displacement. A person, walks along displacement vector A and then along displacement vector B. Her initial, point is the origin, and she would end up at the point at the end of the A + B vector., The sum represents the displacement vector from her initial to final position., To be more specific about the addition process: We start with two vectors, A and B,, both drawn starting at the origin (0, 0). To add them, we move the vector B so it starts, at the head of A. The diagram for Equation 1 shows how the B vector has been moved, so it starts at the head of A. The sum is a vector that starts at the tail of A and ends at, the head of B., In summary, to add two vectors, you:, 1., , Place the tail of the second vector at the head of the first vector. (The order of, addition does not matter, so you can place the tail of the first vector at the head of, the second as well.), , 2., , Draw a vector between the tail end of the first vector and the head of the second, vector. This vector represents the sum of two vectors., , Adding vectors A + B graphically, Move tail of B to head of A, Draw vector from tail of A to head of B, , To emphasize a point: You can think of this as combining a series of vector instructions., If someone says, “Walk positive three in the x direction and then negative two in the y, direction,” you follow one instruction and then the other. This is the equivalent of placing, one vector’s tail at the head of the other. An arrow from where you started to where you, ended represents the resulting vector. Any vector is the vector sum of its rectangular, components., When two vectors are parallel and pointing in the same direction, adding them is, relatively simple: You just combine the two arrows to form a longer arrow. If the vectors, are parallel but pointing in opposite directions, the result is a shorter arrow (three steps, forward plus two steps back equals one step forward)., , Subtracting A í B graphically, Take the opposite of B, , Move it to head of A, To subtract two vectors, take the opposite of the vector that is being subtracted, and, then add. (The opposite or negative of a vector is a vector with the same magnitude but Draw vector from tail of A to head of íB, opposite direction.) This is the same as scalar subtraction (for example 20 í 5 is the, same as 20 + (í5)). To draw the opposite of a vector, draw it with the same length but, the opposite direction. In other words, it starts at the same point but is rotated 180°. The diagram for Equation 2 shows the subtraction of two, vectors., When a vector is added to its opposite, the result is the zero vector, which has zero magnitude and no direction. This is analogous to adding a, scalar number to its opposite, like adding +2 and í2 to get zero., , 3.6 - Adding and subtracting vectors by components, You can combine vectors graphically, but it may be more precise to add up their, components., You perform this operation intuitively outside physics. If you were a dancer or a, cheerleader, you would easily understand the following choreography: “Take two steps, forward, four steps to the right and one step back.” These are vector instructions. You, can add them to determine the overall result. If asked how far forward you are after this, dance move, you would say “one step,” which is two steps forward plus one step back., You realize that your progress forward or back is unaffected by steps to the left or right., You correctly process left/right and forward/back separately. If a physics-oriented dance, instructor asked you to describe the results of your “dancing vector” math, you would, say, “One step forward, four steps to the right.”, You have just learned the basics of vector addition, which is reasonably straightforward: Adding and subtracting vectors, Break the vector into its components and add each component independently. In, by components, physics though, you concern yourself with more than dance steps. You might want to, Add (or subtract) each component, add the vector (20, í40, 60) to (10, 50, 10). Let’s assume the units for both vectors are, separately, meters. As with the dance example, each component is added independently. You add, the first number in each set of parentheses: 20 plus 10 equals 30, so the sum along the, x axis is 30. Then you add í40 and 50 for a total of 10 along the y axis. The sum along the z axis is 60 plus 10, or 70. The vector sum is, (30, 10, 70) meters. If following all this in the text is hard, you can see another problem worked in Example 1 on the right., Although we use displacement vectors in much of this discussion since they may be the most intuitive to understand, it is important to note that, all types of vectors can be added or subtracted. You can add two velocity vectors, two acceleration vectors, two force vectors and so on. As, illustrated in the example problem, where two velocity vectors are added, the process is identical for any type of vector., , 56, , Copyright 2000-2007 Kinetic Books Co. Chapter 03
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Vector subtraction works similarly to addition when you use components. For example,, (5, 3) minus (2, 1) equals 5 minus 2, and 3 minus 1; the result is the vector (3, 2)., , A + B = (Ax + Bx, Ay + By), A í B = (Ax í Bx, Ay í By), A, B = vectors, Ax, Ay = A components, Bx, By = B components, , The boat has the velocity A in, still water. Calculate its velocity, as the sum of A and the velocity, B of the river's current., v=A+B, v = (3, 4) m/s + (2, í1) m/s, v = (3 + 2, 4 + (í1)) m/s, v = (5, 3) m/s, 3.7 - Interactive checkpoint: vector addition, What are the number values of the, constants a and b?, , Answer:, , a=, , ,b=, , Copyright 2000-2007 Kinetic Books Co. Chapter 03, , 57
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3.8 - Multiplying rectangular vectors by a scalar, You can multiply vector quantities by scalar quantities. Let’s say an airplane, as shown, in Concept 1 on the right, travels at a constant velocity represented by the vector (40,, 10) m/s. Let's say you know its current position and want to know where it will be if it, travels for two seconds. Time is a scalar. To calculate the displacement, multiply the, velocity vector by the time., To multiply a vector by a scalar, multiply each component of the vector by the scalar. In, this example, (2 s)(40, 10) m/s = (80, 20) m. This is the plane’s displacement vector, after two seconds of travel., If you wanted the opposite of this vector, you would multiply by negative one. The result, in this case would be (í40, í10) m/s, representing travel at the same speed, but in the, opposite direction., , Multiplying a rectangular vector, by a scalar, Multiply each component by scalar, Positive scalar does not affect direction, , Multiplying a rectangular vector, by a scalar, sr = (srx, sry), s = a scalar, r = a vector, rx, ry = r components, , What is the displacement d of, the plane after 5.0 seconds?, d = (5.0 s)v, d = (5.0 s) (12 m/s, 15 m/s), d = ( (5.0 s)(12 m/s), (5.0 s)(15 m/s) ), d = (60, 75) m, , 58, , Copyright 2000-2007 Kinetic Books Co. Chapter 03
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3.9 - Multiplying polar vectors by a scalar, Multiplying a vector represented in polar notation by a positive scalar requires only one, multiplication operation: Multiply the magnitude of the vector by the scalar. The angle is, unchanged., Let’s say there is a vector of magnitude 50 km with an angle of 30°. You are asked to, multiply it by positive three. This situation is shown in Example 1 to the right. Since you, are multiplying by a positive scalar, the angle stays the same at 30°, and so the answer, is 150 km at 30°., If you multiply a vector by a negative scalar, multiply its magnitude by the absolute, value of the scalar (that is, ignore the negative sign). Then change the direction of the, vector by 180° so that it points in the opposite direction. In polar notation, since the, magnitude is always positive, you add 180° to the vector's angle to take its opposite., The result of multiplying (50 km, 30°) by negative three is (150 km, 210°)., If adding 180° would result in an angle greater than 360°, then subtract 180° instead., For instance, in reversing an angle of 300°, subtract 180° and express the result as, 120° rather than 480°. The two results are identical, but 120° is easier to understand., , Multiplying polar vector by, positive scalar, Multiply vector's magnitude by scalar, Angle unchanged, , Multiplying by negative scalar, Use absolute value and reverse, direction, , su = (su, ș), if s positive, su = (|s|u, ș + 180°), if s negative, s = a scalar, u = a vector, u = magnitude of vector, ș = angle of vector, , Copyright 2000-2007 Kinetic Books Co. Chapter 03, , 59
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What is the displacement vector, if the car travels three times as, far?, su = (su, ș), 3u = ( 3(50 km), 30º), 3u = (150 km, 30º), 3.10 - Gotchas, Stating a value as a scalar when a vector is required. This happens in physics and everyday life as well. You need to use a vector when, direction is required. Throwing a ball up is different than throwing a ball down; taking highway I-5 south is different than taking I-5 north., Vectors always start at the origin. No, they can start at any location., , 3.11 - Summary, A scalar is a quantity, such as time, temperature, or speed, which indicates only, amount., , Polar notation, A vector is a quantity, like velocity or displacement, which has both magnitude and, direction. Vectors are represented by arrows that indicate their direction. The, arrow’s length is proportional to the vector’s magnitude. Vectors are represented, with boldface symbols, and their magnitudes are represented with italic symbols., One way to represent a vector is with polar notation. The direction is indicated by, the angle between the positive x axis and the vector (measured in the, counterclockwise direction). For example, a vector pointing in the negative y, direction would have a direction of 270° in polar notation. The magnitude is, expressed separately. A polar vector is expressed in the form (r, ș) where r is the, magnitude and ș is the direction angle., , v = (v, ș), Rectangular notation, , v = (vx, vy), , A + B = (Ax + Bx, Ay + By), , Another way to represent a vector is by using rectangular notation. The vector’s x, and y components are expressed as an ordered pair of numbers (x, y). The, components of a vector A are also written as Ax and Ay., To add vectors graphically, place the tail of one on the head of the other, then draw a vector that goes from the free tail to the free head: The, new vector is the sum. To subtract, first take the opposite of the vector being subtracted, then add. (The opposite of a vector has the same, magnitude, but it points in the opposite direction.), , 60, , Copyright 2000-2007 Kinetic Books Co. Chapter 03
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Chapter 3 Problems, , Conceptual Problems, C.1, , List three quantities that are represented by vectors., , C.2, , Compare these two vectors: (5, 185°) and the negative of (5, 5°). Are they the same vector? Why or why not?, Yes, , C.3, , No, , An aircraft carrier sails northeast at a speed of 6.0 knots. Its velocity vector is v. What direction and speed would a ship with, velocity vector –v have?, knots, , C.4, , No, , A Boston cab driver picks up a passenger at Fenway Park, drops her off at the Fleet Center. Represent this displacement, with the vector D. What is the displacement vector from the Fleet Center to Fenway Park?, D, , C.6, , Northwest, Northeast, Southwest, Southeast, , Can the same vector have different representations in polar notation that use different angles? Explain., Yes, , C.5, , i., ii., iii., iv., , 2D, , 0, , íD, , Does the multiplication of a scalar and a vector display the commutative property? This property states that the order of, multiplication does not matter. So for example, if the multiplication of a scalar s and a vector r is commutative, then sr = rs for, all values of s and r., Yes, , No, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. Assume that the simulation is, reset before each part and give each answer in the form (x, y). (a) What is the displacement to Ed's Fuel Depot? (b) What is, the displacement to Joe's Diner? (c) What is the displacement to Silver's Gym?, (a) (, , km,, , (b) (, , km,, , km), , (c) (, , km,, , km), , km), , Section 1 - Scalars, 1.1, , The Earth has a mass of 5.97×1024 kg. The Earth's Moon has a mass of 7.35×1022 kg. How many Moons would it take to, have the same mass as the Earth?, , 1.2, , The volume of the Earth's oceans is approximately 1.4×1018 m3. The Earth's radius is 6.4×106 m. What percentage of the, Earth, by volume, is ocean?, %, , 1.3, , Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of, 6.96×108 m. What is the average density of the Sun?, kg/m3, , Section 3 - Polar notation, 3.1, , The tugboat Lawowa is returning to port for the day. It has a speed of 7.00 knots. The heading to the Lawowa's port is 31.0°, west of north. If due east is 0°, what is the tug's heading as a vector in polar notation?, (, , 3.2, , knots,, , °), , An analog clock has stopped. Its hands are stuck displaying the time of 10 o'clock. The hour hand is 5.0 centimeters long,, and the minute hand is 11 centimeters long. Write the position vector of the tip of the hour hand, in polar notation. Consider, 3 o'clock to be 0°, and assume that the center of the clock is the origin., (, , cm,, , °), , Copyright 2007 Kinetic Books Co. Chapter 3 Problems, , 61
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3.3, , What is the polar notation for a vector that points from the origin to the point (0, 3.00)?, ,, , (, , °), , Section 4 - Vector components and rectangular notation, 4.1, , (a) A hotdog vendor named Sam is walking from the southwest corner of Central park to the Empire State Building. He starts, at the intersection of 59th Street and Eighth Avenue, walks 3 blocks east to 59th Street and Fifth Avenue, and then 25 blocks, south to the Empire State Building at the Corner of 34th Street and Fifth Avenue. Write his displacement vector in rectangular, notation with units of "blocks." Orient the axes so positive y is to the north, and positive x is to the east. (b) A stockbroker, named Andrea makes the same trip in a cab that gets lost, and detours 50 blocks south to Washington Square before, reorienting and finally arriving at the Empire State Building. Is her displacement any different?, (a) (, (b), , 4.2, , ,, Yes, , A treasure map you uncovered while vacationing on the Spanish Coast reads as follows: "If me treasure ye wants, me hoard, ye'll have, just follow thee directions these. Step to the south from Brisbain's Mouth, 5 paces through the trees. Then to the, west, 10 paces ye'll quest, with mud as deep as yer knees. Then 3 paces more north, and dig straight down in the Earth, and, me treasure, take it please." What is the displacement vector from Brisbain's Mouth to the spot on the Earth above the, treasure? Consider east the positive x direction and north the positive y direction., (, , 4.3, , 4.4, , ) blocks, , No, , ,, , ) paces, , Write the vectors labeled A, B and C with, rectangular coordinates., (a) A = (, , ,, , ), , (b) B = (, , ,, , ), , (c) C = (, , ,, , ), , Consider these four vectors:, A goes from (0, 0) to (1, 2), B from (1, í2) to (0, 2), C from (í2, í1) to (í3, í3), D from (í3, 1) to (í2, 3), (a) Draw the vectors. Then answer the next two questions. (b) Which two vectors are equal? (c) Which vector is the negative, of the two equal vectors?, (a) Submit anwer on paper., (b), i. A and B, ii. A and C, iii. A and D, iv. B and C, v. B and D, vi. C and D, (c), i. A, ii. B, iii. C, iv. D, , Section 5 - Adding and subtracting vectors graphically, 5.1, , Draw each of the following pairs of vectors on a coordinate system, using separate coordinate systems for parts a and b of, the question. Then, on each coordinate system, also draw the vectors –B, A, (a) A = (0, 5); B = (3, 0), (b) A = (4, 1); B = (2, –3), , 5.2, , 62, , + B, and A – B. Label all your vectors., , The Moon's orbit around the Earth is nearly circular, with an average radius of 3.8×105 kilometers from the Earth's center. It, takes about 28 days for the Moon to complete one revolution around the Earth. (a) Put Earth at the origin of a coordinate, system, and draw labeled vectors to represent the moon's position at 0, 7, 14, 21 and 28 days. (b) On a separate coordinate, system, draw four labeled vectors representing the Moon's displacement from 0 days to 7 days, 7 days to 14 days, 14 days to, 21 days, and 21 days to 28 days. (c) What is the sum of the four displacement vectors you drew in part "b"?, , Copyright 2007 Kinetic Books Co. Chapter 3 Problems
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(a) Submit answer on paper., (b) Submit answer on paper., (c), 5.3, , Consider the following vectors:, A goes from (0, 2) to (4, 2), B from (1, í2) to (2, 1), C from (í1, 0) to (0, 0), D from (í3, í5) to (2, í2), E from (í3, í2) to (í4, í5), F from (í1, 3) to (í3, 0), (a) Draw the vectors. Using your sketch and your knowledge of graphical vector addition and subtraction, which vector listed, above is equal to: (b) A + B? (c) E í F? (d) í(B + C)? (e) Which two vectors sum to zero?, (a) Submit answer on paper., (b), i. A, ii. B, iii. C, iv. D, v. E, vi. F, (c), i. A, ii. B, iii. C, iv. D, v. E, vi. F, (d), i. A, ii. B, iii. C, iv. D, v. E, vi. F, (e), i. A and D, ii. B and C, iii. B and E, iv. B and F, v. C and F, vi. E and F, , 5.4, , The racing yacht America (USA) defeated the Aurora (England) in 1851 to win the 100 Guinea Cup. From the starting buoy, the America's skipper sailed 400 meters at an angle 45° west of north, then 250 meters at an angle 30° east of north, and, finally 350 meters at an angle 60° west of north. Draw the path of the America on a coordinate system as a set of vectors, placed tip to tail. Then draw the total displacement vector., , Section 6 - Adding and subtracting vectors by components, 6.1, , Add the following vectors:, (a) (12, 5) + (6, 3), (b) (í3, 8) + (6, í2), (c) (3, 8, í7) + (7, 2, 17), (d) (a, b, c) + (d, e, f), (a) (, , ,, , (b) (, , ,, , ), , (c) (, , ,, , ,, , ), ), , (d), 6.2, , Solve for the unknown variables:, (a) (a, b) + (3, 3) = (6, 7), (b) (11, c) + (d, 2) = (−12, í3), (c) (5, 5) + (e, 4) = (2, f), (d) ( 4, í3) í (5, g) = (h, 2), (a) a =, , ;b=, , Copyright 2007 Kinetic Books Co. Chapter 3 Problems, , 63
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6.3, , (b) c =, , ;d=, , (c) e =, , ;f=, , (d) g =, , ;h=, , Solve for the unknown variables: (8, 3) + (b, 2) = (4, a)., (a) a =, (b) b =, , 6.4, , Physicists model a magnetic field by assigning to every point in space a vector that represents the strength and direction of, the field at that point. Two magnetic fields that exist in the same region of space may be added as vectors at each point to find, the representation of their combined magnetic field. The "tesla" is the unit of magnetic field strength. At a certain point,, magnet 1 contributes a field of ( í6.4, 6.1, í3.7) tesla and magnet 2 contributes a field of (í1.1, í4.5, 8.6) tesla. What is the, combined magnetic field at this point?, (, , ,, , ,, , ) tesla, , Section 8 - Multiplying rectangular vectors by a scalar, 8.1, , Perform the following calculations., (a) 6(3, í1, 8), (b) í3(í3, 4, í5), (c) ía(a, b, c), (d) í2(a, 5, c) + 6 (3, íb, 2), (a) (, , ,, , ,, , ), , (b) (, , ,, , ,, , ), , (c), (d), 8.2, , Three vectors that are neither parallel nor antiparallel can be arranged to form a triangle if they sum to (0, 0). (a) What vector, forms a triangle with (0, 3) and (3, 0)? (b) If you multiply all three vectors by the scalar 2, do they still form a triangle? (c) What, if you multiply them by the scalar a?, (a) (, , ,, , (b), , Yes, , No, , (c), , Yes, , No, , ), , Section 9 - Multiplying polar vectors by a scalar, 9.1, , 9.2, , Perform the following computations. Express each vector in polar notation with a positive magnitude and an angle between 0°, and 360°., (a) 2(4, 230°), (b) í3(7, 20°), (c) í4(8, 260°), (a) (, , ,, , (b) (, , ,, , °), , (c) (, , ,, , °), , A chimney sweep is climbing a long ladder that leans against the side of a house. If the displacement of her feet from the, base of the ladder is given by ( 2.1 ft, 65°) when she is on the third rung, what is the displacement of her feet from the base, when she has climbed twice as far?, (, , 64, , °), , ft,, , °), , Copyright 2007 Kinetic Books Co. Chapter 3 Problems
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4.0 - Introduction, Imagine that you are standing on the 86th floor observatory of the Empire State, Building, holding a baseball. A friend waits in the street below, ready to catch the, ball. You toss it forward and watch it move in that direction at the same time as it, plummets toward the ground. Although you have not thrown the ball downward at, all, common sense tells you that your friend had better be wearing a well-padded, glove!, When you tossed the ball, you subconsciously split its movement into two, dimensions. You supplied the initial forward velocity that caused the ball to move, out toward the street. You did not have to supply any downward vertical velocity., The force of gravity did that for you, accelerating the ball toward the ground. If you, had wanted to, you could have simply leaned over and dropped the ball off the roof,, supplying no initial velocity at all and allowing gravity to take over., To understand the baseball’s motion, you need to analyze it in two dimensions., Physicists use x and y coordinates to discuss the horizontal and vertical motion of, the ball. In the horizontal direction, along the x axis, you supply the initial forward, velocity to the ball. In the vertical direction, along the y axis, gravity does the work., The ball’s vertical velocity is completely independent of its horizontal velocity. In, fact, the ball will land on the ground at the same time regardless of whether you, drop it straight off the building or hurl it forward at a Randy Johnson-esque 98 miles, per hour., To get a feel for motion in two dimensions, run the simulations on the right. In the, first simulation, you try to drive a race car around a circular track by controlling its x, and y component velocities separately, using the arrow keys on your keyboard. The, right arrow increases the x velocity and the left arrow decreases it. The up arrow, key increases the y velocity, and yes, the down arrow decreases it. Your mission is, to stay on the course and, if possible, complete a lap using these keys., Your car will start moving when you press any of the arrow keys. On the gauges,, you can observe the x and y velocities of your car, as well as its overall speed., Does changing the x velocity affect the y velocity, or vice-versa? How do the two, velocities seem to relate to the overall speed?, There is also a clock, so you can see which among your friends gets the car around, the track in the shortest time. There is no penalty for driving your car off the track,, though striking a wall is not good for your insurance rates. Press RESET to start over. Happy motoring!, In the second simulation, you can experiment with motion in two dimensions by firing the cannon from the castle. The cannon fires the, cannonball horizontally from the top of a tower. You change the horizontal velocity of the cannonball by dragging the head of the arrow. Try to, hit the two haystacks on the plain to see who is hiding inside., As the cannonball moves, look at the gauges in the control panel. One displays the horizontal velocity of the cannonball, its displacement per, unit time along the x axis. The other gauge displays the cannonball’s vertical velocity, its displacement per unit time along the y axis. As you, use the simulation, consider these important questions: Does the cannonball’s horizontal velocity change as it moves through the air? Does its, vertical velocity change? The simulation pauses when a cannonball hits the ground, and the gauges display the values from an instant before, that moment., The simulation also contains a timer that starts when the ball is fired and stops when it hits a haystack or the ground. Note the values in the, timer as you fire shots of varying horizontal velocity. Does the ball stay in the air longer if you increase the horizontal velocity, or does it stay in, the air the same amount of time regardless of that velocity?, The answers to these questions are the keys to understanding what is called projectile motion, motion where the acceleration occurs due to, gravity alone. This chapter will introduce you to motion in two and three dimensions; projectile motion is one example of this type of motion., , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 65
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4.1 - Velocity in two dimensions, Velocity is a vector quantity, meaning it contains two pieces of information: how fast, something is traveling and in which direction. Both are crucial for understanding motion, in multiple dimensions., Consider the car on the track to the right. It starts out traveling parallel to the x axis at a, constant speed. It then reaches a curve and continues to travel at a constant speed, through the curve. Although its speed stays the same, its direction changes. Since, velocity is defined by speed and direction, the change in direction means the car’s, velocity changes., Using vectors to describe the car’s velocity helps to illustrate its change in velocity. The, velocity vector points in the direction of the car’s motion at any moment in time. Initially,, the car moves horizontally, and its velocity vector points to the right, parallel to the x, axis. As the car goes around the curve, the velocity vector starts to point upward as well, as to the right. You see this shown in the illustration for Concept 1., When the car exits the curve, its velocity vector will be straight up, parallel to the y axis., Because the car is moving at a constant speed, the length of the vector stays the same:, The speed, or magnitude of the vector, remains constant. However, the direction of the, vector changes as the car moves around the curve., , Velocity in two dimensions, Velocity has x and y components, Analyze x and y components separately, Two component vectors sum to equal, total velocity, , Like any vector, the velocity vector can be written as the sum of its components, the, velocities along the x and y axes. This is also shown in the Concept 1 illustration. The, gauges display the x and y velocities. If you click on Concept 1 to see the animated, version of the illustration, you will see the gauges constantly changing as the car rounds, the bend. At the moment shown in the illustration, the car is moving at 17 m/s in the, horizontal direction and 10 m/s in the vertical direction. The components of the vector, shown also reflect these values. The horizontal component is longer than the vertical, one., Equations 1 and 2 show equations useful for analyzing the car’s velocity. Equation 1, shows how to break the car’s overall velocity into its components. (These equations, employ the same technique used to break any vector into its components.) The, illustration shows the car’s velocity vector. The angle ș is the angle the velocity vector, makes with the positive x axis. The product of the cosine of that angle and the, magnitude of the car’s velocity (its speed) equals the car’s horizontal velocity, component. The sine of the angle times the speed equals the vertical velocity, component., The first equation in Equation 2 shows how to calculate the car’s average velocity when, its displacement and the elapsed time are known. The displacement ǻr divided by the, elapsed time ǻt equals the average velocity., The equations for determining the average velocity components when the components, of the displacement are known are also shown in Equation 2. Dividing the displacement, along the x axis by the elapsed time yields the horizontal component of the car’s, average velocity. The displacement along the y axis divided by the elapsed time equals, the vertical component of the average velocity. A demonstration of these calculations is, shown in Example 1., , Components of velocity, vx = v cos ș, vy = v sin ș, v = speed, ș = angle with positive x axis, vx = x component of velocity, vy = y component of velocity, , The distinction between average and instantaneous velocity parallels the discussion of these two topics in the study of motion in one, dimension. To determine the instantaneous velocity, ǻr is measured during a very short increment of time and divided by that increment., As with linear motion, the velocity vector points in the direction of motion. On the curved part of the track, the instantaneous velocity vector is, tangent to the curve, since that is the direction of the car’s motion at any instant in time. The average velocity vector points in the same, direction as the displacement vector used to determine its value., , 66, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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Velocity from position, time, , v = velocity, r = position vector, ǻt = elapsed time, ǻx, ǻy = x and y displacements, v becomes instantaneous as ǻt ĺ 0, , What are the x and y, components of the car's average, velocity?, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 67
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4.2 - Acceleration in two dimensions, Analyzing acceleration in two dimensions is analogous to analyzing velocity in two, dimensions. Velocity can change independently in the horizontal and vertical, dimensions. Because acceleration is the change in velocity per unit time, it follows that, acceleration also can change independently in each dimension., The cannonball shown to the right is fired with a horizontal velocity that remains, constant throughout its flight. Constant velocity means zero acceleration. The, cannonball has zero horizontal acceleration., The cannonball starts with zero vertical velocity. Gravity causes its vertical velocity to, become an increasingly negative number as the cannonball accelerates toward the, ground. The vertical acceleration component due to gravity equals í9.80 m/s2., As with velocity, there are several ways to calculate the acceleration and its, components., , Acceleration in two dimensions, , The average acceleration can be calculated using the definition of acceleration, dividing, the change in velocity by the elapsed time. The components of the average acceleration, can be calculated by dividing the changes in the velocity components by the elapsed, time. These equations are shown in Equation 1. Instantaneous acceleration is defined, using the limit as ǻt gets close to zero., , Velocity can vary independently in x, y, dimensions, Change in velocity = acceleration, Acceleration can also vary, independently, , If the overall acceleration is known, in both magnitude and direction, you can calculate, its x and y components by using the cosine and sine of the angle ș that indicates its, direction. These equations are shown in Equation 2., , a = acceleration, v = velocity, ǻt = elapsed time, ǻvx, ǻvy = velocity components, , 68, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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ax = a cos ș, ay = a sin ș, ș = angle with positive x axis, ax = x component of acceleration, ay = y component of acceleration, 4.3 - Projectile motion, , Projectile velocity components, x and y velocity components, ·x velocity constant (ax = 0), ·y velocity changes (ay= í9.80 m/s2), , Projectile motion: Movement determined by an, object’s initial velocity and the constant, acceleration of gravity., The path of a cannonball provides a classic example of projectile motion., The cannonball leaves the cannon with an initial velocity and, ignoring air resistance,, that initial velocity changes during the flight of the cannonball due solely to the, acceleration due to the Earth’s gravity., The cannon shown in the illustrations fires the ball horizontally. After the initial blast, the, cannon no longer exerts any force on the cannonball. The cannonball’s horizontal, velocity does not change until it hits the ground., , Projectile motion, Motion in one dimension independent of, motion in other, , Once the cannonball begins its flight, the force of gravity accelerates it toward the, ground. The force of gravity does not alter the cannonball’s horizontal velocity; it only, affects its vertical velocity, accelerating the cannonball toward the ground. Its y velocity has an increasingly negative value as it moves through, the air., The time it takes for the cannonball to hit the ground is completely unaffected by its horizontal velocity. It makes no difference if the cannonball, flies out of the cannon with a horizontal velocity of 300 m/s or if it drops out of the cannon’s mouth with a horizontal velocity of 0 m/s. In either, case, the cannonball will take the same amount of time to land. Its vertical motion is determined solely by the acceleration due to gravity,, í9.80 m/s2., The equations to the right illustrate how you can use the x and y components of the cannonball’s velocity and acceleration to determine how, long it will take to reach the ground (its flight time) and how far it will travel horizontally (its range). These are standard motion equations, applied in the x and y directions. They hold true when the acceleration is constant, as is the case with projectile motion, where the acceleration, along each dimension is constant., In Equation 1, we show how to determine how long it takes a projectile to reach the ground. This equation holds true when the initial vertical, velocity is zero, as it is when a cannon fires horizontally. To derive the equation, we use a standard linear motion equation applied to the, vertical, or y, dimension. You see that equation in the second line in Equation 1. We substitute zero for the initial vertical velocity and then, solve for t, the elapsed time., , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 69
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Equation 2 shows how to solve for the range. The horizontal displacement of the ball, equals the product of its horizontal velocity and the elapsed time. Since its horizontal, velocity does not change, it equals the initial horizontal velocity. Together, the two, equations describe the flight of a projectile that is fired horizontally., , Projectile flight time, , vyi = initial y velocity, ǻy = vertical displacement, ay = í9.80 m/s2, t = time when projectile hits ground, , Projectile range, ǻx = vxt, ǻx = horizontal displacement, vx = horizontal velocity, , 70, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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4.4 - Sample problem: a horizontal cannon, What horizontal firing (muzzle), velocity splashes the cannonball into, the pond?, , Variables, vertical displacement, , ǻy = í40.0 m, , vertical acceleration, , ay = í9.80 m/s2, , horizontal displacement, , ǻx = 95.0 m, , elapsed time, , t, , horizontal velocity, , vx, , What is the strategy?, 1., , Calculate how long the cannonball remains in the air. This can be done with a standard motion equation., , 2., , Use the elapsed time and the specified horizontal displacement to calculate the required horizontal firing velocity., , Physics principles and equations, , The amount of time the ball takes to fall is independent of its horizontal velocity., Step-by-step solution, We start by determining how long the cannonball is in the air. We can use a linear motion equation to find the time it takes the cannonball to, drop to the ground., , Step, , Reason, , t2, , 1., , ǻy = vyit + ½ ay, , 2., , ǻy = (0)t + ½ ayt2, , linear motion equation, initial vertical velocity zero, , 3., , solve for time, , 4., , enter values, , 5., , t = 2.86 s, , evaluate, , Now that we know the time the cannonball takes to fall to the ground, we can calculate the required horizontal velocity., , Step, , Reason, , 6., , definition of velocity, , 7., , enter values, , 8., , vx = 33.2 m/s divide, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 71
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4.5 - Interactive problem: the monkey and the professor, The monkey at the right has a banana bazooka and plans to shoot a banana at a, hungry physics professor. He has a glove to catch the banana. The professor is, hanging from the tree, and the instant he sees the banana moving, he will drop from, the tree in his eagerness to dine., Can you correctly aim the monkey’s bazooka so that the banana reaches the, professor’s glove as he falls? Should you aim the shot above, below or directly at, the professor? As long as the banana is fired fast enough to reach the professor, before it hits the ground, does its initial speed matter? (Note: The simulation has a, minimum speed so the banana will reach the professor when correctly aimed.), Give it a try in the interactive simulation to the right. No calculations are required to, solve this problem. As you ponder your answer, consider the two key concepts of, projectile motion: (1) all objects accelerate toward the ground at the same rate, and, (2) the horizontal and vertical components of motion are independent. (The effect of, air resistance is ignored.), Aim the banana by dragging the vector arrow at the end of the bazooka. You can, increase the firing speed of the banana by making the arrow longer, and you can change the angle at which the banana is fired by moving the, arrow up or down. Stretching out the vector makes it easy to aim the banana., To shoot the banana, press GO. Press RESET to try again. (Do not worry: We, too, value physics professors, so the professor will emerge, unscathed.), If you have trouble with this problem, review the section on projectile motion., , 4.6 - Interactive checkpoint: golfing, A golfer is on the edge of a 12.5 m, high bluff overlooking the eighteenth, hole, which is located 67.1 m from the, base of the bluff. She launches a, horizontal shot that lands in the hole, on the fly, and the gallery erupts into, cheers. How long was the ball in the, air? What was the ball’s horizontal, velocity? Take upward to be the, positive y direction., , Answer:, , t=, vx =, , s, m/s, , 4.7 - Projectile motion: juggling, Juggling is a form of projectile motion in which the projectiles have initial velocities in, both the vertical and horizontal dimensions. This motion takes more work to analyze, than when a projectile’s initial vertical velocity is zero, as it was with the horizontally, fired cannonball., Jugglers throw balls from one hand to the other and then back. To juggle multiple balls,, the juggler repeats the same simple toss over and over. An experienced juggler’s ability, to make this basic routine seem so effortless stems from the fact that the motion of, each ball is identical. The balls always arrive at the same place for the catch, and in, roughly the same amount of time., A juggler throws each ball with an initial velocity that has both x and y components., Ignoring the effect of air resistance, the x component of the velocity remains constant, as the ball moves in the air from one hand to another., , Projectile motion: y velocity, , y velocity = zero at peak, The initial y velocity is upward, which means it is a positive value. At all times, the ball, accelerates downward at í9.80 m/s2. This means the ball’s velocity decreases as it, Initial y velocity equal but opposite to, rises until it has a vertical velocity of zero. The ball then accelerates back toward the, final y velocity, ground. When the ball plops down into the other hand, the magnitude of the y velocity, will be the same as when the ball was tossed up, but its sign will be reversed. If the ball, is thrown up with an initial y velocity of +5 m/s, it will land with a y velocity of í5 m/s. This symmetry is due to the constant rate of vertical, , 72, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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acceleration caused by gravity., Because the vertical acceleration is constant, the ball takes as much time to reach its, peak of motion as it takes to fall back to the other hand. This also means it has covered, half of the horizontal trip when it is at its peak. The path traced by the ball is a parabola,, a shape symmetrical around its midpoint., We said the juggler chooses the y velocity so that the ball takes the right amount of time, to rise and then fall back to the other hand. What, exactly, is the right amount of time? It, is the time needed for the ball to move the horizontal distance between the juggler’s, hands., If a juggler’s hands are 0.5 meters apart, and the horizontal velocity of the ball is, 0.5 m/s, it will take one second for the ball to move that horizontal distance. (Remember, that the horizontal velocity is constant.) The juggler must throw the ball with enough, vertical velocity to keep it in the air for one second., , Projectile motion: x velocity, x velocity constant, , 4.8 - Sample problem: calculating initial velocity in projectile motion, A juggler throws each ball so it hangs, in the air for 1.20 seconds before, landing in the other hand, 0.750, meters away., What are the initial vertical and, horizontal velocity components?, , Variables, elapsed time, , t = 1.20 s, , horizontal displacement, , ǻx = 0.750 m, , initial vertical velocity, , vy, , horizontal velocity, , vx, , acceleration due to gravity, , ay = í9.80 m/s2, , What is the strategy?, 1., , Use a linear motion equation to determine the initial vertical velocity that will result in a final vertical velocity of 0 m/s after 0.60 s (half the, total time of 1.20 s)., , 2., , Use the definition of velocity to calculate the horizontal velocity, since the displacement and the elapsed time are both provided., , Physics principles and equations, We rely on two concepts. First, the motion of a projectile is symmetrical, so half the time elapses on the way up and the other half on the way, down. Second, the vertical velocity of a projectile is zero at its peak., We also use the two equations listed below., , vyf = vyi + ayt, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 73
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Step-by-step solution, We start by calculating the vertical component of the initial velocity. We use the fact that gravity will slow the ball to a vertical velocity of 0 m/s, at the peak, 0.60 seconds (halfway) into the flight., , Step, , Reason, , 1., , vyf = vyi + ayt, , motion equation, , 2., , 0 = vyi + ayt, , vertical velocity at peak equals zero, , 3., , vyi = –ayt, , rearrange, , 4., , vyi = í(í9.80 m/s2)(0.600 s) enter values, , 5., , vyi = 5.89 m/s, , multiply, , Now we solve for the horizontal velocity component. We could have solved for this first; these steps require no results from the steps above., , Step, , Reason, , 6., , definition of velocity, , 7., , enter values, , 8., , vx = 0.625 m/s, , divide, , From our perspective, the horizontal velocity of the ball when it is going from our left to our right is positive. From the juggler’s, it is negative., , 4.9 - Interactive problem: the monkey and the professor, part II, The monkey is at it again with his banana bazooka. Another hungry professor drops, from the tree the instant the banana leaves the tip of the bazooka. (Are professors, paid enough? Answer: No.), Can you correctly aim the banana so that it hits the professor’s glove as she falls?, Should you aim the shot above, below or directly at the glove? Does the banana’s, initial speed matter? (Note: The simulation has a minimum speed so the banana will, reach the professor when correctly aimed.) Air resistance is ignored in this, simulation., Give it a try in the interactive simulation to the right. No calculations are needed to, solve this problem., To find an answer, think about the key concepts of projectile motion: Objects, accelerate toward the ground at the same rate, and the horizontal and vertical, components of motion are independent., Once you launch the simulation, aim the banana by dragging the vector arrow at the, end of the bazooka. You can increase the firing speed of the banana by making the, arrow longer. Moving the arrow up or down changes the angle at which the banana is fired. Stretching out the vector makes it easy to aim the, banana., To shoot the banana, press GO. Press RESET to try again. Keep trying until the banana reaches the professor!, , 74, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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4.10 - Projectile motion: aiming a cannon, The complexities of correctly aiming artillery pieces have challenged leaders as famed, as Napoleon and President Harry S. Truman. Because a cannon typically fires, projectiles at a particular speed, aiming the cannon to hit a target downfield involves, adjusting the cannon’s angle relative to the ground. If you break the motion into, components, you can determine how far a projectile with a given speed and angle will, travel., To determine when and where a cannonball will land, you must consider horizontal and, vertical motion separately. To start, convert its initial speed and angle into x and y, velocity components. The horizontal velocity will equal the initial speed of the ball, multiplied by the cosine of the angle at which the cannonball is launched. The horizontal, velocity will not change as the cannonball flies toward the target., The initial y velocity equals the initial speed times the sine of the launch angle. The y, velocity is not constant. It changes at the rate of í9.80 m/s2. When the cannonball lands, at the same height at which it was fired, its final y velocity is equal but opposite to its, initial y velocity., , Aiming a projectile, step 1, Start with initial angle, speed, Separate into x and y components, , The initial y velocity of the projectile determines how long it stays in the air. As, mentioned, the cannonball lands with a final y velocity equal to the negative of the initial, y velocity, that is, vyf = ívyi. This means the change in y velocity equals í2vyi., Knowing this, and the value for acceleration due to gravity, enables us to rearrange a, standard motion equation (vf = vi + at) and solve for the elapsed time. The equation for, the flight time of a projectile is the third one in Equation 1., Once you know how long the ball stays in the air, you can determine how far it travels, by multiplying the horizontal velocity by the ball’s flight time. This is the final equation on, the right., You can use these equations to solve projectile motion problems, but understanding the, analysis that led to the equations is more important than knowing the equations. Recall, the basic principles of projectile motion: The x velocity is constant, the y velocity, changes at the rate of í9.80 m/s2, and the projectile’s final y velocity is the opposite of, its initial y velocity., , Aiming a projectile, step 2, Use initial y velocity and ay to calculate, flight time, Use initial x velocity and flight time to, calculate range, , Projectile equations, vx = v cos ș, vy = v sin ș, t = í2vy /ay (same-height landing), ǻx = vxt, vx = x velocity, vy = initial y velocity, t = time projectile is in air, ǻx = horizontal displacement, ay = í9.80 m/s2, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 75
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4.11 - Sample problem: a cannon’s range, How far away is the haystack from, the cannon?, , Draw a diagram, , Variables, speed, , v = 40.0 m/s, , angle, , ș = 60.0°, , initial y velocity, , vyi, , final y velocity, , vyf, , x velocity, , vx, , elapsed time, , t, , horizontal displacement, , ǻx, , acceleration due to gravity, , ay = í9.80 m/s2, , What is the strategy?, 1., , Use trigonometry to determine the x and y components of the cannonball’s initial velocity., , 2., , The final y velocity is the opposite of the initial y velocity. Use that fact and a linear motion equation to determine how long the, cannonball is in the air., , 3., , The x velocity is constant. Rearrange the definition of constant velocity to solve for horizontal displacement (range)., , Physics principles and equations, The projectile’s final y velocity is the opposite of its initial y velocity. The x velocity is constant., We use the following motion equations., , vyf = vyi + ayt, , 76, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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Step-by-step solution, Use trigonometry to determine the x and y components of the initial velocity from the initial speed and the angle., , Step, , Reason, , 1., , vy = v sin ș, , 2., , vy = (40.0 m/s) sin 60.0° enter values, , 3., , vy = 34.6 m/s, , evaluate, , 4., , vx = v cos ș, , trigonometry, , 5., , vx = (40.0 m/s) cos 60.0° enter values, , 6., , vx = 20.0 m/s, , trigonometry, , evaluate, , Now we focus on the vertical dimension of motion, using the initial y velocity to determine the time the cannonball is in the air. We calculated, the initial y velocity in step 3., , Step, , Reason, , 7., , vyf = vyi + ayt, , linear motion equation, , 8., , ívyi = vyi + ayt, , final y velocity is negative of initial y velocity, , 9., , í2vyi = ayt, , rearrange, , 10., , solve for time, , 11., , enter values, , 12. t = 7.07 s, , evaluate, , Now we use the time and the x velocity to solve for the cannonball’s range. We calculated the constant x velocity in step 6., , Step, , Reason, , 13., , definition of velocity, , 14. ǻx = vxt, , rearrange, , 15. ǻx = (20.0 m/s)(7.07 s) enter values, 16. ǻx = 141 m, , multiply, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 77
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4.12 - Interactive checkpoint: clown cannon, A clown in a circus is about to be shot, out of a cannon with a muzzle velocity, of 15.2 m/s, aimed at 52.7° above the, horizontal. How far away should his, fellow clowns position a net to ensure, that he lands unscathed? The net is, at the same height as the mouth of, the cannon., , Answer:, , ǻx =, , m, , 4.13 - Interactive problem: test your juggling!, Much of this chapter focuses on projectile motion: specifically, how objects move in, two dimensions. If you have grasped all the concepts, you can use what you have, learned to make the person at the right juggle., The distance between the juggler’s hands is 0.70 meters and the acceleration due, to gravity is í9.80 m/s2. You have to calculate the initial x and y velocities to send, each ball from one hand to the other. If you do so correctly, he will juggle three balls, at once., There are many possible answers to this problem. A good strategy is to pick an, initial x or y component of the velocity, and then determine the other velocity, component so that the balls, once thrown, will land in the juggler’s opposite hand., You want to pick an initial y velocity above 2.0 m/s to give the juggler time to make, his catch and throw. For similar reasons, you do not want to pick an initial x velocity, that exceeds 2.0 m/s., Make your calculations and then click on the diagram to the right to launch the, simulation. Enter the values you have calculated to the nearest 0.1 m/s and press, the GO button. Do not worry about the timing of the juggler’s throws. They are calculated for you automatically., If you have difficulty with this problem, refer to the sections on projectile motion., , 4.14 - Reference frames, , Reference frame: A coordinate system used to, make observations., The choice of a reference frame determines the perception of motion. A reference, frame is a coordinate system used to make observations. If you stand next to a lab table, and hold out a meter stick, you have established a reference frame for making, observations., The choice of reference frames was a minor issue when we considered juggling: We, chose to measure the horizontal velocity of a ball as you saw it when you stood in front, of the juggler. The coordinate system was established using your position and, orientation, assuming you were stationary relative to the juggler., , Reference frames, System for observing motion, , As the juggler sees the horizontal velocity of the ball, however, it has the same, magnitude you measure, but is opposite in sign. It does so because when you see it, moving from your left to your right, he sees it moving from his right to his left. If you measure the velocity as 1.1 m/s, he measures it as, í1.1 m/s., In the analysis of motion, it is commonly assumed that you, the observer, are standing still. To pursue this further, we ask you to sit or stand, still for a moment. Are you moving? Likely you will answer: “No, you just asked me to be still!”, That response is true for what you are implicitly using as your reference frame, your coordinate system for making measurements. You are, implicitly using the Earth’s surface., But from the perspective of someone watching from the Moon, you are moving due to the Earth’s rotation and orbital motion. Imagine that the, person on the Moon wanted to launch a rocket to pick you up. Unless the person factored in your velocity as the Earth spins about its axis, as, well as the fact that the Earth orbits the Sun and the Moon orbits the Earth, the rocket surely would miss its target. If you truly think you are, , 78, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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stationary here on Earth, you must also conclude that the entire universe revolves, around the Earth (a dubious conclusion, though a common one for centuries)., Reference frames define your perception and measurements of motion. If you are in a, car moving at 80 km/h, another car moving alongside you with the same velocity will, appear to you as if it is not moving at all. As you drive along, objects that you ordinarily, think of as stationary, such as trees, seem to move rapidly past you. On the other hand,, someone sitting in one of the trees would say the tree is stationary and you are the one, moving by., A reference frame is more than just a viewpoint: It is a coordinate system used to make, measurements. For instance, you establish and use a reference frame when you do lab, exercises. Consider making a series of measurements of how long it takes a ball to roll, down a plane. You might say the ball’s starting point is the top of the ramp. Its x position, there is 0.0 meters. You might define the surface of the table as having a y position of, 0.0 meters, and the ball’s initial y position is its height above the table. Typically you, consider the plane and table to be stationary, and the ball to be moving., , Reference frames, Measurements of motion defined by, reference frame, , Two reference frames are shown on the right. One is defined by Joan, the woman, standing at a train station. As the illustration in Concept 1 shows, from Joan’s perspective, she is stationary and the train is moving to the right, at a constant velocity., Another reference frame is defined by the perspective of Ted who is inside the train, and considers the train stationary. This reference frame is, illustrated in Concept 2. Ted in the train perceives himself as stationary, and would see Joan moving backward at a constant velocity. He would, assign Joan the velocity vector shown in the diagram., It is important to note there is no correct reference frame; Joan cannot say her reference frame is better than the reference frame used by Ted., Measurements of velocity and other values made by either observer are equally valid., Reference frames are often chosen for the sake of convenience (choosing the Earth’s surface, not the surface of Jupiter, is a logical choice for, your lab exercises). Once you choose a reference frame, you must use it consistently, making all your measurements using that reference, frame’s coordinate system. You cannot measure a ball’s initial position using the Earth’s surface as a reference frame, and its final position, using the surface of Jupiter, and still easily apply the physics you are learning., , 4.15 - Relative velocity, Observers in reference frames moving past one another may measure different, velocities for the same object. This concept is called relative velocity., In the illustrations to the right, two observers are measuring the velocity of a soccer ball,, but from different vantage points: The man is standing on a moving train, while the, woman is standing on the ground. The man and the woman will measure different, velocities for the soccer ball., Let’s discuss this scenario in more depth. Fred is standing on a train car and kicks a ball, to the right. The train is moving along the track at a constant velocity. The train is Fred’s, reference frame, and, to him, it is stationary. In Fred’s frame of reference, his kick, causes the ball to move at a constant velocity of positive 10 m/s., The train is passing Sarah, who is standing on the ground. Her reference frame is the, ground. From her perspective, the train with the man on it moves by at a constant, velocity of positive 5 m/s., , Observer on train, Measures ball velocity relative to train, , What velocity would Sarah measure for the soccer ball in her reference frame? She, adds the velocity vector of the train, 5 m/s, to the velocity vector of the ball as measured, on the train, 10 m/s. The sum is positive 15 m/s, pointing along the horizontal axis., Summing the velocities determines the velocity as measured by Sarah., Note that there are two different answers for the velocity of one ball. Each answer is, correct in the reference frame of that observer. For someone standing on the ground,, the ball moves at 15 m/s, and for someone standing on the train, it moves at 10 m/s., The equation in Equation 1 shows how to relate the velocity of an object in one, reference frame to the velocity of an object in another frame. The variable vOA is the, velocity vector of the object as measured in reference frame A (which in this diagram is, the ground). The variable vOB is the velocity of an object as measured in reference, frame B (which in this diagram is the train). Finally, the variable vBA is the velocity of, frame B (the train) relative to frame A (the ground). vOA is the vector sum of vOB and, vBA., An important caveat is that this equation can be used to solve relative velocity problems, only when the frames are moving at constant velocity relative to one another. If one or, both frames are accelerating, the equation does not apply., , Observer on ground, Train is moving, Velocity = sum of ball, train velocities, , We use this equation in the example problem on the right. Now Sarah sees the train moving in the opposite direction, at negative 5 m/s. It is, negative because to Sarah, the train is moving to the left along the x axis. Fred is on the train, again kicking the ball from left to right as before., Here he kicks the ball at +5 m/s, as measured in his reference frame. To Sarah, how fast and in what direction is the ball moving now?, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 79
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The answer is that she sees the ball as stationary. The sum of the velocities equals, zero, because it is the sum of +5 m/s (the velocity of the object as measured in, reference frame B) and í5 m/s (the velocity of frame B as measured from frame A)., , Relative velocity equation, vOA = vOB + vBA, vOA = velocity of object measured in, reference frame A, vOB = velocity of object measured in, reference frame B, vBA = velocity of frame B measured in, frame A, , Sarah observes the train moving, to the left at í5 m/s. Fred, on the, train, sees the ball moving to the, right at +5 m/s. What is the ball’s, velocity in Sarah’s reference, frame?, vOA = vOB + vBA, vOA = (5 m/s) + (í5 m/s) = 0 m/s, 4.16 - Gotchas, A ball will land at the same time if you drop it straight down from the top of a building or if you throw it out horizontally. Yes, the ball will hit the, ground at the same time in both cases. Velocity in the y direction is independent of velocity in the x direction., An object has positive velocity along the x and y axes.Along the y axis,it accelerates, has a constant velocity for a while, then accelerates, some more. What happens along the x axis? You have no idea. Information about motion along the y axis tells you nothing about motion along, the x axis, because they can change independently., A projectile has zero acceleration at its peak. No, a projectile has zero y velocity at its peak. Even though it briefly comes to rest in the vertical, dimension, the projectile is always accelerating at í9.80 m/s2due to the force of gravity., , 80, , Copyright 2000-2007 Kinetic Books Co. Chapter 04
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4.17 - Summary, Like any vector, velocity can be broken into its component vectors in the x and y, dimensions using trigonometry. These components sum to equal the velocity vector., The components can also be analyzed separately, reducing a two-dimensional, problem to two separate one-dimensional problems. The same principle applies to, the acceleration vector., Objects that move solely under the influence of gravity are called projectiles. To, analyze projectile motion, consider the motion along each dimension separately., , Horizontal projectile flight time, , Horizontal projectile range, , ǻx = vxt, , Copyright 2000-2007 Kinetic Books Co. Chapter 04, , 81
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Chapter 4 Problems, , Conceptual Problems, C.1, , How can you change only one component of the velocity without changing the speed?, , C.2, , Two girls decide to jump off a diving board. Katherine steps off the diving board. Anna runs straight off the diving board so, that her initial velocity is solely horizontal. They both leave the diving board at the same time. Which one lands in the water, first?, i. Katherine, ii. Anna, iii. They land at the same time, , C.3, , An astronaut throws a baseball horizontally with the same initial velocity: once on the Earth, and once on the Moon. Where, does the ball travel farther? Explain your answer., i. On the Earth, ii. On the Moon, iii. The same distance, , C.4, , What is the name for the shape of a projectile's path?, , C.5, , Sarah is standing on the east bank of a river that runs south to north, and she wants to meet her friends who are on a boat in, the middle of the river, directly west of her. Her friends point the boat east, directly at her, and then turn on the motor. The, boat stays pointed east throughout its journey. (a) At what speed should Sarah walk or run along the bank of the river in order, for her to meet the boat when it touches the bank? (b) In what direction should she move? (c) Does either answer depend on, the speed of the boat relative to the water, which is provided by the motor?, (a), , i., ii., iii., iv., i., ii., iii., , (b), , (c), C.6, , The speed of the river, Zero, Twice the speed of the river, The speed that the boat could travel in still water, North, South, She should stand still, , Yes, , No, , Two students, Jim and Sarah, are walking to different classrooms from the same cafeteria. How does Jim's velocity in, Sarah's reference frame relate to Sarah's velocity in Jim's reference frame?, i., ii., iii., iv., , There is no relationship, They are identical, They are equal but opposite, There is not enough information to answer, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the driving interactive problem from this section to answer the following questions. (a) Does changing, the x velocity affect the y velocity? (b) How does increasing the magnitude of either velocity affect the overall speed?, (a), (b), , 0.2, , 82, , Yes, , No, , i. There is no effect, ii. The speed increases, iii. The speed decreases, , Use the simulation in the cannon interactive in this section to answer the following questions. (a) Does the cannonball's, horizontal velocity change as it moves through the air? (b) Does its vertical velocity change? (c) If you increase the, cannonball's horizontal velocity, does it stay in the air longer, the same amount of time, or shorter? (d) Who is inside the left, haystack?, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems
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(a), , Yes, , No, , (b), , Yes, , No, , (c), , (d), , i., ii., iii., i., ii., iii., , Longer, The same, Shorter, Elmo, Elvis, Marie Curie, , Section 1 - Velocity in two dimensions, 1.1, , A golf ball is launched at a 37.0° angle from the horizontal at an initial velocity of 48.6 m/s. State its initial velocity in, rectangular coordinates., (, , 1.2, , 1.3, , ,, , ) m/s, , The displacement vector for a 15.0 second interval of a jet airplane's flight is ( 2.15e+3, í2430) m. (a) What is the magnitude, of the average velocity? (b) At what angle, measured from the positive x axis, did the airplane fly during this time interval?, (a), , m/s, , (b), , °, , An airplane is flying horizontally at 115 m/s, at a 20.5 degree angle north of east. State its velocity in rectangular coordinates., Assume that north and east are in the positive direction., (, , ,, , ) m/s, , Section 2 - Acceleration in two dimensions, 2.1, , An ice skater starts out traveling with a velocity of (í3.0, í7.0) m/s. He performs a 3.0 second maneuver and ends with a, velocity of (0, 5.0) m/s. (a) What is his average acceleration over this period? (b) A different ice skater starts with the same, initial velocity, accelerates at (1.5, 3.5) m/s2 for 2.0 seconds, and then at (0, 5.0) m/s2 for 1.0 seconds. What is his final, velocity?, , 2.2, , (a) (, , ,, , ) m/s2, , (b) (, , ,, , ) m/s, , A spaceship accelerates at (6.30, 3.50) m/s2 for 6.40 s. Given that its final velocity is (450, 570) m/s, find its initial velocity., ,, , (, 2.3, , An airplane is flying with velocity (152, 0) m/s. It is accelerated in the x direction by its propeller at 8.92 m/s2 and in the, negative y direction by a strong downdraft at 1.21 m/s2. What is the airplane's velocity after 3.10 seconds?, ,, , (, 2.4, , ) m/s, , An electron starts at rest. Gravity accelerates the electron in the negative y direction at 9.80 m/s2 while an electric field, accelerates it in the positive x direction at 3.80×106 m/s2. Find its velocity 2.45 s after it starts to move., ,, , (, 2.5, , ) m/s, , ) m/s, , A particle's initial velocity was (3.2, 4.5) m/s, its final velocity is (2.6, 5.8) m/s, and its constant acceleration was (í2.4, 5.2), m/s2. What was its displacement?, ,, , (, , )m, , Section 3 - Projectile motion, 3.1, , 3.2, , 3.3, , A friend throws a baseball horizontally. He releases it at a height of 2.0 m and it lands 21 m from his front foot, which is, directly below the point at which he released the baseball. (a) How long was it in the air? (b) How fast did he throw it?, (a), , s, , (b), , m/s, , A cannon mounted on a pirate ship fires a cannonball at 125 m/s horizontally, at a height of 17.5 m above the ocean surface., Ignore air resistance. (a) How much time elapses until it splashes into the water? (b) How far from the ship does it land?, (a), , s, , (b), , m, , A juggler throws a ball straight up into the air and it takes 1.20 seconds to reach its peak. How many seconds will it take after, that to fall back into his hand? Assume he throws and catches the ball at the same height., s, , 3.4, , A juggler throws a ball with an initial horizontal velocity of +1.1 m/s and an initial vertical velocity of +5.7 m/s. What is its, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems, , 83
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acceleration at the top of its flight path? Make sure to consider the sign when responding. Consider the upward direction as, positive., m/s2, 3.5, , The muzzle velocity of an armor-piercing round fired from an M1A1 tank is 1770 m/s (nearly 4000 mph or mach 5.2). A tank, is at the top of a cliff and fires a shell horizontally. If the shell lands 6520 m from the base of the cliff, how high is the cliff?, , 3.6, , A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 145 m horizontally before it strikes the, ground. With what velocity did the ball leave the muzzle?, , 3.7, , A box is dropped from a spacecraft moving horizontally at 27.0 m/s at a distance of 155 m above the surface of a moon. The, , m, , m/s, rate of freefall acceleration on this airless moon is 2.79 m/s2. (a) How long does it take for the box to reach the moon's, surface? (b) What is its horizontal displacement during this time? (c) What is its vertical velocity when it strikes the surface?, (d) At what speed does the box strike the moon?, (a), , 3.8, , s, , (b), , m, , (c), , m/s, , (d), , m/s, , Two identical cannons fire cannonballs horizontally with the same initial velocity. One cannon is located on a platform, partway up a tower, and the other is on top, four times higher than the first cannon. How much farther from the tower does the, cannonball from the higher cannon land than the cannonball from the lower cannon?, i., ii., iii., iv., v., , 3.9, , 1, 2, 3, 4, 16, , times farther, , A ball rolls at a constant speed on a level table a height h above the ground. It flies off the table with a horizontal velocity v, and strikes the ground. Choose the correct expression for the horizontal distance it travels, as measured from the table's, edge., , i., ii., iii., iv., , a, b, c, d, , 3.10 An archer wants to determine the speed at which her new bow can fire an arrow. She paces off a distance d from her target, and fires her arrow horizontally. She determines that she launched it at a height h higher than where it strikes the target. She, ignores air resistance. Choose an expression in terms of h, d and g that she can use to determine the initial speed of her, arrow., , 84, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems
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i. a, ii. b, iii. c, 3.11 Randy Johnson throws a baseball horizontally from the top of a building as fast as his 2004 record of 102.0 mph (45.6 m/s)., How much time passes until it moves at an angle 13.0 degrees below the horizontal? Ignore air resistance., s, 3.12 Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?, s, 3.13 You are in a water fight, and your tricky opponent, wishing to stay dry, goes into a nearby building to spray you from above., His first attempt falls one-third of the way from the base of the building to your position. He climbs up to a higher window and, shoots again, and manages to give you a good soaking. In both cases, he aims his water gun parallel to the ground. What is, the ratio of his first height to his second height?, i., ii., iii., iv., v., vi., , 1 to 1, 1 to 2, 1 to 3, 1 to 4, 1 to 9, 1 to 11, , 3.14 You fire a squirt gun horizontally from an open window in a multistory building and make note of where the spray hits the, ground. Then you walk up to a window 5.0 m higher and fire the squirt gun again, discovering that the water goes 1.5 times as, far. Ignore air resistance. How long does the second shot take to hit the ground?, s, , Section 5 - Interactive problem: the monkey and the professor, 5.1, , Using the simulation in the interactive problem in this section, where should the monkey aim the bazooka? Why?, i. Above the professor's glove, ii. At the professor's glove, iii. Below the professor's glove, , Section 7 - Projectile motion: juggling, 7.1, , A juggler throws a ball from height of 0.950 m with a vertical velocity of + 4.25 m/s and misses it on the way down. What is its, velocity when it hits the ground?, m/s, , 7.2, , You are returning a tennis ball that your five-year-old neighbor has accidentally thrown into your yard, but you need to throw it, over a 3.0 m fence. You want to throw it so that it barely clears the fence. You are standing 0.50 m away from the fence and, throw underhanded, releasing the ball at a height of 1.0 m. State the initial velocity vector you would use to accomplish this., (, , ,, , ) m/s, , 7.3, , Your friend has climbed a tree to a height of 6.00 m. You throw a ball vertically up to her and it is traveling at 5.00 m/s when it, reaches her. What was the speed of the ball when it left your hand if you released it at a height of 1.10 m?, , 7.4, , Two people perform the same experiment twice, once on Earth and once on another planet. One observes from the top of a, 40.5 m cliff, and the other stands at ground level. The person at the bottom throws a ball at a speed of 31.0 m/s at a 75.4°, angle above the horizontal. (a) When this experiment is being conducted on Earth, what vertical velocity would you expect, the ball to have when it reaches a height of 40.5 m? (b) On the other planet, the ball is observed to have a vertical velocity of, 24.5 m/s when it reaches that height. Assuming there are no forces besides gravity acting on the ball, what is the acceleration, due to gravity on the other planet? State your answer as a positive value, just as g is stated., , m/s, , 7.5, , (a), , m/s, , (b), , m/s2, , A friend of yours has climbed a tree to a height of 12.0 m, but he forgot his bag lunch and is very hungry. You cannot toss it, straight up to him because there are branches in the way. Instead, you throw the bag from a height of 1.20 m, at a horizontal, distance of 3.50 m from the location of your friend. If you want the bag to have zero vertical velocity when it reaches your, friend, what must the horizontal and vertical velocity components of your throw be? Assume you are throwing the bag in the, positive horizontal direction., (, , ,, , ) m/s, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems, , 85
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Section 9 - Interactive problem: the monkey and the professor, part II, 9.1, , Using the simulation in the interactive problem in this section, where should the monkey aim the bazooka? Why?, i. Above the professor's glove, ii. At the professor's glove, iii. Below the professor's glove, , Section 10 - Projectile motion: aiming a cannon, 10.1 A long jumper travels 8.95 meters during a jump. He moves at 10.8 m/s when starts his leap. At what angle from the, horizontal must he have been moving when he started his jump? You may need the double-angle formula:, 2 sin u cos u = sin (2u), °, 10.2 Your projectile launching system is partially jammed. It can only launch objects with an initial vertical velocity of 42.0 m/s,, though the horizontal component of the velocity can vary. You need your projectile to land 211 m from its launch point. What, horizontal velocity do you need to program into the system?, m/s, 10.3 A professional punter can punt a ball so that its "hang time" is 4.20 seconds and it travels 39.0 m horizontally. State its initial, velocity in rectangular coordinates. Ignore air resistance, and assume the punt receiver catches the ball at the same height at, which the punter kicks it., ,, , (, , ) m/s, , 10.4 A home run just clears a fence 105 m from home plate. The fence is 4.00 m higher than the height at which the batter struck, the ball, and the ball left the bat at a 31.0° angle above the horizontal. At what speed did the ball leave the bat?, m/s, 10.5 You are the stunt director for a testosterone-laden action movie. A car drives up a ramp inclined at 10.0° above the, horizontal, reaching a speed of 40.0 m/s at the end. It will jump a canyon that is 101 meters wide. The lip of the takeoff ramp, is 201 meters above the floor of the canyon. (a) How long will the car take to cross the canyon? (b) What is the maximum, height of the cliff on the other side so that the car lands safely? (c) What angle with the horizontal will the car's velocity make, when it lands on the other side? Assume the height of the other side is the maximum value you just calculated. Express the, angle as a number between í180° and +180°., (a), , s, , (b), , m, , (c), , °, , 10.6 The infamous German "Paris gun" was used to launch a projectile with a flight time of 170 s for a horizontal distance of, 122 km. Based on this information, and ignoring air resistance as well as the curvature of the Earth, calculate (a) the gun's, muzzle speed and (b) the angle, measured above the horizontal, at which it was fired., (a), , m/s, , (b), , °, , 10.7 At its farthest point, the three-point line is 7.24 meters away from the basket in the NBA. A basketball player stands at this, point and releases his shot from a height of 2.05 meters at a 35.0 degree angle. The basket is 3.05 meters off the ground., The player wants the ball to go directly in (no bank shots). At what speed should he throw the ball?, m/s, 10.8 You and your friend are practicing pass plays with a football. You throw the football at a 35.0° angle above the horizontal at, 19.0 m/s. Your friend starts right next to you, and he moves down the field directly away from you at 5.50 m/s. How long after, he starts running should you throw the football? Assume he catches the ball at the same height at which you throw it., s, 10.9 If there were no air resistance, how fast would you have to throw a football at an initial 45° angle in order to complete an 80, meter pass?, m/s, 10.10 A soccer ball is lofted toward the goal from a distance of 9.00 m. It has an initial velocity of 12.7 m/s and is kicked at an angle, of 72.0 degrees above the horizontal. (a) When the ball crosses the goal line, how high will it be above the ground? (b) How, long does it take the ball to reach the goal line? (It will still be in the air.), , 86, , (a), , m, , (b), , s, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems
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Section 13 - Interactive problem: test your juggling!, 13.1 If the juggler in the simulation in the interactive problem in this section wants to throw the balls with a horizontal velocity of, 1.1 m/s, what should the vertical velocity be?, m/s, 13.2 If the juggler in the simulation in the interactive problem in this section wants the balls to remain in the air for 1.4 seconds, (a), what should the vertical velocity be? (b) What should the horizontal velocity be?, (a), , m/s, , (b), , m/s, , Section 15 - Relative velocity, 15.1 Agent Bond is in the middle of one of his trademark, nearly impossible getaways. He is in a convertible driving west at a, speed of 23.0 m/s (this is the reading on his speedometer) on top of a train, heading toward the back end. The train is moving, horizontally, due east at 15.0 m/s. As the convertible goes over the edge of the last train car, Bond jumps off. With what, horizontal velocity relative to the convertible should he jump to hit the ground with a horizontal velocity of 0 m/s, so that he, merely shakes, but does not spill (nor stir) his drink? Consider east to be the positive direction., m/s, 15.2 A boat is moving on Lake Tahoe with a velocity (6.5, 9.5) m/s relative to the water. However, a strong wind is causing the, water on the surface to move with velocity (í4.5, í2.5) m/s relative to the land, and this velocity must be added to the boat's., What is the velocity of the boat, as seen from a skateboarder on the shore with a velocity of (3.5, í0.70) m/s relative to the, land?, (, , ,, , ) m/s, , Copyright 2007 Kinetic Books Co. Chapter 4 Problems, , 87
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5.0 - Introduction, Objects can speed up, slow down, and change direction while they move. In short,, they accelerate., A famous scientist, Sir Isaac Newton, wondered how and why this occurs. Theories, about acceleration existed, but Newton did not find them very convincing. His, skepticism led him to some of the most important discoveries in physics., Before Newton, people who studied motion noted that the objects they observed on, Earth always slowed down. According to their theories, objects possessed an, internal property that caused this acceleration. This belief led them to theorize that a, force was required to keep things moving., This idea seems like common sense. Moving objects do seem to slow down on their, own: a car coasts to a stop, a yo-yo stops spinning, a soccer ball rolls to a halt., Newton, however, rejected this belief, instead suggesting the opposite: The nature, of objects is to continue moving unless some force acts on them. For instance,, Newton would say that a soccer ball stops rolling because of forces like friction and, air resistance, not because of some property of the soccer ball. He would say that if, these forces were not present, the ball would roll and roll and roll. A force (a kick) is, required to start the ball’s motion, and a force such as the frictional force of the, grass is required to stop its motion., Newton proposed several fundamental principles that govern forces and motion., Nearly 300 years later, his insights remain the foundation for the study of forces and, much of motion. This chapter stands as a testament to a brilliant scientist., At the right, you can use a simulation to experience one of Newton’s fundamental, principles: his law relating a net force, mass and acceleration. In the simulation, you, can attempt some of the basic tasks required of a helicopter pilot. To do so, you, control the net force upward on the helicopter. When the helicopter is in the air, the, net force equals the lift force minus its weight. (The lift force is caused by the, interaction of the spinning blades with the air, and is used to propel the helicopter, upward.) The net force, like all forces, is measured in newtons (N)., When the helicopter is in the air, you can set the net force to positive, negative, or, zero values. The net force is negative when the helicopter’s lift force is less than its, weight. When the helicopter is on the ground, there cannot be a negative net force, because the ground opposes the downward force of the helicopter’s weight and, does not allow the helicopter to sink below the Earth’s surface., The simulation starts with the helicopter on the ground and a net force of 0 N. To, increase the net force on the helicopter, press the up arrow key (Ĺ) on your, keyboard; to decrease it, press the down arrow key (Ļ). This net force will continue, to be applied until you change it., To start, apply a positive net force to cause the helicopter to rise off the ground. Next, attempt to have the helicopter reach a constant vertical, velocity. For an optional challenge, have it hover at a constant height of 15 meters, and finally, attempt to land (not crash) the helicopter., Once in the air, you may find that controlling the craft is a little trickier than you anticipated í it may act a little skittish. Welcome to (a) the, challenge of flying a helicopter and (b) Newton’s world., Here are a few hints: Start slowly! Initially, just use small net forces. You can look at the acceleration gauge to see in which direction you are, accelerating. Try to keep your acceleration initially between plus or minus 0.25 m/s2., This simulation is designed to help you experiment with the relationship between force and acceleration. If you find that achieving a constant, velocity or otherwise controlling the helicopter is challenging í read on! You will gain insights as you do., , 5.1 - Force, , Force: Loosely defined as “pushing” or “pulling.”, Your everyday conception of force as pushing or pulling provides a good starting point for explaining what a force is., There are many types of forces. Your initial thoughts may be of forces that require direct contact: pushing a box, hitting a ball, pulling a wagon,, and so on., Some forces, however, can act without direct contact. For example, the gravitational force of the Earth pulls on the Moon even though, hundreds of thousands of kilometers separate the two bodies. The gravitational force of the Moon, in turn, pulls on the Earth., , 88, , Copyright 2007 Kinetic Books Co. Chapter 05
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Electromagnetic forces also do not require direct contact. For instance, two magnets will, attract or repel each other even when they are not touching each other., We have discussed a few forces above, and could continue to discuss more of them:, static friction, kinetic friction, weight, air resistance, electrostatic force, tension, buoyant, force, and so forth. This extensive list gives you a sense of why a general definition of, force is helpful., These varied types of forces do share some essential attributes. Newton observed that, a force, or to be precise, a net force, causes acceleration., All forces are vectors: their direction matters. The weightlifter shown in Concept 1 must, exert an upward force on the barbell in order to accelerate it off the ground. For the, barbell to accelerate upward, the force he exerts must be greater than the downward, force of the Earth’s gravity on the barbell. The net force (the vector sum of all forces on, an object) and the object’s mass determine the direction and amount of acceleration., The SI unit for force is the newton (N). One newton is defined as one kg·m/s2. We will, discuss why this combination of units equals a newton shortly., , Force, “Pushing” or “pulling”, Net force = vector sum of forces, Measured in newtons (N), , We have given examples where a net force causes an object to accelerate. Forces can, 1 N = 1 kg·m/s2, also be in equilibrium (balance), which means there is no net force and no acceleration., When a weightlifter holds a barbell steady over his head after lifting it, his upward force, on the barbell exactly balances the downward gravitational force on it, and the barbell’s acceleration is zero. The net force would also be zero if, he were lifting the barbell at constant velocity., , 5.2 - Newton’s first law, , Newton’s first law: “Every body perseveres in, its state of being at rest or of moving uniformly, straight forward except insofar as it is, compelled to change by forces impressed.”, This translation of Newton’s original definition (Newton wrote it in Latin) may seem, antiquated, but it does state an admirable amount of physics in a single sentence., Today, we are more likely to summarize Newton’s first law as saying that an object, remains at rest, or maintains a constant velocity, unless a net external force acts, upon it. (Newton’s formulation even includes an “insofar” to foreshadow his second, law, which we will discuss shortly.), To state his law another way: An object’s velocity changes í it accelerates í when a net, force acts upon it. In Concept 1, a puck is shown gliding across the ice with nearly, constant velocity because there is little net force acting upon it. The puck that is, stationary in Concept 2 will not move until it is struck by the hockey stick., , Newton’s first law, Objects move at constant velocity, unless acted on by net force, , The hockey stick can cause a great change in the puck’s velocity: a professional’s slap, shot can travel 150 km/hr. Forces also cause things to slow down. As a society, we, spend a fair amount of effort trying to minimize these forces. For example, the grass of, a soccer field is specially cut to reduce the force of friction to ensure that the ball travels, a good distance when passed or shot., Top athletes also know how to reduce air resistance. Tour de France cyclists often bike, single file. The riders who follow the leader encounter less air resistance. Similarly,, downhill ski racers “tuck” their bodies into low, rounded shapes to reduce air resistance,, and they coat the bottoms of their skis with wax compounds to reduce the slowing effect, of the snow’s friction., Newton’s first law states that an object will continue to move “uniformly straight” unless, acted upon by a force. Today we state this as “constant velocity,” since a change in, direction is acceleration as much as a change in speed. In either formulation, the point, is this: Direction matters. An object not only continues at the same speed, it also moves, in the same direction unless a net force acts upon it., , Newton's first law, Objects at rest remain at rest in absence, of net force, , You use this principle every day. Even in as basic a task as writing a note, your fingers apply changing forces to alter the direction of the pen's, motion even as its speed is approximately constant., There is an important fact to note here: Newton’s laws hold true in an inertial reference frame. An object that experiences no net force in an, inertial reference frame moves at a constant velocity. Since we assume that observations are made in such a reference frame, we will be terse, here about what is meant. The surface of the Earth (including your physics lab) approximates an inertial reference frame, certainly closely, enough for the typical classroom lab experiment. (The motion of the Earth makes it less than perfect.), A car rounding a curve provides an example of a non-inertial reference frame. If you decided to conduct your experiments inside such a car,, , Copyright 2007 Kinetic Books Co. Chapter 05, , 89
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Newton’s laws would not apply. Objects might seem to accelerate (a coffee cup sliding along the dashboard, for example) yet you would, observe no net force acting on the cup. However, the nature of observations made in an accelerating reference frame is a topic far removed, from this chapter’s focus, and this marks the end of our discussion of reference frames in this chapter., , 5.3 - Mass, , Mass: A property of an object that determines, how much it will resist a change in velocity., Newton’s second law summarizes the relationship of force, mass and acceleration., Mass is crucial to understanding the second law because an object’s mass determines, how much it resists a change in velocity., More massive objects require more net force to accelerate than less massive objects., An object’s resistance to a change in velocity is called its inertial mass. It requires more, force to accelerate the bus on the right at, say, five m/s2 than the much less massive, bicycle., A common error is to confuse mass and weight. Weight is a force caused by gravity and, is measured in newtons. Mass is an object’s resistance to change in velocity and is, measured in kilograms. An object’s weight can vary: Its weight is greater on Jupiter’s, surface than on Earth’s, since Jupiter’s surface gravity is stronger than Earth’s. In, contrast, the object's mass does not change as it moves from planet to planet. The, kilogram (kg) is the SI unit of mass., , Mass, Measures an object’s resistance to, change in velocity, Measured in kilograms (kg), , 5.4 - Gravitational force: weight, , Weight: The force of gravity on an object., We all experience weight, the force of gravity. On Earth, by far the largest component of, the gravitational force we experience comes from our own planet. To give you a sense, of proportion, the Earth exerts 1600 times more gravitational force on you than does the, Sun. As a practical matter, an object’s weight on Earth is defined as the gravitational, force the Earth exerts on it., Weight is a force; it has both magnitude and direction. At the Earth' surface, the, direction of the force is toward the center of the Earth., The magnitude of weight equals the product of an object’s mass and the rate of freefall, acceleration due to gravity. On Earth, the rate of acceleration g due to gravity is, 9.80 m/s2. The rate of freefall acceleration depends on a planet’s mass and radius, so it, varies from planet to planet. On Jupiter, for instance, gravity exerts more force than on, Earth, which makes for a greater value for freefall acceleration. This means you would, weigh more on Jupiter’s surface than on Earth’s., , Weight, Force of gravity on an object, Direction “down” (toward center of, planet), , Scales, such as the one shown in Concept 1, are used to measure the magnitude of weight. The force of Earth’s gravity pulls Kevin down and, compresses a spring. This scale is calibrated to display the amount of weight in both newtons and pounds, as shown in Equation 1. Forces like, weight are measured in pounds in the British system. One newton equals about 0.225 pounds., A quick word of caution: In everyday conversation, people speak of someone who “weighs 100 kilograms,” but kilograms are units for mass,, not weight. Weight, like any force, is measured in newtons. A person with a mass of 100 kg weighs 980 newtons., , 90, , Copyright 2007 Kinetic Books Co. Chapter 05
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W = mg, W = weight, m = mass, g = freefall acceleration, Units: newtons (N), , What is this person's weight on, Earth?, W = mg, W = (80.0 kg)(9.80 m/s2), W = 784 N, 5.5 - Newton’s second law, , Newton’s second law, Net force equals mass times, acceleration, , Newton's second law: “A change in motion is proportional to the motive force, impressed and takes place along the straight line in which that force is, impressed.”, Newton stated that a change in motion (acceleration) is proportional to force. Today, physicists call this Newton’s second law, and it is stated to, explicitly include mass. Physicists state that acceleration is proportional to the net force on an object and inversely proportional to its, mass., To describe this in the form of an equation: net force equals mass times acceleration, or ȈF = ma. It is the law. The Ȉ notation means the, vector sum of all the forces acting on an object: in other words, the net force. Both the net force and acceleration are vectors that point in the, same direction, and Newton’s formulation stressed this point: “The change in motion…takes place along the straight line in which that force is, , Copyright 2007 Kinetic Books Co. Chapter 05, , 91
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impressed.” The second law explains the units that make up a newton (kg·m/s2); they, are the result of multiplying mass by acceleration., In the illustrations, you see an example of forces and the acceleration caused by the net, force. The woman who stars in these illustrations lifts a suitcase. The weight of the, suitcase opposes this motion. This force points down. Since the force supplied by the, woman is greater than the weight, there is a net force up, which causes the suitcase to, accelerate upward., In Example 1, the woman lifts the suitcase with a force of 158 N upward. The weight of, the suitcase opposes the motion with a downward force of 147 newtons. The two forces, act along a line, so we use the convention that up is positive and down is negative, and, subtract to find the net force. (If both forces were not acting along a line, you would, have to use trigonometry to calculate their components.), The net force is 11 N, upward. The mass of the suitcase is 15 kg. Newton’s second law, can be used to determine the acceleration: It equals the net force divided by the mass., The suitcase accelerates at 0.73 m/s2 in the direction of the net force, upward., , ȈF = ma, ȈF = net force, m = mass, a = acceleration, Units of force: newtons (N, kg·m/s2), , What is the suitcase's, acceleration?, ȈF = ma, F + (ímg) = ma, a = (F ímg)/m, a = (158 N í147 N)/(15 kg), a = 0.73 m/s2 (upward), 5.6 - Sample problem: Rocket Guy, Rocket Guy weighs 905 N and his jet, pack provides 1250 N of thrust,, straight up. What is his acceleration?, , Above you see “Rocket Guy,” a superhero who wears a jet pack. The jet pack provides an upward force on him, while Rocket Guy’s weight, points downward., , 92, , Copyright 2007 Kinetic Books Co. Chapter 05
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Variables, All the forces on Rocket Guy are directed along the y axis., thrust, , FT = 1250 N, , weight, , ímg = í905 N, , mass, , m, , acceleration, , a, , What is the strategy?, 1., , Determine the net force on Rocket Guy., , 2., , Determine Rocket Guy’s mass., , 3., , Use Newton’s second law to find his acceleration., , Physics principles and equations, Newton’s second law, , ȈF= ma, Step-by-step solution, We start by determining the net force on Rocket Guy., , Step, , Reason, , 1., , ȈF = FT + mg, , calculate net vertical force, , 2., , ȈF = FT + (–mg), , apply sign conventions, enter values and add, , 3., Now we find Rocket Guy’s mass., , Step, , 4., , m = weight / g, , Reason, , definition of weight, calculate m, , 5., , Finally we use Newton’s second law to calculate Rocket Guy’s acceleration., , Step, , Reason, , 6., , ȈF = ma, , Newton's second law, , 7., , a = ȈF/m, , solve for a, , 8., , enter values from steps 3 and 5, and divide, , Copyright 2007 Kinetic Books Co. Chapter 05, , 93
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5.7 - Interactive checkpoint: heavy cargo, A helicopter of mass 3770 kg can, create an upward lift force F. When, empty, it can accelerate straight, upward at a maximum of 1.37 m/s2., A careless crewman overloads the, helicopter so that it is just unable to, lift off. What is the mass of the, cargo?, , Answer:, , mc =, , kg, , 5.8 - Interactive checkpoint: pushing a box, Len pushes toward the right on a, 12.0 kg box with a force of magnitude, 31.0 N. Martina applies a 11.0 N force, on the box in the opposite direction., The magnitude of the kinetic friction, force between the box and the very, smooth floor is 4.50 N as the box, slides toward the right. What is the, box’s acceleration?, , Answer:, , a=, , m/s2, , 5.9 - Interactive problem: flying in formation, The simulation on the right will give you some practice with Newton’s second law., Initially, all the space ships have the same velocity. Their pilots want all the ships to, accelerate at 5.15 m/s2. The red ships have a mass of 1.27×104 kg, and the blue, ships, a mass of 1.47×104 kg. You need to set the amount of force supplied by the, ships’ engines so that they accelerate equally. The masses of the ships do not, change significantly as they burn fuel., Apply Newton’s second law to calculate the engine forces needed. The simulation, uses scientific notation; you need to enter three-digit leading values. Enter your, values and press GO to start the simulation. If all the ships accelerate at 5.15 m/s2,, you have succeeded. Press RESET to try again., If you have difficulty solving this problem, review Newton's second law., , 94, , Copyright 2007 Kinetic Books Co. Chapter 05
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5.10 - Newton’s third law, , Newton’s third law, Forces come in pairs, Equal in strength, opposite in direction, The forces act on different objects, , Newton's third law: “To any action there is, always an opposite and equal reaction; in other, words, the actions of two bodies upon each, other are always equal and always opposite in, direction.”, Newton’s third law states that forces come in pairs and that those forces are, equal in magnitude and opposite in direction. When one object exerts a force on, another, the second object exerts a force equal in magnitude but opposite in direction, on the first., For instance, if you push a button, it pushes back on you with the same amount of, force. When someone leans on a wall, it pushes back, as shown in the illustration, above., To illustrate this concept, we use an example often associated with Newton, the falling, apple shown in Example 1. The Earth’s gravitational force pulls an apple toward the, ground and the apple pulls upward on the Earth with an equally strong gravitational, force. These pairs of forces are called action-reaction pairs, and Newton’s third law is, often called the action-reaction law., , Fab = íFba, Force of a on b = opposite of force of b, on a, , If the forces on the apple and the Earth are equal in strength, do they cause them to, accelerate at the same rate? Newton’s second law enables you to answer this question., First, objects accelerate due to a net force, and the force of the apple on the Earth is, minor compared to other forces, such as those of the Moon or Sun. But, even if the, apple were exerting the sole force on the Earth, its acceleration would be very, very, small because of the Earth’s great mass. The forces are equal, but the acceleration for, each body is inversely proportional to its mass., , The weight of the apple is 1.5 N., What force does the apple exert, on the Earth?, 1.5 N upward, 5.11 - Normal force, , Normal force: When two objects are in direct, contact, the force one object exerts in response, to the force exerted by the other. This force is, perpendicular to the objects’ contact surface., The normal force is a force exerted by one object in direct contact with another. The, normal force is a response force, one that appears in response to another force. The, direction of the force is perpendicular to the surfaces in contact. (One meaning of, "normal" is perpendicular.), A normal force is often a response to a gravitational force, as is the case with the block, shown in Concept 1 to the right. The table supports the block by exerting a normal force, upward on it. The normal force is equal in magnitude to the block’s weight but opposite, in direction. The normal force is perpendicular to the surface between the block and the, , Normal force, Occurs with two objects in direct contact, Perpendicular to surface of contact, , Copyright 2007 Kinetic Books Co. Chapter 05, , 95
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table., You experience the normal force as well. The force of gravity pulls you down, and the, normal force of the Earth pushes in the opposite direction. The normal force prevents, you from being pulled to the center of the Earth., , Normal force opposes force, , Let’s consider the direction and the amount of the normal force when you are standing, in your classroom. It is equal in magnitude to the force of gravity on you (your weight), and points in the opposite direction. If the normal force were greater than your weight,, the net force would accelerate you upward (a surprising result), and if it were less, you, would accelerate toward the center of the Earth (equally surprising and likely more, distressing). The two forces are equal in strength and oppositely directed, so the, amount of the normal force is the same as the magnitude of your weight., What is the source of the normal force? The weight of the block causes a slight, deformation in the table, akin to you lying on a mattress and causing the springs to, compress and push back. With a normal force, the deformation occurs at the atomic, level as atoms and molecules attempt to “spring back.”, Normal forces do not just oppose gravity, and they do not have to be directed upward. A, normal force is always perpendicular to the surface where the objects are in contact., When you lean against a wall, the wall applies a normal force on you. In this case, the, normal force opposes your push and is acting horizontally., , What is the direction of the, normal force?, Perpendicular to the surface of the ramp, , We have discussed normal forces that are acting solely vertically or horizontally. The, normal force can also act at an angle, as shown with the block on a ramp in Example 1., The normal force opposes a component of the block’s weight, not the full weight. Why?, Because the normal force is always perpendicular to the contact surface. The normal, force opposes the component of the weight perpendicular to the surface of the ramp., Example 2 makes a similar point. Here again the normal force and weight are not equal, in magnitude. The string pulls up on the block, but not enough to lift it off the surface., Since this reduces the force the block exerts on the table, the amount of the normal, force is correspondingly reduced. The force of the string reduces the net downward, force on the table to 75 N, so the amount of the normal force is 75 N, as well. The, direction of the normal force is upward., , The string supplies an upward, force on the block which is, resting on the table. What is the, normal force of the table on the, block?, ȈF = ma = 0, FN + T + (ímg) = 0, FN + 35 N í 110 N = 0, FN = 75 N (upward), , 5.12 - Tension, , Tension: Force exerted by a string, cord, twine,, rope, chain, cable, etc., In physics textbooks, tension means the pulling force conveyed by a string, rope, chain,, tow-bar, or other form of connection. In this section, we will use a rope to illustrate the, concept of tension., The rope in Concept 1 is shown exerting a force on the block; that force is called, tension. This definition differs slightly from the everyday use of the word tension, which, often refers to forces within a material or object í or a human brain before exams., In physics problems, two assumptions are usually made about the nature of tension., First, the force is transmitted unchanged by the rope. The rope does not stretch or, otherwise diminish the force. Second, the rope is treated as having no mass (it is, massless). This means that when calculating the acceleration of a system, the mass of, the rope can be ignored., , Tension, Force through rope, string, etc., , Example 1 shows how tension forces can be calculated using Newton’s second law. There are two forces acting on the block: its weight and, the tension. The vector sum of those forces, the net force, equals the product of its mass and acceleration. Since the mass and acceleration, are stated, the problem solution shows how the tension can be determined., , 96, , Copyright 2007 Kinetic Books Co. Chapter 05
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What is the amount of tension in, the rope?, ȈF = ma, T + (ímg) = ma, , T = 19 N (upward), 5.13 - Newton's second and third laws, It might seem that Newton’s third law could lead to the conclusion that forces do not, cause acceleration, because for every force there is an equal but opposite force. If for, every force there is an equal but opposite force, how can there be a net non-zero force?, The answer lies in the fact that the forces do not act on the same object. The pair of, forces in an action-reaction pair acts on different objects. In this section, we illustrate, this often confusing concept with an example., Consider the box attached to the rope in Concept 1. We show two pairs of actionreaction forces. Normally, we draw all forces in the same color, but in this illustration,, we draw each pair in a different color. One pair is caused by the force of gravity. The, force of the Earth pulls the box down. In turn, the box exerts an upward gravitational, force of equal strength on the Earth., There is also a pair of forces associated with the rope. The tension of the rope pulls up, on the box. In response, the box pulls down on the rope. These forces are equal but, opposite and form a second action-reaction pair. (Here we only focus on pairs that, include forces acting on the box or caused by the box. We ignore other action-reaction, pairs present in this example, such as the hand pulling on the rope, and the rope pulling, on the hand.), , Action-reaction pairs, Two pairs involving box:, ·Gravity, ·Tension & response, , Now consider only the forces acting on the box. This means we no longer consider the forces the box exerts on the Earth and on the rope. The, two forces on the box are gravity pulling it down and tension pulling it up. In this example, we have chosen to make the force of tension greater, than the weight of the box., The Concept 2 illustration reflects this scenario: The tension vector is longer than the weight vector, and the resulting net force is a vector, upward. Because there is a net upward force on the box, it accelerates in that direction., Now we will clear up another possible misconception: that the weight of an object resting on a surface and the resulting normal force are an, action-reaction pair. They are not. Since they are often equal but opposite, they are easily confused with an action-reaction pair. Consider a, block resting on a table. The action-reaction pair is the Earth pulling the block down and the block pulling the Earth up. It is not the weight of the, block and the normal force., Here is one way to confirm this: Imagine the block is attached to a rope pulling it up so that it just touches the table. The normal force is now, near zero, yet the block’s weight is unchanged. If the weight and the normal force are supposed to be equal but opposite, how could the normal, force all but disappear? The answer is that the action-reaction pair in question is what is stated above: the equal and opposite forces of gravity, between the Earth and the block., , Copyright 2007 Kinetic Books Co. Chapter 05, , 97
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Net force on box, Tension minus weight, Causes acceleration, , 5.14 - Free-body diagrams, , Free-body diagram: A, drawing of the external, forces exerted on an object., Free-body diagrams are used to display multiple, forces acting on an object. In the drawing above, the, free body is a monkey, and the free-body diagram in, Concept 1 shows the forces acting upon the monkey:, the tension forces of the two ropes and the force of, gravity., , A monkey hanging from two ropes., , The diagram only shows the external forces acting on the monkey. There are other, forces present in this configuration, such as forces within the monkey, and forces that, the monkey exerts. Those forces are not shown; a free-body diagram shows just the, forces that act on a single object like the monkey., Although we often draw force vectors where they are applied to an object, in free-body, diagrams it is useful to draw the vectors starting from a single point, typically the origin., This allows the components of the vectors to be more easily analyzed. You see this in, Concept 1., Free-body diagrams are useful in a variety of ways. They can be used to determine the, magnitudes of forces. For instance, if the mass of the monkey and the orientations of, the ropes are known, the tension in each rope can be determined., When forces act along multiple dimensions, the forces and the resulting acceleration, need to be considered independently in each dimension. In the illustration, the monkey, is stationary, hanging from two ropes. Since there is no vertical acceleration, there is no, net force in the vertical dimension. This means the downward force of gravity on the, monkey must equal the upward pull of the ropes., , Free-body diagrams, Shows all external forces acting on body, Often drawn from the origin, , The two ropes also pull horizontally (along the x axis). Because the monkey is not, accelerating horizontally, these horizontal forces must balance as well. By considering, the forces acting in both the horizontal and vertical directions, the tensions of the ropes, can be determined., In Example 1, one of the forces shown is friction, f. Friction acts to oppose motion when, two objects are in contact., , Draw a free-body diagram of the, forces on the box., , 98, , Copyright 2007 Kinetic Books Co. Chapter 05
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Free-body diagram of forces on, box, , 5.15 - Interactive problem: free-body diagram, A rope pulls the block against friction., Draw a free-body diagram. The block, accelerates at 11 m/s2 if the diagram, is correct., , In this section, you practice drawing a free-body diagram. Above, you see the, situation: A block is being pulled horizontally by a rope. It accelerates to the right at, 11 m/s2. In the simulation on the right, the force vectors on the block are drawn, but, each one points in the wrong direction, has the wrong magnitude, or both. We, ignore the force of air resistance in this simulation., Your job is to fix the force vectors. You do this by clicking on the heads of the, vectors and dragging them to point in the correct direction. (To simplify your work,, they “snap” to vertical and horizontal orientations, but you do need to drag them, close before they will snap.) You change both their lengths (which determine their, magnitudes) and their directions with the mouse., The mass of the block is 5.0 kg. The tension force T is 78 N and the force of friction, f is 23 N. The friction force acts opposite to the direction of the motion. Calculate the, magnitudes of the weight mg and the normal force FN to the nearest newton, and, then drag the heads of the vectors to the correct positions, or click on the up and, down arrow buttons, and press GO. If you are correct, the block will accelerate to, the right at 11 m/s2. If not, the block will move based on the net force as determined, by your vectors as well as its mass. Press RESET to try again., There is more than one way to arrange the vectors to create the same acceleration, but there is only one arrangement that agrees with all the, information given., If you have difficulty solving this problem, review the sections on weight and normal force, and the section on free-body diagrams., , 5.16 - Friction, , Friction: A force that resists, the motion of one object, sliding past another., If you push a cardboard box along a wooden floor,, you have to push to overcome the force of friction., This force makes it harder for you to slide the box., The force of friction opposes any force that can cause, one object to slide past another. There are two types, of friction: static and kinetic. These forces are, discussed in more depth in other sections. In this, , Friction between the buffalo's back and the tree scratches an itch., , Copyright 2007 Kinetic Books Co. Chapter 05, , 99
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section, we discuss some general properties of friction., The amount of friction depends on the materials in contact. For example, the box would, slide more easily over ice than wood. Friction is also proportional to the normal force., For a box on the floor, the greater its weight, the greater the normal force, which, increases the force of friction., Humans expend many resources to combat friction. Motor oil, Teflon™, WD-40™, TriFlo™ and many other products are designed to reduce this force. However, friction can, be very useful. Without it, a nail would slip out of a board, the tires of a car would not be, able to “grip” the road, and you would not be able to walk., Friction exists even between seemingly smooth surfaces. Although a surface may, appear smooth, when magnified sufficiently, any surface will look bumpy or rough, as, the illustration in Concept 2 on the right shows. The magnified picture of the “smooth”, crystal reveals its microscopic “rough” texture. Friction is a force caused by the, interaction of molecules in two surfaces., You might think you can defeat friction by creating surfaces that are highly polished., Instead, you may get an effect called cold welding, in which the two highly polished, materials fuse together. Cold welding can be desirable, as when an aluminum, connector is crimped onto a copper wire to create a strong electrical connection., , Friction, Force that opposes “sliding” motion, Varies by materials in contact, Proportional to normal force, , Objects can also move in a fashion that is called slip and slide. They slide for a while,, stick, and then slide some more. This phenomenon accounts for both the horrid noise, generated by fingernails on a chalkboard and the joyous noise of a violin. (Well, joyous, when played by some, chalkboard-like when played by others.), , Friction, Microscopic properties determine friction, force, , 5.17 - Static friction, , Static friction: A force that resists the sliding, motion of two objects that are stationary relative, to one another., Imagine you are pushing a box horizontally but cannot move it due to friction. You are, experiencing a response force called static friction. If you push harder and harder, the, amount of static friction will increase to exactly equal í but not exceed í the amount of, horizontal force you are supplying. For the two surfaces in contact, the friction will, increase up to some maximum amount. If you push hard enough to exceed the, maximum amount of static friction, the box will slide., For instance, let’s say the maximum amount of static friction for a box is 30 newtons. If, you push with a force of 10 newtons, the box does not move. The force of static friction, points in the opposite direction of your force and is 10 newtons as well. If it were less,, the box would slide in the direction you are pushing. If it were greater, the box would, accelerate toward you. The box does not move in either direction, so the friction force is, 10 newtons. If you push with 20 newtons of force, the force of static friction is 20, newtons, for the same reasons., , Static friction, Force opposing sliding when no motion, Balances "pushing" force until object, slides, Maximum static friction proportional to:, ·coefficient of static friction, ·normal force, , You keep pushing until your force is 31 newtons. You have now exceeded the, maximum force of static friction and the box accelerates in the direction of the net force. The box will continue to experience friction once it is, sliding, but this type of friction is called kinetic friction., Static friction occurs when two objects are motionless relative to one another. Often, we want to calculate the maximum amount of static friction, so that we know how much force we will have to apply to get the object to move. The equation in Equation 1 enables you to do so. It depends, on two values. One is the normal force, the perpendicular force between the two surfaces. The second is called the coefficient of static friction., Engineers calculate this coefficient empirically. They place an object (say, a car tire) on top of another surface (perhaps ice) and measure how, hard they need to push before the object starts to move. Coefficients of friction are specific to the two surfaces. Some examples of coefficients, of static friction are shown in the table in Equation 2., You might have noticed a fairly surprising fact: The amount of surface area between the two objects does not enter into the calculation of, maximum static friction. In principle, whether a box of a given mass has a surface area of one square centimeter or one square kilometer, the, maximum amount of static friction is constant. Why? With the greater contact area, the normal and frictional forces per unit area diminish, , 100, , Copyright 2007 Kinetic Books Co. Chapter 05
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proportionally., , fs,max = μsFN, fs,max = maximum static friction, μs = coefficient of static friction, FN = normal force, , Coefficients of static friction, , Anna is pushing but the box, does not move. What is the force, of static friction?, fs = 7 N to the right, , Copyright 2007 Kinetic Books Co. Chapter 05, , 101
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What is the maximum static, friction force? The coefficient of, static friction for these materials, is 0.31., fs,max = μsFN, fs,max = (0.31)(27 N), fs,max = 8.4 N, 5.18 - Kinetic friction, , Kinetic friction: Friction when an object slides, along another., Kinetic friction occurs when two objects slide past each other. The magnitude of kinetic, friction is less than the maximum amount of static friction for the same objects. Some, values for coefficients of kinetic friction are shown in Equation 2 to the right. These are, calculated empirically and do not vary greatly over a reasonable range of velocities., Like static friction, kinetic friction always opposes the direction of motion. It has a, constant value, the product of the normal force and the coefficient of kinetic friction., In Example 1, we state the normal force. Note that the normal force in this case does, not equal the weight; instead, it equals a component of the weight. The other, component of the weight is pulling the block down the plane., , 102, , Kinetic friction, Friction opposing sliding in motion, Force constant as object slides, Proportional to:, ·coefficient of kinetic friction, ·normal force, , Copyright 2007 Kinetic Books Co. Chapter 05
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fk = μkFN, fk = force of kinetic friction, μk = coefficient of kinetic friction, FN = normal force, , Coefficients of kinetic friction, , What is the force of friction?, fk = μkFN, fk = (0.67)(10 N), fk = 6.7 N (pointing up the ramp), , Copyright 2007 Kinetic Books Co. Chapter 05, , 103
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5.19 - Interactive checkpoint: moving the couch, While rearranging your living room,, you push your couch across the floor, at a constant speed with a horizontal, force of 69.0 N. You are using special, pads on the couch legs that help it, slide easier. If the couch has a mass, of 59.5 kg, what is the coefficient of, kinetic friction between the pads and, the floor?, , Answer:, , μk =, , 5.20 - Sample problem: friction and tension, The coefficient of kinetic friction is, 0.200. What is the magnitude of the, tension force in the rope?, , Above, you see a block accelerating to the right due to the tension force applied by a rope. What is the magnitude of tension the rope applies to, the block?, Starting this type of problem with a free-body diagram usually proves helpful., Draw a free-body diagram, , Variables, , x component, , ycomponent, , normal force, , 0, , FN = mg, , acceleration, , a = 2.20 m/s2, , 0, , tension, , T, , 0, , friction force, , ífk, , 0, , mass, , m = 1.60 kg, , coefficient of kinetic friction, , μk = 0.200, , What is the strategy?, 1., , Draw a free-body diagram., , 2., , Find an expression for the net force on the block., , 3., , Substitute the net force into Newton’s second law to find the tension., , Are there any useful relationships?, Since the surface is horizontal, the amount of normal force equals the weight of the block., , 104, , Copyright 2007 Kinetic Books Co. Chapter 05
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Physics principles and equations, Newton’s second law, , ȈF = ma, The magnitude of the force of kinetic friction is found by, , fk =μkFN, Step-by-step solution, We begin by determining the net horizontal force on the block., , Step, , Reason, , 1., , ȈF = T + (–fk), , net horizontal force, , 2., , fk = μkFN, , equation for kinetic friction, , 3., , ȈF = T – μkFN, , substitute equation 2 into 1, , 4., , ȈF = T – μkmg enter value of FN, , Now we substitute the net force just found into Newton’s second law. This allows us to solve for the tension force., , Step, , Reason, , 5., , ȈF = ma, , Newton's second law, , 6., , T – μkmg = ma, , substitute equation 4 into 5, , 7., , T = μkmg + ma, , solve for tension, , 8., , T = (0.200)(1.60 kg)(9.80 m/s2) + (1.60 kg)(2.20 m/s2) enter values, , 9., , T = 6.66 N, , evaluate, , 5.21 - Sample problem: a force at an angle, What is the magnitude of the net, force on the ball along each axis, and, what is the ball's acceleration along, each axis?, , Above, you see a bat hitting a ball at an angle. You are asked to find the net force and the acceleration of the ball along the x and y axes., Draw a free-body diagram, , The forces on the ball are its weight down and the force of the bat at the angle ș to the x axis., , Copyright 2007 Kinetic Books Co. Chapter 05, , 105
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Variables, , x component, , y component, , weight, , 0, , mg sin 270° = í1.40 N, , force, , F cosș, , F sinș, , acceleration, , ax, , ay, , force, , F = 262 N, , angle, , ș = 60.0°, , mass, , m = mg/g = (1.40 N) / (9.80 m/s2) = 0.143 kg, , What is the strategy?, 1., , Draw a free-body diagram., , 2., , Use trigonometry to calculate the net force on the ball along each axis., , 3., , Use Newton’s second law to find the acceleration of the ball along each axis. The mass of the ball is not given, but you can determine it, because you are told its weight. We do this in the variables table., , Physics principles and equations, Newton’s second law, , ȈF = ma, Step-by-step solution, We begin by calculating the net force along the x axis., , Step, , Reason, , 1., , ȈFx = F cos ș, , 2., , ȈFx = (262 N)(cos 60.0°) x component of force, , 3., , ȈFx = 131 N, , net force along x axis, , evaluate, , We next calculate the force along the y axis. In this case, there are two forces to consider., , Step, , 4., , Reason, , ȈFy = F sin ș + (í1.40 N), , net force along y axis, enter values, , 5., 6., , ȈFy = 225 N, , evaluate, , Now we calculate the acceleration along the x axis, using Newton’s second law., , Step, , Reason, , 7., , ȈFx = max, , Newton's second law, , 8., , ax = ȈFx/m, , solve for ax, , 9., , ax = (131 N) / (0.143 kg) enter values from step 3 and table, , 10. ax = 916 m/s2, , division, , We calculate the acceleration along the y axis., , Step, , Reason, , 11. ȈFy = may, , Newton's second law, , 12. ay = ȈFy/m, , solve for ay, , 13. ay = (225 N)/(0.143 kg) enter values from step 6, table, 14. ay = 1570 m/s2, , division, , The acceleration values may seem very large, but this is the acceleration during the brief moment the bat is in contact with the ball., , 106, , Copyright 2007 Kinetic Books Co. Chapter 05
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5.22 - Interactive problem: forces on a sliding block, A rope pulls the block up the ramp., Draw a free-body diagram of the, forces on the block. If the diagram is, correct, the block will accelerate up, the ramp at 4.3 m/s2., , Above, you see an illustration of a block that is being pulled up a ramp by a rope. In, the simulation on the right, the force vectors on the block are drawn, but they are in, the wrong directions, have the wrong magnitudes, or both. Your job is to fix the, force vectors. If you do this correctly, the block will accelerate up the ramp at a rate, of 4.3 m/s2. If not, the block will move due to the net force as determined by your, vectors as well as its mass., The mass of the block is 6.0 kg. The amount of tension from the rope is 78 N and, the coefficient of kinetic friction is 0.45. The angle the ramp makes with the, horizontal is 30°. Calculate (to the nearest newton) the directions and magnitudes, of the weight, normal force and friction force. Drag the head of a vector to set its, magnitude and direction. You can also set the magnitudes in the control panel. The, vectors will “snap” to angles., When you have arranged all the vectors, press the GO button. If your free-body, diagram is accurate, the block will accelerate up the ramp at 4.3 m/s2. Press RESET, to try again., There is more than one way to set the vectors to produce the same acceleration, but only one arrangement agrees with all the information, given. If you have difficulty solving this problem, review the sections on kinetic friction and the normal force, and the sample problem involving, a force at an angle., , 5.23 - Hooke’s law and spring force, You probably already know a few basic things about springs: You stretch them, they, pull back on you. You compress them, they push back., As a physics student, though, you are asked to study springs in a more quantitative, way. Let’s consider the force of a spring using the configuration shown in Concept 1., Initially, no force is applied to the spring, so it is neither stretched nor compressed., When no force is applied, the end of the spring is at a position called the rest point, (sometimes called the equilibrium point)., Then we stretch the spring. In the illustration to the right, the hand pulls to the right, so, the end of the spring moves to the right, away from its rest point, and the spring pulls, back to the left., Hooke’s law is used to determine how much force the spring exerts. It states that the, amount of force is proportional to how far the end of the spring is stretched or, compressed away from its rest point. Stretch the end of the spring twice as far from its, rest point, and the amount of force is doubled., The amount of force is also proportional to a spring constant, which depends on the, construction of the spring. A “stiff” spring has a greater spring constant than one that is, easier to stretch. Stiffer springs can be made from heavier gauge materials. The units, for spring constants are newtons per meter (N/m)., , Spring force, Force exerted by spring depends on:, ·How much it is stretched or, compressed, ·Spring constant, , The equation for Hooke’s law is shown in Equation 1. The spring constant is represented by k. The displacement of the end of the spring is, represented by x. At the rest position, x = 0. When the spring is stretched, the displacement of the end of the spring has a positive x value., When it is compressed, x is negative., Hooke's law calculates the magnitude of the spring force. The equation has a negative sign to indicate that the force of a spring is a restoring, force, which means it acts to restore the end of the spring to its rest point. Stretch a spring and it will pull back toward the rest position;, compress a spring, and it will push back toward the rest position. The direction of the force is the opposite of the direction of the displacement., , Copyright 2007 Kinetic Books Co. Chapter 05, , 107
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Hooke's law, Fs = íkx, Fs = spring force, k = spring constant, x = displacement of end from rest point, , What is the force exerted by the, spring?, Fs = íkx, Fs = í(4.2 N/m)(0.36 m), Fs = í1.5 N (to the left), 5.24 - Air resistance, , Air resistance: A force that opposes motion in, air., If you parachute, or bike or ski, you have experienced air resistance. In each of these, activities, you move through a fluid í air í that resists your motion. As you move, through the air, you collide with the molecules that make up the atmosphere. Although, air is not very dense and the molecules are very small, there are so many of them that, their effects add up to a significant force. The sum of all these collisions is the force, called air resistance., Unlike kinetic friction, air resistance is not constant but increases as the speed of the, object increases. The force created by air resistance is called drag., , Air resistance, Drag force opposes motion in air, , The formula in Equation 1 supplies an approximation of the force of air resistance for, Force increases as speed increases, objects moving at relatively high speeds through air. For instance, it is a relevant, equation for the skysurfer shown in Concept 1, or for an airplane. The resistance is, proportional to the square of the speed and to the cross sectional area of the moving object. (For the skysurfer, the board would constitute the, main part of the cross sectional area.) It is also proportional to an empirically determined constant called the drag coefficient., The shape of an object determines its drag coefficient. A significant change in speed can change the drag coefficient, as well. Aerospace, engineers definitely earn their keep by analyzing air resistance using powerful computers. They also use wind tunnels to check their, computational results., Another interesting implication of the drag force equation is that objects will reach what is called terminal velocity. Terminal velocity is the, maximum speed an object reaches when falling. The drag force increases with speed while the force of gravity is constant; at some point, the, , 108, , Copyright 2007 Kinetic Books Co. Chapter 05
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upward drag force equals the downward force of gravity. When this occurs, there is no, net force and the object ceases to accelerate and maintains a constant speed. The, equation for calculating terminal velocity is shown in Equation 2. It is derived by setting, the drag force equal to the object’s weight and solving for the speed., Research has actually determined that cats reach terminal velocity after falling six, stories. In fact, they tend to slow down after six stories. Here’s why this occurs: The cat, achieves terminal velocity and then relaxes a little, which expands its cross sectional, area and increases its drag force. As a result, it slows down. One has to admire the cat, for relaxing in such a precarious situation (or perhaps doubt its intelligence). If you think, this may be an urban legend, consult the Journal of the American Veterinary, Association, volume 191, page 1399., , Terminal velocity, Drag force equals weight, , Air resistance, FD = ½CȡAv2, FD = drag force, C = drag coefficient for object, ȡ = air density, A = cross-sectional area, v = velocity, , Terminal velocity, , vT = terminal velocity, mg = weight, , Copyright 2007 Kinetic Books Co. Chapter 05, , 109
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Drag coefficients, Based on approximations of shape, , The drag coefficient C is 0.49, and the air density ȡ is 1.1 kg/m3., What is the skydiver’s terminal, velocity?, , vT = 52 m/s, 5.25 - Interactive Problem: Forces in Multiple Dimensions, Now, you will get some additional practice applying Newton’s laws. More, specifically, you will use them in situations where multiple forces are acting on a, single object., If the application of multiple forces results in a net force acting on an object, it, accelerates. On the other hand, if the forces acting on it sum to zero in every, dimension, the result is equilibrium. The object does not accelerate; it either, maintains a constant velocity, or remains stationary. (Forces can also cause an, object to rotate, but rotational motion is a later topic in mechanics.), Equilibrium is an important topic in engineering. The school buildings you study in,, the bridges you travel across í all such structures require careful design to ensure, that they remain in equilibrium., The simulation on the right will help you develop an understanding for how forces in, different directions combine when applied to an object. The 5.0 kg ball has two, forces acting on it, F1 and F2. They act on it as long as the ball is on the screen., You control the direction and magnitude of each force. In the simulation, you set a, force vector's direction and magnitude by dragging its arrowhead; You will notice the angles are restricted to multiples of 90°. You can also, adjust the magnitude of each vector with a controller in the control panel. The net force is shown in the simulation; it is the vector sum of F1, and F2., You can check the box “Display vectors head to tail” if you would like to see them graphically combined in that fashion. Press GO to start the, simulation and set the ball moving in response to the forces on it., Here are some challenges for you. First, set the forces so that the ball does not move at all when you press GO. The individual forces must be, at least 10 newtons, so setting them both to zero is not an option!, Next, hit each of the three animated targets. The center of one is directly to the right of the ball and the center of another is at a 45° angle, , 110, , Copyright 2007 Kinetic Books Co. Chapter 05
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above the horizontal from the ball. Set the individual vectors and press GO to hit the center of each target in turn., The target to the left is at a 150° angle. It is the “extra credit” target. Determining the correct ratio of vectors will require a little thought. We, allow for rounding with this target; if you set one of the vectors to 10 N, you can solve the problem by setting the other one to the appropriate, closest integer value., , 5.26 - Gotchas, An object has a speed of 20 km/h.It swerves to the left but maintains the same speed.Was a force involved? Yes. A change in speed or, direction is acceleration, and acceleration requires a force., An object is moving.A net force must be acting on it. No. Only if the object is accelerating (changing speed or direction) is there a net force., Constant velocity means there is no net force., No acceleration means no forces are present. Close. No acceleration means no net forces. There can be a balanced set of forces and no, acceleration., “I weigh 70 kilograms.” False. Kilograms measure mass, not weight., “I weigh the same on Jupiter as I do on Mars.” Not unless you dieted (lost mass) as you traveled from Jupiter to Mars. Weight is gravitational, force, and Mars exerts less gravitational force., “My mass is the same on Jupiter and Mars.” Yes., The normal force is the response force to gravity. This is too specific of a definition. The normal force appears any time two objects are brought, in contact. It is not limited to gravity. For instance, if you lean against a wall, the force of the wall on you is a normal force. If you stand on the, ground, the normal force of the ground is a response force to gravity., “I push against a wall with a force of five newtons. The wall pushes back with the same force.” Close, but it is better to say, “The same amount, (magnitude) of force but in the opposite direction.”, “I pull on the Earth with the same amount of gravitational force that the Earth exerts on me.”True. You are an action-reaction pair., , Copyright 2007 Kinetic Books Co. Chapter 05, , 111
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5.27 - Summary, Force, and Newton's laws which describe force, are fundamental concepts in the, study of physics. Force can be described as a push or pull. It is a vector quantity, that is measured in newtons (1 N = 1 kg·m/s2). Net force is the vector sum of all the, external forces on an object., Free-body diagrams depict all the external forces on an object. Even though the, forces may act on different parts of the object, free-body diagrams are drawn so that, the forces are shown as being applied at a single point., Newton's first law states that an object maintains a constant velocity (including, remaining at rest) until a net force acts upon it., , weight = mg, Newton’s second law, , ȈF = ma, Newton’s third law, , Mass is the property of an object that determines its resistance to a change in, velocity, and it is a scalar, measured in kilograms. Mass should not be confused, with weight, which is a force caused by gravity, directed toward the center of the, Earth., , Fab = –Fba, , Newton's second law states that the net force on an object is equal to its mass times, its acceleration., , fs,max = μsFN, , Newton's third law states that the forces that two bodies exert on each other are, always equal in magnitude and opposite in direction., The normal force is a force that occurs when two objects are in direct contact. It is, always directed perpendicular to the surface of contact., Tension is a force exerted by a means of connection such as a rope, and the, tension force always pulls on the bodies to which the rope is attached., Friction is a force that resists the sliding motion of two objects in direct contact. It is, proportional to the magnitude of the normal force and varies according to the, composition of the objects., , Static friction, , Kinetic friction, , f k = μ kF N, Hooke’s law, , Fs = –kx, , Static friction is the term for friction when there is no relative motion between two objects. It balances any applied pushing force that tends to, slide the body, up to a maximum determined by the normal force and the coefficient of static friction between the two objects, μs. If the applied, pushing force is greater than the maximum static friction force, then the object will move., Once an object is in motion, kinetic friction applies. The force of kinetic friction is determined by the magnitude of the normal force multiplied by, , μk, the coefficient of kinetic friction between the two objects., Hooke's law describes the force that a spring exerts when stretched or compressed away from its equilibrium position. The force increases, linearly with the displacement from the equilibrium position. The equation for Hooke’s law includes k, the spring constant; a value that depends, on the particular spring. The negative sign indicates that the spring force is a restoring force that points in the opposite direction as the, displacement, that is, it resists both stretching and compression., Air resistance, or drag, is a force that opposes motion through a fluid such as air. Drag increases as speed increases. Terminal velocity is, reached when the drag force on a falling object equals its weight, so that it ceases to accelerate., , 112, , Copyright 2007 Kinetic Books Co. Chapter 05
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Chapter 5 Problems, , Conceptual Problems, C.1, , Can an object move if there is no net force on it? Explain., , C.2, , Suppose you apply a force of 1 N to block A and a force of 2 N to block B. Does it follow that block B has twice the, acceleration of block A? Justify your answer using Newton's second law., , C.3, , When a brick rests on a flat, stationary, horizontal table, there is an upward normal force on it from the table. Explain why the, brick does not accelerate upward in response to this force., , C.4, , A rocket in space can change course with its engines. Since in empty space there is nothing for the exhaust gases to push, on, how can it accelerate?, , C.5, , Blocks 1 and 2, and 2 and 3 are connected by, two identical thin wires. All three blocks are, resting on a frictionless table. Block 1 is pulled, by a constant force and all three blocks, accelerate equally in a line, with block 1, leading. Are the tensions in the two wires the, same or different? If the tensions are different,, which has the larger magnitude? Why?, , Yes, , Yes, , No, , No, , i. Tensions are the same, ii. Greater between blocks 1 and 2, iii. Greater between blocks 2 and 3, C.6, , Two blocks of different mass are connected by a massless rope which goes over a massless, frictionless pulley. The rope is, free to move, and both of the blocks hang vertically. What is the magnitude of the tension in the rope?, i., ii., iii., iv., v., , The weight of the heavier block, The weight of the lighter block, Their combined weight, A value between the two weights, zero, , C.7, , Why is the frictional force proportional to the normal force, and not weight?, , C.8, , A college rower can easily push a small car along a flat road, but she cannot lift the car in the air. Since the mass of the car is, constant, how can you explain this discrepancy?, , C.9, , Without friction, you would not be able to walk along a level sidewalk. Why? (Imagine being stranded in the middle of an ice, rink, wearing shoes made of ice.), , C.10 If an acrobat who weighs 800 N is clinging to a vertical pole using only his hands, neither moving up nor down, can we, determine the coefficient of static friction between his hands and the pole? Explain your answer., Yes, , No, , C.11 State two reasons why it is easier to push a heavy object down a hill than it is to push that same object across a flat,, horizontal surface., C.12 One end of a spring is attached firmly to a wall, and a block is attached to the other end. When the spring is fully compressed,, it exerts a force F on the block, and when the spring is fully extended, the force it exerts on the block is íF. What is the force, of the spring on the block at (a) equilibrium (neither compressed nor stretched), (b) halfway between maximum stretch and, equilibrium, and (c) halfway between maximum compression and equilibrium? Carefully consider the signs in your answer,, which indicate direction, and express your answers in terms of F., , Copyright 2007 Kinetic Books Co. Chapter 5 Problems, , 113
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(a), , i., ii., iii., iv., v., i., ii., iii., iv., v., i., ii., iii., iv., v., , (b), , (c), , F, -F, 0, -F/2, F/2, F, -F, 0, -F/2, F/2, F, -F, 0, -F/2, F/2, , C.13 From the example of the falling cat, we see that the cross-sectional area of a falling object affects its terminal velocity. Does, an object's mass also affect its terminal velocity? Why or why not?, Yes, , No, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following question. If the net force on the helicopter, is zero, what must the helicopter be doing?, i., ii., iii., iv., , Rising, Falling, Staying still, It can be doing any of these, , Section 2 - Newton’s first law, 2.1, , An airplane of mass 2867 kg flies at a constant horizontal velocity. The force of air resistance on it is 2225 N. What is the net, force on the plane (magnitude and direction)?, i., ii., iii., iv., , 2.2, , 0 N (direction does not matter), 2225 N opposing the motion, 642 N at 30° below horizontal, It is impossible to say, , Three children are pulling on a toy that has a mass of 1.25 kg. Child A pulls with force (5.00, 6.00, 2.50) N. Child B pulls with, force (0.00, 9.20, 2.40) N. With what force must Child C pull for the toy to remain stationary?, (, , ,, , ,, , )N, , Section 4 - Gravitational force: weight, 4.1, , A weightlifter can exert an upward force of 3750 N. If a dumbbell has a mass of 225 kg, what is the maximum number of, dumbbells this weightlifter could hold simultaneously if he were on the Moon? (The Moon's acceleration due to gravity is, approximately 0.166 times freefall acceleration on Earth.) The weightlifter cannot pick up a fraction of a dumbbell, so make, sure your answer is an integer., , 4.2, , (a) How much does a 70.0 kg person weigh on the Earth? (b) How much would she weigh on the Moon (gmoon = 0.166g)? (c), , dumbells, How much would she weigh on a neutron star where gstar = 1.43×1011g?, (a), , 4.3, , 114, , N, , (b), , N, , (c), , N, , A dog weighs 47.0 pounds on Earth. (a) What is its weight in newtons? (One newton equals 0.225 pounds.) (b) What is its, mass in kilograms?, (a), , N, , (b), , kg, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems
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4.4, , A dog on Earth weighs 136 N. The same dog weighs 154 N on Neptune. What is the acceleration due to gravity on Neptune?, m/s2, , Section 5 - Newton’s second law, 5.1, , When empty, a particular helicopter of mass 3770 kg can accelerate straight upward at a maximum acceleration of 1.34 m/s2., A careless crewman overloads the helicopter so that it is just unable to lift off. What is the mass of the cargo?, kg, , 5.2, , 5.3, , A 0.125 kg frozen hamburger patty has two forces acting on it that determine its horizontal motion. A 2.30 N force pushes it to, the left, and a 0.800 N force pushes it to the right. (a) Taking right to be positive, what is the net force acting on it? (b) What is, its acceleration?, (a), , N, , (b), , m/s2, , In the illustration, you see the graph of an, object's acceleration over time. (a) At what, moment is it experiencing the most positive, force? (b) The most negative force? (c) Zero, force?, (a), , 5.4, , s, , (b), , s, , (c), , s, , The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes, another boat to accelerate at 0.125 m/s2. (a) What is the mass of the second boat? (b) One of the boats is now loaded on the, other, and the same net force is applied to this combined mass. What acceleration does it cause?, (a), , kg, , (b), , m/s2, , 5.5, , A flea has a mass of 4.9×10í7 kg. When a flea jumps, its rear legs act like catapults, accelerating it at 2400 m/s2. What force, do the flea's legs have to exert on the ground for a flea to accelerate at this rate?, , 5.6, , An extreme amusement park ride accelerates its riders upward from rest to 50.0 m/s in 7.00 seconds. Ignoring air resistance,, what average upward force does the seat exert on a rider who weighs 1120 N?, , 5.7, , A giant excavator (used in road construction) can apply a maximum vertical force of 2.25×105 N. If it can vertically accelerate, a load of dirt at 0.200 m/s2, what is the mass of that load? Ignore the mass of the excavator itself., , N, , N, , kg, 5.8, , 5.9, , In moving a standard computer mouse, a user applies a horizontal force of 6.00×10í2 N. The mouse has a mass of 125 g. (a), What is the acceleration of the mouse? Ignore forces like friction that oppose its motion. (b) Assuming it starts from rest, what, is its speed after moving 0.159 m across a mouse pad?, (a), , m/s2, , (b), , m/s, , A solar sail is used to propel a spacecraft. It uses the pressure (force per unit area) of sunlight instead of wind. Assume the, sail and its spacecraft have a mass of 245 kg. If the sail has an area of 62,500 m2 and achieves a velocity of 8.93 m/s in 12.0, hours starting from rest, what pressure does light of the Sun exert on the sail? To simplify the problem, ignore other forces, acting on the spacecraft and assume the pressure is constant even as its distance from the Sun increases., N/m2, , 5.10 A tennis player strikes a tennis ball of mass 56.7 g when it is at the top of the toss, accelerating it to 68.0 m/s in a distance of, 0.0250 m. What is the average force the player exerts on the ball? Ignore any other forces acting on the ball., N, 5.11 There are two forces acting on a box of golf balls, F1 and F2. The mass of the box is 0.750 kg. When the forces act in the, same direction, they cause an acceleration of 0.450 m/s2. When they oppose one another, the box accelerates at 0.240 m/s2, in the direction of F2. (a) What is the magnitude of F1? (b) What is the magnitude of F2?, (a), , N, , (b), , N, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems, , 115
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5.12 A chain of roller coaster cars moving horizontally comes to an abrupt stop and the passengers are accelerated by their safety, harnesses. In one particular car in the chain, the car has a mass M = 122 kg, the first passenger has a mass m1 = 55.2 kg,, and the second passenger has a mass m2 = 68.8 kg. If the chain of cars slows from 26.5 m/s to a stop in 4.73 s, calculate the, average magnitude of force exerted by their safety harnesses on (a) the first passenger and (b) the second passenger., (a), , N, , (b), , N, , 5.13 Cedar Point's Top Thrill Dragster Strata-Coaster in Ohio, the fastest amusement park ride in the world as of 2004, can, accelerate its riders from rest to 193 km/h in 4.00 seconds. (a) What is the magnitude of the average acceleration of a rider?, (b) What is the average net force on a 45.0 kg rider during these 4.00 seconds? (c) Do people really pay money for this?, m/s2, , (a), (b), (c), , N, Yes, , No, , 5.14 The leader is the weakest part of a fly-fishing line. A given leader can withstand 19 N of force. A trout when caught will, accelerate, taking advantage of slack in the line, and some trout are strong enough to snap the line. Assume that with the line, taut and the rod unable to flex further, a 1.3 kg trout is just able to snap this leader. How much time would it take this trout to, accelerate from rest to 5.0 m/s if it were free of the line? Note: Trout can reach speeds like this in this interval of time., s, 5.15 A 7.6 kg chair is pushed across a frictionless floor with a force of 42 N that is applied at an angle of 22° downward from the, horizontal. What is the magnitude of the acceleration of the chair?, m/s2, , Section 9 - Interactive problem: flying in formation, 9.1, , Using the simulation in the interactive problem in this section, (a) what is the force required for the red ships to accelerate at, the desired magnitude? (b) What force is required for the blue ships?, (a), , N, , (b), , N, , Section 10 - Newton’s third law, 10.1 A 75.0 kg man sits on a massless cart that is on a horizontal surface. The cart is initially stationary and it can move without, friction or air resistance. The man throws a 5.00 kg stone in the positive direction, applying a force to it so that it has, acceleration +3.50 m/s2 as measured by a nearby observer on the ground. What is the man's acceleration during the throw,, as seen by the same observer? Be careful to use correct signs., m/s2, 10.2 Two motionless ice skaters face each other and put their palms together. One skater pushes the other away using a constant, force for 0.80 s. The second skater, who is pushed, has a mass of 110 kg and moves off with a velocity of í1.2 m/s relative to, the rink. If the first skater has a mass of 45 kg, what is her velocity relative to the rink after the push? (Consider any forces, other than the push acting on the skaters as negligible.), m/s, , Section 11 - Normal force, 11.1 A cup and saucer rest on a table top. The cup has mass 0.176 kg and the saucer 0.165 kg. Calculate the magnitude of the, normal force (a) the saucer exerts on the cup and (b) the table exerts on the saucer., (a), , N, , (b), , N, , 11.2 Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of, mass 18.0 kg sits on top of it and a 8 kg block sits on top of the middle block. A downward vertical force of 170 N is applied to, the top block. What is the magnitude of the normal force exerted by the bottom block on the middle block?, N, 11.3 A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal, force that the slide exerts on the child? (b) At what angle from the horizontal is this force directed? State your answer as a, number between 0 and 90°., (a), , N, , (b), , °, , 11.4 A 6.00 kg box is resting on a table. You push down on the box with a force of 8.00 N. What is the magnitude of the normal, , 116, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems
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force of the table on the block?, N, , Section 12 - Tension, 12.1 An ice rescue team pulls a stranded hiker off a frozen lake by throwing him a rope and pulling him horizontally across the, essentially frictionless ice with a constant force. The hiker weighs 1040 N, and accelerates across the ice at 1.10 m/s2. What, is the magnitude of the tension in the rope? (Ignore the mass of the rope.), N, 12.2 During recess, Maria, who has mass 27.0 kg, hangs motionless on the monkey bars, with both hands gripping a horizontal, bar. Assume her arms are vertical and evenly support her body. What is the tension in each of her arms?, N, , Section 14 - Free-body diagrams, 14.1 A tugboat is towing an oil tanker on a straight section of a river. The current in the river applies a force on the tanker that is, one-half the magnitude of the force that the tugboat applies. Draw a free-body diagram for the tanker when the force provided, by the tugboat is directed (a) straight upstream (b) directly cross-stream, and (c) at a 30° angle to upstream. In your drawing,, let the current flow in the "left" direction, and have the tugboat pull "up" when applying a cross-stream force. Label the force, vectors., 14.2 A person lifts a 3.60 kg textbook (remember when they were made of paper and so heavy?) with a 52.0 N force at a 60.0°, angle from the horizontal. (a) Draw a free-body diagram of the forces acting on the book, ignoring air resistance. Label the, forces. (b) Draw a free-body diagram showing the net force acting on the book., 14.3 A 17.0 N force F acting on a 4.00 kg block is directed at 30.0° from the horizontal, parallel to the surface of a frictionless, ramp. Draw a free-body diagram of the forces acting on the block, including the normal force, and label the forces. Make sure, the vectors are roughly proportional to the forces!, 14.4 An apple is resting on a table. Draw free-body diagrams for the apple, table and the Earth., , Section 15 - Interactive problem: free-body diagram, 15.1 Using simulation in the interactive problem in this section, what are the direction and magnitude of (a) the weight, (b) the, normal force, (c) the tension, and (d) the frictional force that meet the stated requirements and give the desired acceleration?, (a), , N,, , (b), , N,, , (c), , N,, , (d), , N,, , i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., , Up, Down, Left, Right, Up, Down, Left, Right, Up, Down, Left, Right, Up, Down, Left, Right, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems, , 117
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Section 17 - Static friction, 17.1 A piece of steel is held firmly in the jaws of a, vise. A force larger than 3350 N will cause the, piece of steel to start to move out of the vise. If, the coefficient of friction between the steel and, each of the jaws of the vise is 0.825 and each, jaw applies an equal force, what is the, magnitude of the normal force exerted on the, steel by each jaw?, N, , 17.2 A wooden block of mass 29.0 kg sits on a horizontal table. A wire of negligible mass is attached to the right side of the block, and goes over a pulley (also of negligible mass, and frictionless), where it is allowed to dangle vertically. When a mass of, 15.5 kg is attached to the dangling wire, the block on the table just barely starts to slide. What is the coefficient of static friction, between the block and the table?, , 17.3 The Occupational Safety and Health Administration (OSHA) suggests a minimum coefficient of static friction of μs = 0.50 for, floors. If Ethan, who has mass of 53 kg, stands passively, how much horizontal force can be applied on him before he will slip, on a floor with OSHA's minimum coefficient of static friction?, N, 17.4 A block of mass m sits on top of a larger block of mass 2m, which in turn sits on a flat, frictionless table. The coefficient of, static friction between the two blocks is μs.What is the largest possible horizontal acceleration you can give the bottom block, without the top block slipping?, , μsg/2, μ sg, 2 μ sg, , Section 18 - Kinetic friction, 18.1 A 1.0 kg brick is pushed against a vertical wall by a horizontal force of 24 N. If μs = 0.80 and μk = 0.70 what is the, acceleration of the brick?, m/s2, 18.2 A firefighter whose weight is 812 N is sliding down a vertical pole, her speed increasing at the rate of 1.45 m/s2. Gravity and, friction are the two significant forces acting on her. What is the magnitude of the frictional force?, N, 18.3 A plastic box of mass 1.1 kg slides along a horizontal table. Its initial speed is 3.9 m/s, and the force of kinetic friction, opposes its motion, causing it to stop after 3.1 s. What is the coefficient of kinetic friction between the block and the table?, , 18.4 An old car is traveling down a long, straight, dry road at 25.0 m/s when the driver slams on the brakes, locking the wheels., The car comes to a complete stop after sliding 215 m in a straight line. If the car has a mass of 755 kg, what is the coefficient, of kinetic friction between the tires and the road?, , 18.5 A rescue worker pulls an injured skier lying on a toboggan (with a combined mass of 127 kg) across flat snow at a constant, speed. A 2.43 m rope is attached to the toboggan at ground level, and the rescuer holds the rope taut at shoulder level. If the, rescuer's shoulders are 1.65 m above the ground, and the tension in the rope is 148 N, what is the coefficient of kinetic friction, between the toboggan and the snow?, , 118, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems
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Section 22 - Interactive problem: forces on a sliding block, 22.1 Using the simulation in the interactive problem in this section, what are the direction and magnitude of (a) the weight, (b) the, normal force, (c) the frictional force, and (d) the tension that meet the stated requirements and give the desired acceleration?, (a), , N,, , (b), , N,, , (c), , N,, , (d), , N,, , i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., , Straight up, Straight down, Up the plane, To the right, Straight up, Straight down, Up and to the left, To the right, Down the plane, Up the plane, To the right, To the left, Down the plane, Up the plane, Perpendicular to the plane, Straight down, , Section 23 - Hooke’s law and spring force, 23.1 A 10.0 kg mass is placed on a frictionless, horizontal surface. The mass is connected to the end of a horizontal compressed, spring which has a spring constant 339 N/m. When the spring is released, the mass has an initial, positive acceleration of, 10.2 m/s2. What was the displacement of the spring, as measured from equilibrium, before the block was released? Watch the, sign of your answer., m, 23.2 Consider a large spring, hanging vertically, with spring constant k = 3220 N/m. If the spring is stretched 25.0 cm from, equilibrium and a block is attached to the end, the block stays still, neither accelerating upward nor downward. What is the, mass of the block?, kg, 23.3 A spring with spring constant k = 15.0 N/m hangs vertically from the ceiling. A 1.20 kg mass is attached to the bottom end of, the spring, and allowed to hang freely until it becomes stationary. Then, the mass is pulled downward 10.0 cm from its resting, position and released. At the moment of its release, what is (a) the magnitude of the mass's acceleration and (b) the, direction? Ignore the mass of the spring., (a), , m/s2, , (b), , i. Downward, ii. Upward, , 23.4 A 5.00 kg wood cube rests on a frictionless horizontal table. It has two springs attached to it on opposite faces. The spring on, the left has a spring constant of 55.0 N/m, and the spring on the right has a spring constant of 111 N/m. Both springs are, initially in their equilibrium positions (neither compressed nor stretched). The block is moved toward the left 10.0 cm,, compressing the left spring and stretching the right spring. (a) Calculate the resulting net force on the block. (b) Calculate the, the initial acceleration of the block when it is released. Use the convention that to the right is positive and to the left is, negative., (a), , N, , (b), , m/s2, , 23.5 A man steps on his bathroom scale and obtains a reading of 243 lb. The spring in the scale is compressed by a displacement, of í0.0590 inches. Calculate the value of its spring constant in (a) pounds per inch (b) newtons per meter., (a), , lbs/in, , (b), , N/m, , Section 24 - Air resistance, 24.1 A parachutist and her parachuting equipment have a combined mass of 101 kg. Her terminal velocity is 5.30 m/s with the, parachute open. Her parachute has a cross-sectional area of 35.8 m2. The density of air at that altitude is 1.23 kg/m3. What is, the drag cofficient of the parachutist with her parachute?, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems, , 119
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24.2 A tennis ball of mass 57.0 g is dropped from the observation deck of the Empire State building (369 m). The tennis ball has a, cross-sectional area of 3.50×10-3 m2 and a drag coefficient of 0.600. Using 1.23 kg/m3 for the density of air, (a) what is the, speed of the ball when it hits the ground? (b) What would be the final speed of the ball if you did not include air resistance in, your calculations? Remember to use the appropriate sign when answering the question., (a), , m/s, , (b), , m/s, , Section 25 - Interactive Problem: Forces in Multiple Dimensions, 25.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) If F1 is set to 12 N directly, to the left, what should F2 be set to so that the ball does not move when you press GO? (b) If F1 is set to 10 N directly to the, left, what should F2 be set to so that the ball hits the target directly to the right of the ball? (c) If F1 is set to 10 N straight up,, what should F2 be set to so that the ball hits the target that is up and to the right of the ball?, (a), , (b), , (c), , 120, , N,, , i. Straight up, ii. Directly to the right, iii. Straight down, iv. Directly to the left, i. Any magnitude less than 10 N, ,, ii. Exactly 10 N, iii. Any magnitude greater than 10 N, N,, , i., ii., iii., iv., , i., ii., iii., iv., , Straight up, Directly to the right, Straight down, Directly to the left, , Straight up, Directly to the right, Straight down, Directly to the left, , Copyright 2007 Kinetic Books Co. Chapter 5 Problems
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6.0 - Introduction, The use of energy has played an important role in defining much of human history., Fire warmed and protected our ancestors. Coal powered the Industrial Revolution., Gasoline enabled the proliferation of the automobile, and electricity led to indoor, lighting, then radio, television and the computer. The enormous energy unleashed, by splitting the atom was a major factor in ending World War II. Today, businesses, involved in technology or media may garner more newspaper headlines, but energy, is a larger industry., Humankind has long studied how to harness and transform energy. Early machines, used the energy of flowing water to set wheels spinning to mill grain. Machines, designed during the Industrial Revolution used energy unleashed by burning coal to, create the steam that drove textile looms and locomotives. Today, we still use these, same energy sources í water and coal í but often we transform the energy into, electric energy., Scientists continue to study energy sources and ways to store energy. Today,, environmental concerns have led to increased research in areas including atomic, fusion and hydrogen fuel cells. Even as scientists are working to develop new, energy technologies, there is renewed interest in some ancient energy sources: the Sun and the wind. They too can provide clean energy via, photovoltaic cells and wind turbines., Why is energy so important? Because humankind uses it to do work. It no longer requires as much human labor to plow fields, to travel, or to, entertain ourselves. We can tap into other energy sources to serve those needs., This chapter is an introduction to work and energy. It appears in the mechanics section of the textbook, because we focus here on what is, called mechanical energy, energy arising from the motion of particles and objects, and energy due to the force of gravity. Work and energy also, are major topics in thermodynamics, a topic covered later. Thermodynamics adds the topic of heat to the discussion. We will only mention heat, briefly in this chapter., Whatever the source and ultimate use of energy, certain fundamental principles always apply. This chapter begins your study of those, principles, and the simulation to the right is your first opportunity to experiment with them., Your mission in the simulation is to get the car over the hill on the right and around a curve that is beyond the hill. You do this by dragging the, car up the hill on the left and releasing it. If you do not drag it high enough, it will fail to make it over the hill. If you drag it too high, it will fly off, the curve after the hill. The height of the car is shown in a gauge in the simulation., Only the force of gravity is factored into this simulation; the forces of friction and air resistance are ignored. In this chapter, we consider only the, kinetic energy due to the object moving as a whole and ignore rotational energy, such as the energy of the car wheels due to their rotational, motion. (Taxes, title and dealer prep are also not factored into the simulation; contact your local dealership for any other additional restrictions, or limitations.), Make some predictions before you try the simulation. If you release the car at a higher point, will its speed at the bottom of the hill be greater,, the same or less? How high will you have to drag the car to have it just reach the summit of the other hill: to the same height, higher or lower?, You can use PAUSE to see the car’s speed more readily at any point., When you use this simulation, you are experimenting with some of the key principles of this chapter. You are doing work on the car as you drag, it up the hill, and that increases the car’s energy. That energy, called potential energy, is transformed into kinetic energy as the car moves, down the hill. Energy is conserved as the car moves down the hill. It may change forms from potential energy to kinetic energy, but as the car, moves on the track, its total energy remains constant., , 6.1 - Work, , Work: The product of displacement and the, force in the direction of displacement., You may think of work as homework, or as labor done to earn money, or as exercise in, a demanding workout., But physicists have a different definition of work. To them, work equals the component, of force exerted on an object along the direction of the object’s displacement, times the, object’s displacement., When the force on an object is in the same direction as the displacement, the, magnitude of the force and the object’s displacement can be multiplied together to, calculate the work done by the force. In Concept 1, a woman is shown pushing a crate, so that all her force is applied in the same direction as the crate’s motion., , Work, Product of force and displacement, , Copyright 2007 Kinetic Books Co. Chapter 06, , 121
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All the force need not be in the direction of the displacement. When the force and, displacement vectors are not in the same direction, only the component of the force in, the direction of the displacement contributes to work. Consider the woman pulling the, crate at an angle with a handle, as shown in Concept 2. Again, the crate slides along, the ground. The component of the force perpendicular to the displacement contributes, nothing to the work because there is no motion up or down., Perhaps subconsciously, you may have applied this concept. When you push on a, heavy object that is low to the floor, like a sofa, it is difficult to slide it if you are mostly, pushing down on top of it. Instead, you bend low so that more of your force is, horizontal, parallel to the desired motion., The equation on the right is used to calculate how much work is done by a force. It has, notation that may be new to you. The equation states that the work done equals the “dot, product” of the force and displacement vectors (the name comes from the dot between, the F and the ǻx)., , Force at angle to displacement, Only force component along, displacement contributes to work, , The equation is also expressed in a fashion that you will find useful: (F cos ș)ǻx. The, angle ș is the angle between the force and displacement vectors when they are placed, tail to tail. The vectors and the angle are shown in Equation 1., By multiplying the amount of force by cos ș, you calculate the component of the force, parallel to the displacement. You may recall other cases in which you used the cosine, or sine of an angle to calculate a component of a vector. In this case, you are, calculating the component of one vector that is parallel to another., The equation to the right is for a constant or average force. If the force varies as the, motion occurs, then you have to break the motion into smaller intervals within which the, force is constant in order to calculate the total work., The everyday use of the word “work” can lead you astray. In physics, if there is no, displacement, there is no work. Suppose the woman on the right huffed and puffed and, pushed the crate as hard as she could for ten minutes, but it did not move. She would, certainly believe she had done work. She would be exhausted. But a physicist would, say she has done zero work on the crate because it did not move. No displacement, means no work, regardless of how much force is exerted., Work can be a positive or negative value. Positive work occurs when the force and, displacement vectors point in the same direction. Negative work occurs when the force, and displacement vectors point in opposite directions. If you kick a stationary soccer, ball, propelling it downfield, you have done positive work on the ball because the force, and the displacement are in the same direction., When a goalie catches a kicked ball, negative work is done by the force from the, goalie’s hands on the ball. The force on the ball is in the opposite direction of the ball’s, displacement, with the result that the ball slows down., , W = F · ǻx = (F cos ș)ǻx, W = work, F = force, ǻx = displacement, ș = angle between force and, displacement, Unit: joule (J), , The sign of work can be calculated with the equation to the right. When force and, displacement point in the same direction, the angle between them is 0°, and the cosine, of 0° is positive one. When force and displacement point in opposite directions, the, angle between the vectors is 180°, and the cosine of 180° is negative one. This, mathematically confirms the points made above: Force in the direction of motion results, in positive work; force opposing the motion results in negative work., Work is a scalar quantity, which means it has magnitude but no direction. The joule is, the unit for work. The units that make up the joule are kg·m2/s2 and come from, multiplying the unit for force (kg·m/s2) by the unit for displacement (m)., If several forces act on an object, each of them can do work on the object. You can, calculate the net work done on the object by all the forces by calculating the net force, and using the equation in Equation 1., , How much work does the woman, do on the crate?, W = (F cos ș)ǻx, W = (120 N)(cos 0°)(3.0 m), W = (120 N)(1)(3.0 m) = 360 J, , 122, , Copyright 2007 Kinetic Books Co. Chapter 06
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Now the woman is pulling the, crate at an angle. If she does the, same amount of work as before,, how much force must the woman, exert?, W = (F cos ș)ǻx, F = W/ (cos ș)ǻx, F = (360 J)/ (cos 39°)(3.0 m), F = 150 N, 6.2 - Interactive checkpoint: work, Sally does 401 J of work moving a, couch 1.30 meters. If she applies a, constant force at an angle of 22.0° to, the horizontal as shown, what is the, magnitude of this force?, , Answer:, , F=, , N, , 6.3 - Energy, Before delving into some specific forms of energy, in this section we address the, general topic of energy. Although it is a very important concept in physics, and an, important topic in general, energy is notoriously hard to define., Why? There are several reasons. Many forms of energy exist: electric, atomic,, chemical, kinetic, potential, and so on. Finding a definition that fits all these forms is, challenging. You may associate energy with motion, but not all forms of energy involve, motion. A very important class of energy, potential energy, is based on the position or, configuration of objects, not their motion., Energy is a property of an object, or of a system of objects. However, unlike many other, properties covered so far in this textbook, is hard to observe and measure directly. You, can measure most forces, such as the force of a spring. You can see speed and decide, which of two objects is moving faster. You can use a stopwatch to measure time., Quantifying energy is more elusive, because energy depends on multiple factors, such, as an object’s mass and the square of its speed, or the mass and positions of a system, of objects., , Energy, Is changed by work, , Despite these caveats, there are important principles that concern all forms of energy. First, there is a relationship between work and energy., For instance, if you do work by kicking a stationary soccer ball, you increase a form of its energy called kinetic energy, the energy of motion., Second, energy can transfer between objects. When a cue ball in the game of pool strikes another ball, the cue ball slows or stops, and the, other ball begins to roll. The cue ball’s loss of energy is the other ball’s gain., , Copyright 2007 Kinetic Books Co. Chapter 06, , 123
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Third, energy can change forms. When water falls over a dam, its energy of position, becomes the energy of motion (kinetic energy). The kinetic energy from the moving, water can cause a turbine to spin in a dam, generating electric energy. If that electricity, is used to power a blender to make a milkshake, the energy is transformed again, this, time into the rotational kinetic energy of the blender’s spinning blades., In Concept 1, an archer does work by applying a force to pull a bowstring. This work, increases the elastic potential energy of the bow. When the string is released, it, accelerates the arrow, transferring and transforming the bow’s elastic potential energy, into the kinetic energy of the arrow., One can trace the history of the energy in the bow and arrow example much farther, back. Maybe the chemical energy in the archer that was used by his muscles to stretch, the bow came from the chemical energy of a hamburger, and the cow acquired that, energy by digesting plants, which got energy via photosynthesis by tapping, electromagnetic energy, which came from nuclear reactions in the Sun. We could go, on, but you get the idea., , Energy, Transfers between objects, Exists in many forms, , Energy is a scalar. Objects can have more or less energy, and some forms of energy, can be positive or negative, but energy does not have a direction, only a value. The joule is the unit for energy, just as it is for work. The fact, that work and energy share the same unit is another indication that a fundamental relationship exists between them., , 6.4 - Kinetic energy, , Kinetic energy: The energy of motion., Physicists describe the energy of objects in motion using the concept of kinetic energy (, KE). Kinetic energy equals one-half an object’s mass times the square of its speed., To the right is an arrow in motion. The archer has released the bowstring, causing the, arrow to fly forward. A fundamental property of the arrow changes when it goes from, motionless to moving: It gains kinetic energy., The kinetic energy of an object increases with mass and the square of speed. A, 74,000 kg locomotive barreling along at 40 m/s has four times as much kinetic energy, as when it is going 20 m/s, and about five million times the kinetic energy of a 6-kg, bowling ball rolling at 2 m/s., With kinetic energy, only the magnitude of the velocity (the speed) matters, not, direction. The locomotive, whether heading east or west, north or south, has the same, kinetic energy., , Kinetic energy, Energy of motion, Proportional to mass, square of speed, , Objects never have negative kinetic energy, only zero or positive kinetic energy. Why?, Kinetic energy is a function of the speed squared and the square of a value is never, negative., Because it is a type of energy, the unit for kinetic energy is the joule, which is one kg·, (m/s)2. This is the product of the units for mass and the square of the units for velocity., , KE = ½ mv2, KE = kinetic energy, m = mass, v = speed, Unit: joule (J), , 124, , Copyright 2007 Kinetic Books Co. Chapter 06
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What is the kinetic energy of the, arrow?, KE = ½ mv2, KE = ½(0.015 kg)(5.0 m/s)2, KE = 0.19 J, 6.5 - Work-kinetic energy theorem, , Work-kinetic energy theorem: The net work, done on a particle equals its change in kinetic, energy., Consider the foot kicking the soccer ball in Concept 1. We want to relate the work done, by the force exerted by the foot on the ball to the ball’s change in kinetic energy. To, focus solely on the work done by the foot, we ignore other forces acting on the ball,, such as friction., Initially, the ball is stationary. It has zero kinetic energy because it has zero speed. The, foot applies a force to the ball as it moves through a short displacement. This force, accelerates the ball. The ball now has a speed greater than zero, which means it has, kinetic energy. The work-kinetic energy theorem states that the work done by the foot, on the ball equals the change in the ball’s kinetic energy. In this example, the work is, positive (the force is in the direction of the displacement) so the work increases the, kinetic energy of the ball., , Work done on a particle, Net work equals change in kinetic, energy, Positive work on object increases its KE, , As shown in Concept 2, a goalie catches a ball kicked directly at her. The goalie’s, hands apply a force to the ball, slowing it. The force on the ball is opposite the ball’s, displacement, which means the work is negative. The negative work done on the ball, slows and then stops it, reducing its kinetic energy to zero. Again, the work equals the, change in energy; in this case, negative work on the ball decreases its energy., In the scenarios described here, the ball is the object to which a force is applied. But, you can also think of the soccer ball doing work. The ball applies a force on the goalie,, causing the goalie’s hands to move backward. The ball does positive work on the goalie, because the force it applies is in the direction of the displacement of the goalie’s hands., When stated precisely, which is always worthwhile, the work-kinetic energy theorem is, defined to apply to a particle: The net work done on a particle equals the change in its, KE. A particle is a small, indivisible point of mass that does not rotate, deform, and so, on. Various properties of the particle can be observed at any point in time, and by, recording those properties at any instant, its state can be defined., , Negative work on object, Decreases object’s kinetic energy, , The only form of energy that a single particle can possess is kinetic energy. Stating the work-kinetic energy theorem for a particle means that, the work contributes solely to the change in the particle’s kinetic energy. A soccer ball is not a particle. When you kick a soccer ball, the, surface of the ball deforms, the air particles inside move faster, and so forth., Having said this, we (and others) apply the work-kinetic energy theorem to objects such as soccer balls. A textbook filled solely with particles, would be a drab textbook indeed. We simplify the situation, modeling the ball as a particle, so that we can apply the work-kinetic energy, theorem. We can always make it more complicated (have the ball rotate or lift off the ground, so rotational KE and gravitational potential, energy become factors), but the work-kinetic energy theorem provides an essential starting point., For instance, in Example 1, we first calculate the work done on the ball by the foot. We then use the work-kinetic energy theorem to equate the, work to the change in KE of the ball. Using the definition of KE, we can calculate the ball’s speed immediately after being kicked., It is important that the theorem applies to the net work done on an object. Here, we ignore the force of friction, but if it were being considered,, we would have to first calculate the net force being applied on the ball in order to consider the net work that is done on it., , Copyright 2007 Kinetic Books Co. Chapter 06, , 125
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Work-kinetic energy theorem, W = ǻKE, W = net work, KE = kinetic energy, , What is the soccer ball’s speed, immediately after being kicked?, Its mass is 0.42 kg., W = F · ǻx, W = (240 N) (0.20 m) = 48 J, W = ǻKE = 48 J, KE = ½ mv2 = 48 J, v2 = 2(48 J)/0.42 kg, v = 15 m/s, 6.6 - Sample problem: work-kinetic energy theorem, Four bobsledders push their 235 kg, sled with a constant force, moving it, from rest to a speed of 10.0 m/s along, a flat, 50.0-meter-long icy track., Ignoring friction and air resistance,, what force does the team exert on the, sled?, , We assume here that all the work done by the athletes goes to increasing the kinetic energy of the sled., , 126, , Copyright 2007 Kinetic Books Co. Chapter 06
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Variables, mass of sled, , m = 235 kg, , displacement, , ǻx = 50.0 m, , sled’s initial speed, , vi = 0 m/s, , sled’s final speed, , vf = 10.0 m/s, , work, , W, , force, , F, , What is the strategy?, 1., , Calculate the change in kinetic energy of the sled, using the sled’s mass and its initial and final speeds., , 2., , Use the work-kinetic energy theorem and the definition of work to find the force exerted on the sled., , Physics principles and equations, The definition of work, applied when the force is in the direction of the displacement, , W = Fǻx, The definition of kinetic energy, , KE =½ mv2, The work-kinetic energy theorem, , W = ǻKE, Step-by-step solution, Start by calculating the change in kinetic energy of the sled., , Step, , Reason, , 1., , ǻKE = KEf – KEi, , definition of change in kinetic energy, , 2., , ǻKE = ½mvf2 – ½mvi2, , definition of kinetic energy, , 3., , ǻKE = ½m(vf2 – vi2), , factor, enter values, , 4., 5., , ǻKE = 11,800 J, , solve, , Use the work-kinetic energy theorem to find the work done on the sled. Then, use the definition of work to determine how much force was, exerted on the sled., , Step, , Reason, , 6., , W = ǻKE, , work-kinetic energy theorem, , 7., , W = (F cos ș)ǻx, , definition of work, , 8., , (F cos ș)ǻx = ǻKE, , set two work equations equal, , 9., , F cos ș = ǻKE/ǻx, , rearrange, , 10. F = ǻKE/ǻx, , force in direction of displacement, , 11. F = 11,800 J/50.0 m enter values, 12. F = 236 N, , solve, , Copyright 2007 Kinetic Books Co. Chapter 06, , 127
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6.7 - Interactive problem: work-kinetic energy theorem, In this simulation, you are a skier and your challenge is to do the correct amount of, work to build up enough energy to soar over the canyon and land near the lip of the, slope on the right., You, a 50.0 kg skier, have a flat 12.0 meter long runway leading up to the lip of the, canyon. In that stretch, you must apply a force such that at the end of the, straightaway, you are traveling with a speed of 8.00 m/s. Any slower, and your jump, will fall short. Any faster, and you will overshoot., How much force must you apply, in newtons, over the 12.0 meter flat stretch?, Ignore other forces like friction and air resistance., Enter the force, to the nearest newton, in the entry box and press GO to check your, result., If you have trouble with this problem, review the section on the work-kinetic energy, theorem. (If you want to, you can check your answer using a linear motion equation, and Newton’s second law.), , 6.8 - Interactive checkpoint: a spaceship, A 45,500 kg spaceship is far from any, significant source of gravity. It, accelerates at a constant rate from, 13,100 m/s to 15,700 m/s over a, distance of 2560 km. What is the, magnitude of the force on the ship, due to the action of its engines? Use, equations involving work and energy, to solve the problem, and assume, that the mass is constant., , Answer:, , F=, , N, , 6.9 - Power, , Power: Work divided by time; also the rate of, energy output or consumption., The definition of work í the dot product of force and displacement í does not say, anything about how long it takes for the work to occur. It might take a second, or a year,, or any interval of time. Power adds the concept of time to the topics of work and energy., Power equals the amount of work divided by the time it took to do the work. You see, this expressed as an equation in Equation 1., One reason we care about power is because more power means that work can be, accomplished faster. Would you rather have a car that accelerated you from zero to 100, kilometers per hour in five seconds, or five minutes? (Some of us have owned cars of, the latter type.), , Power, Rate of work, , The unit of power is the watt (W), which equals one joule per second. It is a scalar unit., Power can also be expressed as the rate of change of energy. For instance, a 100 megawatt power plant supplies 100 million joules of energy, to the electric grid every second., Sometimes power is expressed in terms of an older unit, the horsepower. This unit comes from the days when scientists sought to establish a, standard for how much work a horse could do in a set amount of time. They then compared the power of early engines to the power of a horse., One horsepower equals 550 foot-pounds/second, which is the same as 746 watts., We still measure the power of cars in horsepower. For instance, a 300-horsepower Porsche is more powerful than a 135-horsepower Toyota., The Porsche’s engine is capable of doing more work in a given period of time than the Toyota’s., , 128, , Copyright 2007 Kinetic Books Co. Chapter 06
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Equation 2 shows two other useful equations for power. Power can be expressed as the, rate of change of energy, as you see in the first equation in Equation 2. Sometimes this, is stated as “energy consumption”, as in a 100-watt light bulb “consumes” 100 joules of, energy each second. In other words, the light bulb converts 100 joules of electrical, energy each second into other forms of energy, such as light and heat., The companies that provide electrical power to homes measure each household’s, energy consumption in kilowatt·hours. You can check that this is a unit of energy. The, companies multiply power (thousands of joules per second, or kilowatts) by time, (hours). The result is that one kilowatt·hour equals 60 kilojoules, a unit of energy., The second equation in Equation 2 shows that when there is a constant force in the, direction of an object’s displacement, the power can be measured as the product of, the force and the velocity. This equation can be derived from the definition of work:, W = Fǻx. Dividing both sides of that equation by time yields power on the left (work, divided by time). On the right side, dividing displacement by time yields velocity., As with other rates of change, such as velocity or acceleration, we can consider, average or instantaneous power. Average power is the total amount of work done over, a period of time, divided by that time. Instantaneous power has the same definition, but, the time interval must be a brief instant (more precisely, it is defined as the limit of the, average power as the time interval approaches zero)., , W = work, ǻt = time, Units: watts (W), , Other power equations, , E = energy, F = force, , Copyright 2007 Kinetic Books Co. Chapter 06, , 129
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Applying a force of 2.0×105 N, the, tugboat moves the log boom, 1.0 kilometer in 15 minutes. What, is the tugboat’s average power?, W = Fǻx, W = Fǻx = (2.0×105 N)(1.0×103 m), W = 2.0×108 J, , 6.10 - Potential energy, , Potential energy: Energy related to the positions, of and forces between the objects that make up a, system., Although the paint bucket in Concept 1 is not moving, it makes up part of a system that, has a form of energy called potential energy. In general, potential energy is the energy, due to the configuration of objects that exert forces on one other., In this section, we focus on one form of potential energy, gravitational potential energy., The paint bucket and Earth make up a system that has this form of potential energy., , Potential energy, A system is some “chunk” of the universe that you wish to study, such as the bucket, and the Earth. You can imagine a boundary like a bubble surrounding the system,, Energy of position or configuration, separating it from the rest of the universe. The particles within a system can interact, with one another via internal forces or fields. Particles outside the system can interact with the system via external forces or fields., Gravitational potential energy is due to the gravitational force between the bucket and Earth. As the bucket is raised or lowered, its change in, potential energy (ǻPE) equals the magnitude of its weight, mg, times its vertical displacement, ǻh. (We follow the common convention of, using ǻh for change in height, instead of ǻy.) The weight is the amount of force exerted on the bucket by the Earth (and vice versa). This, formula is shown in Equation 1., A change in PE can be positive or negative. The magnitude of weight is a positive value, but change in height can be positive (when the, bucket moves up) or negative (when it moves down)., To define a system’s PE, we must define a configuration at which the system has zero PE. Unlike kinetic energy, where zero KE has a natural, value (when an object’s speed is zero), the configuration with zero PE is defined by you, the physicist., In the diagrams to the right, it is convenient to say the system has zero PE when the bucket is on the Earth’s surface. This convention means, its PE equals its weight times its height above the ground, mgh. Only the bucket’s distance above the Earth, h, matters here; if the bucket, moves left or right, its PE does not change., In Example 1, we calculate the paint bucket’s gravitational potential energy as it sits on the scaffolding, four meters above the ground., There are other types of potential energy. One you will frequently encounter is elastic potential energy, which is the energy stored in a, compressed or stretched object such as a spring. As you may recall, this form of energy was present in the bow that was used to fire an arrow., , 130, , Copyright 2007 Kinetic Books Co. Chapter 06
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Change in gravitational potential, energy, ǻPE = mgǻh, PE = potential energy, mg = object’s weight, ǻh = vertical displacement, , Gravitational potential energy, PE = mgh, PE = 0 when h = 0, , What is the bucket’s, gravitational potential energy?, PE = mgh, PE = (2.00 kg)(9.80 m/s2)(4.00 m), PE = 78.4 J, , Copyright 2007 Kinetic Books Co. Chapter 06, , 131
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6.11 - Work and gravitational potential energy, Potential energy is the energy of a system due to forces between the particles or, objects that make up the system. It can be related to the amount of work done on a, system by an external force. We will use gravitational force and gravitational potential, energy as an example of this general principle. Our discussion applies to what are, called conservative forces, a type of force we will later discuss in more detail., The system we consider consists of two objects, the bucket and the Earth, illustrated to, the right. The painter applies an external force to this system (via a rope) when she, raises or lowers the bucket. The bucket starts at rest on the ground, and she raises it up, and places it on the scaffolding. That means the work she does as she moves the, bucket from its initial to its final position changes only its gravitational PE. The system’s, kinetic energy is zero at the beginning and the end of this process., As she raises the bucket, the painter does work on it. She pulls the bucket up against, the force of gravity, which is equal in magnitude to the bucket’s weight, mg. She pulls in, the direction of the bucket’s displacement, ǻh. The work equals the force multiplied by, the displacement: mgǻh. The paint bucket’s change in gravitational potential energy, also equals mgǻh. The analysis lets us reach an important conclusion: The work done, on the system, against gravity, equals the system’s increase in gravitational potential, energy., , Work and potential energy, Work equals change in energy, , Earlier, we stated the work-kinetic energy theorem: The net work done on a particle, equals its change in kinetic energy. Here, where there is no change in kinetic energy,, we state that the work done on a system equals its change in potential energy., Are we confused? No. Work performed on a system can change its mechanical energy,, which consists of its kinetic energy and its potential energy. Either or both of these, forms of energy can change when work is applied to the system., As the painter does work against the force of gravity, the force of gravity itself is also, doing work. The work done by gravity is the negative of the work done by the painter., This means the work done by gravity is also the negative of the change in potential, energy, as seen in Equation 2., , Work done against gravity, , Imagine that the painter drops the bucket from the scaffolding. Only the force of gravity, does work on the bucket as it falls. The system has more potential energy when the, bucket is at the top of the scaffolding than when it is at the bottom, so the work done by, gravity has lowered the system’s PE: the change in PE due to the work done by gravity, is negative., , W = work done against gravity, , W = ǻPE, , PE = potential energy of system, , Work done by gravity, W = íǻPE, W = work done by gravity, PE = potential energy of system, , 132, , Copyright 2007 Kinetic Books Co. Chapter 06
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6.12 - Sample problem: potential energy and Niagara Falls, In its natural state, an average of, 5.71×106 kg of water flowed per, second over Niagara Falls, falling, 51.0 m. If all the work done by gravity, could be converted into electric power, as the water fell to the bottom, how, much power would the falls generate?, , Variables, height of falls, , h = 51.0 m, , magnitude of acceleration due to gravity, , g = 9.80 m/s2, , potential energy, , PE, , mass of water over falls per unit time, , m/t = 5.71×106 kg/s, , power, , P, , work done by gravity, , W, , What is the strategy?, 1., , Use the definition of power as the rate of work done to define an equation for the power of the falls., , 2., , Use the fact that work done by gravity equals the negative of the change in gravitational potential energy to solve for the power., , Physics principles and equations, Power is the rate at which work is performed., , Change in gravitational PE, , ǻPE = mgǻh, Work done by gravity, , W = íǻPE, Step-by-step solution, We start with the definition of power í work done per unit time í and then substitute in the definition of work done by gravity and the definition, of gravitational potential energy to solve the problem., , Step, , Reason, , 1., , power equation, , 2., , work done by gravity lowers PE, , 3., , definition of gravitational potential energy, , 4., , enter values for g and h, , 5., , enter value for m/t, , 6., , P = 2.85×109 W, , solve, , This is the theoretical maximum power that could be generated. A real power plant cannot be 100% efficient., , Copyright 2007 Kinetic Books Co. Chapter 06, , 133
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6.13 - Interactive checkpoint: an elevator, An elevator with a mass of 515 kg is, being pulled up a shaft at constant, velocity. It takes the elevator 3.00, seconds to travel from floor two to, floor three, a distance of 4.50 m., What is the average power of the, elevator motor during this time?, Neglect friction., , Answer:, , P=, , W, , 6.14 - Work and energy, We have discussed work on a particle increasing its KE, and work on a system, increasing its PE. Now we discuss what happens when work increases both forms of, mechanical energy., Because we are considering only KE and PE in this chapter, we can say the net work, done on an object equals the change in the sum of its KE and PE. Positive work done, on an object increases its energy; negative work decreases its energy., Let’s also consider what happens when an object does work, and how that affects the, object’s energy. Consider a soccer ball slamming into the hands of a goalie. The ball is, doing work, forcing the goalie’s hands backwards. The ball slows down; its energy, decreases. Work done by an object decreases its energy. At the same time, this work, on the goalie increases her energy. Work has transferred energy from one system (the, ball) to another (the goalie)., We will use the scenario in Example 1 to show how both an object’s KE and PE can, change when work is done on it. A cannon shoots a 3.20 kg cannonball straight up. The, barrel of the cannon is 2.00 m long, and it exerts an average force of 6,250 N while the, cannonball is in the cannon. We will ignore air resistance. Can we determine the, cannonball’s velocity when it has traveled 125 meters upward?, , Work and energy, Work on system equals its change in, total energy, , As you may suspect, the answer is “yes”., The cannon does 12,500 J of work on the cannonball, the product of the force (6,250 N), and the displacement (2.00 m). (We assume the cannon does no work on the, cannonball after it leaves the cannon.), At a height of 125 meters, the cannonball’s increase in PE equals mgǻh, or 3,920 J., Since a total of 12,500 J of work was done on the ball, the rest of the work must have, gone into raising the cannonball’s KE: The change in KE is 8,580 J., Applying the definition of kinetic energy, we determine that its velocity at 125 m is, 73.2 m/s. We could further analyze the cannonball’s trip if we were so inclined. At the, peak of its trip, all of its energy is potential since its velocity (and KE) there are zero., The PE at the top is 12,500 J. Again applying the formula mgǻh , we can determine, that its peak height above the cannon is about 399 m., , The cannon supplies 6,250 N of, force along its 2.00 m barrel., How much work does the, cannon do on the cannonball?, W = (F cos ș)ǻx = Fǻx, W = (6250 N)(2.00 m) = 12,500 J, , 134, , Copyright 2007 Kinetic Books Co. Chapter 06
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What is the cannonball’s velocity, at 125 m? Its mass is 3.20 kg., W = ǻPE + ǻKE, W = mgǻh + ǻKE, 12,500 J = (3.20 kg)(9.80 m/s2)(125 m), + ǻKE, ǻKE = 8,580 J, ½ mv2 = 8,580 J, v2 = 2(8,580 J)/(3.20 kg), v = 73.2 m/s, 6.15 - Conservative and non-conservative forces, Earlier, when discussing potential energy, we mentioned that we would explain, conservative forces later. The concept of potential energy only applies to conservative, forces., Gravity is an example of a conservative force. It is conservative because the total work, it does on an object that starts and finishes at the same point is zero. For example, if a, 20 kg barbell is raised 2.0 meters, gravity does í40 J of work, and when the barbell is, lowered 2.0 meters back to its initial position, gravity does +40 J of work. When the, barbell is returned to its initial position, the sum of the work done by gravity on the, barbell equals zero., You can confirm this by considering the barbell’s gravitational potential energy. Since, that equals mgh, it is the same at the beginning and end because the height is the, same. Since there is no change in gravitational PE, there is no work done by gravity on, the barbell., We illustrate this with the roller coaster shown in Concept 1. For now, we ignore other, forces, such as friction, and consider gravity as the only force doing work on the roller, coaster car. When the roller coaster car goes down a hill, gravity does positive work., When the roller coaster car goes up a hill, gravity does negative work. The sum of the, work done by gravity on this journey equals zero., , Conservative force, A force that does no work on closed, path, , When a roller coaster car makes such a trip, the roller coaster car travels on what is, called a closed path, a trip that starts and stops at the same point. Given a slight push, at the top of the hill, the roller coaster would make endless trips around the roller, coaster track., Kinetic friction and air resistance are two examples of non-conservative forces. These, forces oppose motion, whatever its direction. Friction and air resistance do negative, work on the roller coaster car, slowing it regardless of whether it is going uphill or, downhill., , Non-conservative force, We show non-conservative forces at work in Concept 2. The roller coaster glides down, the hill, but it does not return to its initial position because kinetic friction and air, A force that does work on closed, resistance dissipate some of its energy as it goes around the track. The presence of, these forces dictates that net work must be done on the roller coaster car by some other, force to return it to its initial position. A mechanism such as a motorized pulley system can accomplish this., , path, , A way to differentiate between conservative and non-conservative forces is to ask: Does the amount of work done by the force depend on the, path?, Consider only the force of gravity, a conservative force, as it acts on the skier shown in Concept 3. When considering the work done by gravity,, it does not matter in terms of work and energy whether the skier goes down the longer, zigzag route (path A), or the straight route (path B). The, work done by gravity is the same along either path. All that matters are the locations of the initial and final points of the path. The conservative, , Copyright 2007 Kinetic Books Co. Chapter 06, , 135
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force is path independent., However, with non-conservative forces, the path does influence the amount of work, done by a force. Consider the skier in the context of kinetic friction, a non-conservative, force. The amount of force of kinetic friction is the same along either route, but it acts, along a greater distance if the skier chooses the longer route. The amount of work done, by the force of kinetic friction increases with the path length. A non-conservative force is, path dependent., , Effect of path on work and, energy, Conservative forces: work does not, depend on path, Non-conservative forces: work depends, on path, , 6.16 - Conservation of energy, , Conservation of energy: The total energy in an, isolated system remains constant., Energy never disappears. It only changes form and transfers between objects., To illustrate this principle, we use the boy to the right who is swinging on a rope. We, consider his mechanical energy, the sum of his kinetic and gravitational potential, energy. When he jumps from the riverbank and swings toward the water, his, gravitational potential energy becomes kinetic energy. The decrease in gravitational PE, (shown in the gauge labeled PE in Concept 1) is matched by an increase in his kinetic, energy (shown in the gauge labeled KE). Ignoring air resistance or any other nonconservative forces, the sum of KE and PE is a constant at any point. The total amount, of energy (labeled TE, for total energy) stays the same. The total energy is conserved;, its amount does not change., , Conservation of energy, Total energy in isolated system stays, constant, , The law of conservation of energy applies to an isolated system. An isolated system is, one that has no interactions with its environment. The particles within the system may, interact with one another, but no net external force or field acts on an isolated system., Only external forces can change the total energy of a system. If a giant spring lifts a car,, you can say the spring has increased the energy of the car. In this case, you are, considering the spring as supplying an external force and not as part of the system. If, you include the spring in the system, the increase in the energy of the car is matched by, a decrease in the potential energy contained of the spring, and the total energy of the, system remains the same. For the law of conservation of energy to apply, there can be, no non-conservative forces like friction within the system., The law of conservation of energy can be expressed mathematically, as shown in, Equation 1. The equation states that an isolated system’s total energy at any final point, in time is the same as its total energy at an initial point in time. When considering, mechanical energy, we can state that the sum of the kinetic and potential energies at, some final moment equals the sum of the kinetic and potential energies at an initial, moment., , Conservation of energy, PE transforms to KE ..., , In the case of the boy on the rope, if you know his mass and height on the riverbank,, you can calculate his gravitational potential energy. In this example, rather than saying, his PE equals zero on the ground, we say it equals zero at the bottom of the arc. This, simplifies matters. Using the law of conservation of energy, you can then determine, what his kinetic energy, and therefore his speed, will be when he reaches the bottom of, the arc, nearest to the water, since at that point all his energy is kinetic., Let’s leave the boy swinging for a while and switch to another example: You drop a, weight. When the weight hits the ground it will stop moving. At this point, the weight has, neither kinetic energy nor potential energy because it has no motion and its height off, the Earth’s surface is zero. Does the law of conservation of energy still hold true?, Yes, it does, although we need to broaden the forms of energy included in the, discussion. With careful observation you might note that the ground shakes as the, weight hits it (more energy of motion). The weight and the ground heat up a bit (thermal, energy). The list can continue: energy of the motion of flying dirt, the energy of sound, , 136, , ... but total energy remains the same, , Copyright 2007 Kinetic Books Co. Chapter 06
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and so on. The amount of mechanical energy does decline, but when you include all, forms of energy, the overall energy stays constant., There is a caveat to the law of conservation of energy. Albert Einstein demonstrated, that there is a relationship between mass and energy. Mass can be converted into, energy, as it is inside the Sun or a nuclear reactor, and energy can be converted into, mass. It is the sum of mass and energy that remains constant. Our current focus is on, much less extreme situations., Using the principle of conservation of energy can have many practical benefits, as, automotive engineers are now demonstrating. When it comes to energy and cars, the, focus is often on how to cause the car to accelerate, how fast they will reach say a, speed of 100 km/h., Of course, cars also need to slow down, a task assigned to the brakes. As conventional, cars brake, the energy is typically dissipated as heat as the brake pads rub on the, rotors. Innovative new cars, called hybrids, now capture some of the kinetic energy and, convert it to chemical energy stored in batteries or mechanical energy stored in, flywheels. The engine then recycles that energy back into kinetic energy when the car, needs to accelerate, saving gasoline., , Ef = Ei, PEf + KEf = PEi + KEi, E = total energy, KE = kinetic energy, PE = potential energy, , 6.17 - Sample problem: conservation of energy, Sam is at the peak of his jump., Calculate Sam's speed when he, reaches the trampoline's surface., , Sam is jumping up and down on a trampoline. He bounces to a maximum height of 0.25 m above the surface of the trampoline. How fast will, he be traveling when he hits the trampoline? We define Sam’s potential energy at the surface of the trampoline to be zero., Variables, Sam’s height at peak, , h = 0.25 m, , Sam’s speed at peak, , vpeak = 0 m/s, , Sam’s speed at bottom, , v, , What is the strategy?, 1., , Use the law of conservation energy, to state that Sam’s total energy at the peak of his jump is the same as his total energy at the, surface of the trampoline. Simplify this equation, using the facts that his kinetic energy is zero at the peak, and his potential energy is, zero at the surface of the trampoline., , 2., , Solve the resulting equation for his speed at the bottom., , Physics principles and equations, The definition of gravitational potential energy, , PE = mgh, The definition of kinetic energy, , KE = ½ mv2, Total energy is conserved in this isolated system., , Ef = Ei, , Copyright 2007 Kinetic Books Co. Chapter 06, , 137
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Step-by-step solution, We start by stating the law of conservation of energy in equation form, and then adapting it to fit the specifics of Sam’s trampoline jump., , Step, , Reason, , 1., , Ef = Ei, , 2., , KEf + PEf = KEi + PEi energy is mechanical energy, , 3., , KEf + 0 = 0 + PEi, , enter values, , 4., , KEf = PEi, , simplify, , law of conservation of energy, , Now we have a simpler equation for Sam’s energy. The task now is to solve it for the one unknown variable, his speed at the surface of the, trampoline., , Step, , Reason, , 5., , KEf = PEi, , state equation again, , 6., , ½ mv2 = mgh, , definitions of KE, PE, , 7., , solve for v, , 8., , enter values, , 9., , v = 2.2 m/s, , evaluate, , 6.18 - Interactive checkpoint: conservation of energy, A boy releases a pork chop on a, rope. The chop is moving at a speed, of 8.52 m/s at the bottom of its swing., How much higher than this point is, the point from which the pork chop is, released? Assume that it has no initial, speed when starting its swing., , Answer:, , h=, , 138, , m, , Copyright 2007 Kinetic Books Co. Chapter 06
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6.19 - Interactive problem: conservation of energy, The law of conservation of energy states that the total energy in an isolated system, remains constant. In the simulation on the right, you can use this law and your, knowledge of potential and kinetic energies to help a soapbox derby car make a, jump., A soapbox derby car has no engine. It gains speed as it rolls down a hill. You can, drag the car to any point on the hill. A gauge will display the car’s height above the, ground. Release the mouse button and the car will fly down the hill., In this interactive, if the car is traveling 12.5 m/s at the bottom of the ramp, it will, successfully make the jump through the hoop. Too slow and it will fall short; too fast, and it will overshoot., You can use the law of conservation of energy to figure out the vertical position, needed for the car to nail the jump., , 6.20 - Friction and conservation of energy, In this section, we show how two principles we have discussed can be combined to, solve a typical problem. We will use the principle of conservation of energy and how, work done by an external force affects the total energy of a system to determine the, effect of friction on a block sliding down a plane., Suppose the 1.00 kg block shown to the right slides down an inclined wooden plane., Since the block is released from rest, it has no initial velocity. It loses 2.00 meters in, height as it slides, and it slides 6.00 meters along the surface of the inclined plane. The, force of kinetic friction is 2.00 N. You want to know the block’s speed when it reaches, the bottom position., To solve this problem, we start by applying the principle of conservation of energy. The, block’s initial energy is all potential, equal to the product of its mass, g and its height, (mgh). At a height of 2.00 meters, the block’s PE equals 19.6 J. The potential energy, will be zero when the block reaches the bottom of the plane. Ignoring friction, the PE of, the block at the top equals its KE at the bottom., , Effect of non-conservative forces, Reduce object’s energy, , Now we will factor in friction. The force of friction opposes the block’s motion down the, inclined plane. The work it does is negative, and that work reduces the energy of the, block. We calculate the work done by friction on the block as the force of friction times, the displacement along the plane, which equals í12.0 J. The block’s energy at the top, (19.6 J) plus the í12.0 J means the block has 7.6 J of kinetic energy at the bottom., Using the definition of kinetic energy, we can conclude that the 1.00 kg block is moving, at 3.90 m/s., You can also calculate the effect of friction by determining how fast the block would be, traveling if there were no friction. All 19.6 J of PE would convert to KE, yielding a, speed of 6.26 m/s. Friction reduces the speed of the block by approximately 38%., In general, non-conservative forces like friction and air resistance are dissipative forces:, They reduce the energy of a system. They do negative work since they act opposite the, direction of motion. (There are a few cases where they do positive work, such as when, the force of friction causes something to move, as when you step on a moving, sidewalk.), , Wnc = Ef í Ei, Wnc = work by non-conservative force, Ef = final energy, Ei = initial energy, , Copyright 2007 Kinetic Books Co. Chapter 06, , 139
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What is the block’s kinetic, energy at the bottom?, Wnc = Ef í Ei, (F cos ș)ǻx = KEf íPEi, KEf = PEi + (F cos ș ǻx), KEf = (mgh) + (F cos ș ǻx), KEf = (1.00 kg × 9.80 m/s2 × 2.00 m) +, (2.00 N × cos 180° × 6.00 m), KEf = 19.6 J + (í12.0 J), KEf = 7.60 J, 6.21 - Gotchas, You are asked to push two wheelbarrows up a hill. One wheelbarrow is empty, and you are able to push it up the hill in one minute. The other, is filled with huge rocks, and even after you push it for an hour, you cannot budge it. In which case do you do more work on the wheelbarrow?, You do more work on the empty wheelbarrow because it is the only one that moves. You do no work on the wheelbarrow filled with rocks, because you do not move it; its displacement is zero., Two 1/4 kg cheeseburgers are moving in opposite directions. One is rising at three m/s, the other is falling at three m/s. Which has more kinetic, energy? They are the same. The direction of velocity does not matter for kinetic energy; only the magnitude of velocity (the speed) matters., You start on a beach. You go to the moon. You come back. You go to Hollywood. You then go to the summit of Mount Everest. Have you, increased your gravitational potential energy more than if you had climbed to this summit without all the other side trips? No, the increase in, energy is the same. In both cases, the increase in gravitational potential energy equals your mass times the height of Mount Everest times g., You are holding a cup of coffee at your desk, a half-meter above the floor. You extend your arm laterally out a nearby third-story window so, that the cup is suspended 10 meters above the ground. Have you increased the cup’s gravitational potential energy? No. Assume you choose, the floor as the zero potential energy point. The potential energy is the same because the cup remains the same distance above the floor., An apple falls to the ground.Because the Earth’s gravity did work on it, the apple’s energy has increased. It depends on how you define “the, system.” If you said the apple had gravitational potential energy, then you are including the Earth as part of the system. In this case, the Earth’s, gravity is not an external force. Energy is conserved: The system’s decreased PE is matched by an increase in KE. You could also say “the, system” is solely the apple. In that case, it has no PE, since PE requires the presence of a force between objects in a system, and this system, has only one object. Gravity is now an external force acting on the apple, and it does increase the apple’s KE. This is not, perhaps, the typical, way to think about it, but it is valid., It is impossible for a system to have negative PE. Wrong: Systems can have negative PE. For instance, if you define the system to have zero, PE when an object is at the Earth’s surface, the object has negative PE when it is below the surface. Its PE has decreased from zero, so it, must be negative. Negative PE is common in some topics, such as orbital motion., , 140, , Copyright 2007 Kinetic Books Co. Chapter 06
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6.22 - Summary, Work is the product of the force on an object and its displacement in the direction of, that force. It is a scalar quantity with units of joules (1 J = 1 kg·m2/s2)., Work and several other scalar quantities can be computed by taking the dot product, of two vectors. The dot product is a scalar equal to the product of the magnitudes of, the two vectors and the cosine of the angle between them. Loosely, it tells you how, much of one vector is in the direction of another., Energy is a property of an object or a system. It has units of joules and is a scalar, quantity. Energy can transfer between objects and change forms. Work on an object, or system will change its energy., , W = F · ǻx = (F cos ș)ǻx, KE = ½ mv2, W = ǻKE, for a particle, , One form of energy is kinetic energy. It is the energy possessed by objects in, motion and is proportional to the object's mass and the square of its speed., The work-kinetic energy theorem states that the work done on a particle or an object, modeled as a particle is equal to its change in kinetic energy. Positive work, increases the energy, while negative work decreases it., Power is work divided by time. The unit of power is the watt (1 W = 1 J/s), a scalar, quantity. It is often expressed as a rate of energy consumption or output. For, example, a 100-watt light bulb converts 100 joules of electrical energy per second, into light and heat., , ǻPE = mgǻh, Ef = Ei, , Another form of energy is potential energy. It is the energy related to the positions of the objects in a system and the forces between them., Gravitational potential energy is an object's potential energy due to its position relative to a body such as the Earth., Forces can be classified as conservative or non-conservative. An object acted upon only by conservative forces, such as gravitational and, spring forces, requires no net work to return to its original position. An object acted upon by non-conservative forces, such as kinetic friction,, will not return to its initial position without additional work being done on it., When only conservative forces are present, the work to move an object between two points does not depend on the path taken. The work is, path independent. When non-conservative forces are acting, the work does depend on the path taken, and the work is path dependent., The law of conservation of energy states that the total energy in an isolated system remains constant, though energy may change form or be, transferred from object to object within the system., Mechanical energy is conserved only when there are no non-conservative forces acting in the system. When a non-conservative force such as, friction is present, the mechanical energy of the system decreases. The law of conservation of energy still holds, but we have not yet learned to, account for the other forms into which the mechanical energy might be transformed, such as thermal (heat) energy., , Copyright 2007 Kinetic Books Co. Chapter 06, , 141
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Chapter 6 Problems, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) If you release the car at a, higher point, will its speed at the bottom of the hill be greater, the same, or less? (b) How high will you have to drag the car to, have it just reach the summit of the other hill: to the same height, higher or lower?, (a), , (b), , i., ii., iii., i., ii., iii., , Greater, The same, Less, Higher than the other hill, To the same height as the other hill, Lower than the other hill, , Section 1 - Work, 1.1, , An airline pilot pulls her 12.0 kg rollaboard suitcase along the ground with a force of 25.0 N for 10.0 meters. The handle she, pulls on makes an angle of 36.5 degrees with the horizontal. How much work does she do over the ten-meter distance?, J, , 1.2, , A parent pushes a baby stroller from home to daycare along a level road with a force of 34 N directed at an angle of 30°, below the horizontal. If daycare is 0.83 km from home, how much work is done by the parent?, , 1.3, , A horizontal net force of 75.5 N is exerted on a 47.2 kg sofa, causing it to slide 2.40 meters along the ground. How much, work does the force do?, , J, , J, 1.4, , Charlie pulls horizontally to the right on a wagon with a force of 37.2 N. Sara pulls horizontally to the left with a force of, 22.4 N. How much work is done on the wagon after it has moved 2.50 meters to the right?, J, , Section 4 - Kinetic energy, 4.1, , You are about shoot two identical cannonballs straight up into the air. The first cannonball has 7.0 times as much initial, velocity as the second. How many times higher will the first cannonball go compared to the second?, , 4.2, , What is the change in kinetic energy of a baseball as it accelerates from rest to 45.0 m/s? The mass of a baseball is 145, grams., , times higher, , J, 4.3, , A bullet of mass 10.8 g leaves a gun barrel with a velocity of 511 m/s. What is the bullet's kinetic energy?, , 4.4, , A 0.50 kg cream pie strikes a circus clown in the face at a speed of 5.00 m/s and stops. What is the change in kinetic energy, of the pie?, , J, , J, , Section 5 - Work-kinetic energy theorem, 5.1, , A net force of 1.6×10í15 N acts on an electron over a displacement of 5.0 cm, in the same direction as the net force. (a) What, is the change in kinetic energy of the electron? (b) If the electron was initially at rest, what is the speed of the electron? An, electron has a mass of 9.1×10í31 kg., , 5.2, , 142, , (a), , J, , (b), , m/s, , A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10í27 kg.) (b), Assume the net braking force acting on it has magnitude 8.01×10í16 N and is directed opposite to its initial velocity. Over what, distance must the force be applied? Watch your negative signs in this problem., (a), , J, , (b), , m, , Copyright 2007 Kinetic Books Co. Chapter 6 Problems
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5.3, , A hockey stick applies a constant force over a distance of 0.121 m to an initially stationary puck, of mass 152 g. The puck, moves with a speed of 51.0 m/s. With what force did the hockey stick strike the puck?, N, , 5.4, , 5.5, , 5.6, , At the Aircraft Landing Dynamics Facility located at NASA's Langley Research Center in Virginia, a water-jet nozzle propels a, 46,000 kg sled from zero to 400 km/hour in 2.5 seconds. (This sled is equipped with tires that are being tested for the space, shuttle program, which are then slammed into the ground to see how they hold up.) (a) Assuming a constant acceleration for, the sled, what distance does it travel during the speeding-up phase? (b) What is the net work done on the sled during the time, interval?, (a), , m, , (b), , J, , A 25.0 kg projectile is fired by accelerating it with an electromagnetic rail gun on the Earth's surface. The rail makes a 30.0, degree angle with the horizontal, and the gun applies a 1250 N force on the projectile for a distance of 7.50 m along the rail., (a) Ignoring air resistance and friction, what is the net work done on the projectile, by all the forces acting on it, as it moves, 7.50 m along the rail? (b) Assuming it started at rest, what is its speed after it has moved the 7.50 m?, (a), , J, , (b), , m/s, , The force of gravity acts on a 1250 kg probe in outer space. It accelerates the probe from a speed of 225 m/s to 227 m/s over, a distance of 6750 m. How much work does gravity do on the probe?, J, , Section 7 - Interactive problem: work-kinetic energy theorem, 7.1, , Use the information given in the interactive problem in this section to answer the following question. What force is required for, the skier to make the jump? Test your answer using the simulation., N, , Section 9 - Power, 9.1, , How much power would be required to hoist a 48 kg couch up to a 22 m high balcony in 5.0 seconds? Assume it starts and, ends at rest., , 9.2, , Stuntman's Freefall, a ride at Six Flags Great Adventure in New Jersey, stands 39.6 meters high. Ignoring the force of, , W, friction, what is the minimum power rating of the motor that raises the 1.20×105 kg ride from the ground to the top in 10.0, seconds at a constant velocity?, W, 9.3, , Lori, who loves to ski, has rigged up a rope tow to pull herself up a local hill that is inclined at an angle of 30.0 degrees from, the horizontal. The motor works against a retarding frictional force of 100 N. If Lori has a mass of 60.5 kg, and the power of, the motor is 1350 W, at what speed can the motor pull her up the hill?, m/s, , 9.4, , A power plant supplies 1,100 megawatts of power to the electric grid. How many joules of energy does it supply each, second?, J, , 9.5, , How much work does a 5.00 horsepower outboard motor do in one minute? State your answer in joules., J, , 9.6, , The Porsche® 911 GT3 has a 380 hp engine and a mass of 1.4×103 kg. The car can accelerate from 0 to 100 km/h in 4.3, seconds. What percentage of the power supplied by the engine goes into making the car move? Assume that the car's, acceleration is constant and that there are 746 Watts/hp., %, , Section 10 - Potential energy, 10.1 You get paid $1.25 per joule of work. How much do you charge for moving a 10.0 kg box up to a shelf that is 1.50 meters off, the ground?, $, , Copyright 2007 Kinetic Books Co. Chapter 6 Problems, , 143
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Section 11 - Work and gravitational potential energy, 11.1 You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a, height of 1.85 m to 1.50 m. What is its change in potential energy?, J, 11.2 The reservoir behind the Grand Coulee dam in the state of Washington holds water with a total gravitational potential energy, of 1×1016 J, where the reference point for zero potential energy is taken as the height of the base of the dam. (a) Suppose, that the dam released all its water, which flowed to form a still pool at the base of the dam. What would be the change in the, gravitational potential energy of the Earth-dam-water system? (b) What work was done by the gravitational force? (Part of this, work is ordinarily used to turn electric generators.) (c) How much work would it take to pump all the water back up into the, reservoir?, (a), , J, , (b), , J, , (c), , J, , 11.3 What is the change in the gravitational potential energy of a Boeing 767 jet as it soars from the runway up to a cruising, altitude of 10.2 km? Assume its mass is a constant 2.04×105 kg., J, 11.4 A weightlifter raises a 115 kg weight from the ground to a height of 1.95 m in 1.25 seconds. What is the average power of this, maneuver?, W, 11.5 A small domestic elevator has a mass of 454 kg and can ascend at a rate of 0.180 m/s. What is the average power that must, be supplied for the elevator to move at this rate?, W, , Section 14 - Work and energy, 14.1 A firm bills at the rate of $1.00 per 125 J of work. It bills you $45.00 for carrying a sofa up some stairs. Their workers moved, the sofa up 3 flights of stairs, with each flight being 4.50 meters high. What is the mass of the sofa?, kg, 14.2 An engine supplies an upward force of 9.00 N to an initially stationary toy rocket, of mass 54.0 g, for a distance of 25.0 m., The rocket rises to a height of 339 meters before falling back to the ground. What was the magnitude of the average force of, air resistance on the rocket during the upward trip?, N, 14.3 On an airless moon, you drop a golf ball of mass 106 g out of a skyscraper window. After it has fallen 125 meters, it is moving, at 19.0 m/s. What is the rate of freefall acceleration on this moon?, m/s2, 14.4 You are pulling your sister on a sled to the top of a 16.0 m high, frictionless hill with a 10.0° incline. Your sister and the sled, have a total mass of 50.0 kg. You pull the sled, starting from rest, with a constant force of 127 N at an angle of 45.0° to the, hill. If you pull from the bottom to the top, what will the speed of the sled be when you reach the top?, m/s, , Section 16 - Conservation of energy, 16.1 A large block of ice is moving down the hill toward you at 25.0 m/s. Its mass is 125 kg. It is sliding down a slope that makes a, 30.0 degree angle with the horizontal. In short: Think avalanche. Assume the block started stationary and moves down the hill, with zero friction. How many meters has it been sliding?, m, , Section 19 - Interactive problem: conservation of energy, 19.1 Use the information given in the interactive problem in this section to answer the following question. What initial height is, required for the soapbox car to make it through the hoop? Test your answer using the simulation., m, , 144, , Copyright 2007 Kinetic Books Co. Chapter 6 Problems
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7.0 - Introduction, “The more things change, the more they stay the same” is a well-known French, saying., However, though witty and perhaps true for many matters on which the French have, great expertise, this saying is simply not good physics., Instead, a physicist would say: “Things stay the same, period. That is, unless acted, upon by a net force.” Perhaps a little less joie de vivre than your average, Frenchman, but nonetheless the key to understanding momentum., What we now call momentum, Newton referred to as “quantity of motion.” The linear, momentum of an object equals the product of its mass and velocity. (In this chapter,, we focus on linear momentum. Angular momentum, or momentum due to rotation,, is a topic in another chapter.) Momentum is a useful concept when applied to, collisions, a subject that can be a lot of fun. In a collision, two or more objects exert, forces on each other for a brief instant of time, and these forces are significantly, greater than any other forces they may experience during the collision., At the right is a simulation í a variation of shuffleboard í that you can use to begin, your study of momentum and collisions. You can set the initial velocity for both the blue and the red pucks and use these velocity settings to, cause them to collide. The blue puck has a mass of 1.0 kg, and the red puck a mass of 2.0 kg. The shuffleboard has no friction, but the pucks, stop moving when they fall off the edge. Their momenta and velocities are displayed in output gauges., Using the simulation, answer these questions. First, is it possible to have negative momentum? If so, how can you achieve it? Second, does, the collision of the pucks affect the sum of their velocities? In other words, does the sum of their velocities remain constant? Third, does the, collision affect the sum of their momenta? Remember to consider positive and negative signs when summing these values. Press PAUSE, before and after the collisions so you can read the necessary data. For an optional challenge: Does the collision conserve the total kinetic, energy of the pucks? If so, the collision is called an elastic collision. If it reduces the kinetic energy, the collision is called an inelastic collision., , 7.1 - Momentum, , Momentum (linear):Mass times velocity., An object’s linear momentum equals the product of its mass and its velocity. A fast, moving locomotive has greater momentum than a slowly moving ping-pong ball., The units for momentum are kilogram·meters/second (kg·m/s). A ping-pong ball with a, mass of 2.5 grams moving at 1.0 m/s has a momentum of 0.0025 kg·m/s. A 100,000 kg, locomotive moving at 5 m/s has a momentum of 5×105 kg·m/s., Momentum is a vector quantity. The momentum vector points in the same direction as, the velocity vector. This means that if two identical locomotives are moving at the same, speed and one is heading east and the other west, they will have equal but opposite, momenta, since they have equal but oppositely directed velocities., , Momentum, Moving objects have momentum, Momentum increases with mass,, velocity, , p = mv, , Copyright 2007 Kinetic Books Co. Chapter 07, , 145
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p = momentum, m = mass, v = velocity, Momentum same direction as velocity, Units: kg·m/s, , What is the toy truck’s, momentum?, p = mv, p = (0.14 kg)(1.2 m/s), p = 0.17 kg·m/s to the right, 7.2 - Momentum and Newton's second law, , Momentum and Newton's second, law, , ȈF = net force, p = momentum, t = time, Although you started your study of physics with velocity and acceleration, early physicists such as Newton focused much of their attention on, momentum., The equation in Equation 1 may remind you of Newton's second law. In fact, it is equivalent to the second law (as we show below) and Newton, stated his law in this form. This equation is useful when you know the change in the momentum of an object (or, equivalently for an object of, constant mass, the change in velocity). This equation as stated assumes an average or constant external net force., , 146, , Copyright 2007 Kinetic Books Co. Chapter 07
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Below, we show that this equation is equivalent to the more familiar version of Newton's second law based on mass and acceleration. We state, that version of Newton’s law, and then use the definition to restate the law as you see it here., , Step, , 1., , 2., , Reason, , Newton’s second law; definition of acceleration, , p = mv, , definition of momentum, , 3., , divide equation 2 by ǻt, , 4., , substitute equation 3 into equation 1, , 7.3 - Impulse, , Impulse, Average force times elapsed time, Change in momentum, , Impulse:Change in momentum., In the prior section, we stated that net force equals change in momentum per unit time., We can rearrange this equation and state that the change in momentum equals the, product of the average force and the elapsed time, which is shown in Equation 1., The change in momentum is called the impulse of the force, and is represented by J., Impulse is a vector, has the same units as momentum, and points in the same direction, as the change in momentum and as the force. The relationship shown is called the, impulse-momentum theorem., In this section, we focus on the case when a force is applied for a brief interval of time,, and is stated or approximated as an average force. This is a common and important, way of applying the concept of impulse., A net force is required to accelerate an object, changing its velocity and its momentum., The greater the net force, or the longer the interval of time it is applied, the more the, object's momentum changes, which is the same as saying the impulse increases., Engineers apply this concept to the systems they design. For instance, a cannon barrel, is long so that the cannonball is exposed to the force of the explosive charge longer,, which causes the cannonball to experience a greater impulse, and a greater change in, momentum., Even though a longer barrel allows the force to be applied for a longer time interval, it is, still brief. Measuring a rapidly changing force over such an interval may be difficult, so, the force is often modeled as an average force. For example, when a baseball player, swings a bat and hits the ball, the duration of the collision can be as short as 1/1000th of, a second and the force averages in the thousands of newtons., , J = impulse, , t = time, p = momentum, Units of impulse: kg·m/s, , The brief but large force that the bat exerts on the ball is called an impulsive force. When analyzing a collision like this, we ignore other forces, (like gravity) that are acting upon the ball because their effect is minimal during this brief period of time., In the illustration for Equation 1, you see a force that varies with time (the curve) and the average of that force (the straight dashed line). The, area under the curve and the area of the rectangle both equal the impulse, since both equal the product of force and time., The nature of impulse explains why coaches teach athletes like long jumpers, cyclists, skiers and martial artists to relax when they land or fall,, and why padded mats and sand pits are used. In Example 1 on the right, we calculate the (one-dimensional) impulse experienced by a long, jumper on landing in the sand pit, from her change in momentum., , Copyright 2007 Kinetic Books Co. Chapter 07, , 147
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Her impulse (change in momentum) is the same, however long it takes her to stop when, she hits the ground. In the example, that impulse is í530 kg·m/s., Why does she want to extend the time of her landing? She wants to make this time as, long as possible (by landing in the sand, by flexing her knees), since it means the, collision lasts longer. Since impulse equals force multiplied by elapsed time, the, average force required to produce the change in momentum decreases as the time, increases. The reduced average force lessens the chance of injury. Padded mats are, another application of this concept: The impulse of landing is the same on a padded or, unpadded floor, but a mat increases the duration of a landing and reduces its average, force., There are also numerous applications of this principle outside of sports. For example,, cars have “crumple zones” designed into them that collapse upon impact, extending the, duration of the impulse during a collision and reducing the average force., , The long jumper's speed just, before landing is 7.8 m/s. What is, the impulse of her landing?, J = pf ípi, J = mvf ímvi, J = 0.0 í (68 kg)(7.8 m/s), J = í530 kg·m/s, , 7.4 - Physics at play: hitting a baseball, When a professional baseball player swings a bat and hits a ball square on, he will, dramatically change its velocity in a millisecond. A fastball can approach the plate at, around 95 miles per hour, and in a line drive shot, the ball can leave the bat in roughly, the opposite direction at about 110 miles per hour, a change of about 200 mph in about, a millisecond., The bat exerts force on the baseball in the very brief period of time they are in contact., The amount of force varies over this brief interval, as the graph to the right reflects., At the moment of contact, the bat and ball are moving toward each other. The force on, the ball increases as they come together and the ball compresses against the bat. The, force applied to the ball during the time it is in contact with the bat is responsible for the, ball’s change in momentum., How long the bat stays in contact with the ball is much easier to measure than the, average force the bat exerts on the ball, but by applying the concept of impulse, that, force can be calculated. Impulse equals both the average force times the elapsed time, and the change in momentum. Since the velocities of the baseball can be observed, (say, with a radar gun), and the baseball’s mass is known, its change in momentum can, be calculated, as we do in Example 1. The time of the collision can be observed using, stroboscopic photography and other techniques. This leaves one variable í average, force í and we solve for that in the example problem., , Baseballs, bats and impulse, Force applied over time changes, momentum, Impulse = change in momentum, Impulse = average force × elapsed time, , The average force equals 2.5×104 N. A barrier that stops a car moving at 20 miles per, hour in half a second exerts a comparable average amount of force., , The ball arrives at 40 m/s and, leaves at 49 m/s in the opposite, direction. The contact time is, 5.0×10í4 s. What is the average, force on the ball?, J = Favg ǻt = ǻp = mǻv, Favg = mǻv/ǻt, , 148, , Copyright 2007 Kinetic Books Co. Chapter 07
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Favg = 2.5×104 N, 7.5 - Conservation of momentum, , Conservation of momentum, No net external force on system:, Total momentum is conserved, , Conservation of momentum: The total, momentum of an isolated system is constant., Momentum is conserved in an isolated system. An isolated system is one that does not, interact with its environment. Momentum can transfer from object to object within this, system but the vector sum of the momenta of all the objects remains constant., Excluding deep space, it can be difficult to find locations where a system has no, interaction with its environment. This makes it useful to state that momentum is, conserved in a system that has no net force acting on it., We will use pool balls on a pool table to discuss the conservation of momentum. To put, it more formally, the pool balls are the system., In pool, a player begins play by striking the white cue ball. To use the language of, physics, a player causes there to be a net external force acting on the cue ball., Once the cue ball has been struck, it may collide with another ball, and more collisions, may ensue. However, there is no net external force acting on the balls after the cue ball, has been struck (ignoring friction which we will treat as negligible). The normal force of, the table balances the force of gravity on the balls. Since there is now no net external, force acting on the balls, the total amount of their momentum remains constant. When, they collide, the balls exert forces on one another, but this is a force internal to the, system, and does not change its total momentum., , pi1 + pi2 +…+ pin = pf1 + pf2 +…+ pfn, pi1, pi2, …, pin = initial momenta, pf1, pf2, …, pfn = final momenta, , In the scenario you see illustrated above, the cue ball and another ball are shown, before and after a collision. The cue ball initially has positive momentum since it is, moving to the right. The ball it is aimed at is initially stationary and has zero momentum., When the cue ball strikes its target, the cue ball slows down and the other ball speeds, up. In fact, the cue ball may stop moving. You see this shown on the right side of the, illustration above. During the collision, momentum transfers from one ball to another., The law of conservation of momentum states that the combined momentum of both, remains constant: One ball’s loss equals the other ball’s gain., A rifle also provides a notable example of the conservation of momentum. Before it is, fired, the initial momentum of a rifle and the bullet it fires are both zero (since neither, has any velocity). When the rifle is fired, the bullet moves in one direction and the rifle, recoils in the opposite direction. The bullet and the rifle each now have nonzero, momentum, but the vector sum of their momenta must remain at zero., Two factors account for this. First, the rifle and the bullet are moving in opposite, directions. In the case of the rifle and bullet, all the motion takes place along a line, so, we can use positive and negative to indicate direction. Let’s assume the bullet has, positive velocity; since the rifle moves (recoils) in the opposite direction, it has negative, velocity. The momentum vector of each object points in the same direction as its, velocity vector. This means the bullet has positive momentum while the rifle has, negative momentum., , The balls have the same mass., The cue ball strikes the, stationary yellow ball head on,, and stops. What is the yellow, ball’s resulting velocity?, pi,cue + pi,yel = pf,cue + pf,yel, mvi,cue + mvi,yel = mvf,cue + mvf,yel, vi,cue + vi,yel = vf,cue + vf,yel, vi,cue + vi,yel = vf,cue + vf,yel, 3.1 m/s + 0.0 m/s = 0.0 m/s + vf,yel, , vf,yel = 3.1 m/s to the right, Second, for momentum to be conserved, the sum of these momenta must equal zero,, since the sum was zero before the rifle was fired. The amount of momentum of the, faster moving but less massive bullet equals the amount of momentum of the more massive but slower moving rifle. When the two are added, together, the total momentum continues to equal zero., , Copyright 2007 Kinetic Books Co. Chapter 07, , 149
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7.6 - Derivation: conservation of momentum from Newton’s laws, , Conservation of momentum, m1vi1 + m2vi2 = m1vf1 + m2vf2, p = mv = momentum, m1, m2 = masses of objects, vi1, vi2 = initial velocities, vf1, vf2 = final velocities, Newton formulated many of his laws concerning motion using the concept of momentum, although today his laws are stated in terms of force, and acceleration. The law of conservation of momentum can be derived from his second and third laws., The derivation uses a collision between two balls of masses m1 and m2 with velocities v1 and v2. You see the collision illustrated above, along, with the conservation of momentum equation we will prove for this situation., To derive the equation, we consider the forces on the balls during their collision. During the time, and F2 on each other. (F1 is the force on ball 1 and F2 the force on ball 2.), , ǻt of the collision, the balls exert forces F1, , Diagram, , This diagram shows the forces on the balls during the collision., Variables, duration of collision, , ǻt, ball 1, , ball 2, , force on ball, , F1, , F2, , mass, , m1, , m2, , acceleration, , a1, , a2, , initial velocity, , vi1, , vi2, , final velocity, , vf1, , vf2, , Strategy, 1., , Use Newton’s third law: The forces will be equal but opposite., , 2., , Use Newton’s second law, F, , 3., , Use the definition that expresses acceleration in terms of change in velocity. This will result in an equation that contains momentum (, mv) terms., , = ma, to determine the acceleration of the balls., , Physics principles and equations, In addition to Newton’s laws cited above, we will use the definition of acceleration., , a = ǻv/ǻt, , 150, , Copyright 2007 Kinetic Books Co. Chapter 07
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Step-by-step derivation, In these first steps, we use Newton’s third law followed by his second law., , Step, , Reason, , 1., , F1 = –F2, , Newton's third law, , 2., , m1a1 = –m2a2, , Newton's second law, definition of acceleration, , 3., , In the next steps, we apply the definition of the change ǻv in velocity. After some algebraic simplification we obtain the result we want: The, sum of the initial momenta equals the sum of the final momenta., , Step, , Reason, , 4., , definition of change in velocity, , 5., , m1vf1 – m1vi1 = –m2vf2 + m2vi2, , multiply both sides by ǻt, , 6., , m1vi1 + m2vi2 = m1vf1 + m2vf2, , rearrange, , 7.7 - Interactive checkpoint: astronaut, The 55.0 kg astronaut is stationary in, the spaceship’s reference frame. She, wants to move at 0.500 m/s to the, left. She is holding a 4.00 kg bag of, dehydrated astronaut chow. At what, velocity must she throw the bag to, achieve her desired velocity?, (Assume the positive direction is to, the right.), , Answer:, , vfb =, , m/s, , 7.8 - Collisions, , Elastic collision, Kinetic energy is conserved, , Elastic collision: The kinetic energy of the system is unchanged by the, collision., Inelastic collision: The kinetic energy of the system is changed by the, collision., In a collision, one moving object briefly strikes another object. During the collision, the forces the objects exert on each other are much greater, , Copyright 2007 Kinetic Books Co. Chapter 07, , 151
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than the net effect of other forces acting on them, so we may ignore these other forces., Elastic and inelastic are two terms used to define types of collisions. These types of, collisions differ in whether the total amount of kinetic energy in the system stays, constant or is reduced by the collision. In any collision, the system’s total amount of, energy must be the same before and after, because the law of conservation of energy, must be obeyed. But in an inelastic collision, some of the kinetic energy is transformed, by the collision into other types of energy, so the total kinetic energy decreases., For example, a car crash often results in dents. This means some kinetic energy, compresses the car permanently; other KE becomes thermal energy, sound energy, and so on. This means that an inelastic collision reduces the total amount of KE., In contrast, the total kinetic energy is the same before and after an elastic collision., None of the kinetic energy is transformed into other forms of energy. The game of pool, provides a good example of nearly elastic collisions. The collisions between balls are, almost completely elastic and little kinetic energy is lost when they collide., , Inelastic collision, Kinetic energy is not conserved, , In both elastic and inelastic collisions occurring within an isolated system, momentum is, conserved. This important principle enables you to analyze any collision., We will mention a third type of collision briefly here: explosive collisions, such as what, occurs when a bomb explodes. In this type of collision, the kinetic energy is greater, after the collision than before. However, since momentum is conserved, the explosion, does not change the total momentum of the constituents of the bomb., , Either type of collision, Momentum is conserved, , 7.9 - Sample problem: elastic collision in one dimension, The small purple ball strikes the, stationary green ball in an elastic, collision. What are the final velocities, of the two balls?, , The picture and text above pose a classic physics problem. Two balls collide in an elastic collision. The balls collide head on, so the second, ball moves away along the same line as the path of the first ball. The balls’ masses and initial velocities are given. You are asked to calculate, their velocities after the collision. The strategy for solving this problem relies on the fact that both the momentum and kinetic energy remain, unchanged., Variables, ball 1 (purple), , ball 2 (green), , mass, , m1 = 2.0 kg, , m2 = 3.0 kg, , initial velocity, , vi1 = 5.0 m/s, , vi2 = 0 m/s, , final velocity, , vf1, , vf2, , What is the strategy?, 1., , Set the momentum before the collision equal to the momentum after the collision., , 2., , Set the kinetic energy before the collision equal to the kinetic energy after the collision., , 3., , Use algebra to solve two equations with two unknowns., , Physics principles and equations, Since problems like this one often ask for values after a collision, it is convenient to state the following conservation equations with the final, values on the left., Conservation of momentum, , m1vf1 + m2vf2 = m1vi1 + m2vi2, , 152, , Copyright 2007 Kinetic Books Co. Chapter 07
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Conservation of kinetic energy, , ½ m1vf12 + ½ m2vf22 = ½ m1vi12 + ½ m2vi22, Step-by-step solution, First, we use the conservation of momentum to find an equation where the only unknown values are the two final velocities. Since all the, motion takes place on a horizontal line, we use sign to indicate direction., , Step, , Reason, , 1., , conservation of momentum, , 2., , enter values, , 3., , vf1 = (–1.5)vf2 + 5.0, , solve for vf1, , The conservation of kinetic energy gives us another equation with these two unknowns., , Step, , Reason, , 4., , elastic collision: KE conserved, , 5., , simplify, , 6., , enter values, , 7., , re-arrange as quadratic equation, , We substitute the expression for the first ball's final velocity found in equation 3 into the quadratic equation, and solve. This gives us the second, ball's final velocity. Then we use equation 3 again to find the first ball's final velocity. One velocity is negative, and one positive í one ball, moves to the left after the collision, the other to the right., , Step, , Reason, , 8., , substitute equation 4 into equation 8, , 9., , simplify, , 10. vf2 = 4.0 m/s (to the right), , solve equation, , 11., , use equation 3 to find vf1, , A quick check shows that the total momentum both before and after the collision is 10 kg·m/s. The kinetic energy is 25 J in both cases. This, verifies that we did the computations correctly., There is a second solution to this problem: You can see that vf2 = 0 is also a solution to the quadratic equation in step 9, and then using step, 3, you see that vf1 would equal 5.0 m/s. This solution satisfies the conditions that momentum and KE are conserved, and it describes what, happens if the balls do not collide. In other words, the purple ball passes by the green ball without striking it., , Copyright 2007 Kinetic Books Co. Chapter 07, , 153
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7.10 - Interactive checkpoint: another one dimensional collision problem, Two balls move toward each other, and collide head-on in an elastic, collision. What is the mass and final, velocity of the green ball?, , Answer:, , m2 =, , kg, , vf2 =, , m/s, , 7.11 - Physics at play: clicky-clack balls, , Clicky-clack balls, Momentum conserved, Elastic collision: kinetic energy, conserved, , The law of conservation of momentum and the nature of elastic collisions underlie the functioning of a desktop toy: a set of balls of equal mass, hanging from strings. This toy is shown in the photograph above., You may have seen these toys in action but if you have not, imagine pulling back one ball and releasing it toward the pack of balls. (Click on, Concept 1 to launch a video.) The one ball strikes the pack, stops, and a ball on the far side flies up, comes back down, and strikes the pack., The ball you initially released now flies back up again, returns to strike the pack, and so forth. The motion continues in this pattern for quite a, while., Interestingly, if you pick up two balls and release them, then two balls on the far side of the pack will fly off, resulting in a pattern of two balls, moving. This pattern obeys the principle of the conservation of momentum as well as the definition of an elastic collision: kinetic energy, remains constant. Other scenarios that on the surface might seem plausible fail to meet both criteria. For instance, if one ball moved off the, pack at twice the speed of the two balls striking, momentum would be conserved, but the KE of the system would increase, since KE is a, function of the square of velocity. Doubling the velocity of one ball quadruples its KE. One ball leaving at twice the velocity would have twice, the combined KE of the two balls that struck the pack., However, it turns out this is not the only solution which obeys the conservation of momentum and of kinetic energy. For instance, the striking, ball could rebound at less than its initial speed, and the remaining four balls could move in the other direction as a group. With certain speeds, for the rebounding ball and the pack of four balls, this would provide a solution that would obey both principles. Why the balls behave exactly, as they do has inspired plenty of discussion in physics journals., , 154, , Copyright 2007 Kinetic Books Co. Chapter 07
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7.12 - Interactive problem: shuffleboard collisions, The simulation at the right shows a variation of the game of shuffleboard. The red, puck has an initial velocity of í2.0 m/s. You want to set the initial velocity of the blue, puck so that after the two pucks collide head on in an elastic collision, the red puck, moves with a velocity of +2.0 m/s. This will cause the red puck to stop at the scoring, line, since the friction in the green area on the right side of the surface will cause it, to slow down and perhaps stop., The blue puck has a mass of 1.0 kilograms. The red puck’s mass is 3.0 kilograms., Use the fact that the collision is elastic to calculate and enter the initial velocity of, the blue puck to the nearest 0.1 m/s and press GO to see the results. Press RESET, to try again., , 7.13 - Inelastic collisions, , Inelastic collisions, Collision reduces total KE, Momentum conserved, Completely inelastic collisions:, ·Objects "stick together", ·Have common velocity after collision, , Inelastic collision: The collision results in a, decrease in the system’s total kinetic energy., In an inelastic collision, momentum is conserved. But kinetic energy is not conserved., In inelastic collisions, kinetic energy transforms into other forms of energy. The kinetic, energy after an inelastic collision is less than the kinetic energy before the collision., When one boxcar rolls and connects with another, as shown above, some of the kinetic, energy of the moving car transforms into elastic potential energy, thermal energy and so, forth. This means the kinetic energy of the system of the two boxcars decreases,, making this an inelastic collision., A completely inelastic collision is one in which two objects “stick together” after they, collide, so they have a common final velocity. Since they may still be moving,, completely inelastic does not mean there is zero kinetic energy after the collision. For, instance, after the boxcars connect, the two “stick together” and move as one unit. In, this case, the train combination continues to move after the collision, so they still have, kinetic energy, although less than before the collision., As with elastic collisions, we assume the collision occurs in an isolated system, with no, net external forces present. You can think of elastic collisions and completely inelastic, collisions as the extreme cases of collisions. Kinetic energy is not reduced at all in an, elastic collision. In a completely inelastic collision, the total amount of kinetic energy, after the collision is reduced as much as it can be, consistent with the conservation of, momentum., , Completely inelastic collision, , vf = common final velocity, m1, m2 = masses, vi1, vi2 = initial velocities, , In Equation 1, you see an equation to calculate the final velocity of two objects, like the snowballs shown, after a completely inelastic collision., This equation is derived below. The derivation hinges on the two objects having a common velocity after the collision., , Copyright 2007 Kinetic Books Co. Chapter 07, , 155
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In this derivation, two objects, with masses m1 and m2, collide in a completely inelastic, collision., , Step, , 1., , vf1 = vf2 = vf, , Reason, , final velocities equal, , 2., , conservation of momentum, , 3., , factor out vf, , 4., , divide, , Example 1 on the right applies this equation to a collision seen on many fall weekends:, a football tackle., , The players collide head-on at, the velocities shown. What is, their common velocity after this, completely inelastic collision?, , vf = 0.65 m/s to the right, 7.14 - Center of mass, , Center of mass, “Average” location of mass, , Center of mass: Average location of mass. An, object can be treated as though all its mass were, located at this point., The center of mass is useful when considering the motion of a complex object, or, system of objects. You can simplify the analysis of motion of such an object, or system, of objects, by determining its center of mass. An object can be treated as though all its, mass is located at this point. For instance, you could consider the force of a weightlifter, lifting the barbell pictured above as though she applied all the force at the center of, mass of the bar, and determine the acceleration of the center of mass., You may react: “But we have been doing this in many sections of this book,” and yes,, implicitly we have been. If we asked earlier how much force was required to accelerate, this barbell, we assumed that the force was applied at the center of mass, rather than at, one end of the barbell, which would cause it to rotate., , Center of mass, At geometric center of uniform,, symmetric objects, , In this section, we focus on how to calculate the center of mass of a system of objects. Consider the barbell above. Its center of mass is on the, rod that connects the two balls, nearer the ball labeled “Work,” because that ball is more massive., When an object is symmetrical and made of a uniform material, such as a solid sphere of steel, the center of mass is at its geometric center., So for a sphere, cube or other symmetrical shape made of a uniform material, you can use your sense of geometry and decide where the, center of the object is. That point will be the center of the mass. (We can relax the condition of uniformity if an object is composed of different, parts, but each one of them is symmetrical, like a golf ball made of different substances in spherically symmetrical layers.), The center of mass does not have to lie inside the object. For example, the center of mass of a doughnut lies in the middle of its hole., The equation to the right can be used to calculate the overall center of mass of a set of objects whose individual centers of mass lie along a, line. To use the equation, place the center of mass of each object on the x axis. It helps to choose for the origin a point where one of the, centers of mass is located, since this will simplify the calculation. Then, multiply the mass of each object times its center’s x position and divide, the sum of these products by the sum of the masses. The resulting value is the x position of the center of mass of the set of objects., , 156, , Copyright 2007 Kinetic Books Co. Chapter 07
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If the objects do not conveniently lie along a line, you can calculate the x and y, positions of the center of mass by applying the equation in each dimension separately., The result is the x, y position of the system’s center of mass., , xCM = x position of center of mass, mi = mass of object i, xi = x position of object i, , What is the location of the center, of mass?, , xCM = 32/5.0, xCM = 6.4 m, 7.15 - Center of mass and motion, , Center of mass and motion, Laws of mechanics apply to center of, mass, Shifting center of mass creates "floating", illusion, , Above, you see a ballet dancer performing a grand jeté, a “great leap.” When a ballet dancer performs this leap well, she seems to float, through the air. In fact, if you track the dancer’s motion by noting the successive positions of her head, you can see that its path is nearly, horizontal. She seems to be defying the law of gravity. This seeming physics impossibility is explained by considering the dancer’s center of, mass. An object (or a system of objects) can be analyzed by considering the motion of its center of mass., Look carefully at the locations of the dancer’s center of mass in the diagram. The center of mass follows the parabolic path of projectile motion., To achieve the illusion of floating í moving horizontally í the dancer alters the location of her center of mass relative to her body as she, performs the jump. As she reaches the peak of her leap, she raises her legs, which places her center of mass nearer to her head. This, , Copyright 2007 Kinetic Books Co. Chapter 07, , 157
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decreases the distance from the top of her head to her center of mass, so her head, does not rise as high as it would otherwise. This allows her head to move in a straight, line while her center of mass moves in the mandatory parabolic projectile arc., At the right, we use another example to make a similar point. A cannonball explodes in, midair. Although the two resulting fragments move in different directions, the center of, mass continues along the same trajectory the cannonball would have followed had it not, exploded. The two fragments have different masses. The path of the center of mass is, closer to the path of the more massive fragment, as you might expect., , Center of mass and motion, Center of mass follows projectile path, , 7.16 - Interactive summary problem: types of collisions, On the right is a simulation featuring three collisions. Each collision is classified as, one of the following: an elastic collision, a completely inelastic collision, an inelastic, (but not completely inelastic) one, or an impossible collision that violates the laws of, physics. The colliding disks all have the same mass, and there is no friction. Each, disk on the left has an initial velocity of 1.00 m/s. The disks on the right have an, initial velocity of í0.60 m/s., Press GO to watch the collisions. Use the PAUSE button to stop the action after the, collisions and record data, then make whatever calculations you need to classify, each collision using the choices in the drop-down controls labeled “Collision type.”, Press RESET if you want to start the simulation from the beginning., If you have difficulty with this, review the sections on elastic and inelastic collisions., , 7.17 - Gotchas, One object has a mass of 1 kg and a speed of 2 m/s, and another object has a mass of 2 kg and a speed of 1 m/s.The two objects have, identical momenta. Only if they are moving in the same direction. You can say they have equal magnitudes of momentum, but momentum is a, vector, so direction matters. Consider what happens if they collide. The result will be different depending on whether they are moving in the, same or opposite directions., In inelastic collisions, momentum is not conserved. No. Kinetic energy decreases, but momentum is conserved., Two objects are propelled by equal constant forces, and the second one is exposed to its force for three times as long.The second object’s, change in momentum must be greater than the first’s. This is true. It experienced a greater impulse, and impulse equals the change in, momentum., , 158, , Copyright 2007 Kinetic Books Co. Chapter 07
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7.18 - Summary, An object's momentum is the product of its mass and velocity. It is a vector quantity, with units of kg·m/s., Like energy, momentum is conserved in an isolated system. If no net external force, acts on a system, its total momentum is constant., , p = mv, , A change in momentum is called impulse. It is a vector with the same units as, momentum. Impulse can be calculated as the difference between the final and initial, momenta, or as an average applied force times the duration of the force., The conservation of momentum is useful in analyzing collisions between objects,, since the total momentum of the objects involved must be the same before and after, the collision. In an elastic collision, kinetic energy is also conserved. In an inelastic, collision, the kinetic energy is reduced during the collision as some or all of it is, converted into other forms of energy., A collision is completely inelastic if the kinetic energy is reduced as much as, possible, consistent with the conservation of momentum. The two objects "stick, together" after the collision., , Conservation of momentum, , pi1 + pi2 +…+ pin = pf1 + pf2 +…+ pfn, Center of mass, , The center of mass of an object (or system of objects) is the average location of the, object's (or system's) mass. For a uniform object, this is the object's geometric, center. For more complicated objects and systems, center of mass equations must, be applied., Moving objects behave as if all their mass were concentrated at their center of mass. For example, a hammer thrown into the air may rotate as, it falls, but its center of mass will follow the parabolic path followed by any projectile., , Copyright 2007 Kinetic Books Co. Chapter 07, , 159
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Chapter 7 Problems, , Conceptual Problems, C.1, , Two balls of equal mass move at the same speed in different directions. Are their momenta equal? Explain., , C.2, , (a) If a particle that is moving has the same momentum and the same kinetic energy as another, must their masses and, velocities be equal? If so, explain why, and if not, give a counterexample. (b) Describe the properties of two particles that, have the same momentum, but different kinetic energies., , Yes, , (a), , No, , Yes, , No, , (b), C.3, , Two balls are moving in the same direction. Ball A has half the mass of ball B, and is moving at twice its speed. (a) Which ball, has the greater momentum? (b) Which ball has greater kinetic energy?, (a), , (b), , i., ii., iii., iv., i., ii., iii., iv., , A's momentum is greater, B's momentum is greater, They have the same momentum, You cannot tell, A's kinetic energy is greater, B's kinetic energy is greater, They have the same kinetic energy, You cannot tell, , C.4, , Two salamanders have the same mass. Their momenta (considered as signed quantities) are equal in magnitude but, opposite in sign. Describe the relationship of their velocities., , C.5, , A large truck collides with a small car. How does the magnitude of the impulse experienced by the truck compare to that, experienced by the car?, i. Truck impulse greater, ii. Truck impulse equal, iii. Truck impulse less, , C.6, , Object A is moving when it has a head-on collision with stationary object B. No external forces act on the objects. Which of, the following situations are possible after the collision? Check all that are possible., A and B move in the same direction, A and B move in opposite directions, A moves and B is stationary, A is stationary and B moves, A and B are both stationary, , C.7, , A cannonball is on track to hit a distant target when, at the top of its flight, it unexpectedly explodes into two pieces that fly out, horizontally. You find one piece of the cannonball, and you know the target location. What physics principle would you apply, to find the other piece? Explain., i. Conservation of momentum, ii. Impulse, iii. Center of mass, , C.8, , An object can have a center of mass that does not lie within the object itself. Give examples of two such objects., , Section Problems, Section 0 - Introduction, 0.1, , Use the information given in the interactive problem in this section to answer the following questions. (a) Is it possible to have, negative momentum? (b) Does the sum of the pucks' velocities remain constant before and after the collision? (c) Does the, sum of the pucks' momenta remain constant?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , Section 1 - Momentum, 1.1, , 160, , Belle is playing tennis. The mass of the ball is 0.0567 kg and its speed after she hits it is 22.8 m/s. What is the magnitude of, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems
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the momentum of the ball?, kg · m/s, 1.2, , A sport utility vehicle is travelling at a speed of 15.3 m/s. If its momentum has a magnitude of 32,800 kg·m/s, what is the, SUV's mass?, , 1.3, , A truck with a mass of 2200 kg is moving at a constant 13 m/s. A 920 kg car with the same momentum as the truck passes it, at a constant speed. How fast is the car moving?, , kg, , m/s, 1.4, , At a circus animal training facility, a monkey rides a miniature motorscooter at a speed of 7.0 m/s. The monkey and scooter, together have a mass of 29 kg. Meanwhile, a chimpanzee on roller skates with a total mass of 44 kg moves at a speed of 1.5, meters per second. The magnitude of the momentum of the monkey plus scooter is how many times the magnitude of the, momentum of the chimpanzee plus skates?, , 1.5, , A net force of 30 N is applied to a 10 kg object, which starts at rest. What is the magnitude of its momentum after 3.0, seconds?, kg · m/s, , Section 2 - Momentum and Newton's second law, 2.1, , A golden retriever is sitting in a park when it sees a squirrel. The dog starts running, exerting a constant horizontal force of, 89 N against the ground for 3.2 seconds. What is the magnitude of the dog's change in momentum?, kg · m/s, , Section 3 - Impulse, 3.1, , A 1400 kg car traveling in the positive direction takes 10.5 seconds to slow from 25.0 meters per second to 12.0 meters per, second. What is the average force on the car during this time?, , 3.2, , A cue stick applies an average force of 66 N to a stationary 0.17 kg cue ball for 0.0012 s. What is the magnitude of the, impulse on the cue ball?, , 3.3, , A baseball arrives at home plate at a speed of 43.3 m/s. The batter hits the ball along the same line straight back to the, , N, , kg · m/s, pitcher at 68.4 m/s. The baseball has a mass of 0.145 kg and the bat is in contact with the ball for 6.28×10í4 s. What is the, magnitude of the average force on the ball from the bat?, N, 3.4, , 3.5, , A steel ball with mass 0.347 kg falls onto a hard floor and bounces. Its speed just before hitting the floor is 23.6 m/s and its, speed just after bouncing is 12.7 m/s. (a) What is the magnitude of the impulse of the ball? (b) If the ball is in contact with the, floor for 0.0139 s, what is the magnitude of the average force exerted on the ball by the floor?, (a), , kg · m/s, , (b), , N, , The net force on a baby stroller increases at a constant rate from 0 N to 88 N, over a period of 0.075 s. The mass of the baby, stroller is 7.4 kg. It starts stationary; what is its speed after the 0.075 s the force is applied?, m/s, , 3.6, , 3.7, , A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also, stated that air bags have been confirmed as killing 242 people, and they stress that seat belts are estimated to save 11,000, lives a year.) Assume that a car crashes and has come to a stop when the air bag inflates, causing a 75.0 kg person moving, forward at 15.0 m/s to stop moving in 0.0250 seconds. (a) What is the magnitude of the person's impulse? (b) What is the, magnitude of the average force the airbag exerts on the person?, (a), , kg · m/s, , (b), , N, , Imagine you are a NASA engineer, and you are asked to design an airbag to protect the Mars Pathfinder from its impact with, the Martian plain when it lands. Your system can allow an average force of up to 53,000 N on the spacecraft without damage., Your Pathfinder mock-up has a mass of 540 kg. If the spacecraft will strike the planet at 24 m/s, what is the minimum time for, your airbag system to bring Pathfinder to rest so that the average force will not exceed 53,000 N?, s, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems, , 161
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3.8, , The graph shows the net force applied on a, 0.15 kg object over a 3.0 s time interval. (a), What is the average force applied to the object, over the 3.0 seconds? (b) What is the, impulse? (c) What is its change in velocity?, (a), , 3.9, , N, , (b), , kg · m/s, , (c), , m/s, , A ball is traveling horizontally over a volleyball net when a player "spikes" it, driving it straight down to the ground. The ball's, mass is 0.22 kg, its speed before being hit is 6.4 m/s and its speed immediately after the spike is 21 m/s. What is the, magnitude of the impulse from the spike?, kg · m/s, , 3.10 For a movie scene, an 85.0 kg stunt double falls 12.0 m from a building onto a large inflated landing pad. After touching the, landing pad surface, it takes her 0.468 s to come to a stop. What is the magnitude of the average net force on her as the, landing pad stops her?, N, 3.11 A relative of yours belly flops from a height of 2.50 m (ouch!) and stops moving after descending 0.500 m underwater. Her, mass is 62.5 kg. (a) What is her speed when she strikes the water? Ignore air resistance. (b) What is the magnitude of her, impulse between when she hit the water, and when she stopped? (c) What was the magnitude of her acceleration in the pool?, Assume that it is constant. (d) How long was she in the water before she stopped moving? (e) What was the magnitude of the, average net force exerted on her after she hit the water until she stopped? (f) Do you think this hurt?, (a), , m/s, , (b), , kg · m/s, , (c), , m/s2, , (d), , s, N, , (e), (f), , Yes, , No, , 3.12 Nitrogen gas molecules, which have mass 4.65×10í26 kg, are striking a vertical container wall at a horizontal velocity of, positive 440 m/s. 5.00×1021 molecules strike the wall each second. Assume the collisions are perfectly elastic, so each, particle rebounds off the wall in the opposite direction but at the same speed. (a) What is the change in momentum of each, particle? (b) What is the average force of the particles on the wall?, (a), , kg · m/s, , (b), , N, , 3.13 A pickup truck has a mass of 2400 kg when empty. It is driven onto a scale, and sand is poured in at the rate of 150 kg/s from, a height of 3.0 m above the truck bed. At the instant when 440 kg of sand have already been added, what weight does the, scale report?, N, , Section 5 - Conservation of momentum, 5.1, , A probe in deep space is infested with alien bugs and must be blown apart so that the icky creatures perish in the interstellar, vacuum. The craft is at rest when its self-destruction device is detonated, and the craft explodes into two pieces. The first, piece, with a mass of 6.00×108 kg, flies away in a positive direction with a speed of 210 m/s. The second piece has a mass of, 1.00×108 kg and flies off in the opposite direction. What is the velocity of the second piece after the explosion?, m/s, , 5.2, , A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it, encounters a lost 19.5 kg scuba tank. Thinking the tank is a meal, it has lunch. Assuming momentum is conserved in the, collision, what is the velocity of the shark immediately after it swallows the tank?, m/s, , 5.3, , A rifle fires a bullet of mass 0.0350 kg which leaves the barrel with a positive velocity of 304 m/s. The mass of the rifle and, bullet is 3.31 kg. At what velocity does the rifle recoil?, m/s, , 5.4, , 162, , A cat stands on a skateboard that moves without friction along a level road at a constant velocity of 2.00 m/s. She is carrying, a number of books. She wishes to stop, and does so by hurling a 1.20 kg book horizontally forward at a speed of 15.0 m/s, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems
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with respect to the ground. (a) What is the total mass of the cat, the skateboard, and any remaining books? (b) What mass, book must she now throw at 15.0 m/s with respect to the ground to move at í2.00 m/s?, , 5.5, , (a), , kg, , (b), , kg, , Stevie stands on a rolling platform designed for moving heavy objects. The platform has mass of 76 kg and is on a flat floor,, supported by rolling wheels that can be considered to be frictionless. Stevie's mass is 43 kg. The platform and Stevie are, stationary when she begins walking at a constant velocity of +1.2 m/s relative to the platform. (a) What is the platform's, velocity relative to the floor? (b) What is Stevie's velocity relative to the floor?, (a), , m/s, , (b), , m/s, , Section 8 - Collisions, 8.1, , A steel ball of mass 0.76 kg strikes a brick wall, in an elastic collision. Incoming, it strikes the, wall moving 63° directly above a line normal, (perpendicular) to the wall. It rebounds off the, wall at an angle of 63° directly below the, normal line. The ball's speed is 8.4 m/s, immediately before and immediately after the, collision, which lasts 0.18 s. What is the, magnitude of the average force exerted by the, ball on the wall?, N, , 8.2, , A quarterback is standing stationary waiting to make a pass when he is tackled from behind by a linebacker moving at, 4.75 m/s. The linebacker holds onto the quarterback and they move together in the same direction as the linebacker was, moving, at 2.60 m/s. If the linebacker's mass is 143 kg, what is the quarterback's mass?, kg, , Section 9 - Sample problem: elastic collision in one dimension, 9.1, , 9.2, , 9.3, , 9.4, , Two balls collide in a head-on elastic collision and rebound in opposite directions. One ball has velocity 1.2 m/s before the, collision, and í2.3 m/s after. The other ball has a mass of 1.1 kg and a velocity of í4.2 m/s before the collision. (a) What is the, mass of the first ball? (b) What is the velocity of the second ball after the collision?, (a), , kg, , (b), , m/s, , Two identical balls collide head-on in an elastic collision and rebound in opposite directions. The first ball has speed 2.3 m/s, before the collision and 1.7 m/s after. What is the speed of the second ball (a) before and (b) after the collision?, (a), , m/s, , (b), , m/s, , A 4.3 kg block slides along a frictionless surface at a constant velocity of 6.2 m/s in the positive direction and collides head-on, with a stationary block in an elastic collision. After the collision, the stationary block moves at 4.3 m/s. (a) What is the mass of, the initially stationary block? (b) What is the velocity of the first block after the collision?, (a), , kg, , (b), , m/s, , Ball A, which has mass 23 kg, is moving horizontally left to right at +7.8 m/s when it overtakes the 67 kg ball B, which is, moving left to right at +4.7 m/s, and they collide elastically. (a) What is ball A's velocity after the collision? (b) What is ball B's, velocity?, (a), , m/s, , (b), , m/s, , Section 12 - Interactive problem: shuffleboard collisions, 12.1 Use the information given in the interactive problem in this section to determine the initial velocity of the blue puck that will, cause the red puck to stop at the scoring line. Test your answer using the simulation., m/s, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems, , 163
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Section 13 - Inelastic collisions, 13.1 Ball A has mass 5.0 kg and is moving at í3.2 m/s when it strikes stationary ball B, which has mass 3.9 kg, in a head-on, collision. If the collision is elastic, what is the velocity of (a) ball A, and (b) ball B after the collision? (c) If the collision is, completely inelastic, what is the common velocity of balls A and B?, (a), , m/s, , (b), , m/s, , (c), , m/s, , 13.2 During a snowball fight, two snowballs travelling towards each other collide head-on. The first is moving east at a speed of, 16.1 m/s and has a mass of 0.450 kg. The second is moving west at 13.5 m/s. When the snowballs collide, they stick together, and travel west at 3.50 meters per second. What is the mass of the second snowball?, kg, 13.3 Three railroad cars, each with mass 2.3×104 kg, are moving on the same track. One moves north at 18 m/s, another moves, south at 12 m/s, and third car between these two moves south at 6 m/s. When the three cars collide, they couple together and, move with a common velocity. What is their velocity after they couple?, m/s, 13.4 A 110 kg quarterback is running the ball downfield at 4.5 m/s in the positive direction when he is tackled head-on by a 150 kg, linebacker moving at í3.8 m/s. Assume the collision is completely inelastic. (a) What is the velocity of the players just after the, tackle? (b) What is the kinetic energy of the system consisting of both players before the collision? (c) What is the kinetic, energy of the system consisting of both players after the collision?, (a), , m/s, , (b), , J, , (c), , J, , 13.5 Two clay balls of the same mass stick together in an completely inelastic collision. Before the collision, one travels at 5.6 m/s, and the other at 7.8 m/s, and their paths of motion are perpendicular. If the mass of each ball is 0.21 kg, what is the, magnitude of the momentum of the combined balls after the collision?, kg · m/s, 13.6 A large flat 3.5 kg boogie board is resting on the beach. Jessica, whose mass is 55 kg, runs at a constant horizontal velocity, of 2.8 m/s. While running, she jumps on the board, and the two of them move together across the beach. (a) What is the, speed of the board (with Jessica on it) just after she jumps on? (b) If the board and Jessica slide 7.7 m before coming to a, stop, what is the coefficient of kinetic friction between the board and the beach?, (a), , m/s, , (b), 13.7 A 6.0 kg ball A and a 5.0 kg ball B move directly toward each other in a head-on collision, then move in opposite directions, away from the site of the collision. Ball A has velocity 4.1 m/s before the collision and í1.1 m/s after, and ball B has velocity, í2.9 m/s before the collision. (a) What is ball B's velocity after the collision? (b) Is this an elastic collision?, (a), (b), , m/s, Yes, , No, , Section 14 - Center of mass, 14.1 How far is the center of mass of the Earth-Moon system from the center of the Earth? The Earth's mass is 5.97×1024 kg, the, Moon's mass is 7.4×1022 kg, and the distance between their centers is 3.8×108 m., m, 14.2 Four particles are postioned at the corners of a square that is 4.0 m on each side. One corner of the square is at the origin,, one on the positive x axis, one in the first quadrant and one on the positive y axis. Starting at the origin, going clockwise, the, particles have masses 2.3 kg, 1.4 kg, 3.7 kg, and 2.9 kg. What is the location of the center of mass of the system of particles?, (, , 164, , ,, , )m, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems
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14.3 The uniform sheet of siding shown has a, centrally-located doorway of width 2.0 m and, height 6.0 m cut out of it. The sheet, with the, doorway hole, has a mass of 264 kg. (a) What, is the x coordinate of the center of mass? (b), What is the y coordinate of the center of, mass? Assume the lower left hand corner of, the siding is at (0, 0)., (a), , m, , (b), , m, , Section 16 - Interactive summary problem: types of collisions, 16.1 Use the information given in the interactive problem in this section to determine the collision type for (a) collision A, (b), collision B, and (c) collision C. Test your answer using the simulation., (a), , (b), , (c), , i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., , Elastic, Completely inelastic, Inelastic, Impossible, Elastic, Completely inelastic, Inelastic, Impossible, Elastic, Completely inelastic, Inelastic, Impossible, , Copyright 2007 Kinetic Books Co. Chapter 7 Problems, , 165
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8.0 - Introduction, A child riding on a carousel, you riding on a Ferris wheel: Both are examples of, uniform circular motion. When the carousel or Ferris wheel reaches a constant rate, of rotation, the rider moves in a circle at a constant speed. In physics, this is called, uniform circular motion., Developing an understanding of uniform circular motion requires you to recall the, distinction between speed and velocity. Speed is the magnitude, or how fast an, object moves, while velocity includes both magnitude and direction. For example,, consider the car in the graphic on the right. Even as it moves around the curve at a, constant speed, its velocity constantly changes as its direction changes. A change, in velocity is called acceleration, and the acceleration of a car due to its change in, direction as it moves around a curve is called centripetal acceleration., Although the car moves at a constant speed as it moves around the curve, it is, accelerating. This is a case where the everyday use of a word í acceleration í and, its use in physics differ. A non-physicist would likely say: If a car moves around a, curve at a constant speed, it is not accelerating. But a physicist would say: It most, certainly is accelerating because its direction is changing. She could even point out,, as we will discuss later, that a net external force is being applied on the car, so the car must be accelerating., Uniform circular motion begins the study of rotational motion. As with linear motion, you begin with concepts such as velocity and acceleration, and then move on to topics such as energy and momentum. As you progress, you will discover that much of what you have learned about, these topics in earlier lessons will apply to circular motion., In the simulation shown to the right, the car moves around the track at a constant speed. The red velocity vector represents the direction and, magnitude of the car’s instantaneous velocity., The simulation has gauges for the x and y components of the car's velocity. Note how they change as the car travels around the track. These, changes are reflected in the centripetal acceleration of the car. You can also have the car move at different constant speeds, and read the, corresponding centripetal acceleration in the appropriate gauge. Is the centripetal acceleration of the car higher when it is moving faster? Note:, If you go too fast, you can spin off the track. Happy motoring!, , 8.1 - Uniform circular motion, , Uniform circular motion: Movement in a circle, at a constant speed., The toy train on the right moves on a circular track in uniform circular motion. The, identical lengths of the velocity vectors in the diagram indicate a constant magnitude of, velocity í a constant speed. When an object is moving in uniform circular motion, its, speed is uniform (constant) and its path is circular., The train does not have constant velocity; in fact, its velocity is constantly changing., Why? As you can also see in the diagram to the right, the direction of the velocity vector, changes as the train moves around the track. A change in the direction of velocity, means a change in velocity. The velocity vector is tangent to the circle at every instant, because the train’s displacement is tangent to the circle during every small interval of, time., , Uniform circular motion, , Motion in a circle with constant speed, ·Velocity changes!, Uniform circular motion is important in physics. For instance, a satellite in a circular orbit Instantaneous velocity always tangent, around the Earth moves in uniform circular motion., , 166, , Copyright 2000-2007 Kinetic Books Co. Chapter 08
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8.2 - Period, , Period: The amount of time it takes for an, object to return to the same position., The concept of period is useful in analyzing motion that repeats itself. We use the, example of the toy train shown in Concept 1 to illustrate a period. The train moves, around a circular track at a constant rate, which is to say in uniform circular motion. It, returns to the same position on the track after equal intervals of time. The period, measures how long it takes the train to complete one revolution. In this example, it, takes the train six seconds to make a complete lap around the track., When an object like a train moves in uniform circular motion, that motion is often, described in terms of the period. Many other types of motion can be discussed using, the notion of a period, as well. For example, the Earth follows an elliptical path as it, moves around the Sun, and its period is called a year. A metronome is designed to, have a constant period that provides musicians with a source of rhythm., , Period, Time to complete one revolution, , The equation on the right enables you to calculate the period of an object moving in, uniform circular motion. The period is the circumference of the circle, 2ʌr, divided by, the object’s speed. To put it more simply, it is distance divided by speed., , Period for uniform circular, motion, , T = period, r = radius, v = speed, , What is the period of the train?, , Copyright 2000-2007 Kinetic Books Co. Chapter 08, , 167
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8.3 - Centripetal acceleration, , Centripetal acceleration: The centrally directed, acceleration of an object due to its circular, motion., An object moving in uniform circular motion constantly accelerates because its direction, (and therefore its velocity) constantly changes. This type of acceleration is called, centripetal acceleration. Any object moving along a circular path has centripetal, acceleration., In Concept 1 at the right is a vector analysis of centripetal acceleration that uses a toy, train as an example of an object moving along a circular path. As the drawing indicates,, the train’s velocity is tangent to the circle., In uniform circular motion, the acceleration vector always points toward the center of the, circle, perpendicular to the velocity vector. In other words, the object accelerates toward, the center. This can be proven by considering the change in the velocity vector over a, short period of time and using a geometric argument (an argument that is not shown, here)., , Centripetal acceleration, Acceleration due to change in direction, in circular motion, In uniform circular motion, acceleration:, ·Has constant magnitude, ·Points toward center, , The equation for calculating centripetal acceleration is shown in Equation 1 on the right., The magnitude of centripetal acceleration equals the speed squared divided by the, radius. Since both the speed and the radius are constant in uniform circular motion, the, magnitude of the centripetal acceleration is also constant., With uniform circular motion, the only acceleration is centripetal acceleration, but for, circular motion in general, there may be both centripetal acceleration, which changes, the object’s direction, and acceleration in the direction of the object’s motion (tangential, acceleration), which changes its speed. If you ride on a Ferris wheel which is starting, up, rotating faster and faster, you are experiencing both centripetal and tangential, acceleration. For now, we focus on uniform circular motion and centripetal acceleration,, leaving tangential acceleration as another topic., , ac = centripetal acceleration, v = speed, r = radius, , What is the centripetal, acceleration of the train?, , Accelerates toward center, , 168, , Copyright 2000-2007 Kinetic Books Co. Chapter 08
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8.4 - Interactive problems: racing in circles, The two simulations in this section let you experience uniform circular motion and, centripetal acceleration as you race your car against the computer’s., In the first simulation, you race a car around a circular track. Both your car and the, computer’s move around the loop at constant speeds. You control the speed of the, blue car. Halfway around the track, you encounter an oil slick. If the centripetal, acceleration of your car is greater than 3.92 m/s2 at this point, it will leave the track, and you will lose. The radius of the circle is 21.0 m., To win the race, set the centripetal acceleration equal to 3.92 m/s2 in the centripetal, acceleration equation, solve for the velocity, and then round down the velocity to, the nearest 0.1 m/s; this is a value that will keep your car on the track and beat the, computer car. Enter this value using the controls in the simulation. Press GO to start, the simulation and test your calculations., In the second simulation, the track consists of two half-circle curves connected by a, straight section. Your blue car runs the entire race at the speed that you set for it., You want to set this speed to just keep the car on the track. The first curve has a, radius of 14.0 meters; the second, 8.50 meters. On either curve, if the centripetal, acceleration of your car exceeds 9.95 m/s2, its tires will lose traction on the curve,, causing it to leave the track. If your car moves at the fastest speed possible without, leaving the track, it will win. Again, calculate the speed of the blue car on each curve, but using a centripetal acceleration value of 9.95 m/s2, and round down to the, nearest 0.1 m/s. Since the car will go the same speed on both curves, you need to, decide which curve determines your maximum speed. Enter this value, then press, GO., If you have difficulty solving these interactive problems, review the equation relating, centripetal acceleration, circle radius, and speed., , 8.5 - Newton's second law and centripetal forces, If you hold onto the string of a yo-yo and twirl it in a circle overhead, as illustrated in, Concept 1, you know you must hold the string firmly or the yo-yo will fly away from you., This is true even when the toy moves at a constant speed. A force must be applied to, keep the yo-yo moving in a circle., A force is required because the yo-yo is accelerating. Its change in direction means its, velocity is changing. Using Newton’s second law, F = ma, we can calculate the amount, of force as the product of the object’s mass and its centripetal acceleration. That, equation is shown in Equation 1. It applies to any object moving in uniform circular, motion. The force, called a centripetal force, points in the same direction as the, acceleration, toward the center of the circle., The term “centripetal” describes any force that causes circular motion. A centripetal, force is not a new type of force. It can be the force of tension exerted by a string, as in, the yo-yo example, or it can be the force of friction, such as when a car goes around a, curve on a level road. It can also be a normal force; for example, the walls of a clothes, dryer supply a normal force that keeps the clothes moving in a circle, while the holes in, those walls allow water to “spin out” of the fabric. Or, as in the case of the motorcycle, rider in Example 1, the centripetal force can be a combination of forces, such as the, normal force from the wall and the force of friction., , Forces and centripetal, acceleration, Force causes circular motion, Directed toward center, Any force can be centripetal, , Sometimes the source of a centripetal force is easily seen, as with a string or the walls of a dryer. Sometimes that force is invisible: The force, of gravity cannot be directly seen, but it keeps the Earth in its orbit around the Sun. The centripetal force can also be quite subtle, such as, when an airplane tilts or banks; the air passing over the plane’s angled wings creates a force inward. In each of these examples, a force, causes the object to accelerate toward the center of its circular path., Identifying the force or forces that create the centripetal acceleration is a key step in solving many problems involving circular motion., , Copyright 2000-2007 Kinetic Books Co. Chapter 08, , 169
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F = net force, m = mass, v = speed, r = radius, , A daredevil bike rider goes, around a circular track. The bike, and rider together have the mass, shown. What is the centripetal, force on them?, , F = (180 kg)(25 m/s)2/15 m, F = (180)(625)/15, F = 7500 N, Force directed toward center, 8.6 - Artificial gravity, When a spacecraft is far away from the Earth and any, other massive body, the force of gravity is near zero., As a result of this lack of gravitational force, the, astronauts and their equipment float in space., Although perhaps amusing to watch and experience,, floating presents an unusual challenge because, humans are accustomed to working in environments, with enough gravity to keep them, and their, equipment, anchored to the floor. (Note: This same, feeling of weightlessness occurs in a spacecraft, orbiting the Earth, but the cause of the apparent, weightlessness in this case arises from the free-fall, motion of the craft and the astronaut, not from a lack, of gravitational force.), , 170, , A rotating space station provides an illusion of gravity., , Copyright 2000-2007 Kinetic Books Co. Chapter 08
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A number of science fiction books and films have featured spacecraft that rotate very, slowly as they travel through the universe. Arthur C. Clarke’s science fiction novel, Rendezvous with Rama is set in a massive rotating spacecraft, as is part of the movie, 2001, which is also based on his work. This rotation supplies artificial gravity í the, illusion of gravity í to the astronauts and their equipment. Artificial gravity has effects, similar to true gravity, and as a result can mislead the people riding in such machines to, believe they are experiencing true gravitational force., Why does the rotation of a spacecraft produce the sensation of gravity? Consider what, happens when an airplane takes off from a runway: You feel a force pulling you back, into your seat, as if the force of gravity were increasing. The force of gravity has not, been significantly altered (in fact, it decreases a bit as you gain elevation). However,, while the airplane accelerates upward, you feel a greater normal force pushing up from, your seat, and you may interpret this subconsciously as increased gravity., A roughly analogous situation occurs on a rotating spaceship. The astronauts are, rotating in uniform circular motion. The outside wall of the station (the floor, from the, astronauts’ perspective) provides the centrally directed normal force that is the, centripetal force. This force keeps the astronauts moving in a circle. From the, astronauts’ perspective, this force is upwards, and they relate it to the upward normal, force of the ground they feel when standing on the Earth. On Earth, the normal force is, equal but opposite to the force of gravity. Because they typically associate the normal, force with gravity, the astronauts may erroneously perceive this force from the, spacecraft floor as being caused by some form of artificial or simulated gravity., , Artificial gravity, Space station rotates, Floor of craft provides centripetal force, Person (incorrectly) assumes normal, force counters force of gravity, , Artificial gravity is a pseudo, or fictitious, force. The astronauts assume it exists because, of the normal force. The perception of this fictitious force is a function of the, acceleration of the astronauts as they move in a circle. It would disappear if the, spacecraft stopped rotating., Although discussed as the realm of science fiction, real-world carnival rides (like the, “Gravitron”) use this effect. Riders are placed next to the wall of a cylinder. The cylinder, then is rotated at a high speed and the floor (or seats) below the riders is lowered. The, walls of the cylinder supply a normal force and the force of friction keeps the riders from, slipping down., , To simulate Earth's gravity, what, should the radius of the space, station be?, , Copyright 2000-2007 Kinetic Books Co. Chapter 08, , 171
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8.7 - Interactive summary problem: race curves, In the simulation on the right, you are asked to race a truck on an S-shaped track, against the computer. This time, the first curve is covered with snow and you are, racing against a snowmobile. As you go around the track, the static friction between, the tires of your truck and the snow or pavement provides the centripetal force. If, you go too fast, you will exceed the maximum force of friction and your truck will, leave the track. If you go as fast as you can without sliding, you will beat the, snowmobile., The snowmobile runs the entire race at its maximum speed. The blue truck, negotiates each curve at a constant speed, but these speeds must be different for, you to win the race. You set the speed of the blue truck on each curve., Straightaway sections are located at the start of the race and between the two, curves. The simulation will automatically supply the acceleration you need on the, straightaway sections., The blue truck has a mass of 1,800 kg. The first curve is icy, and the coefficient of, static friction of the truck on this curve is 0.51. (The snowmobile has a greater, coefficient thanks to its snow-happy treads.) On the second curve, the coefficient of, static friction is 0.84. The radius of the first curve is 13 m, and the second curve is 11 m. Set the speed of the blue truck on each curve as fast, as it can go without sliding off the track, and you will win., You set the speed in increments of 0.1 m/s in the simulation. If you need to round a value after your calculations, make sure you round down to, the nearest 0.1 m/s. (If you round up, you will be exceeding the maximum safe speed.) Press GO to begin the race, and RESET if you need to, try again., If you have difficulty with this problem, you may want to review the section on static friction in a previous chapter and the section on centripetal, acceleration in this chapter., , 8.8 - Gotchas, A car is moving around a circular track at a constant speed of 20 km/h. This means its velocity is constant, as well. Wrong. The car’s velocity, changes because its direction changes as it moves., Since an object moving in uniform circular motion is constantly changing direction, it is hard at any point in time to know the direction of its, velocity and the direction of its acceleration. This is not true. The velocity vector is always tangent to the circle at the location of the object., Centripetal acceleration always points toward the center of the circle., No force is required for an object to move in uniform circular motion. After all, its speed is constant. Yes, but its velocity is changing due to its, change in direction, which means it is accelerating. By Newton’s second law, this means there must be a net force causing this acceleration., Centripetal force is another type of force. No, rather it is a way to describe what a force is “doing.” The normal force, gravity, tension í each of, these forces can be a centripetal force if it is causing an object to move in uniform circular motion., , 8.9 - Summary, Uniform circular motion is movement in a circle at a constant speed. But while, speed is constant in this type of motion, velocity is not. Since instantaneous velocity, in uniform circular motion is always tangent to the circle, its direction changes as the, object's position changes., The period is the time it takes an object in uniform circular motion to complete one, revolution of the circle., Since the velocity of an object moving in uniform circular motion changes, it is, accelerating. The acceleration due to its change in direction is called centripetal, acceleration. For uniform circular motion, the acceleration vector has a constant, magnitude and always points toward the center of the circle., Newton's second law can be applied to an object in uniform circular motion. The net, force causing centripetal acceleration is called a centripetal force. Like centripetal, acceleration, it is directed toward the center of the circle., A centripetal force is not a new type of force; rather, it describes a role that is played by one or more forces in the situation, since there must be, some force that is changing the velocity of the object. For example, the force of gravity keeps the Moon in a roughly circular orbit around the, Earth, while the normal force of the road and the force of friction combine to keep a car in circular motion around a banked curve., , 172, , Copyright 2000-2007 Kinetic Books Co. Chapter 08
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Chapter 8 Problems, , Conceptual Problems, C.1, , A string's tension force supplies the centripetal force needed to keep a yo-yo whirling in a circle. (a) What force supplies the, centripetal force keeping a satellite in uniform circular motion around the Earth? (b) What kinds of forces keep a roller coaster, held to a looping track?, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Is the centripetal, acceleration of the car higher when it is moving faster? (b) If the speed of the car remains constant, do the x and y, components of the car's velocity change as the car goes around the track?, (a), , Yes, , No, , (b), , Yes, , No, , Section 2 - Period, 2.1, , Jupiter's distance from the Sun is 7.78×1011 meters and it takes 3.74×108 seconds to complete one revolution of the Sun in, its roughly circular orbit. What is Jupiter's speed?, , 2.2, , Saturn travels at an average speed of 9.66×103 m/s around the Sun in a roughly circular orbit. Its distance from the Sun is, 1.43×1012 m. How long (in seconds) is a "year" on Saturn?, , m/s, , s, 2.3, , Mars travels at an average speed of 2.41×104 m/s around the Sun, and takes 5.94×107 s to complete one revolution. How far, is Mars from the Sun?, m, , 2.4, , Long-playing vinyl records, still used by club DJs, are 12 inches in diameter and are played at 33 1/3 revolutions per minute., What is the speed (in m/s) of a point on the edge on such a record?, m/s, , Section 3 - Centripetal acceleration, 3.1, , A runner rounds a circular curve of radius 24.0 m at a constant speed of 5.25 m/s. What is the magnitude of the runner's, centripetal acceleration?, m/s2, , 3.2, , 3.3, , 3.4, , In a carnival ride, passengers are rotated at a constant speed in a seat at the end of a long horizontal arm. The arm is 8.30 m, long, and the period of rotation is 4.00 s. (a) What is the magnitude of the centripetal acceleration experienced by a rider? (b), State the acceleration in "gee's," that is, as a multiple of the gravitational acceleration constant g., (a), , m/s2, , (b), , g, , Consider the radius of the Earth to be 6.38×106 m. What is the magnitude of the centripetal acceleration experienced by a, person (a) at the equator and (b) at the North Pole due to the Earth's rotation?, (a), , m/s2, , (b), , m/s2, , When tires are installed or reinstalled on a car, they are usually first balanced on a device that spins them to see if they, wobble. A tire with a radius of 0.380 m is rotated on a tire balancing device at exactly 460 revolutions per minute. A small, stone is embedded in the tread of the tire. What is the magnitude of the centripetal acceleration experienced by the stone?, m/s2, , 3.5, , A toy airplane connected by a guideline to the top of a flagpole flies in a circle at a constant speed. If the plane takes 4.5 s to, complete one loop, and the radius of the circular path is 11 m, what is the magnitude of the plane's centripetal acceleration?, m/s2, , Copyright 2007 Kinetic Books Co. Chapter 8 Problems, , 173
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Section 4 - Interactive problems: racing in circles, 4.1, , Use the information given in the first interactive problem in this section to calculate the initial speed that keeps the blue car on, the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s. Test your answer using the, simulation., m/s, , 4.2, , Use the simulation in the second interactive problem in this section to calculate the initial speed that keeps the blue car on, the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s., m/s, , Section 5 - Newton's second law and centripetal forces, 5.1, , An astronaut in training rides in a seat that is moved in uniform circular motion by a radial arm 5.10 meters long. If her speed, is 15.0 m/s, what is the centripetal force on her in "G's," where one G equals her weight on the Earth?, "G's", , 5.2, , A ball with mass 0.48 kg moves at a constant speed. A centripetal force of 23 N acts on the ball, causing it to move in a circle, with radius 1.7 m. What is the speed of the ball?, m/s, , 5.3, , A bee loaded with pollen flies in a circular path at a constant speed of 3.20 m/s. If the mass of the bee is 133 mg and the, radius of its path is 11.0 m, what is the magnitude of the centripetal force?, , 5.4, , Fifteen clowns are late to a party. They jump into their sporty coupe and start driving. Eventually they come to a level curve,, with a radius of 27.5 meters. What is the top speed at which they can drive successfully around the curve? The coefficient of, static friction between the car's tires and the road is 0.800., , 5.5, , You are playing tetherball with a friend and hit, the ball so that it begins to travel in a circular, horizontal path. If the ball is 1.2 meters from, the pole, has a speed of 3.7 m/s, a mass of, 0.42 kilograms, and its (weightless) rope, makes a 49° angle with the pole, find the, tension force that the rope exerts on the ball, just after you hit it., , N, , m/s, , N, , 5.6, , A car with mass 1600 kg drives around a flat circular track of radius 28.0 m. The coefficient of friction between the car tires, and the track is 0.830. How fast can the car go around the track without losing traction?, m/s, , Section 6 - Artificial gravity, 6.1, , A rotating space station has radius 1.31e+3 m, measured from the center of rotation to the outer deck where the crew lives., What should the period of rotation be if the crew is to feel that they weigh one-half their Earth weight?, s, , Section 7 - Interactive summary problem: race curves, 7.1, , 174, , Use the simulation in the interactive problem in this section to calculate the speed for (a) the first curve and (b) the second, curve that keeps the blue truck on the track and wins the race. For safety, round your answer down to the nearest 0.1 m/s., (a), , m/s, , (b), , m/s, , Copyright 2007 Kinetic Books Co. Chapter 8 Problems
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9.0 - Introduction, If you feel as though you spend your life spinning around in circles, you may be, pleased to know that an entire branch of physics is dedicated to studying that kind, of motion. This chapter is for you! More seriously, this chapter discusses motion that, consists of rotation about a fixed axis. This is called pure rotational motion. There, are many examples of pure rotational motion: a spinning Ferris wheel, a roulette, wheel, or a music CD are three instances of this type of motion., In this chapter, you will learn about rotational displacement, rotational velocity, and, rotational acceleration: the fundamental elements of what is called, rotationalkinematics., The simulation on the right features the “Angular Surge,” an amusement park ride, you will be asked to operate in order to gain insight into rotational kinematics. The, ride has a rotating arm with a “rocket” where passengers sit. You can move the, rocket closer to or farther from the center by setting the distance in the simulation., You can also change the rocket’s period, which is the amount of time it takes to, complete one revolution., By changing these parameters, you affect two values you see displayed in gauges: the rocket’s angular velocity and its linear speed. The, rocket’s angular velocity is the change per second in the angle of the ride’s arm, measured from its initial position. Its units are radians per, second. For instance, if the rocket completes one revolution in one second, its angular velocity is 2ʌ radians (360°) per second., This simulation has no specific goal for you to achieve, although you may notice that you can definitely have an impact on the passengers!, What you should observe is this: How do changes in the period affect the angular velocity? The linear speed? And how does a change in the, distance from the center (the radius of the rocket’s motion) affect those values, if at all? Can you determine how to maximize the linear speed, of the rocket?, To run the ride, you start the simulation, set the values mentioned above, and press GO. You can change the settings while the ride is in, motion., , 9.1 - Angular position, , Angular position: The amount of rotation from a, reference position, described with a positive or, negative angle., When an object such as a bicycle wheel rotates about its axis, it is useful to describe, this motion using the concept of angular position. Instead of being specified with a linear, coordinate such as x, as linear position is, angular position is stated as an angle., In Concept 1, we use the location of a bicycle wheel’s valve to illustrate angular, position. The valve starts at the 3 o’clock position (on the positive x axis), which is zero, radians by convention. As the illustration shows, the wheel has rotated one-eighth of a, turn, or ʌ/4 radians (45°), in a counterclockwise direction away from the reference, position. In other words, angular position is measured from the positive x axis., , Angular position, , Rotation from 3 o’clock position, ·Counterclockwise rotation: positive, Note that this description of the wheel’s position used radians, not degrees; this is, ·Clockwise rotation: negative, because radians are typically used to describe angular position. The two lines we use to Units are radians, measure the angle radiate from the point about which the wheel rotates., The axis of rotation is a line also used to describe an object’s rotation. It passes through the wheel’s center, since the wheel rotates about that, point, and it is perpendicular to the wheel. The axis is assumed to be stationary, and the wheel is assumed to be rigid and to maintain a, constant shape. Analyzing an object that changes shape as it rotates, such as a piece of soft clay, is beyond the scope of this textbook. We are, concerned with the wheel’s rotational motion here: its motion around a fixed axis. Its linear motion when moving along the ground is another, topic., As mentioned, angular position is typically measured with radians (rad) instead of degrees. The formula that defines the radian measure of an, angle is shown in Equation 1. The angle in radians equals the arc length s divided by the radius r. As you may recall, 2ʌ radians equals one, revolution around a circle, or 360°. One radian equals about 57.3°. To convert radians to degrees, multiply by the conversion factor 360°/2ʌ., To convert degrees to radians, multiply by the reciprocal: 2ʌ/360°. The Greek letter ș (theta) is used to represent angular position., The angular position of zero radians is defined to be at 3 o’clock, which is to say along a horizontal line pointing to the right. Let’s now consider, what happens when the wheel rotates a quarter turn counterclockwise, moving the valve from the 3 o’clock position to 12 o’clock. A quarter, turn is ʌ/2 rad (or 90°). The valve’s angular position when it moves a quarter turn counterclockwise is ʌ/2 rad. By convention, angular position, increases with counterclockwise motion., , Copyright 2007 Kinetic Books Co. Chapter 09, , 175
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The valve can be placed in the same angular position, ʌ/2 rad, by rotating the wheel in, the other direction, by rotating it clockwise three quarters of a turn. By convention,, angular position decreases with clockwise motion, so this rotation would be described, as an angular position of í3ʌ/2 rad., An angular position can be greater than 2ʌrad. An angular position of 3ʌ rad, represents one and a half counterclockwise revolutions. The valve would be at 9 o’clock, in that position., , Radian measure, , ș = angle in radians, s = arc length, r = radius, , What is the arc length?, , s = rș = (0.35 m)(ʌ/3 rad), s = 0.37 m, 9.2 - Angular displacement, , Angular displacement: Change in angular, position., In Concept 1 you see a pizza topped with a single mushroom (we are not going back to, that pizzeria!). We use a mushroom to make the rotational motion of the pizza easier to, see. As the pizza rotates, its angular position changes. This change in angular position, is called angular displacement., To calculate angular displacement, you subtract the initial angular position from the final, position. For instance, the mushroom in the Equation illustration moves from ʌ/2 rad to, ʌ rad, a displacement of ʌ/2 rad. As you can see in this example, angular displacement, in the counterclockwise direction is positive., Revolution is a common term in the study of rotational motion. It means one complete, rotational cycle, with the object starting and returning to the same position. One, counterclockwise revolution equals 2ʌ radians of angular displacement., , Angular displacement, Change in angular position, Counterclockwise rotation is positive, , The angular displacement is the total angle “swept out” during rotational motion from an initial to a final position. If the pizza turns, counterclockwise three complete revolutions, its angular displacement is 6ʌradians., The definition of angular displacement resembles that of linear displacement. However, the discussion above points out a difference. A, mushroom that makes a complete revolution has an angular displacement of 2ʌ rad. On the other hand, its linear displacement equals zero,, , 176, , Copyright 2007 Kinetic Books Co. Chapter 09
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since it starts and stops at the same point., , ǻș = șf í și, ǻș = angular displacement, șf = final angular position, și = initial angular position, Units: radians (rad), , At 12:10, the initial angular, position of the minute hand is, ʌ / 6. After 15 minutes have, passed, what is the minute, hand’s angular displacement?, , 9.3 - Angular velocity, , Angular velocity: Angular displacement per unit, time., In Concept 1, a ball attached to a string is shown moving counterclockwise around a, circle. Every four seconds, it completes one revolution of the circle. Its angular velocity, is the angular displacement 2ʌ radians (one revolution) divided by four seconds, or ʌ/2, rad/s. The Greek letter Ȧ (omega) represents angular velocity., As is the case with linear velocity, angular velocity can be discussed in terms of, average and instantaneous velocity. Average angular velocity equals the total angular, displacement divided by the elapsed time. This is shown in the first equation in Equation, 1., Instantaneous angular velocity refers to the angular velocity at a precise moment in, time. It equals the limit of the average velocity as the increment of time approaches, zero. This is shown in the second equation in Equation 1., , Angular velocity, Angular displacement per unit time, , Copyright 2007 Kinetic Books Co. Chapter 09, , 177
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The sign of angular velocity follows that of angular displacement: positive for, counterclockwise rotation and negative for clockwise rotation. The magnitude (absolute, value) of angular velocity is angular speed., , Ȧ = instantaneous angular velocity, ș = angular position, ǻș = angular displacement, ǻt = elapsed time, Units: rad/s, , The motorcycle rider takes two, seconds for a counterclockwise, lap around the track. What is her, average angular velocity?, , 178, , Copyright 2007 Kinetic Books Co. Chapter 09
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9.4 - Angular acceleration, , Angular acceleration: The change in angular, velocity per unit time., By now, you might be experiencing a little déjà vu in this realm of angular motion., Angular velocity equals angular displacement per unit time, but if you drop the word, “angular” you are stating that velocity equals displacement per unit time, an equation, that should be familiar to you from your study of linear motion., So it is with angular acceleration. Angular acceleration equals the change in angular, velocity divided by the elapsed time. The toy train shown in Concept 1 is experiencing, angular acceleration. This is reflected in the increasing separation between the images, you see. Its angular velocity is becoming increasingly negative since it is moving in the, clockwise direction. It is moving faster and faster in the negative angular direction., , Angular acceleration, Change in angular velocity per unit time, , Average angular acceleration equals the change in angular velocity divided by the, elapsed time. The instantaneous angular acceleration equals the limit of this ratio as the, increment of time approaches zero. These two equations are shown in Equation 1 to, the right. The Greek letter Į (alpha) is used to represent angular acceleration., With rotational kinematics, we often pose problems in which the angular acceleration is, constant; this helps to simplify the mathematics involved in solving problems. We made, similar use of constant acceleration for the same reason in the linear motion chapter., , Į = instantaneous angular acceleration, Ȧ = angular velocity, ǻt = elapsed time, Units: rad/s2, , The toy train starts from rest and, reaches the angular velocity, shown in 5.0 seconds. What is, its average angular acceleration?, , Copyright 2007 Kinetic Books Co. Chapter 09, , 179
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9.5 - Sample problem: a clock, Over the course of 1.00 hour, what is, (a) the angular displacement, (b) the, angular velocity and (c) the angular, acceleration of the minute hand?, , Think about the movement of the minute hand over the course of an hour. Be sure to consider the direction!, Variables, elapsed time, , ǻt= 1.00 h, , angular displacement, , ǻș, , angular velocity, , Ȧ, , angular acceleration, , Į, , What is the strategy?, 1., , Calculate the angular displacement., , 2., , Convert the elapsed time to seconds., , 3., , Use the angular displacement and time to determine the angular velocity and angular acceleration., , Physics principles and equations, Definition of angular velocity, , Definition of angular acceleration, , Step-by-step solution, We start by calculating the angular displacement of the minute hand over 1.00 hour. We then calculate the angular velocity., , Step, , 1., , ǻș = –2ʌ rad, , Reason, , minute hand travels clockwise one revolution, , 2., , convert to seconds, , 3., , definition of angular velocity, , 4., , substitute, , 5., , Ȧ = –1.75×10í3 rad/s, , evaluate, , The angular displacement is calculated in step 1, and the angular velocity in step 5. Since the angular velocity is constant, the angular, acceleration is zero., , 180, , Copyright 2007 Kinetic Books Co. Chapter 09
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9.6 - Interactive checkpoint: a potter’s wheel, At a particular instant, a potter's, wheel rotates clockwise at 12.0 rad/s;, 2.50 seconds later, it rotates at 8.50, rad/s clockwise. Find its average, angular acceleration during the, elapsed time., , Answer:, , =, , rad/s2, , 9.7 - Tangential velocity, , Tangential velocity: The instantaneous linear, velocity of a point on a rotating object., Concepts such as angular displacement and angular velocity are useful tools for, analyzing rotational motion. However, they do not provide the complete picture., Consider the salt and pepper shakers rotating on the lazy Susan shown to the right. The, containers have the same angular velocity because they are on the same rotating, surface and complete a revolution in the same amount of time., However, at any instant, they have different linear speeds and velocities. Why? They, are located at different distances from the axis of rotation (the center of the lazy Susan),, which means they move along circular paths with different radii. The circular path of the, outer shaker is longer, so it moves farther than the inner one in the same amount of, time. At any instant, its linear speed is greater. Because the direction of motion of an, object moving in a circle is always tangent to the circle, the object’s linear velocity is, called its tangential velocity., , Tangential velocity, Linear velocity at an instant, ·Magnitude: magnitude of linear velocity, ·Direction: tangent to circle, , To reinforce the distinction between linear and angular velocity, consider what happens if you decide to run around a track. Let’s say you are, asked to run one lap around a circular track in one minute flat. Your angular velocity is 2ʌ radians per minute., Could you do this if the track had a radius of 10 meters? The answer is yes. The circumference of that track is 2ʌr, which equals approximately, 63 meters. Your pace would be that distance divided by 60 seconds, which works out to an easy stroll of about 1.05 m/s (3.78 km/h)., What if the track had a radius of 100 meters? In this case, the one-minute accomplishment would require the speed of a world-class sprinter, capable of averaging more than 10 m/s. (If the math ran right past you, note that we are again multiplying the radius by 2ʌ to calculate the, circumference and dividing by 60 seconds to calculate the tangential velocity.) Even though the angular velocity is the same in both cases, 2ʌ, radians per minute, the tangential speed changes with the radius., As you see in Equation 1, tangential speed equals the product of the distance to the axis of rotation, r, and the angular velocity, Ȧ. The units, for tangential velocity are meters per second. The direction of the velocity is always tangent to the path of the object., Confirming the direction of tangential velocity can be accomplished using an easy home experiment. Let’s say you put a dish on a lazy Susan, and then spin the lazy Susan faster and faster. Initially, the dish moves in a circle, constrained by static friction. At some point, though, it will fly, off. The dish will always depart in a straight line, tangent to the circle at its point of departure., The tangential speed equation can also be used to restate the equation for centripetal acceleration in terms of angular velocity. Centripetal, acceleration equals v2/r. Since v, , = rȦ, centripetal acceleration also equals Ȧ2r., , We derive the equation for tangential speed using the diagram below., , Copyright 2007 Kinetic Books Co. Chapter 09, , 181
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To understand the derivation, you must recall that the arc length ǻs (the distance along, the circular path) equals the angular displacement ǻș in radians times the radius r., Also recall that the instantaneous speed vT equals the displacement divided by the, elapsed time for a very small increment of time. Combining these two facts, and the, definition of angular velocity, yields the equation for tangential speed., , Step, , 1., , Reason, , definition of instantaneous, velocity, , 2., , ǻs = rǻș, , definition of radian measure, substitute equation 2 into, , 3., , equation 1, 4., , vT = rȦ, , definition of angular velocity, , vT = rȦ, vT = tangential speed, r = distance to axis, Ȧ = angular velocity, Direction: tangent to circle, , At the instant shown, what is the, salt shaker's tangential velocity?, vT = rȦ, vT = (0.25 m)(ʌ/2 rad/s), vT = 0.39 m/s, pointing down, 9.8 - Tangential acceleration, , Tangential acceleration: A vector tangent to the, circular path whose magnitude is the rate of, change of tangential speed., As discussed earlier, an object moving in a circle at a constant speed is accelerating, because its direction is constantly changing. This is called centripetal acceleration., Now consider the mushroom on the pizza to the right. Let’s say the pizza has a positive, angular acceleration. Since it is rotating faster and faster, its angular velocity is, increasing. Since tangential speed is the product of the radius and the angular velocity,, the magnitude of its tangential velocity is also increasing., The magnitude of the tangential acceleration vector equals the rate of change of, tangential speed. The tangential acceleration vector is always parallel to the linear, velocity vector. When the object is speeding up, it points in the same direction as the, tangential velocity vector; when the object is slowing down, tangential acceleration, points in the opposite direction., , Tangential acceleration, Rate of change of tangential speed, Increases with distance from center, Direction of vector is tangent to circle, , Since the centripetal acceleration vector always points toward the center, the centripetal and tangential acceleration vectors are perpendicular, to each other. An object’s overall acceleration is the sum of the two vectors. To put it another way: The centripetal and tangential acceleration, are perpendicular components of the object's overall acceleration., Like tangential velocity, tangential acceleration increases with the distance from the axis of rotation. Consider again the pizza and its toppings, in Concept 1. Imagine that the pizza started stationary and it now has positive angular acceleration. Since tangential velocity is proportional to, , 182, , Copyright 2007 Kinetic Books Co. Chapter 09
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radius, at any moment in time the mushroom near the outer edge of the pizza has, greater tangential velocity than the piece of pepperoni closer to the center. Since the, mushroom’s change in tangential velocity is greater, it must have accelerated at a, greater rate., Tangential acceleration can be calculated as the product of the radius and the angular, acceleration. This relationship is stated in Equation 1. The units for tangential, acceleration are meters per second squared, the same as for linear acceleration. Note, that it only makes sense to calculate the tangential acceleration for an object (or really a, point) on the pizza. You cannot speak of the tangential acceleration of the entire pizza, because it includes points that are at different distances from its center and have, different rates of tangential acceleration., Because it is easy to confuse angular and linear motion, we will now review a few, fundamental relationships., An object rotating at a constant angular velocity has zero angular acceleration and zero, tangential acceleration. An example of this is a car driving around a circular track at a, constant speed, perhaps at 100 km/hr. This means the car completes a lap at a, constant rate, so its angular velocity is constant. A constant angular velocity means, zero angular acceleration. Since the angular acceleration is zero, so is the tangential, acceleration., In contrast, the car’s linear (or tangential) velocity is changing since it changes direction, as it moves along the circular path. This accounts for the car’s centripetal acceleration,, which equals its speed squared divided by the radius of the track. The direction of, centripetal acceleration is always toward the center of the circle., , aT = rĮ, aT = tangential acceleration, r = distance to axis, Į = angular acceleration, Direction: tangent to circle, , Now imagine that the car speeds up as it circles the track. It now completes a lap more, quickly, so its angular velocity is increasing, which means it has positive angular, acceleration (when it is moving counterclockwise; it is negative in the other direction)., The car now has tangential acceleration (its linear speed is changing), and this can be, calculated by multiplying its angular acceleration by the track’s radius., The equation for tangential acceleration is derived below from the equations for, tangential velocity and angular acceleration. We begin with the basic definition of linear, acceleration and substitute the tangential velocity equation. The result is an expression, which contains the definition of angular acceleration. We replace this expression with Į,, angular acceleration, which yields the equation we desire., , Step, , 1., , Reason, , definition of linear, acceleration, , 2., , ǻvT = rǻȦ, , tangential velocity equation, , What is the tangential, acceleration of the mushroom, slice at this instant?, aT = rĮ, aT = (ʌ/10 rad/s2)(0.15 m), aT = 0.047 m/s2, pointing down, , substitute equation 2 into, , 3., , equation 1, 4., , aT = rĮ, , definition of angular, acceleration, , Copyright 2007 Kinetic Books Co. Chapter 09, , 183
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9.9 - Interactive checkpoint: a marching band, The performers in a marching band, move in straight rows, maintaining, constant side-to-side spacing, between them. Each row sweeps 90°, through a circular arc when the band, turns a corner. The radii of the paths, followed by the marchers at the inner, and outer ends of a row are 1.50 m, and 7.50 m. If the innermost marcher, in a row moves at 0.350 m/s, what is, the speed of the outermost marcher?, , Answer:, , vout =, , m/s, , 9.10 - Gotchas, A potter’s wheel rotates.A location farther from the axis will have a greater angular velocity than one closer to the axis.Wrong. They all have the, same angular displacement over time, which means they have the same angular velocity, as well. In contrast, they do have different linear, (tangential) velocities., A point on a wheel rotates from 12 o’clock to 3 o’clock, so its angular displacement is 90 degrees, correct? No. This would be one definite error, and one “units police” error. The displacement is negative because clockwise motion is negative. And, using radians is preferable and, sometimes essential in the study of angular motion, so the angular displacement should be stated as íʌ/2 radians., , 9.11 - Summary, Rotational kinematics applies many of the ideas of linear motion to rotational, motion., Angular position is described by an angle ș, measured from the positive x axis., Radians are the typical units., Angular displacement is a change ǻș in angular position. By convention, the, counterclockwise direction is positive., , ǻș = șf – și, , Angular velocity is the angular displacement per unit time. It is represented by Ȧ, and has units of radians per second., Angular acceleration is the change in angular velocity per unit time. It is represented, by Į and has units of radians per second squared., As with linear motion, physicists define instantaneous and average angular velocity, and angular acceleration. Instantaneous and average are defined in ways, analogous to those used in the study of linear motion., The linear velocity of a point on a rotating object is called its tangential velocity,, because it is always directed tangent to its circular path. Any two points on a rigid, rotating object have the same angular velocity, but do not have the same tangential, velocity unless they are the same distance from the rotational axis. Tangential, speed increases as the distance from the axis of rotation increases., , vT = rȦ, , aT = rĮ, Tangential acceleration is the change in tangential speed per unit time. Its, magnitude increases as the radius increases. Its direction is the same as the, tangential velocity if the object is speeding up, and in the opposite direction as the velocity if it is slowing down., , 184, , Copyright 2007 Kinetic Books Co. Chapter 09
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Chapter 9 Problems, , Conceptual Problems, C.1, , Is it possible for a rotating object to have increasing angular speed and negative angular acceleration? Explain your answer., , C.2, , Order these three cities from smallest to largest tangential velocity due to the rotation of the Earth: Washington, DC, USA;, Havana, Cuba; Ottawa, Canada., , Yes, , smallest:, , middle:, , largest:, , No, , i. Havana, ii. Ottawa, iii. Washington, i. Havana, ii. Ottawa, iii. Washington, i. Havana, ii. Ottawa, iii. Washington, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) If you increase the period,, will the angular velocity increase, decrease or stay the same? (b) If you increase the period, will the linear speed increase,, decrease or stay the same? (c) If you increase the distance from the center, will the angular velocity increase, decrease or, stay the same? (d) If you increase the distance from the center, will the linear speed increase, decrease or stay the same?, (a), , i., ii., iii., i., ii., iii., i., ii., iii., i., ii., iii., , (b), , (c), , (d), , 0.2, , Increase, Stay the same, Decrease, Increase, Stay the same, Decrease, Increase, Stay the same, Decrease, Increase, Stay the same, Decrease, , Using the simulation in the interactive problem in this section and referring to your answers to the previous problem, what is, the best way to maximize the linear speed of the rocket? Test your answer using the simulation., i., ii., iii., iv., , Maximize both the period and the distance from the center, Maximize the period and minimize the distance from the center, Minimize both the period and the distance from the center, Minimize the period and maximize the distance from the center, , Section 1 - Angular position, 1.1, , Two cars are traveling around a circular track. The angle between them, from the center of the circle, is 55° and the track has, a radius of 50 m. How far apart are the two cars, as measured around the curve of the track?, m, , 1.2, , Glenn starts his day by walking around a circular track with radius 48 m for 15 minutes. First he walks in a counterclockwise, direction for 1000 meters, then he walks clockwise until the 15 minutes are up. This morning, his clockwise walk is 880 meters, long. When he ends his walk, what is his angular position with respect to where he starts?, rad, , Section 2 - Angular displacement, 2.1, , A dancer completes 2.2 revolutions in a pirouette. What is her angular displacement?, rad, , Copyright 2007 Kinetic Books Co. Chapter 9 Problems, , 185
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2.2, , What is the angular displacement in radians that the minute hand of a watch moves through from 3:15 A.M. to 7:30 P.M. the, same day? Express your answer to the nearest whole radian., rad, , 2.3, , What is the angular displacement in radians of the Earth around the Sun in one hour? Assume the orbit is circular and takes, exactly 365 days in a counterclockwise direction, as viewed from above the North Pole., , 2.4, , The radius of the tires on your car is 0.33 m. You drive 1600 m in a straight line. What is the angular displacement of a point, on the outer rim of a tire, around the center of the tire, during this trip? Assume the tire rotates in the counterclockwise, direction., , 2.5, , A heavy vault door is shut. The angular position of the door from t = 0 to the time the door is shut is given by ș(t) = 0.125t 2,, where ș is in radians. (a) The door is completely shut at ș = ʌ/2 radians. At what time does this occur? (b) What is the angular, displacement of the door between t = 0.52 s and t = 1.67 s? (c) What is the door's average angular velocity between t = 1.50 s, and t = 2.50 s?, , rad, , rad, , (a), , s, , (b), , rad, , (c), , rad/s, , Section 3 - Angular velocity, 3.1, , A hamster runs in its wheel for 2.7 hours every night. If the wheel has a 6.8 cm radius and its average angular velocity is 3.0, radians per second, how far does the hamster run in one night?, m, , 3.2, , An LP record rotates at 33 1/3 rpm (revolutions per minute) and is 12.0 inches in diameter. What is the angular velocity in, rad/s for a fly sitting on the outer edge of an LP rotating in a clockwise direction?, , 3.3, , What is the average angular velocity of the Earth around the sun? Assume a circular counterclockwise orbit, and 365 days in, a year., , rad/s, , rad/s, 3.4, , A car starts a race on a circular track and completes the first three laps in a counterclockwise direction in 618 seconds,, finishing with an angular velocity of 0.0103 rad/s. What is the car's average angular velocity for the first three laps?, rad/s, , 3.5, , Your bicycle tires have a radius of 0.33 m. It takes you 850 seconds to ride 14 times counterclockwise around a circular track, of radius 73 m at constant speed. (a) What is the angular velocity of the bicycle around the track? (b) What is the magnitude, of the angular velocity of a tire around its axis? (That is, don't worry about whether the tire's rotation is clockwise or, counterclockwise.), (a), , rad/s, , (b), , rad/s, , Section 4 - Angular acceleration, 4.1, , The blades of a fan rotate clockwise at í225 rad/s at medium speed, and í355 rad/s at high speed. If it takes 4.65 seconds to, get from medium to high speed, what is the average angular acceleration of the fan blades during this time?, rad/s2, , 4.2, , The blades of a kitchen blender rotate counterclockwise at 2.2×104 rpm (revolutions per minute) at top speed. It takes the, blender 2.1 seconds to reach this top speed after being turned on. What is the average angular acceleration of the blades?, rad/s2, , 4.3, , A potter's wheel is rotating at 3.2 rad/s in a counterclockwise direction when the potter turns it off and lets it slow to a stop., This takes 26 seconds. What is the average angular acceleration of the wheel during this time?, rad/s2, , Section 7 - Tangential velocity, 7.1, , 186, , How might a magician make the Statue of Liberty disappear? Imagine that you are sitting with some spectators on a circular, platform that, unknown to all of you, can rotate very slowly. It is evening, and you can see the Statue of Liberty a short, distance away between two tall brightly lit columns at the rim of the platform. A large curtain can be drawn between the, , Copyright 2007 Kinetic Books Co. Chapter 9 Problems
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columns to temporarily hide the statue. The magician closes the curtain, then rotates the platform through an angle of just, 0.170 radians so the statue is hidden behind one of the columns when the curtain is opened. (a) If the platform rotation takes, 24.0 seconds, what is the average angular speed required? (b) You are sitting 4.00 m from the center of rotation while the, platform is rotating. What is the centripetal acceleration required to move you along the circular arc? (c) Calculate the, centripetal acceleration as a fraction of g. You could be unaware of the rotation, especially if you were distracted., , 7.2, , (a), , rad/s, , (b), , m/s2 (c), , An old-fashioned LP record rotates at 33 1/3 rpm (revolutions per minute) and is 12 inches in diameter. A "single" rotates at, 45 rpm and is 7.0 inches in diameter. If a fly sits on the edge of an LP and then on the edge of a single, on which will the fly, experience the greater tangential speed?, On the LP, , 7.3, , g, , On the 45, , A computer hard drive disk with a diameter of 3.5 inches rotates at 7200 rpm. The "read head" is positioned exactly halfway, from the axis of rotation to the outer edge of the disk. What is the tangential speed in m/s of a point on the disk under the read, head?, m/s, , 7.4, , 7.5, , A radio-controlled toy car has a top speed of 7.90 m/s. You tether it to a pole with a rigid horizontal rod and let it drive in a, circle at top speed. (a) If the rod is 1.80 m long, how long does it take the car to complete one revolution? (b) What is the, angular velocity?, (a), , s, , (b), , rad/s, , You accelerate your car from rest at a constant rate down a straight road, and reach 22.0 m/s in 111 s. The tires on your car, have radius 0.320 m. Assuming the tires rotate in a counterclockwise direction, what is the angular acceleration of the tires?, rad/s2, , 7.6, , When a compact disk is played, the angular velocity varies so that the tangential speed of the area being read by the player, is constant. If the angular velocity of the CD when the player is reading at a distance of 3.00 cm from the center is 3.51, revolutions per second, what is the angular velocity when the player is reading at a distance of 4.00 cm from the center?, rev/s, , Section 8 - Tangential acceleration, 8.1, , A whirling device is launched spinning counterclockwise at 35 rad/s. It slows down with a constant angular acceleration and, stops after 16 seconds. If the radius of the device is 0.038 m, what is the magnitude of the tangential acceleration of a point, on the edge of the device?, m/s2, , 8.2, , Two go-karts race around a course that has concentric circular tracks. The radius of the inner track is 15.0 m, and the radius, of the outer track is 19.0 m. The go-karts start from rest at the same angular position and time, and move at the same, constant angular acceleration. The race ends in a tie after one complete lap, which takes 21.5 seconds. (a) What is the, common angular acceleration of the carts? (b) What is the tangential acceleration of the inner cart? (c) What is the tangential, acceleration of the outer cart?, (a), , rad/s2, , (b), , m/s2, , (c), , m/s2, , Copyright 2007 Kinetic Books Co. Chapter 9 Problems, , 187
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10.0 - Introduction, In the study of rotational kinematics, you analyze the motion of a rotating object by, determining such properties as its angular displacement, angular velocity or angular, acceleration. In this chapter, you explore the origins of rotational motion by studying, rotational dynamics., At the right is a simulation that lets you conduct some experiments in the arena of, rotational dynamics. In it, you play the role of King Kong, and your mission is to, save the day, namely, the bananas on the truck. The bridge is initially open, and the, truck loaded with bananas is heading toward it. You must rotate the bridge to a, closed position. You determine where on the bridge you push and with how much, force. If you cause the bridge to rotate too slowly, it will not close in time, and the, truck will fall into the river. If you accelerate the bridge at too great a rate, the bridge, will smash through the pilings., In this simulation, you are experimenting with torque, the rotational analog to force., A net force causes linear acceleration, and a net torque causes angular, acceleration. The greater the torque you apply on the bridge, the greater the angular, acceleration of the bridge. You control two of the elements that determine torque:, the amount of force and how far it is applied from the axis of rotation. The third factor, the angle at which the force is applied, is a constant 90°, in this simulation., Try pushing with the same amount of force at different points on the bridge. Is the angular acceleration the same or different? Where do you, push to create the maximum torque and angular acceleration? Select a combination of force and location that swings the bridge closed before, the truck arrives, but not so hard that the pilings get smashed., , 10.1 - Torque, , Torque: A force that causes or opposes rotation., A net force causes linear acceleration: a change in the linear velocity of an object. A net, torque causes angular acceleration: a change in the angular velocity. For instance, if, you push hard on a wrench like the one shown in Concept 1, you will start it and the nut, rotating., We will use a wrench that is loosening a nut as our setting to explain the concept of, torque in more detail. In this section, we discuss two of the factors that determine the, amount of torque. One factor is how much force F is exerted and the other is the, distance r between the axis of rotation and the location where the force is applied. We, assume in this section that the force is applied perpendicularly to the line from the axis, of rotation and the location where the force is applied. (If this description seems cryptic,, look at Concept 1, where the force is being applied in this manner.), When the force is applied as stated above, the torque equals the product of the force F, and the distance r. In Equation 1, we state this as an equation. The Greek letter IJ (tau), represents torque., , Torque, Causes or opposes rotation, Increases with:, ·amount of force, ·distance from axis to point of force, , Your practical experience should confirm that the torque increases with the amount of force and the distance from the axis of rotation. If you, are trying to remove a “frozen” nut, you either push harder or you get a longer wrench so you can apply the force at a greater distance., The location of a doorknob is another classic example of factoring in where force is applied. A torque is required to start a door rotating. The, doorknob is placed far from the axis of rotation at the hinges so that the force applied to opening the door results in as much torque as, possible. If you doubt this, try opening a door by pushing near its hinges., The wrench and nut scenario demonstrates another aspect of torque. The angular acceleration of the nut is due to a net torque. Let's say the, nut in Concept 1 is stuck: the force of static friction between it and the bolt creates a torque that opposes the torque caused by the force of the, hand. If the hand pushes hard enough and at a great enough distance from the nut, the torque it causes will exceed that caused by the force of, static friction, and the nut will accelerate and begin rotating. The torque caused by the force of kinetic friction will continue to oppose the, motion., A net torque can cause an object to start rotating clockwise or counterclockwise. By convention, a torque that would cause counterclockwise, rotation is a positive torque. A negative torque causes clockwise rotation. In Example 1, the torque caused by the hand on the wrench is, positive, and the torque caused by friction between the nut and bolt is negative., The unit for torque is the newton-meter (N·m). You might notice that work and energy are also measured using newton-meters, or, equivalently,, joules. Work (and energy) and torque are different, however, and to emphasize that difference, the term "joule" is not used when discussing, torque, but only when analyzing work or energy., , 188, , Copyright 2007 Kinetic Books Co. Chapter 10
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For force applied perpendicularly, IJ = rF, IJ = magnitude of torque, r = distance from axis to force, F = force, Counterclockwise +, clockwise í, Units: newton-meters (N·m), , The hand applies a force of 34 N, as shown. Static friction creates, an opposing torque of 8.5 N·m., Does the nut rotate?, IJ = rF, IJ = (0.25 m)(34 N), IJ = 8.5 N·m, No, the nut does not rotate, 10.2 - Torque, moment of inertia and angular acceleration, , Moment of inertia: The measure of resistance to, angular acceleration., An object’s moment of inertia is the measure of its resistance to a change in its angular, velocity. It is analogous to mass for linear motion; a more massive object requires more, net force to accelerate at a given rate than a less massive object. Similarly, an object, with a greater moment of inertia requires more net torque to angularly accelerate at a, given rate than an object with a lesser moment of inertia. For example, it takes more, torque to accelerate a Ferris wheel than it does a bicycle wheel, for the same rate of, acceleration., To state this as an equation: The net torque equals the moment of inertia times the, angular acceleration. This equation, ȈIJ = IĮ, resembles Newton’s second law,, ȈF = ma. We sometimes refer to this equation as Newton’s second law for rotation., The moment of inertia is measured in kilogram·meters squared (kg·m2). Like mass, the, moment of inertia is always a positive quantity., , Torque and moment of inertia, Net torque = moment of inertia ×, angular acceleration, , Copyright 2007 Kinetic Books Co. Chapter 10, , 189
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We show how the moment of inertia of an object could be experimentally determined in, Example 1. A block, attached to a massless rope, is causing a pulley to accelerate. The, angular acceleration and the net torque are stated in the problem. (The net torque could, be determined by multiplying the tension by the radius of the pulley, keeping in mind, that the tension is less than the weight of the block since the block accelerates, downward.) With these facts known, the moment of inertia of the pulley can be, determined., , ȈIJ = IĮ, ȈIJ = net torque, I = moment of inertia, Į = angular acceleration, Units for I: kg·m2, , What is the moment of inertia of, the pulley?, ȈIJ = IĮ, I = ȈIJ/Į, I = (55 N·m)/(22 rad/s2), I = 2.5 kg·m2, 10.3 - Calculating the moment of inertia, If you were asked whether the same amount of torque would cause a greater angular, acceleration with a Ferris wheel or a bicycle wheel, you would likely answer: the bicycle, wheel. The greater mass of the Ferris wheel means it has a greater moment of inertia. It, accelerates less with a given torque., But more than the amount of mass is required to determine the moment of inertia; the, distribution of the mass also matters. Consider the case of a boy sitting on a seesaw., When he sits close to the axis of rotation, it takes a certain amount of torque to cause, him to have a given rate of angular acceleration. When he sits farther away, it takes, more torque to create the same rate of acceleration. Even though the boy’s (and the, seesaw's) mass stays constant, he can increase the system’s moment of inertia by, sitting farther away from the axis., When a rigid object or system of particles rotates about a fixed axis, each particle in the, object contributes to its moment of inertia. The formula in Equation 1 to the right shows, how to calculate the moment of inertia. The moment equals the sum of each particle’s, mass times the square of its distance from the axis of rotation., , Moment of inertia, Sum of each particle’s, ·Mass times its, ·Distance squared from the axis, , A single object often has a different moment of inertia when its axis of rotation changes., For instance, if you rotate a baton around its center, it has a smaller moment of inertia than if you rotate it around one of its ends. The baton is, harder to accelerate when rotated around an end. Why is this the case? When the baton rotates around an end, more of its mass on average is, farther away from the axis of rotation than when it rotates around its center., If the mass of a system is concentrated at a few points, we can calculate its moment of inertia using multiplication and addition. You see this in, , 190, , Copyright 2007 Kinetic Books Co. Chapter 10
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Example 1, where the mass of the object is concentrated in two balls at the ends of the, rod. The moment of inertia of the rod is very small compared to that of the balls, and we, do not include it in our calculations. We also consider each ball to be concentrated at its, own center of mass when measuring its distance from the axis of rotation (marked by, the ×). This is a reasonable approximation when the size of an object is small relative to, its distance from the axis., Not all situations lend themselves to such simplifications. For instance, let’s assume we, want to calculate the moment of inertia of a CD spinning about its center. In this case, the mass is uniformly distributed across the entire CD. In such a case, we need to use, calculus to sum up the contribution that each particle of mass makes to the moment, or, we must take advantage of a table that tells us the moment of inertia for a disk rotating, around its center., , I = Ȉmr2, I = moment of inertia, m = mass of a particle, r = distance of particle from axis, Units: kg·m2, , What is the system's moment of, inertia? Ignore the rod's mass., I = Ȉmr2 = m1r12 + m2r22, I = (2.3 kg)(1.3 m) 2 + (1.2 kg)(1.1 m)2, I = 5.3 kg·m2, 10.4 - A table of moments of inertia, , Moments of inertia, Cylinders, , Sets of objects are shown in the illustrations above and to the right. Above each object is a description of it and its axis of rotation. Below each, object is a formula for calculating its moment of inertia, I., The variable M represents the object’s mass. It is assumed that the mass is distributed uniformly throughout each object., If you look at the formulas in each table, they will confirm an important principle underlying moments of inertia: The distribution of the mass, relative to the axis of rotation matters. For instance, consider the equations for the hollow and solid spheres, each of which is rotating about an, axis through its center. A hollow sphere with the same mass and radius as a solid sphere has a greater moment of inertia. Why? Because the, mass of the hollow sphere is on average farther from its axis of rotation than that of the solid sphere., Note also that the moment of inertia for an object depends on the location of the axis of rotation. The same object will have different moments, of inertia when rotated around differing axes. As shown on the right, a thin rod rotated around its center has one-fourth the moment of inertia as, the same rod rotated around one end. Again, the difference is due to the distribution of mass relative to the axis of rotation. On average, the, mass of the rod is further away from the axis when it is rotated around one end., , Copyright 2007 Kinetic Books Co. Chapter 10, , 191
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Moments of inertia, Spheres, , Slabs, , Thin rods, , 10.5 - Sample problem: a seesaw, The seesaw plank is horizontal. Its, mass is 36.5 kg, and it is 4.40 m long., What is the initial angular acceleration, of this system?, , The axis of rotation is the point where the fulcrum touches the midpoint of the plank. The plank itself creates no net torque since it is balanced, at its middle. For every particle at a given distance from the axis that creates a clockwise torque, there is a matching particle at the same, distance creating a counterclockwise torque. However, the plank does factor into the moment of inertia., , 192, , Copyright 2007 Kinetic Books Co. Chapter 10
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Draw a diagram, , Variables, mass of seesaw plank, , mS = 36.5 kg, , seesaw plank’s moment of inertia, , IS, , girl, , boy, , mass, , mG = 42.8 kg, , mB = 55.2 kg, , distance from axis, , rG = 2.13 m, , rB = 1.51 m, , moment of inertia, , IG, , IB, , What is the strategy?, 1., , Calculate the moment of inertia of the system: the sum of the moments for the children, and the moment of the plank., , 2., , Calculate the net torque by summing the torques created by each child. The torques of the left and right sides of the plank cancel, so, you do not have to consider them., , 3., , Divide the net torque by the moment of inertia to determine the initial angular acceleration., , Physics principles and equations, We will use the definitions of torque and moment of inertia., , IJ = rF sin ș, I = Ȉmr2, To calculate the moments of inertia of the children, we consider the mass of each to be concentrated at one point., The plank can be considered as a slab rotating on an axis parallel to an edge through the center, with moment of inertia, , The equation relating net torque and moment of inertia is Newton's second law for rotation,, , ȈIJ = IĮ, Step-by-step solution, First, we add the moments of inertia for the two children and the seesaw plank. The sum of these values equals the system's moment of inertia., , Step, , Reason, , 1., , I = IG + IB + IS, , total moment is sum, , 2., , I = mGrG2 + mBrB2 + IS, , definition of moment of inertia, , 3., , moment of slab, , 4., , enter values, , 5., , I = 379 kg·m2, , evaluate, , Copyright 2007 Kinetic Books Co. Chapter 10, , 193
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The children create torques, and to calculate the net torque, we sum their torques, being careful about signs. The plank creates no net torque, since its midpoint is at the fulcrum., , Step, , Reason, , 6., , ȈIJ = IJG + IJB, , net torque equals sum of torques, , 7., , ȈIJ = mGgrG + (ímBgrB), , equation for torque, enter values, , 8., 9., , ȈIJ = 76.6 N·m, , evaluate, , Now we use the values we calculated for the net torque and the moment of inertia to calculate the angular acceleration., , Step, , Reason, , 10. Į =ȈIJ / I, , Newton’s second law for rotation, , 11. Į = (76.6 N·m)/(379 kg·m2), , substitute equations 5 and 9 into equation 10, , 12. Į = 0.202 rad/s2 (counterclockwise) solve for Į, Because various quantities change, such as the angle between the direction of each child’s weight and the seesaw, the angular acceleration, changes as the seesaw rotates. This is why we asked for the initial angular acceleration., , 10.6 - Interactive problem: close the bridge, Once again, you are King Kong, and your task is to close the bridge you see on the, right in order to save an invaluable load of bananas (well, invaluable to you at least)., Here, we ask you to be a more precise gorilla than you may have been in the, introductory exercise., To close the bridge quickly enough to save the fruit without breaking off the bumper, pilings, you need to apply a torque so that the bridge’s angular acceleration is, ʌ/16.0 rad/s2. The moment of inertia of the rotating part of the bridge is, 45,400,000 kg·m2., Two trucks are parked on this part of the bridge, and you must include them when, you calculate the total moment of inertia. Each truck has a mass of 4160 kg; the, midpoint of one is 20.0 m and the midpoint of the other is 30.0 m from the pivot (axis, of rotation) of the swinging bridge. The trucks will increase the bridge's moment of, inertia. To solve the problem, consider all the mass of each truck to be concentrated, at its midpoint., You apply your force 35.0 m from the pivot and your force is perpendicular to the, rotating component of the bridge. Enter the amount of force you wish to apply to the nearest 0.01×105 N and press GO to start the simulation., Press RESET if you need to try again., If you have difficulty solving this problem, review the sections on calculating the moment of inertia, and the relation between torque, angular, acceleration, and moment of inertia., , 10.7 - Physics at work: flywheels, Flywheels are rotating objects used to store energy, as rotational kinetic energy. Recently, environmental, and other concerns have caused flywheels to receive, increased attention. Many of these new flywheels, serve as mechanical “batteries,” replacing traditional, electric batteries., Why the interest? Traditional chemical batteries,, rechargeable or not, have a shorter total life span, than flywheels and can cause environmental, problems when disposed of incorrectly. On the other, hand, flywheels cost more to produce than traditional, batteries, and their ability to function in demanding, situations is unproven., , This advanced flywheel is being developed by NASA as a, source of stored energy for use by satellites and spacecraft., , Flywheels can be powered by “waste” energy. For instance, when a bus slows down, its brakes warm up. The bus’s kinetic energy becomes, thermal energy, which the vehicle cannot re-use efficiently. Some buses now use a flywheel to convert a portion of that linear kinetic energy, into rotational energy, and then later transform that rotational energy back into linear kinetic energy as the bus speeds up., , 194, , Copyright 2007 Kinetic Books Co. Chapter 10
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Flywheels can receive power from more traditional sources, as well. For instance,, uninterruptible power sources (UPS) for computers use rechargeable batteries to keep, computers powered during short-term power outages. Flywheels are being considered, as an alternative to chemical batteries in these systems. Less traditional sources can, also supply energy to a flywheel: NASA uses solar power to energize flywheels in, space., The equation for rotational KE is shown to the right. The moment of inertia and, maximum angular velocity determine how much energy a flywheel can store. The, moment of inertia, in turn, is a function of the mass and its distance (squared) from the, axis of rotation., In Concept 1, you see a traditional flywheel. It is large, massive, and constructed with, most of its mass at the outer rim, giving it a large moment of inertia and allowing it to, store large amounts of rotational KE. In Concept 2, you see a modern flywheel, which, is much smaller and less massive, but capable of rotating with a far greater angular, velocity. Flywheels in these systems can rotate at 60,000 revolutions per minute, (6238 rad/s). Air drag and friction losses are greatly reduced by enclosing the flywheel, in a near vacuum and by employing magnetic bearings., , Flywheels, Spinning objects “store” rotational KE, Energy depends on, ·angular velocity, ·moment of inertia (mass, radius), , Flywheels, Serve as mechanical batteries, , Flywheel energy, KE = ½IȦ2, KE = kinetic energy (rotational), I = moment of inertia, Ȧ = angular velocity, , Copyright 2007 Kinetic Books Co. Chapter 10, , 195
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10.8 - Angular momentum of a particle in circular motion, The concepts of linear momentum and conservation of linear momentum prove very, useful in understanding phenomena such as collisions. Angular momentum is the, rotational analog of linear momentum, and it too proves quite useful in certain settings., For instance, we can use the concept of angular momentum to analyze an ice skater’s, graceful spins., In this section, we focus on the angular momentum of a single particle revolving in a, circle. Angular momentum is always calculated using a point called the origin. With, circular motion, the simple and intuitive choice for the origin is the center of the circle,, and that is the point we will use here. The letter L represents angular momentum., As with linear momentum, angular momentum is proportional to mass and velocity., However, with rotational motion, the distance of the particle from the origin must be, taken into account, as well. With circular motion, the amount of angular momentum, equals the product of mass, speed and the radius of the circle: mvr. Another way to, state the same thing is to say that the amount of angular momentum equals the linear, momentum mv times the radius r., , Angular momentum of a particle, Proportional to mass, speed, and, distance from origin, , Like linear momentum, angular momentum is a vector. When the motion is, counterclockwise, by convention, the vector is positive. The angular momentum of, clockwise motion is negative. The units for angular momentum are kilogram-meter2 per, second (kg·m2/s)., , L = mvr, L = angular momentum, m= mass, v = speed, r = distance from origin (radius), Counterclockwise +, clockwise í, Units: kg·m2/s, , How much angular momentum, does the engine have?, L = ímvr, L = í(0.15 kg)(1.1 m/s)(0.50 m), L = í0.083 kg·m2/s, , 196, , Copyright 2007 Kinetic Books Co. Chapter 10
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10.9 - Angular momentum of a rigid body, On the right, you see a familiar sight: a rotating compact disc. In the prior section, we, defined the angular momentum of a single particle as the product of its mass, speed, and radial distance from the axis of revolution. A CD is more complex than that. It, consists of many particles rotating at different distances from a common axis of rotation., The CD is rigid, which means the particles all rotate with the same angular velocity, and, each remains at a constant radial distance from the axis., We can determine the angular momentum of the CD by summing the angular momenta, of all the particles that make it up. The resulting sum can be expressed concisely using, the concept of moment of inertia. The magnitude of the angular momentum of the CD, equals the product of its moment of inertia, I, and its angular velocity, Ȧ., We derive this formula for calculating angular momentum below. In Equation 1, you see, one of the rotating particles drawn, with its mass, velocity and radius indicated., Variables, mass of a particle, , mi, , tangential (linear) speed of a particle, , vi, , radius of a particle, , ri, , angular momentum of particle, , Li, , angular momentum of CD, , L, , angular velocity of CD, , Ȧ, , moment of inertia of CD, , I, , Angular momentum of a rigid, body, Product of moment of inertia, angular, velocity, , Strategy, 1., , Express the angular momentum of the CD as the sum of the angular momenta of, all the particles of mass that compose it., , 2., , Replace the speed of each particle with the angular velocity of the CD times the, radial distance of the particle from the axis of rotation., , 3., , Express the sum in concise form using the moment of inertia of the CD., , L = IȦ, , Physics principles and equations, , L = angular momentum, , The angular momentum of a particle in circular motion, , I = moment of inertia, Ȧ = angular velocity, , L = mvr, We will use the equation that relates tangential speed and angular velocity., , v = rȦ, The formula for the moment of inertia of a rotating body, , I = Ȉmiri2, Step-by-step derivation, First, we express the angular momentum of the CD as the sum of the angular momenta, of the particles that make it up., , Step, , Reason, , 1., , Li = miviri, , 2., , L = Ȉmiviri angular momentum of object is sum of particles, , definition of angular momentum, , We now express the speed of the ith particle as its radius times the constant angular, velocity Ȧ, which we then factor out of the sum. The angular velocity is the same for all, particles in a rigid body., , Step, , How much angular momentum, does the skater have?, L = IȦ, L = (1.4 kg·m2)(21 rad/s), L = 29 kg·m2/s, , Reason, , 3., , vi = riȦ, , 4., , L = Ȉmi(riȦ)ri substitute equation 3 into equation 2, , 5., , L = (Ȉmiri2)Ȧ factor out Ȧ, , tangential speed and angular velocity, , Copyright 2007 Kinetic Books Co. Chapter 10, , 197
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In the final steps we express the above result concisely, replacing the sum in the last equation by the single quantity I., , Step, , Reason, , Ȉmiri2, , 6., , I=, , 7., , L = IȦ, , moment of inertia, substitute equation 6 into equation 5, , 10.10 - Torque and angular momentum, A net force changes an object’s velocity, which means its linear momentum changes as, well. Similarly, a net torque changes a rotating object’s angular velocity, and this, changes its angular momentum., As Equation 1 shows, the product of torque and an interval of time equals the change in, angular momentum. This equation is analogous to the equation from linear dynamics, stating that the impulse (the product of average force and elapsed time) equals the, change in linear momentum., In the illustration to the right, we show a satellite rotating at a constant angular velocity, and then firing its thruster rockets, which changes its angular momentum. The thrusters, apply a constant torque IJ to the satellite for an elapsed time ǻt. In Example 1, the, satellite’s change in angular momentum is calculated using the equation. (The change, in the satellite’s mass and moment of inertia resulting from the expelled fuel are small, enough to be ignored.), , Torque and angular momentum, IJǻt = ǻL, IJ = torque, ǻt = time interval, ǻL = change in angular momentum, , The rockets provide 56 N·m of, torque for 3.0 s. What is the, amount of change in the, satellite's angular momentum?, IJǻt = ǻL, (56 N·m)(3.0 s) = ǻL, ǻL = 168 kg·m2/s, 10.11 - Conservation of angular momentum, , Angular momentum conserved, No external torque, Angular momentum is constant, , Linear momentum is conserved when there is no external net force acting on a system. Similarly, angular momentum is conserved when there, is no net external torque. To put it another way, if there is no net external torque, the initial angular momentum equals the final angular, momentum. This is stated in Equation 1., The principle of conservation of linear momentum is often applied to collisions, and the masses of the colliding objects are assumed to remain, constant. However, with angular momentum, we often examine what occurs when the moment of inertia of a body changes. Since angular, momentum equals the product of the moment of inertia and angular velocity, if one of these properties changes, the other must as well for the, angular momentum to stay the same. This principle is used both in classroom demonstrations and in the world of sports. In a common, classroom demonstration, a student is set rotating on a stool. The student holds weights in each hand, and as she pushes them away from her, body, she slows down. In doing so, she demonstrates the conservation of angular momentum: As her moment of inertia increases, her angular, , 198, , Copyright 2007 Kinetic Books Co. Chapter 10
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velocity decreases. In contrast, pulling the weights close in to her body decreases her, moment of inertia and increases her angular velocity., Ice skaters apply this principle skillfully. When they wish to spin rapidly, they wrap their, arms tightly around their bodies. They decrease their moment of inertia to increase their, angular velocity. You can see images of a skater applying this principle to the right and, above., , Conservation of angular, momentum, Li = Lf, IiȦi = IfȦf, L = angular momentum, I = moment of inertia, Ȧ = angular velocity, , The skater pulls in her arms,, cutting her moment of inertia in, half. How much does her angular, velocity change?, IiȦi = IfȦf, IiȦi = (½Ii)Ȧf, Ȧf = 2Ȧi (it doubles), 10.12 - Gotchas, A torque is a force. No, it is not. A net torque causes angular acceleration. It requires a force., A torque that causes counterclockwise acceleration is a positive torque. Yes, and a torque that causes clockwise acceleration is negative., I have a baseball bat. I shave off some weight from the handle and put it on the head of the bat.A baseball player thinks I have changed the, bat’s moment of inertia. Is he right? Yes. A baseball player swings from the handle, so you have increased the amount of mass at the farthest, distance, changing the bat’s moment. The player will find it harder to apply angular acceleration to the bat., A skater begins to rotate more slowly, so his angular momentum must be changing. Not necessarily. An external torque is required to change, the angular momentum. The slower rotation could instead be caused by the skater altering his moment of inertia, perhaps by moving his hands, farther from his body. On the other hand, if he digs the tip of a skate into the ice, that torque would reduce his angular momentum., , Copyright 2007 Kinetic Books Co. Chapter 10, , 199
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10.13 - Summary, Torque is a force that causes rotation. Torque is a vector quantity with units N·m., An object's moment of inertia is a measure of its resistance to angular acceleration,, just as an object's mass is a measure of its resistance to linear acceleration. The, moment of inertia is measured in kg·m2 and depends not only upon an object’s, mass, but also on the distribution of that mass around the axis of rotation. The, farther the distribution of the mass from the axis, the greater the moment of inertia., Another linear analogy applies: Just as Newton's second law states that net force, equals mass times linear acceleration, the net torque on an object equals its, moment of inertia times its angular acceleration., Rotational kinetic energy depends upon the moment of inertia and the angular, velocity. Mechanical devices called flywheels can store rotational kinetic energy., Angular momentum is the rotational analog to linear momentum. Its units are, kg·m2/s. The magnitude of the angular momentum of an object in circular motion is, the product of its mass, tangential velocity, and the radius of its path. The angular, momentum of a rigid rotating body equals its moment of inertia multiplied by its, angular velocity., Just as a change in linear momentum (impulse) is equal to a force times its, duration, a change in angular momentum is equal to a torque times its duration., , Torque, , IJ = rF, Newton's second law for rotation, , ȈIJ = IĮ, Moment of inertia, , I = Ȉmr2, Angular momentum, , L = IȦ, IJǻt = ǻL, , Angular momentum is conserved in the absence of a net torque on the system., , Conservation of angular momentum, , L i = Lf, I i Ȧ i = I fȦ f, , 200, , Copyright 2007 Kinetic Books Co. Chapter 10
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Chapter 10 Problems, , Conceptual Problems, C.1, , C.2, , (a) Can an object have an angular velocity if there is no net torque acting on it? (b) Can an object have a net torque acting on, it if it has zero angular velocity?, (a), , Yes, , No, , (b), , Yes, , No, , On which of the following does the moment of inertia depend?, Angular velocity, Angular momentum, Shape of the object, Location of axis of rotation, Mass, Linear velocity, , C.3, , An object of fixed mass and rigid shape has one unique value for its moment of inertia. True or false? Explain your answer., True, , False, , C.4, , Compared to a solid sphere, will a hollow spherical shell (like a basketball) of the same mass and radius have a greater or a, lesser moment of inertia for rotations about an axis passing through the center? Explain your answer., , C.5, , Scoring in the sport of gymnastics has much to do with the difficulty of the skills performed, with higher scores awarded to, more difficult tricks. The position of a gymnast's body in a trick determines the difficulty. There are three positions to hold the, body in a flip: the tuck, in which the legs are bent and tucked into the body; the pike, in which the body is bent at the hips but, the legs remain straight; and the layout, in which the whole body is straight. A layout receives a higher score than a pike, which receives a higher score than a tuck. Use the principles of rotational dynamics to describe why the position of the body, affects the difficulty of a flip., , Greater, , Lesser, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Does changing the distance, from the axis of rotation affect the angular acceleration? (b) Where should you push to create the maximum torque and, angular acceleration?, (a), (b), , Yes, , No, , i. Farthest from the axis of rotation, ii. In the middle of the bridge, iii. Closest to the axis of rotation, , Section 1 - Torque, 1.1, , The wheel on a car is held in place by four nuts. Each nut should be tightened to 94.0 N·m of torque to be secure. If you have, a wrench with a handle that is 0.250 m long, what minimum force do you need to exert perpendicular to the end of the wrench, to tighten a nut correctly?, N, , 1.2, , A 1.1 kg birdfeeder hangs from a horizontal tree branch. The birdfeeder is attached to the branch at a point that is 1.1 m from, the trunk. What is the amount of torque exerted by the birdfeeder on the branch? The origin is at the pivot point, where the, branch attaches to the trunk., N·m, , 1.3, , Bob and Ray push on a door from opposite sides. They both push perpendicular to the door. Bob pushes 0.63 m from the, door hinge with a force of 89 N. Ray pushes 0.57 m from the door hinge with a force of 98 N, in a manner that tends to turn, the door in a clockwise direction. What is the net torque on the door?, N·m, , Section 2 - Torque, moment of inertia and angular acceleration, 2.1, , The pulley shown in the illustration has a radius of 2.70 m and a moment of inertia of 39.0 kg·m2. The hanging mass is, , Copyright 2007 Kinetic Books Co. Chapter 10 Problems, , 201
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4.20 kg and it exerts a force tangent to the, edge of the pulley. What is the angular, acceleration of the pulley?, rad/s2, , 2.2, , A string is wound tight around the spindle of a top, and then pulled to spin the top. While it is pulled, the string exerts a, constant torque of 0.150 N·m on the top. In the first 0.220 s of its motion, the top reaches an angular velocity of 12.0 rad/s., What is the moment of inertia of the top?, kg · m2, , 2.3, , A wheel with a radius of 0.71 m is mounted, horizontally on a frictionless vertical axis. A toy, rocket is attached to the edge of the wheel, so, that it makes a 72° angle with a radius line, as, shown. The moment of inertia of the wheel, and rocket is 0.14 kg·m2. If the rocket starts, from rest and fires for 8.4 seconds with a, constant force of 1.3 N, what is the angular, velocity of the wheel when the rocket stops, firing?, rad/s, , Section 3 - Calculating the moment of inertia, 3.1, , Four small balls are arranged at the corners of a rigid metal square with sides of length 3.0 m. An axis of rotation in the plane, of the square passes through the center of the square, and is parallel to two sides of the square. On one side of the axis, the, two balls have masses 1.8 kg and 2.3 kg; on the other side, 1.5 kg and 2.7 kg. The mass of the square is negligible compared, to the mass of the balls. What is the moment of inertia of the system for this axis?, kg · m2, , 3.2, , For the same arrangement as in the previous problem, what is the moment of inertia of the system of balls for the axis that, is perpendicular to the plane of the square, and passes through its center point?, kg · m2, , 3.3, , Four balls are connected by a straight rod. One end of the rod is painted blue. The first ball has mass 1.0 kg and is 1.0 m, from the blue end of the rod, the second ball has mass 2.0 kg and is 2.0 m from the blue end, and so on for the other two, balls. The mass of the rod is negligible compared to the mass of the balls. What is the moment of inertia of this system for an, axis of rotation perpendicular to the rod and touching the blue end?, kg · m2, , 3.4, , Four small balls of identical mass 2.36 kg are, arranged in a rigid structure as a regular, tetrahedron. (A regular tetrahedron has four, faces, each of which is an equilateral triangle.), Each edge of the tetrahedron has length, 3.20 m. What is the moment of inertia of the, system, for an axis of rotation passing, perpendicularly through the center of one of, the faces of the tetrahedron?, kg · m2, , Section 4 - A table of moments of inertia, 4.1, , A tire of mass 1.3 kg and radius 0.34 m experiences a constant net torque of 2.1 N·m. Treat the tire as though all of its mass, is concentrated at its rim. How long does it take for the tire to reach an angular speed of 18 rad/s from a standing stop?, s, , 4.2, , 202, , A hollow glass holiday ornament in the shape of a sphere is suspended on a string that forms an axis of rotation through the, , Copyright 2007 Kinetic Books Co. Chapter 10 Problems
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ornament's center. If the ornament has mass 0.0074 kg and radius 0.036 m, what is its moment of inertia for an axis that, passes through its center?, kg · m2, 4.3, , Three glass panels, each 2.7 m tall and 1.6 m, wide and with mass 23 kg, are joined together, to make a rotating door. The axis of rotation is, the line where the panels join together. A, person pushes perpendicular to a panel at its, outer edge with a force of 73 N. What is the, angular acceleration of the door?, rad/s2, , 4.4, , A solid cylinder 0.230 m long and with a radius, of 0.0730 m is rotated around an axis through, its middle and parallel to its ends, as shown. A, constant net torque of 1.20 N·m is applied to, the cylinder, resulting in an angular, acceleration of 23.0 rad/s2. What is the mass, of the cylinder?, kg, , 4.5, , Two golf balls are glued together and rotated about an axis through the point where they are joined. The axis is tangent to, both golf balls. The mass of a golf ball is 0.046 kg and its radius is 0.021 m. What is the moment of inertia of the pair of golf, balls around the chosen axis?, kg · m2, , 4.6, , Bob and Ray push an ordinary door from opposite sides. Both of them push perpendicular to the door. Bob pushes 0.670 m, from the hinge with a force of 121 N. Ray pushes 0.582 m from the hinge with force of 132 N. Consider the door as a slab,, with height 2.03 m, width 0.813 m, and mass 13.6 kg. What is the magnitude of the angular acceleration of the door?, rad/s2, , 4.7, , A potter's wheel consists of a uniform concrete disk, 0.035 m thick and with a radius of 0.42 m. It has a mass of 48 kg. The, wheel is rotating freely with an angular velocity of 9.6 rad/s when the potter stops it by pressing a wood block against the edge, of the wheel, directing a force of 65 N on a line toward the center of the wheel. If the wheel stops in 7.2 seconds, what is the, coefficient of friction between the block and the wheel?, , 4.8, , Two thin, square slabs of metal, each with side length of 0.30 m and mass 0.29 kg, are welded together in a T shape and, rotated on an axis through their line of intersection. What is the moment of inertia of the T?, kg · m2, , Section 6 - Interactive problem: close the bridge, 6.1, , Use the simulation in the interactive problem in this section to answer the following question. What is the force required to, close the bridge with the desired angular acceleration? Test your answer using the simulation., N, , Section 8 - Angular momentum of a particle in circular motion, 8.1, , What is the magnitude of angular momentum of a 1070 kg car going around a circular curve with a 15.0 m radius at, 12.0 m/s? Assume the origin is at the center of the curve's arc., kg · m2 /s, , 8.2, , Calculate the magnitude of the angular momentum of the Earth around the Sun, using the Sun as the origin. The Earth's, mass is 5.97×1024 kg and its roughly circular orbit has a radius of 1.50×1011 m. Use a 365-day year with exactly 24 hours in, each day., kg · m2 /s, , Copyright 2007 Kinetic Books Co. Chapter 10 Problems, , 203
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Section 9 - Angular momentum of a rigid body, 9.1, , A thin rod 2.60 m long with mass 3.80 kg is rotated counterclockwise about an axis through its midpoint. It completes 3.70, revolutions every second. What is the magnitude of its angular momentum?, kg · m2 /s, , 9.2, , A 0.16 meter long, 0.15 kg thin rigid rod has a small 0.22 kg mass stuck on one of its ends and a small 0.080 kg mass stuck, on the other end. The rod rotates at 1.7 rad/s through its physical center without friction. What is the magnitude of the angular, momentum of the system taking the center of the rod as the origin? Treat the masses on the ends as point masses., kg · m2 /s, , 9.3, , A solid cylinder is rotated counterclockwise around an axis at its base (one circular end). The axis lies in the plane of the, base, and passes through the center of the circle. The cylinder is 3.60 m long and has a radius of 0.870 m. Its mass is, 2.30×104 kg and the center of the base opposite the axis has a tangential speed of 12.0 m/s. What is the cylinder's angular, momentum?, kg · m2 /s, , Section 10 - Torque and angular momentum, 10.1 An electric drill delivers a net torque of 15.0 N·m to a buffing wheel used to polish a car. The buffing wheel has a moment of, inertia of 2.30×10í3 kg·m2. At 0.0220 s after the drill is turned on, what is the angular velocity of the buffing wheel?, rad/s, 10.2 A merry-go-round is 18.0 m in diameter and has a mass, unloaded, of 48,100 kg. It is fairly uniform in structure and can be, considered to be a solid cylindrical disk. The merry-go-round carries 28 passengers, who have an average mass of 68.5 kg, and all sit at a distance of 8.75 m from the center. The fully-loaded merry-go-round takes 48.0 s to reach an angular velocity, of 0.650 rad/s. (a) What is the moment of inertia of the loaded merry-go-round? (b) What constant net torque is applied to the, merry-go-round to reach this working speed?, (a), , kg · m2 (b), , N·m, , 10.3 A string is wound around the edge of a solid 1.60 kg disk with a 0.130 m radius. The disk is initially at rest when the string is, pulled, applying a force of 6.50 N in the plane of the disk and tangent to its edge. If the force is applied for 1.90 seconds, what, is the magnitude of its final angular velocity?, rad/s, , Section 11 - Conservation of angular momentum, 11.1 A 1.6 kg disk with radius 0.63 m is rotating freely at 55 rad/s around an axis perpendicular to its center. A second disk that is, not rotating is dropped onto the first disk so that their centers align, and they stick together. The mass of the second disk is, 0.45 kg and its radius is 0.38 m. What is the angular velocity of the two disks combined?, rad/s, 11.2 Two astronauts in deep space are connected by a 22 m rope, and rotate at an angular velocity of 0.48 rad/s around their, center of mass. The mass of each astronaut, including spacesuit, is 97 kg, and the rope has negligible mass. One astronaut, pulls on the rope, shortening it to 14 m. What is the resulting angular velocity of the astronauts?, rad/s, 11.3 A 27.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to, the edge of the merry-go-round, what is the new angular velocity of the system? Model the merry-go-round as a solid disk., rad/s, 11.4 A puck with mass 0.28 kg moves in a circle at, the end of a string on a frictionless table, with, radius 0.75 m. The string goes through a hole, in the table, and you hold the other end of the, string. The puck is rotating at an angular, velocity of 18 rad/s when you pull the string to, reduce the radius of the puck's travel to, 0.55 m. Consider the puck to be a point mass., What is the new angular velocity of the puck?, rad/s, , 204, , Copyright 2007 Kinetic Books Co. Chapter 10 Problems
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11.0 - Introduction, Although much of physics focuses on motion and, change, the topic of how things stay the same í, equilibrium í also merits study. Bridges spanning, rivers, skyscrapers standing tall… None of these, would be possible without engineers having achieved, a keen understanding of the conditions required for, equilibrium. In this chapter, we focus on static, equilibrium. To do so, we must consider both forces, and torques, since for an object to be in static, equilibrium, the net force and net torque on it must, both equal zero., A tightrope walker balances forces and torques to maintain equilibrium., The forces and masses involved in equilibrium can be, stupendous. The Brooklyn Bridge was the, engineering marvel of its day; it gracefully spans the East River between Brooklyn and Manhattan. The anchorages at the ends of the bridge, each have a mass of almost 55 million kilograms, while the suspended superstructure between the anchorages has a mass of 6 million, kilograms. Supporting those 6 million kilograms are four cables of 787,000 kilograms apiece. A five-year construction effort resulted in the, largest suspension bridge of its time, and one that over 200,000 vehicles pass over daily., , Engineers who design structures such as bridges must concern themselves with forces that cause even a material like steel to change shape,, to lengthen or contract. If the material returns to its initial dimensions when the force is removed, it is called elastic., , 11.1 - Static equilibrium, , Static equilibrium: No net torque, no net force, and no motion., In California, equilibrium is achieved either by renouncing one’s possessions, moving to, a commune and selecting a guru, or by becoming extremely rich, moving to Malibu and, choosing a personal trainer., In physics, static equilibrium also requires a threefold path. First, there is no net force, acting on the body. Second, there is no net torque on it about any axis of rotation., Finally, in the case of static equilibrium, there is no motion. An object moving with a, constant linear and rotational velocity is also in equilibrium, but not in static equilibrium., Let’s see how we can apply these concepts to the seesaw at the right. There are two, children of different weights on the seesaw. They have adjusted their positions so that, the seesaw is stationary in the position you now see. (We will only concern ourselves, with the weights of the children, and will ignore the weight of the seesaw.), In Equation 2, we examine the torques. The fulcrum is the axis of rotation. Since the, system is stationary, there is no angular acceleration, which means there is no net, torque., , Static equilibrium, ȈFx = 0, ȈFy = 0, Net force along each axis is zero, , Let’s consider the torques in more detail. They must sum to zero since the net torque, equals zero. We choose to use an axis of rotation that passes through the point where, the fulcrum touches the seesaw. The normal force of the fulcrum creates no torque, because its distance to this axis is zero. The boy exerts a clockwise (negative) torque., The girl exerts a counterclockwise (positive) torque. Since the girl weighs less than the, boy, she sits farther from the fulcrum to make their torques equal but opposite. In sum,, there are no net forces, no net torques, and the system is not moving: It is in static, equilibrium., Note that in analyzing the seesaw, we used the fulcrum as the axis of rotation. This, seems natural, since it is the point about which the seesaw rotates when the children, are “seesawing.” However, in problems you will encounter later, it is not always so easy, to determine the axis of rotation. In those cases, it is helpful to remember that if the net, torque is zero about one axis, it will be zero about any axis, so the choice of axis is up, to you. However, this trick only works for cases when the net torque is zero. In general,, the torque depends on one’s choice of axis., , Static equilibrium, ȈIJ = 0, Net torque is zero, , Copyright 2000-2007 Kinetic Books Co. Chapter 11, , 205
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11.2 - Sample problem: a witch and a duck balance, The witch and the duck are balanced, on the scale. The duck weighs 44.5, N. What is the witch’s weight?, , Since the system is stationary, it is in static equilibrium. This means there is no net torque and no net force. We use the pivot point of the scale, as the axis of rotation. We start by drawing a diagram for the problem. (The diagrams in this section are not drawn to scale; we also are not, really sure witches exist, but then we do assume objects to be massless and frictionless and so on, so who are we to complain?) We ignore the, masses of the various parts of the balance in this problem., Draw a Diagram, , Variables, Sign is important with torques: the duck’s torque is in the positive (counterclockwise) direction while the witch’s torque is in the negative, (clockwise) direction. To calculate the magnitude of each torque, we can multiply the force by the lever arm, since the two are perpendicular., , Forces, , x, , y, , weight, duck, , 0N, , ímdg = í44.5 N, , weight, witch, , 0N, , ímwg, , tension, , 0N, , T, , Torques, , lever arm (m), , torque (N·m), , weight, duck, , 1.65, , (1.65)(44.5), , weight, witch, , 0.183, , í(0.183)( mwg), , tension, , 0, , 0, , What is the strategy?, 1., , Draw a free-body diagram that shows the forces on the balance beam., , 2., , Place the axis of rotation at the location of an unknown force (the tension). This simplifies solving the problem. There is no need to, calculate the amount of this force since a force applied at the axis of rotation does not create a torque., , 3., , Use the fact that there is no net torque to solve the problem. The only unknown in this equation is the weight of the witch., , Physics Principles and Equations, There is no net torque since there is no angular acceleration., , ȈIJ = 0, The weights are perpendicular to the beam, so we calculate the torques they create using, , IJ = rF, A force (like tension) applied at the axis of rotation creates no torque., , 206, , Copyright 2000-2007 Kinetic Books Co. Chapter 11
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Step-by-step solution, We use the fact that the torques sum to zero to solve this problem. There are only two torques in the problem due to the location of the axis of, rotation: The tension creates no torque., , Step, , 1., , Reason, , torque of witch + torque of duck = 0, , no net torque, substitute values, , 2., 3., , mwg = (1.65 m)(44.5 N) / (0.183 m), , solve for mwg, , 4., , mwg = 401 N, , evaluate, , 11.3 - Center of gravity, , Center of gravity: The force, of gravity effectively acts at a, single point of an object, called the center of gravity., The concept of center of gravity complements the, concept of center of mass. When working with torque, and equilibrium problems, the concept of center of, gravity is highly useful., , The center of gravity of a barbell., , Consider the barbell shown above. The sphere on the, left is heavier than the one on the right. Because the spheres are not equal in weight, if, you hold the barbell exactly in the center, the force of gravity will create a torque that, causes the barbell to rotate. If you hold it at its center of gravity however, which is closer, to the left ball than to the right, there will be no net torque and no rotation., When a body is symmetric and uniform, you can calculate its center of gravity by, locating its geometric center. Let’s consider the barbell for a moment as three distinct, objects: the two balls and the bar. Because each of the balls on the barbell is a uniform, sphere, the geometric center of each coincides with its center of gravity. Similarly, the, center of gravity of the bar connecting the two spheres is at its midpoint., When we consider the entire barbell, however, the situation gets more complicated. To, calculate the center of gravity of this entire system, you use the equation to the right., This equation applies for any group of masses distributed along a straight line. To apply, the equation, pick any point (typically, at one end of the line) as the origin and measure, the distance to each mass from that point. (With a symmetric, uniform object like a ball,, you measure from the origin to its geometric center.), , Center of gravity, One point where weight effectively acts, Can be found by "dangling" object twice, , Then, multiply each distance by the corresponding weight, add the results, and divide, that sum by the sum of the weights. The result is the distance from the origin you, selected to the center of gravity of the system. The center of gravity of an object does, not have to be within the mass of the object: For example, the center of gravity of a, doughnut is in its hole., If you have studied the center of mass, you may think the two concepts seem, equivalent. They are. When g is constant across an object, its center of mass is the, same as its center of gravity. Unless the object is enormous (or near a black hole where, the force of gravity changes greatly with location), a constant g is a good assumption., You can empirically determine the center of gravity of any object by dangling it. In, Concept 1, you see the center of gravity of a painter’s palette being determined by, dangling. To find the center of gravity of an object using this method, hang (dangle) the, object from a point and allow it to move until it naturally stops and rests in a state of, equilibrium. The center of gravity lies directly below the point where the object is, suspended, so you can draw an imaginary line through the object straight down from, the point of suspension. The object is then dangled again, and you draw another line, down from the suspension point. Since both of these lines go through the center of, gravity, the center of gravity is the point where the lines intersect., , xCG = x position of center of gravity, wi = weight of object i, xi = x position of object i, , Copyright 2000-2007 Kinetic Books Co. Chapter 11, , 207
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Where is the center of gravity?, , Put the origin at center of gold ball:, xCG =, , xCG = (8.4 + 29.9) / 66, xCG = 0.58 m from center of goldcolored ball, 11.4 - Interactive problem: achieve equilibrium, In the simulation on the right, you are asked to apply three forces to a rod so that it, will be in static equilibrium. Two of the forces are given to you and you have to, calculate the magnitude, position, and direction of the third force. If you do this, correctly, when you press GO, the rod will not move., The rod is 2.00 meters long, and is horizontal. A force of 323 N is applied to the left, end, straight up. A force of 627 N is applied to the right end, also straight up. You, are asked to apply a force to the rod that will balance these two forces and keep it in, static equilibrium., Here is a free-body diagram of the situation. We have not drawn the third force, where it should be!, , After you calculate the third vector's magnitude, position and direction, follow these steps to set up the simulation., 1., , Adjust the rod length so it is 2.00 m., , 2., , Drag the axis of rotation to an appropriate position., , 3., , Apply all the forces. Drag a force vector by its tail from the control panel and attach the tail to the rod. You can then move the tail of the, vector along the rod to the correct position, and drag the head of the vector to change its length and angle., , The control panel will show you the force's magnitude, direction and distance to the axis of rotation. The vector whose properties are being, displayed has its head in blue., When you have the simulation set up, press GO. If everything is set up correctly, the rod will be in equilibrium and will not move. Press RESET, if you need to make any adjustments. If you have trouble, refer to the section on static equilibrium in this chapter, and the section on torque in, the Rotational Dynamics chapter., After you solve this interactive problem, consider the following additional challenge. What do you think will happen in the simulation if you, change the position of the axis of rotation? Make a guess, and test your hypothesis with the simulation., , 208, , Copyright 2000-2007 Kinetic Books Co. Chapter 11
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11.5 - Elasticity, , Elastic: An elastic object returns to its original, dimensions when a deforming force is removed., In much of physics, the object being analyzed is assumed to be rigid and its dimensions, are unchanging. For instance, if a homework problems asks: “What is the effect of a net, force on a car?” and you answer, “The net force on the car caused a dent in its fender, and enraged the owner,” then you are thinking a little too much outside the box. The, question anticipates that you will apply Newton’s second law, not someone’s insurance, policy., However, objects do stretch or compress when an external force is applied to them., They may return to their original dimensions when that force is removed: Objects with, this property are called elastic. The external force causes the bonds between the, molecules that make up the material to stretch or compress, and when that force is, removed, the molecules can “spring back” to their initial configuration. The extent to, which the dimensions of an object change in response to a deforming force is a function, of the object’s original dimensions, the material that makes it up, and the nature of the, force that is applied., , Elastic, Shape changes under force, Elastic: when force is removed, original, shape is restored, , It is also possible to stretch or compress an object to the point where it is unable to spring back. When this happens, the object is said to be, deformed., It can require a great deal of force to cause significant stretching. For instance, if you hang a 2000 kg object, like a midsize car, at the end of a, two-meter long steel bar with a radius of 0.1 meters, the bar will stretch only about 6 × 10í6 meters., There are various ways to change the dimensions of an object. For instance, it can be stretched or compressed along a line, like a vertical steel, column supporting an overhead weight. Or, an object might experience compressive forces from all dimensions, like a ball submerged in water., Calculating the change of dimensions is a slightly different exercise in each case., , 11.6 - Stress and strain, , Stress and strain, Stress: force per unit area, Strain: fractional change in dimension, due to stress, Modulus of elasticity: relates stress,, strain for given material, , Stress: External force applied per unit area., Strain: Fractional change in dimension due to, stress., Above, you see a machine that tests the behavior of materials under stress. Stress is, the external force applied per unit area that causes deformation of an object. The, machine above stretches the rod, increasing its length. Although we introduce the topic, of stress and strain by discussing changes in length, stress can alter the dimensions of, objects in other fashions as well., Strain measures the fractional change in an object’s dimensions: the object’s change, in dimension divided by its original dimension. This means that strain is dimensionless., For instance, if a force stretches a two-meter rod by 0.001 meters, the resulting strain is, 0.001 meters divided by two meters, which is 0.0005., , Stress-strain graphs, Show behavior of material, ·proportionality limit, ·elastic limit, ·rupture point, , Stress measures the force applied per unit area. If a rod is being stretched, the area, equals the surface area of the end of the rod, called its cross-sectional area. If the force is being applied over the entirety of an object, such as, a ball submerged in water, then the area is the entire surface area of the ball., For a range of stresses starting at zero, the strain of a material (fractional change in size) is linearly proportional to the stress on it. When the, stress ends, the material will resume its original shape: That is, it is elastic. The modulus of elasticity is a proportionality constant, the ratio of, , Copyright 2000-2007 Kinetic Books Co. Chapter 11, , 209
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stress to strain. It differs by material, and depends upon what kind of stress is applied., Values for modulus of elasticity apply only for a range of stress values. Here, we specifically use changes in length to illustrate this point. If you, stress an object beyond its material’s proportionality limit, the strain ceases to be linearly proportional to stress. A “mild steel” rod will exhibit a, linear relation between stress and strain for a stress on it up to about 230,000,000 N/m2. This is its proportionality limit., Beyond a further point called the elastic limit, the object becomes permanently deformed and will not return to its original shape when the, stress ceases to be applied. You have exceeded its yield strength. At some point, you pull so far it ruptures. At that rupture point, you have, exceeded the material’s ultimate strength., Concept 2 shows a stress-strain graph typical of a material like soft steel (a type of steel that is easily cut or bent) or copper. Graphs like these, are commonly used by engineers. They are created with testing machines similar to the one above by carefully applying force to a material, over time. The graph plots the minimum stress required to achieve a certain amount of strain, which in this case is measured as the, lengthening of a rod of the material. (We say “minimum stress” because the strain may vary depending on whether a given amount of stress is, applied suddenly or is slowly increased to the same value.), You may notice that the graph has an interesting property: Near the end, the curve flattens and its slope decreases. Less stress is required to, generate a certain strain. The material has become ductile, or stretchy like taffy, and it is easier to stretch than it was before. Soft steel and, copper are ductile. Other materials will reach the rupture point without becoming stretchy; they are called brittle. Concrete and glass are two, brittle materials, and hardened steel is more brittle than soft steel., , 11.7 - Tensile stress, , Tensile stress: A stress that stretches., On the right, you see a rod being stretched. Tensile forces cause materials to lengthen., Tensile stress is the force per cross-sectional unit area of the object. Here, the crosssectional area equals the surface area of the end of the rod. The strain is measured as, the rod’s change in length divided by its initial length., Young’s modulus equals tensile stress divided by strain. The letter Y denotes Young’s, modulus, which is measured in units of newtons per square meter. The value for, Young’s modulus for various materials is shown in Concept 2. Note the scale used in, the table: billions of newtons per square meter., To correctly apply the equation in Equation 1 on the right, you must be careful with the, definitions of stress and strain. First, stress is force per unit area. Second, strain is the, fractional change in size. The relevant area for a rod in calculating tensile stress is the, area of its end, as the diagrams on the right reflect., The equation on the right can be used for compression as well as expansion. When a, material is compressed, ǻL is the decrease in length. For some materials, Young’s, modulus is roughly the same for compressing (shortening) as for stretching, so you can, use the same modulus when a compressive force is applied., , Tensile stress, Causes stretching/compression along a, line, Stress: force per cross-section unit area, Strain: fractional change in length, , In addition to supplying the values for Young’s modulus, we supply values for the yield, strength (elastic limit) of a few of the materials. The table can give you a sense of why, certain materials are used in certain settings. Steel, for example, has both a high, Young’s modulus and high yield strength. This means it requires a lot of stress to, stretch steel elastically, as well as a lot of stress to deform it permanently., Some materials have different yield strengths for compression and tension. Bone, for, instance, resists compressive forces better than tensile forces. The value listed in the, table is for compressive forces., , Young's modulus, Relates stress, strain, , 210, , Copyright 2000-2007 Kinetic Books Co. Chapter 11
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F = force, A = cross-sectional area, Y = Young’s modulus (N/m2), ǻL = change in length, L i= initial length, , How much does this aluminum, rod stretch under the force?, , ǻL = 1.5×10í4 m, , Copyright 2000-2007 Kinetic Books Co. Chapter 11, , 211
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11.8 - Volume stress, , Volume stress: A stress that, acts on the entire surface of, an object, changing its, volume., Tensile stress results in a change along a single, dimension of an object. A volume stress is one that, exposes the entire surface of an object to a force. The, force is assumed to be perpendicular to the surface, and uniform at all points., One way to exert volume stress on an object is to, submerge it in a fluid (a liquid or a gas). For example,, submersible craft that visit the wreck of the Titanic, travel four kilometers below the surface, experiencing, a huge amount of volume stress on their hulls., , The effect of volume stress on a Styrofoam cup submerged 7875 feet underwater, (with an un-stressed cup also shown for comparison). The stress on the cup, exceeded its elastic limit: It is permanently deformed., , In all cases, stress is force per unit area, which is also the definition of pressure. With, volume stresses, the term pressure is used explicitly, since the pressure of fluids is a, commonly measured property., Volume strain is measured as a fractional change in the volume of an object. The, modulus of elasticity that relates volume stress and strain is called the bulk modulus,, and is represented with the letter B., The equation in Equation 1 states that the change in pressure equals the bulk modulus, times the strain. The negative sign means that an increase in pressure results in a, decrease in volume. Unlike the equation for tensile stress, this equation does not have, an explicit term for area, because the pressure term already takes this factor into, account. Notice that the equation is stated in terms of the change in pressure., At the right is a table that lists values for the bulk modulus for some materials. To give, you a sense of the deformation, the increase in pressure at 100 meters depth of water,, as compared to the surface, is about 1.0×106 N/m2. The volume of water will be, reduced by 0.043% at this depth; steel, only 0.00063%., , Volume stress, Pressure of fluids alters volume, Stress: pressure (force per unit area), Strain: fractional change in volume, , At a depth of 11 km, approximately the maximum depth of the Earth’s oceans, the, increased pressure is 1.1×108 N/m2. At this depth, a steel ball with a radius of 1.0 meter, will compress to a radius of 0.997 m., , Bulk modulus, Relates stress, strain, , 212, , Copyright 2000-2007 Kinetic Books Co. Chapter 11
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P = pressure (force per unit area), B = bulk modulus (N/m2), V = volume, Pressure units: pascals (Pa = N/m2), , The increase in water pressure is, 3.2×103 Pa. The balloon’s initial, volume was 0.50 m3. What is it, now?, , ǻV = í0.011 m3, Vf = Vi + ǻV = 0.50 í 0.011 m3, Vf = 0.49 m3, , Copyright 2000-2007 Kinetic Books Co. Chapter 11, , 213
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11.9 - Interactive checkpoint: stress and strain, A 3.50 meter-long rod, composed of a, titanium alloy, has a cross-sectional, area of 4.00×10í4 m2. It increases in, length by 0.0164 m under a force of, 2.00×105 N. What is the stress on the, rod? What is the strain? What is, Young's modulus for this titanium, alloy?, , Answer:, , stress =, , N/m2, , strain =, Y=, , N/m2, , 11.10 - Gotchas, Stress equals force. No, stress is always a measure of force per unit surface area., A force stretches a rod by 0.01 meters, so the strain is 0.01 meters. No, strain is always the fractional change (a dimensionless ratio). To, calculate the strain, you need to divide this change in length by the initial length of the rod. This is not stated here, so you cannot determine the, strain without more information., , 11.11 - Summary, An object is in equilibrium when there are no net forces or torques on it. Static, equilibrium is a special case where there is also no motion., , Static equilibrium, The center of gravity is the average location of the weight of an object. The force of, gravity effectively acts on the object at this point. The concept of center of gravity is, very similar to the concept of center of mass. The two locations differ only when an, object is so large that the pull of gravity varies across it., Elasticity refers to an object's shape changing when forces are applied to it. Objects, are called elastic when they return to their original shape as forces are removed., Related to elasticity are stress and strain. Stress is the force applied to an object per, unit area. The area used to determine stress depends on how the force is applied., Strain is the fractional change in an object's dimensions due to stress., , ȈFx = 0, ȈFy = 0, ȈIJ = 0, Tensile stress, , Volume stress, , The relationship between stress and strain is determined by a material's modulus of, elasticity up to the proportionality limit. An object will become permanently deformed, if it is stressed past its elastic limit. It will finally break at its rupture point. Ductile, materials are easily deformed, while brittle materials tend to break rather than, stretch., Tensile stress is the application of stress causing stretching or compression along a line. Strain under tensile stress is measured as a fractional, change in length and stress is measured as the force per unit cross-sectional area. Young's modulus is the tensile stress on a material divided, by its strain., Volume stress acts over the entire surface of an object. The amount of volume stress is calculated as the force per unit surface area, while the, strain is the fractional change in volume. The bulk modulus relates volume stress and strain., , 214, , Copyright 2000-2007 Kinetic Books Co. Chapter 11
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Chapter 11 Problems, , Conceptual Problems, C.1, , The sketch shows a rod divided into five equal, parts. The rod has negligible mass and a fixed, pivot at point c. An upward force of magnitude, F is applied at point a, and an identical force is, applied at point f, as shown. At what point on, the rod could you apply a third force of the, same magnitude F, perpendicular to the rod,, either upward or downward, so that the net, torque on the rod is zero? Check all of the, points for which this is possible., a, b, c, d, e, f, , C.2, , Use principles of equilibrium to describe how, the gymnast is able to hold his body in this, position., , C.3, , The Space Needle in Seattle, Washington is over 500 feet tall, but the center of gravity of the complete structure is only 5 feet, off the ground. This is achieved with a huge concrete foundation that weighs more than the Space Needle itself. What is the, advantage of having a center of gravity so low?, , C.4, , (a) Which point is the proportionality limit? (b), Which point is the rupture point? (c) Which, point is the elastic limit?, (a), , (b), , (c), , i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., , Point 1, Point 2, Point 3, Point 4, Point 5, Point 6, Point 1, Point 2, Point 3, Point 4, Point 5, Point 6, Point 1, Point 2, Point 3, Point 4, Point 5, Point 6, , Copyright 2007 Kinetic Books Co. Chapter 11 Problems, , 215
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C.5, , Suppose you have four rods made of four different materials-aluminum, copper, steel and titanium. They all have the same, dimensions, and the same force is applied to each. Order the rods from smallest to largest by how much each stretches., i., ii., iii., iv., , Aluminum, Copper, Steel, Titanium, , ,, , i., ii., iii., iv., , Aluminum, Copper, Steel, Titanium, , ,, , i., ii., iii., iv., , Aluminum, Copper, Steel, Titanium, , ,, , i., ii., iii., iv., , Aluminum, Copper, Steel, Titanium, , Section Problems, Section 1 - Static equilibrium, 1.1, , Two children sit 2.60 m apart on a very low-mass horizontal seesaw with a movable fulcrum. The child on the left has a mass, of 29.0 kg, and the child on the right has a mass of 38.0 kg. At what distance, as measured from the child on the left, must the, fulcrum be placed in order for them to balance?, m, , 1.2, , A 3.6-meter long horizontal plank is held up by two supports. One support is at the left end, and the other is 0.80 m from the, right end. The plank has uniform density and has a mass of 40 kg. How close can a 70 kg person stand to the unsupported, end before causing the plank to rotate?, m, , 1.3, , 1.4, , The mass of the sign shown is 28.5 kg. Find, the weight supported by (a) the left support, and (b) the right support., (a), , N, , (b), , N, , A horizontal log just barely spans a river, and its ends rest on opposite banks. The log is uniform and weighs 2400 N. If a, person who weighs 840 N stands one fourth of the way across the log from the left end, how much weight does the bank, under the right end of the log support?, N, , 1.5, , A person who weighs 620 N stands at x = 5.00 m, right on the end of a long horizontal diving board that weighs 350 N. The, diving board is held up by two supports, one at its left end at x = 0, and one at the point x = 2.00 m. (a) What is the force, exerted on the support at x = 0? (b) What is the force acting on the other support? (Use positive to indicate an upward force,, negative for a downward force.), (a), , N, , (b), , N, , 1.6, , Two brothers, Jimmy and Robbie, sit 3.00 m apart on a horizontal seesaw with its fulcrum exactly midway between them., Jimmy sits on the left side, and his mass is 42.5 kg. Robbie's mass is 36.5 kg. Their sister Betty sits at the exact point on the, seesaw so that the entire system is balanced. If Betty is 29.8 kg, at what location should she sit? Take the fulcrum to be the, origin, and right to be positive. Assume that the mass of the seesaw is negligible., , 1.7, , If a cargo plane is improperly loaded, the plane can tip up onto its tail while it rests on the runway (this has actually, , m, happened). Suppose a plane is 45.0 m long, and weighs 1.20×106 N. The center of mass of the plane is located 21.0 m from, the nose. The nose wheel located 3.50 m from the nose and the main wheels are 25.0 m from the nose. Cargo is loaded into, the back end of the plane. If the center of mass of the cargo is located 40.0 m from the nose of the plane, what is the, maximum weight of the cargo that can be put in the plane without tipping it over (that is, so that the plane remains horizontal)?, N, , 216, , Copyright 2007 Kinetic Books Co. Chapter 11 Problems
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1.8, , 1.9, , The sketch shows a mobile in equilibrium., Each of the rods is 0.16 m long, and each, hangs from a supporting string that is attached, one fourth of the way across it. The mass of, each rod is 0.10 kg. The mass of the strings, connecting the blocks to the rods is negligible., What is the mass of (a) block A? (b) block B?, (a), , kg, , (b), , kg, , Two identical 1.80 m long boards just barely, balance on the edge of a table, as shown in, the figure. What is the distance x?, m, , Section 3 - Center of gravity, 3.1, , A weightlifter has been given a barbell to lift. One end has a mass of 5.5 kg while the other end has a mass of 4.7 kg. The bar, is 0.20 m long. (Consider the bar to be massless, and assume that the masses are thin disks, so that their centers of mass, are at the ends of the bar.) How far from the heavier end should she hold the bar so that the weight feels balanced?, m, , 3.2, , A 0.65 m rod with uniform mass distribution runs along the x axis with its left end at the origin. A 1.8 m rod with uniform mass, distribution runs along the y axis with its top end at the origin. Find the coordinates of the center of gravity for this system., (, , 3.3, , 3.4, , ,, , )m, , A length of uniform wire is cut and bent into, the shapes shown. Find the location of the, center of gravity of each shape. In each, instance, consider the corners of the shape to, be located at integer coordinates., (a) (, , ,, , (b) (, , ,, , ), ), , (c) (, , ,, , ), , (d) (, , ,, , ), , Three beetles stand on a grid. Two beetles have the same weight, W, and the third beetle weighs 2W. (a) The lighter beetles, are located at (1.00, 0) and (0, 2.00), and the heavier beetle is at (3.00, 1.00). Find the coordinates of the center of gravity of, the beetles. (b) If the heavy beetle moves to (1.00, 1.00), what is the new location of the center of gravity?, (a) (, , ,, , ), , (b) (, , ,, , ), , 3.5, , A woman with weight 637 N lies on a bed of nails. The bed has a weight of 735 N and a length of 1.72 m. The bed is held up, by two supports, one at the head and one at the foot. Underneath each support is a scale. When the woman lies in the bed,, the scale at the foot reads 712 N. How far is the center of gravity of the system from the foot of the bed of nails?, , 3.6, , (a) An empty delivery truck weighs 5.20×105 N. Of this weight, 3.20×105 N is on the front wheels. The distance between the, front axle and the back axle is 4.10 m. How far is the center of gravity of the truck from the front wheels? (b) Now the delivery, , m, , truck is loaded with a 2.40×105 N shipment, 2.60 m from the front wheels. Now how far is the center of gravity from the front, wheels?, , 3.7, , (a), , m, , (b), , m, , A skateboarder stands on a skateboard so that 62% of her weight is located on the front wheels. If the distance between the, , Copyright 2007 Kinetic Books Co. Chapter 11 Problems, , 217
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wheels is 0.85 m, how far is her center of gravity from the front wheels? Ignore the weight of this rather long skateboard., m, 3.8, , The figure shows a uniform disk with a circular, hole cut out of it. The disk has a radius of, 4.00 m. The hole's center is at 2.00 m, and its, radius is 1.00 m. What is the x-coordinate of, the center of gravity of the system?, m, , Section 4 - Interactive problem: achieve equilibrium, 4.1, , Use the data and the diagram in the interactive problem in this section to answer the following question. If the pivot is placed, at the left end of the rod, what is (a) the magnitude and (b) direction of the unknown force? (c) How far should it be placed, from the pivot? Use the simulation to test your answer., (a), (b), , (c), , N, i., ii., iii., iv., , Positive x, Negative x, Positive y, Negative y, m, , Section 6 - Stress and strain, 6.1, , (a) Suzy the elephant weighs 65,000 N. The cross-sectional area of each foot is 0.10 m2. When she stands on all fours, what, is the average stress on her feet? (b) Suzanne the stockbroker weighs 610 N. She wears a pair of high-heeled shoes whose, heels each have a cross-sectional area of 1.5×10í4 m2 in contact with the floor. If she stands with all her weight on her heels,, what is the stress on the heels of the shoes?, (a), , N/m2, , (b), , N/m2, , Section 7 - Tensile stress, 7.1, , A 85.0 kg window washer hangs down the side of a building from a rope with a cross-sectional area of 4.00×10í4 m2. If the, rope stretches 0.740 cm when it is let out 7.50 m, how much will it stretch when it is let out 26.0 m?, m, , 7.2, , A building with a weight of 4.10e+7 N is built on a concrete foundation with an area of 1300 m2 and an (uncompressed), height of 2.82 m. By what vertical distance does the foundation compress?, , 7.3, , Many gyms have an exercise machine called a vertical leg press. To use this machine, you lie on your back and press a, weight upward with your feet until your legs are perpendicular to the ground. Find the amount the femur bone compresses, when a person with a 0.385 m femur lifts 525 N with a vertical leg press, using one leg, until the leg is fully extended straight, , m, , up. Assume that the cross sectional area of the femur is 5.30×10í4 m2., m, 7.4, , (a) An engineer is designing a bridge with four supports made of cylinders of concrete. The cylinders are 3.5 m tall, and they, cannot compress by more than 1.6×10–4 m. If the bridge must be able to hold 7.5×106 N, what should be the radius of the, cylinders? (b) Suppose the restriction on the amount of compression is removed. If the yield strength of the concrete is, , 5.7×106 N/m2, what is the smallest radius that will prevent the concrete from deforming under the weight?, , 7.5, , (a), , m, , (b), , m, , A climber with a mass of 75 kg is attached to a 6.5 m rope with a cross sectional area of 3.5×10í4 m2. When the climber, hangs from the rope, it stretches 3.4×10í3 m. What is the Young's modulus of the rope?, N/m2, , 7.6, , A helicopter is working to bring water to a forest fire. A steel cable hangs from the helicopter with a large bucket on the end., The cable is 8.5 m long and has a cross-sectional area of 4.0×10í4 m2. Together, the bucket and the water inside it weigh, , 218, , Copyright 2007 Kinetic Books Co. Chapter 11 Problems
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3.0×104 N. (a) How much will the cable stretch when the helicopter is hovering? (b) If the yield strength of the steel in the, cable is 2.1×108 N/m2, at what rate can the helicopter accelerate upward before the cable deforms permanently?, , 7.7, , (a), , m, , (b), , m/s2, , Two rods with identical dimensions are placed end to end to form a new rod. If one of the smaller rods is aluminum alloy and, the other is titanium, what is the effective Young's modulus of the large rod?, N/m2, , Section 8 - Volume stress, 8.1, , The deepest point in the seven seas is the Marianas Trench in the Pacific Ocean. The pressure in the deepest parts of the, Marianas Trench is, 1.1×108 Pa. Pressure at the surface of the ocean is 1.0×105 Pa. If a mass of salt water has a volume of 1.6 m3 at the surface, of the ocean, what will be its volume at the bottom of the Marianas Trench?, m3, , 8.2, , If a spherical glass marble has a radius of 0.00656 m at 1.02×105 Pa, at what pressure will it have a radius of 0.00650 m?, Pa, , 8.3, , The pressure on the surface of Venus is about 9.0×106 Pa, and the pressure on the surface of Earth is 1.0×105 Pa. What, would be the volume strain on a solid stainless steel sphere if it were moved from Earth to Venus?, , 8.4, , A sealed, expandable plastic bag is filled with equal volumes of saltwater and mercury. The volume of the bag is 0.120 m3 at, a pressure of 1.01×105 N/m2. What is the change in volume of the bag when the pressure increases to 1.09×105 N/m2?, Assume that the bag exerts no force., m3, , Copyright 2007 Kinetic Books Co. Chapter 11 Problems, , 219
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12.0 - Introduction, The topic of gravity has had a starring role in some of the most famous tales in the, history of physics. Galileo Galilei was studying the acceleration due to the Earth’s, gravity when he dropped two balls from the Leaning Tower of Pisa. A theory to, explain the force of gravity came to Isaac Newton shortly after an apple fell from a, tree and knocked him in the head. Although historians doubt whether these events, actually occurred, the stories have come to symbolize how a simple experiment or a, sudden moment of insight can lead to important and lasting scientific progress., Galileo is often said to have dropped two balls of different masses from the tower so, that he could see if they would land at the same time. Most scientists of his era, would have predicted that the heavier ball would hit the ground first. Instead, the, balls landed at the same instant, showing that the acceleration due to gravity is a, constant for differing masses. While it is doubtful that Galileo actually dropped balls, off that particular tower, his writings show that he performed many experiments, studying the acceleration due to gravity., In another famous story, Newton formed his theory of gravity after an apple fell and, hit him on the head. At least one person (the daughter of French philosopher, Voltaire) said that Newton mentioned that watching a falling apple helped him to, comprehend gravity. Falling apple or no, his theory was not the result of a, momentary insight; Newton pondered the topic of gravity for decades, relying on the, observations of contemporary astronomers to inform his thinking. Still, the image of, a scientist deriving a powerful scientific theory from a simple physical event has, intrigued people for generations., The interactive simulations on the right will help you begin your study of gravity., Interactive 1 permits you to experiment with the gravitational forces between, objects. In its control panel, you will see five point masses. There are three identical, red masses of mass m. The green mass is twice as massive as the red, and the, blue mass is four times as massive., You can start your experimentation by dragging two masses onto the screen. The, purple vectors represent the gravitational forces between them. You can move a, mass around the screen and see how the gravitational forces change. What is the, relationship between the magnitude of the forces and the distances between the, masses?, Experiment with different masses. Drag out a red mass and a green mass. Do you expect the force between these two masses to be smaller or, larger than it would be between two red masses situated the same distance apart? Make a prediction and test it. You can also drag three or, more masses onto the screen to see the gravitational force vectors between multiple bodies. There is yet more to do: You can also press GO, and see how the gravitational forces cause the masses to move., The other major topic of this chapter is orbital motion. The force of gravity keeps bodies in orbit. Interactive 2 is a reproduction of the inner part, of our solar system. It shows the Sun, fixed at the center of the screen, along with the four planets closest to it: Mercury, Venus, Earth and, Mars., Press GO to watch the planets orbit about the Sun. You can experiment with our solar system by changing the position of the planets prior to, pressing GO, or by altering the Sun’s mass as the planets orbit., In the initial configuration of this system, the period of each planet’s orbit (the time it takes to complete one revolution around the Sun) is, proportional to its actual orbital period. Throughout this chapter, as in this simulation, we will often not draw diagrams to scale, and will speed, up time. If we drew diagrams to scale, many of the bodies would be so small you could not see them, and taking 365 days to show the Earth, completing one revolution would be asking a bit much of you., , 220, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.1 - Newton’s law of gravitation, , Newton’s law of gravitation: The attractive, force of gravity between two particles is, proportional to the product of their masses and, inversely proportional to the square of the, distance between them., Newton’s law of gravitation states that there is a force between every pair of particles, of, any mass, in the universe. This force is called the gravitational force, and it causes, objects to attract one another. The force does not require direct contact. The Earth, attracts the Sun, and the Sun attracts the Earth, yet 1.5×1011 meters of distance, separate the two., The strength of the gravitational force increases with the masses of the objects, and, weakens proportionally to the square of the distance between them. Two masses exert, equal but opposite attractive forces on each other. The forces act on a line between the, two objects. The magnitude of the force is calculated using the equation on the right., The symbol G in the equation is the gravitational constant., , Newton’s law of gravitation, Gravitational force, ·Proportional to masses of bodies, ·Inversely proportional to square of, distance, , It took Newton about 20 years and some false starts before he arrived at the, relationship between force, distance and mass. Later scientists established the value for, G, which equals 6.674 2×10í11 N·m2/kg2. This small value means that a large amount, of mass is required to exert a significant gravitational force., The example problems on the right provide some sense of the magnitude of the, gravitational force. First, we calculate the gravitational force the Earth exerts on the, Moon (and the Moon exerts on the Earth). Although separated by a vast distance (on, average, their midpoints are separated by about 384,000,000 meters), the Earth and, the Moon are massive enough that the force between them is enormous: 1.98×1020 N., In the second example problem, we calculate the gravitational force between an 1100kg car and a 2200-kg truck parked 15 meters apart. The force is 0.00000072 N. When, you press a button on a telephone, you press with a force of about one newton,, 1,400,000 times greater than this force., , F = force of gravity, M, m = masses of objects, r = distance between objects, G = gravitational constant, G = 6.67×10í11 N·m2/kg2, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 221
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What is the gravitational force, between the Earth and the, Moon?, , F = 1.98×1020 N, Each body attracts the other, , How much gravitational force do, the car and the truck exert on, each other?, , F = 7.2×10í7 N, , 222, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.2 - G and g, Newton’s law of gravity includes the gravitational constant G., In this section, we discuss the relationship between G and g, the rate of freefall, acceleration in a vacuum near the Earth’s surface., The value of g used in this textbook is 9.80 m/s2, an average value that varies slightly, by location on the Earth. Both a 10-kg object and a 100-kg object will accelerate toward, the ground at 9.80 m/s2. The rate of freefall acceleration does not vary with mass., Newton’s law of gravitation, however, states that the Earth exerts a stronger, gravitational force on the more massive object. If the force on the more massive object, is greater, why does gravity cause both objects to accelerate at the same rate?, The answer becomes clear when Newton’s second law of motion, F = ma, is applied., The acceleration of an object is proportional to the force acting on it, divided by the, object’s mass. The Earth exerts ten times the force on the 100-kg object that it does on, the 10-kg object. But that tenfold greater force is acting on an object with a mass ten, times greater, meaning the object has ten times more resistance to acceleration. The, result is that the mass term cancels out and both objects accelerate toward the center, of the Earth at the same rate, g., , G and g, G = gravitational constant everywhere in, universe, g = freefall acceleration at Earth’s, surface, , If the gravitational constant G and the Earth’s mass and radius are known, then the, acceleration g of an object at the Earth’s surface can be calculated. We show this, calculation in the following steps. The distance r used below is the average distance, from the surface to the center of the Earth, that is, the Earth’s average radius. We treat, the Earth as a particle, acting as though all of its mass is at its center., Variables, gravitational constant, , G = 6.67×10í11 N·m2/kg2, , mass of the Earth, , M = 5.97×1024 kg, , mass of object, , m, , Earth-object distance, , r = 6.38×106 m, , acceleration of object, , g, , Strategy, 1., , Set the expressions for force from Newton’s second law and his law of gravitation, equal to each other., , 2., , Solve for the acceleration of the object, and evaluate it using known values for, the other quantities in the equation., , Physics principles and equations, , g = free fall acceleration at Earth’s, surface, G = gravitational constant, M = mass of Earth, , We will use Newton’s second law of motion and his law of gravitation. In this case, the, acceleration is g., , r = distance to center of Earth, , Step-by-step derivation, Here we use two of Newton’s laws, his second law (F = ma) and his law of gravitation (F = GMm/r 2). We use g for the acceleration instead, of a, because they are equal. We set the right sides of the two equations equal and solve for g., , Step, , Reason, , 1., , Newton’s laws, , 2., , simplify, , 3., , substitute known values, , 4., , g = 9.78 m/s2, , evaluate, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 223
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Our calculations show that g equals 9.78 m/s2. The value for g varies by location on the Earth for reasons you will learn about later., In the steps above, the value for G, the gravitational constant, is used to calculate g, with the mass of the Earth given in the problem. However,, if g and G are both known, then the mass of the Earth can be calculated, a calculation performed by the English physicist Henry Cavendish in, the late 18th century., Physicists believe G is the same everywhere in the universe, and that it has not changed since the Big Bang some 13 billion years ago. There, is a caveat to this statement: some research indicates the value of G may change when objects are extremely close to each other., , 12.3 - Shell theorem, , Shell theorem: The force of gravity outside a, sphere can be calculated by treating the sphere’s, mass as if it were concentrated at the center., Newton’s law of gravitation requires that the distance between two particles be known in, order to calculate the force of gravity between them. But applying this to large bodies, such as planets may seem quite daunting. How can we calculate the force between the, Earth and the Moon? Do we have to determine the forces between all the particles that, compose the Earth and the Moon in order to find the overall gravitational force between, them?, Fortunately, there is an easier way. Newton showed that we can assume the mass of, each body is concentrated at its center., Newton proved mathematically that a uniform sphere attracts an object outside the, sphere as though all of its mass were concentrated at a point at the sphere’s center., Scientists call this the shell theorem. (The word “shell” refers to thin shells that together, make up the sphere and which are used to mathematically prove the theorem.), , The shell theorem, Consider sphere’s mass to be, concentrated at center, ·r is distance between centers of, spheres, , Consider the groundhog on the Earth’s surface shown to the right. Because the Earth is, approximately spherical and the matter that makes up the planet is distributed in a, spherically symmetrical fashion, the shell theorem can be applied to it. To use Newton’s, law of gravitation, three values are required: the masses of two objects and the distance, between them. The mass of the groundhog is 5.00 kg, and the Earth’s mass is, 5.97×1024 kg. The distance between the groundhog and the center of the Earth is the, Earth’s radius, which averages 6.38×106 meters., In the example problem to the right, we use Newton’s law of gravitation to calculate the, gravitational force exerted on the groundhog by the Earth. The force equals the, groundhog’s weight (mg), as it should., , How much gravitational force, does the Earth exert on the, groundhog?, , F = 48.9 N, , 224, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.4 - Shell theorem: inside the sphere, Another section discussed how to calculate the force of gravity exerted on an object on, the surface of a sphere (a groundhog on the Earth). Now imagine a groundhog, burrowing halfway to the center of the Earth, as shown to the right. For the purposes of, calculating the force of gravity, what is the distance between the groundhog and the, Earth? And what mass should be used for the Earth in the equation?, The first question is easier: The distance used to calculate gravity’s force remains the, distance between the groundhog and the Earth’s center. Determining what mass to use, is trickier. We use the sphere defined by the groundhog’s position, as shown to the, right. The mass inside this new sphere, and the mass of the groundhog, are used to, calculate the gravitational force. (Again, we assume that the Earth’s mass is, symmetrically distributed.), If the groundhog is 10 meters from the Earth’s center, the mass enclosed in a sphere, with a radius of 10 meters is used in Newton’s equation. If the animal moves to the, center of the Earth, then the radius of the sphere is zero. At the center, no mass is, enclosed, meaning there is no net force of gravity. The groundhog is perhaps feeling a, little claustrophobic and warm, but is effectively weightless at the Earth’s center., , Inside a sphere, To calculate gravitational force, ·Use mass inside the new shell, ·r is distance between object, sphere’s, center, , The volume of a sphere is proportional to the cube of the radius, as the equation to the, right shows. If the groundhog burrows halfway to the center of the Earth, then the, sphere encloses one-eighth the volume of the Earth and one-eighth the Earth’s mass., Let’s place the groundhog at the Earth’s center and have him burrow back to the, planet’s surface. The gravitational force on him increases linearly as he moves back out, to the Earth’s surface. Why? The force increases proportionally to the mass enclosed, by the sphere, which means it increases as the cube of his distance from the center., But the force also decreases as the square of the distance. When the cube of a quantity, is divided by its square, the result is a linear relationship., If the groundhog moves back to the Earth’s surface and then somehow moves above, the surface (perhaps he boards a plane and flies to an altitude of 10,000 meters), the, force again is inversely proportional to the square of the groundhog’s distance from the, Earth’s center. Since the mass of the sphere defined by his position no longer varies,, the force is computed using the full mass of the Earth, the mass of the groundhog, and, the distance between their centers., , Volume of a sphere, , V = volume, r = radius, 12.5 - Sample problem: gravitational force inside the Earth, What is the gravitational force on the, groundhog after it has burrowed, halfway to the center of the Earth?, , Assume the Earth’s density is uniform. The Earth’s mass and radius are given in the table of variables below., Variables, gravitational force, , F, , mass of Earth, , ME = 5.97×1024 kg, , radius of Earth, , rE = 6.38×106 m, , mass of inner sphere, , Ms, , distance between groundhog and Earth’s, center, , rs= 3.19×106 m, , mass of groundhog, , m = 5.00 kg, , gravitational constant, , G = 6.67×10í11 N·m2/kg2, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 225
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What is the strategy?, 1., , Use the shell theorem. Compute the ratio of the volume of the Earth to the volume of the sphere defined by the groundhog’s current, position. Use this comparison to calculate the mass of the inside sphere., , 2., , Use the mass of the inside sphere, the mass of the groundhog and the distance between the groundhog and the center of the Earth to, calculate the gravitational force., , Physics principles and equations, Newton’s law of gravitation, , Mathematics principles, The equation for the volume of a sphere is, , Step-by-step solution, First, we calculate the mass enclosed by the sphere., , Step, , Reason, , 1., , ratio of masses proportional to ratio of volumes, , 2., , rearrange, , 3., , volume of a sphere, , 4., , cancel common factors, , 5., , enter values, , 6., , Ms = 7.46×1023 kg, , evaluate, , Now that we know the mass of the inner sphere, the shell theorem states that we can use it to calculate the gravitational force on the, groundhog using Newton’s law of gravitation., , Step, , Reason, , 7., , law of gravitation, , 8., , enter values, , 9., , F = 24.4 N, , solve for the force, , This calculation also confirms a point made previously: Inside the Earth, the groundhog’s weight increases linearly with distance from the, planet’s center. At half the distance from the center, his weight is half his weight at the surface., , 226, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.6 - Earth’s composition and g, The force of gravity differs slightly at different locations on the Earth. These variations, mean that g, the acceleration due to the force of gravity, also differs by location. Why do, the force of gravity and g vary?, First, the surface of the Earth can be slightly below sea level (in Death Valley, for, example), and it can be more than 8000 meters above it (on peaks such as Everest and, K2). Compared to an object at sea level, an object at the summit of Everest is 0.14%, farther from the center of the Earth. This greater distance to the Earth’s center means, less force (Newton’s law of gravitation), which in turn reduces acceleration (Newton’s, second law). If you summit Mt. Everest and jump with joy, the force of gravity will, accelerate you toward the ground about 0.03 m/s2 slower than if you were jumping at, sea level., Second, the Earth has a paunch of sorts: It bulges at the equator and slims down at the, poles, making its radius at the equator about 21 km greater than at the poles. This is, shown in an exaggerated form in the illustration for Concept 3. The bulge is caused by, the Earth’s rotation and the fact that it is not entirely rigid. This bulge means that at the, equator, an object is farther from the Earth’s center than it would be at the poles and,, again, greater distance means less force and less acceleration., , Value of g, about 9.80 m/s2 at sea level, , Finally, the density of the planet also varies. The Earth consists of a jumble of rocks,, minerals, metals and water. It is denser in some regions than in others. The presence of, materials such as iron that are denser than the average increases the local gravitational, force by a slight amount. Geologists use gravity gradiometers to measure the Earth’s, density. Variations in the density can be used to prospect for oil or to analyze seismic, faults., , Value for g varies:, Due to altitude, , Value for g varies:, Because the Earth is not a perfect, sphere, , Value for g varies:, Because the Earth is not uniformly, dense, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 227
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12.7 - Newton’s cannon, In addition to noting that the Earth exerts a force on an apple, Newton also pondered, why the Moon circles the Earth. He posed a fundamental question: Is the force the, Earth exerts on the Moon the same type of force that it exerts on an apple? He, answered yes, and his correct answer would forever change humanity’s understanding, of the universe., Comparing the orbit of the distant Moon to the fall of a nearby apple required great, intellectual courage. Although the motion of the Moon overhead and the fall of the apple, may not seem to resemble one another, Newton concluded that the same force dictates, the motion of both, leading him to propose new ways to think about the Moon’s orbit., To explain orbital motion, Newton conducted a thought experiment: What would happen, if you used a very powerful cannon to fire a stone from the top of a very tall mountain?, He knew the stone would obey the basic precepts of projectile motion, as shown in the, diagrams to the right., But, if the stone were fired fast enough, could it just keep going, never touching the, ground? (Factors such as air resistance, the Earth’s rotation, and other mountains that, might block the stone are ignored in Newton’s thought experiment.), Newton concluded that the stone would not return to the Earth if fired fast enough. As, he wrote in his work, Principia, published in 1686:, , Newton’s cannon, Newton imagined a powerful cannon, The faster the projectile, the farther it, travels, At ~8,000 m/s, projectile never touches, ground, , “ ... the greater the velocity with which [a stone] is projected, the farther it goes, before it falls to the earth. We may therefore suppose the velocity to be so, increased, that it would describe an arc of 1, 2, 5, 10, 100, 1000 miles before it, arrived at earth, till at last, exceeding the limits of the earth, it should pass into, space without touching.”, Newton correctly theorized that objects in orbit í moons, planets and, today, artificial, satellites í are in effect projectiles that are falling around a central body but moving fast, enough that they never strike the ground. He could use his theory of gravity and his, knowledge of circular motion to explain orbits. (In this section, we focus exclusively on, circular orbits, although orbits can be elliptical, as well.), Why is it that the stone does not return to the Earth when it is fired fast enough? Why, can it remain in orbit, forever circling the Earth, as shown to the right?, First, consider what happens when a cannon fires a cannonball horizontally from a, mountain at a relatively slow speed, say 100 m/s. In the vertical direction, the, cannonball accelerates at g toward the ground. In the horizontal direction, the ball, continues to move at 100 m/s until it hits the ground. The force of gravity pulls the ball, down, but there is neither a force nor a change in speed in the horizontal direction, (assuming no air resistance)., , Close-up of Newton’s cannon, At high speeds, Earth’s curvature affects, whether projectile lands, At ~8,000 m/s, ground curves away at, same rate that object falls, , Now imagine that the cannonball is fired much faster. If the Earth were flat, at some, point the ball would collide with the ground. But the Earth is a sphere. Its approximate, curvature is such that it loses five meters for every 8000 horizontal meters, as shown in, Concept 2. At the proper horizontal (or more properly, tangential) velocity, the, cannonball moves in an endless circle around the planet. For every 8000 meters it, moves forward, it falls 5 meters due to gravity, resulting in a circle that wraps around the, globe., In this way, satellites in orbit actually are falling around the Earth. The reason, astronauts in a space shuttle orbiting close to the Earth can float about the cabin is not, because gravity is no longer acting on them (the Earth exerts a force of gravity on, them), but rather because they are projectiles in freefall., , 228, , Objects in orbit, Move fast enough to never hit the, ground, Continually fall toward the ground,, pulled by force of gravity, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.8 - Interactive problem: Newton’s cannon, Imagine that you are Isaac Newton. It is the year 1680, and you are staring up at the, heavens. You see the Moon passing overhead., You think: Perhaps the motion of the Moon is related to the motion of Earth-bound, objects, such as projectiles. Suppose you threw a stone very, very fast. Is there a, speed at which the stone, instead of falling back to the Earth, would instead circle, the planet, passing around it in orbit like the Moon?, You devise an experiment in your head, a type of experiment called a thought, experiment. A thought experiment is a way physicists can test or explain valuable, concepts even though they cannot actually perform the experiment. You ask: “What, if I had an extremely powerful cannon mounted atop a mountain. Could I fire a stone, so fast it would never hit the ground?”, Try Newton’s cannon in the simulation to the right. You control the initial speed of, the cannonball by clicking the up and down buttons in the control panel. The cannon, fires horizontally, tangent to the surface of the Earth. See if you can put the stone, into orbit around the Earth. You can create a nearly perfect circular orbit, as well as, orbits that are elliptical., (In this simulation we ignore the rotation of the Earth, as Newton did in his thought experiment. When an actual satellite is launched, it is fired in, the same direction as the Earth’s rotation to take advantage of the tangential velocity provided by the spinning Earth.), , 12.9 - Circular orbits, The Moon orbits the Earth, the Earth orbits the Sun, and today artificial satellites are, propelled into space and orbit above the Earth’s surface. (We will use the term satellite, for any body that orbits another body.), These satellites move at great speeds. The Earth’s orbital speed around the Sun, averages about 30,000 m/s (that is about 67,000 miles per hour!) A communications, satellite in circular orbit 250 km above the surface of the Earth moves at 7800 m/s., In this section, we focus on circular orbits. Most planets orbit the Sun in roughly circular, paths, and artificial satellites typically travel in circular orbits around the Earth as well., The force of gravity is the centripetal force that along with a tangential velocity keeps, the body moving in a circle. We use an equation for centripetal force on the right to, derive the relationship between the mass of the body being orbited, orbital radius, and, satellite speed. As shown in Equation 1, we first set the centripetal force equal to the, gravitational force and then we solve for speed., This equation has an interesting implication: The mass of the satellite has no effect on, its orbital speed. The speed of an object in a circular orbit around a body with mass M, is determined solely by the orbital radius, since M and G are constant. Satellite speed, and radius are linked in circular orbits. A satellite cannot increase or decrease its speed, and stay in the same circular orbit. A change in speed must result in a change in orbital, radius, and vice versa., , Circular orbits, Satellites in circular orbit have constant, speed, , At the same orbital radius, the speed of a satellite increases with the square root of the, mass of the body being orbited. A satellite in a circular orbit around Jupiter would have, to move much faster than it would if it were in an orbit of the same radius around the, Earth., , Orbital speed and radius, Satellite speed and radius are linked, ·The smaller the orbit, the greater the, speed, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 229
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Speed in circular orbit, , m = mass of satellite, v = satellite speed, G = gravitational constant, M = mass of body being orbited, r = orbital radius, 12.10 - Sample problem: speed of an orbiting satellite, What is the speed of a satellite in, circular orbit 500 kilometers above, the Earth's surface?, , The illustration above shows a satellite in circular orbit 500 km above the Earth’s surface. The Earth’s radius is stated in the variables table., Variables, satellite speed, , v, , satellite orbital radius, , r, , satellite height, , h = 500 km, , Earth’s radius, , rE = 6.38×106 m, , Earth’s mass, , ME = 5.97×1024 kg, , gravitational constant, , G = 6.67×10í11 N·m2/kg2, , What is the strategy?, 1., , Determine the satellite’s orbital radius, which is its distance from the center of the body being orbited., , 2., , Use the orbital speed equation to determine the satellite’s speed., , Physics principles and equations, The equation for orbital speed is, , 230, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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Step-by-step solution, We start by determining the satellite’s orbital radius. Careful: this is not the satellite’s height above the surface of the Earth, but its distance, from the center of the planet., , Step, , Reason, , 1., , r = rE + h, , 2., , r = 6.38×106 m + 5.00×105 m enter values, , 3., , r = 6.88×106 m, , equation for satellite’s orbital radius, , add, , Now we apply the equation for orbital speed., , Step, , Reason, , 4., , orbital speed equation, , 5., , enter values, , 6., , v = 7610 m/s, , evaluate, , 12.11 - Interactive problem: intercept the orbiting satellite, In this simulation, your mission is to send up a rocket to intercept a rogue satellite, broadcasting endless Barney® reruns. You can accomplish this by putting the, rocket into an orbit with the same radius and speed as that of the satellite, but, traveling in the opposite direction. The resulting collision will destroy the satellite, (and your rocket, but that is a small price to pay to save the world’s sanity)., The rogue satellite is moving in a counterclockwise circular orbit 40,000 kilometers, (4.00×107 m) above the center of the Earth. Your rocket will automatically move to, the same radius and will move in the correct direction., You must do a calculation to determine the proper speed to achieve a circular orbit, at that radius. You will need to know the mass of the Earth, which is 5.97×1024 kg., Enter the speed (to the nearest 10 m/s) in the control panel and press GO. Your, rocket will rise from the surface of the Earth to the same orbital radius as the, satellite, and then go into orbit with the speed specified. You do not have to worry, about how the rocket gets to the orbit; you just need to set the speed once the, rocket is at the radius of the satellite. If you fail to destroy the satellite, check your, calculations, press RESET and try a new value., , 12.12 - Interactive problem: dock with an orbiting space station, In this simulation, you are the pilot of an orbiting spacecraft, and your mission is to, dock with a space station. As shown in the diagram to the right, your ship is initially, orbiting in the same circular orbit as the space station. However, it is on the far side, of the Earth from the space station., In order to dock, your ship must be in the same orbit as the space station, and it, must touch the space station. To dock, your speed and radius must be very close to, that of the space station. A high speed collision does not equate to docking!, You have two buttons to control your ship. The “Forward thrust” button fires rockets, out the back of the ship, accelerating your spacecraft in the direction of its current, motion. The other button, labeled “Reverse thrust,” fires retrorockets in the opposite, direction. To use more “professional” terms, the forward thrust is called prograde, and the reverse thrust is called retrograde., Using these two controls, can you figure out how to dock with the space station?, To assist in your efforts, the current orbital paths for both ships are drawn on the, screen. Your rocket’s path is drawn in yellow, and the space station’s is drawn in, white., This simulation requires no direct mathematical calculations, but some thought and experimentation are necessary. If you get too far off track,, you may want to press RESET and try again from the beginning., , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 231
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There are a few things you may find worth observing. You will learn more about them when you study orbital energy. First, what is the change, in the speed of the rocket after firing its rear (Forward thrust) engine? What happens to its speed after a few moments? If you qualitatively, consider the total energy of the ship, can you explain you what is going on? You may want to consider it akin to what happens if you throw a, ball straight up into the sky., If you cannot dock but are able to leave and return to the initial circular orbit, you can consider your mission achieved., , 12.13 - Kepler’s first law, , The law of orbits: Planets move in elliptical, orbits around the Sun., The reason Newton’s comparison of the Moon’s motion to the motion of an apple was, so surprising was that many in his era believed the orbits of the planets and stars were, “divine circles:” arcs across the cosmic sky that defied scientific explanation. Newton, used the fact that the force of gravity decreases with the square of the distance to, explain the geometry of orbits., Scientists had been proposing theories about the nature of orbits for centuries before, Newton stated his law of gravitation. Numerous theories held that all bodies circle the, Earth, but subsequent observations began to point to the truth: the Earth and other, planets orbit the Sun. The conclusion was controversial; in 1633 the Catholic Church, forced Galileo to repudiate his writings that implied this conclusion., , Kepler’s first law, Planets move in elliptical orbits with the, Sun at one focus, , Even earlier, in 1609, the astronomer and astrologer Johannes Kepler began to, propose what are now three basic laws of astronomy. He developed these laws through, careful mathematical analysis, relying on the detailed observations of his mentor, Tycho, Brahe, a talented and committed observational astronomer., Kepler and Brahe formed one of the most productive teams in the history of astronomy., Brahe had constructed a state of the art observatory on an island off the coast of, Denmark. “State of the art” is always a relative term í the telescope had not yet been, invented, and Brahe might well have traded his large observatory for a good pair of, current day binoculars. However, Brahe’s records of years of observations allowed, Kepler, with his keen mathematical insight, to derive the fundamental laws of planetary, motion. He accomplished this decades before Newton published his law of gravitation., Kepler, using Brahe’s observations of Mars, demonstrated what is now known as, Kepler’s first law. This law states that all the planets move in elliptical orbits, with the, Sun at one focus of the ellipse., , Solar system, Most planetary orbits are nearly circular, , The planets in the solar system all move in elliptical motion. The distinctly elliptical orbit, of Pluto is shown to the right, with the Sun located at one focus of the ellipse. Had Kepler been able to observe Pluto, the elliptical nature of, orbits would have been more obvious., The other planets in the solar system, some of which he could see, have orbits that are very close to circular. (If any of them moved in a, perfectly circular orbit, they would still be moving in an ellipse, since a circle is an ellipse with both foci at its center.) Some of the orbits of these, other planets are shown in Concept 2., , 12.14 - More on ellipses and orbits, The ellipse shape is fundamental to orbits and can be described by two quantities: the, semimajor axis a and the eccentricity e. Understanding these properties of an ellipse, proves useful in the study of elliptical orbits., The semimajor axis, represented by a, is one-half the width of the ellipse at its widest,, as shown in Concept 1. You can calculate the semimajor axis by averaging the, maximum and minimum orbital radii, as shown in Equation 1., The eccentricity is a measure of the elongation of an ellipse, or how much it deviates, from being circular. (The word eccentric comes from “ex-centric,” or off-center.), Mathematically, it is the ratio of the distance d between the ellipse’s center and one, focus to the length a of its semimajor axis. You can see both these lengths in Equation, 2. Since a circle’s foci are at its center, d for a circle equals zero, which means its, eccentricity equals zero., , Elliptical orbits, , Pluto has the most eccentric orbit in our solar system, with an eccentricity of 0.25, as, calculated on the right. By comparison, the eccentricity of the Earth’s orbit is 0.0167., Most of the planets in our solar system have nearly circular orbits., , Semimajor axis: one half width of orbit, at widest, Eccentricity: elongation of orbit, , Comets have extremely eccentric orbits. This means their distance from the Sun at the, aphelion, the point when they are farthest away, is much larger than their distance at the perihelion, the point when they are closest to the Sun., , 232, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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(Both terms come from the Greek, “far from the Sun” and “near the Sun” respectively.), Halley’s comet has an eccentricity of 0.97. We show the comet’s orbit in Example 1, but, for visual clarity it is not drawn to scale. The comet’s orbit is so eccentric that its, maximum distance from the Sun is 70 times greater than its minimum distance. In, Example 1, you calculate the perihelion of this object in AU. The AU (astronomical unit), is a unit of measurement used in planetary astronomy. It is equal to the average radius, of the Earth’s orbit around the Sun: about 1.50×1011 meters., , Semimajor axis, , a = semimajor axis, rmin = minimum orbital distance, rmax = maximum orbital distance, , Eccentricity, , e = eccentricity, d = distance from center to focus, a = semimajor axis, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 233
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What is the perihelion distance, of Halley’s Comet? The comet’s, semimajor axis is 17.8 AU., , rmin = 2a írmax, rmin = 2(17.8 AU) í 35.1 AU, rmin = 0.5 AU, , What is the eccentricity of, Pluto’s orbit?, , e = 0.25, 12.15 - Kepler’s second law, , The law of areas: An orbiting body sweeps out, equal areas in equal amounts of time., In his second law, Kepler used a geometrical technique to show that the speed of an, orbiting planet is related to its distance from the Sun. (We use the example of a planet, and the Sun; this law applies equally well for a satellite orbiting the Earth, or for Halley’s, comet orbiting the Sun.), Kepler used the concept of a line connecting the planet to the Sun, moving like a, second hand on a watch. As shown to the right, the line “sweeps out” slices of area over, time. His second law states that the planet sweeps out an equal area in an equal, amount of time in any part of an orbit. In an elliptical orbit, planets move slowest when, they are farthest from the Sun and move fastest when they are closest to the Sun., , Kepler’s second law, Planets in orbit sweep out equal areas, in equal times, , Kepler established his second law nearly a century before Newton proposed his theory, of gravitation. Although Kepler did not know that gravity varied with the inverse square, of the distance, using Brahe’s data and his own keen quantitative insights he determined a key aspect of elliptical planetary motion., , 234, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.16 - Kepler’s third law, , The law of periods: The square of the period of, an orbit is proportional to the cube of the, semimajor axis of the orbit., Kepler’s third law, proposed in 1619, states that the period of an orbit around a central, body is a function of the semimajor axis of the orbit and the mass of the central body., The semimajor axis a is one half the width of the orbit at its widest. In a circular orbit,, the semimajor axis is the same as the radius r of the orbit., We illustrate this in Concept 1 using the Earth and the Sun. Given the scale of, illustrations in this section, the Earth’s nearly circular orbit appears as a circle., The length of the Earth’s period í a year, the time required to complete a revolution, about the Sun í is solely a function of the mass of the Sun and the distance a shown in, Concept 2., , Orbital period, Time of one revolution, , Kepler’s third law states that the square of the period is proportional to the cube of the, semimajor axis, and inversely proportional to the mass of the central body. The law is, shown in Equation 1. For the equation to hold true, the mass of the central body must, be much greater than that of the satellite., This law has an interesting implication: The square of the period divided by the cube of, the semimajor axis has the same value for all the bodies orbiting the Sun. In our solar, system, that ratio equals about 3×10í34 years2/meters3 (where “years” are Earth years), or 3×10í19 s2/m3. This is demonstrated in the graph in Concept 3. The horizontal and, vertical scales of the coordinate system are logarithmic, with semimajor axis measured, in AU and period measured in Earth years., , Kepler’s third law, Square of orbital period proportional to, cube of semimajor axis, , Graph of Kepler’s third law, Orbital size versus period for planets, orbiting Sun, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 235
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T = satellite period (in seconds), G = gravitational constant, M = mass of central body, a = semimajor axis, 12.17 - Orbits and energy, Satellites have both kinetic energy and potential energy. The KE and PE of a satellite, in elliptical orbit both change as it moves around its orbit. This is shown in Concept 1., Energy gauges track the satellite’s changing PE and KE. When the satellite is closer to, the body it orbits, Kepler’s second law states that it moves faster, and greater speed, means greater KE., The PE of the system is less when the satellite is closer to the body it orbits. When, discussing gravitational potential energy, we must choose a reference point that has, zero PE. For orbiting bodies, that reference point is usually defined as infinite, separation. As two bodies approach each other from infinity, potential energy decreases, and becomes increasingly negative as the value declines from zero., Concept 2 shows that even while the PE and KE change continuously in an elliptical, orbit, the total energy TE stays constant. Because there are no external forces acting, on the system consisting of the satellite and the body it orbits, nor any internal, dissipative forces, its total mechanical energy must be conserved. Any increase in, kinetic energy is matched by an equivalent loss in potential energy, and vice versa., In a circular orbit, a satellite’s speed is constant and its distance from the central body, remains the same, as shown in Concept 3. This means that both its kinetic and potential, energies are constant., , Orbital energy, Satellites have kinetic and potential, energy, Since PE = 0 at infinite distance, PE, always negative, , The total energy of a satellite increases with the radius (in the case of circular orbits) or, the semimajor axis (in the case of elliptical orbits). Moving a satellite into a larger orbit, requires energy; the source of that energy for a satellite might be the chemical energy, present in its rocket fuel., Equations used to determine the potential and kinetic energies and the total energy of a, satellite in circular orbit are shown in Equation 1. These equations can be used to, determine the energy required to boost a satellite from one circular orbit to another with, a different radius. The KE equation can be derived from the equation for the velocity of, a satellite. The PE equation holds true for any two bodies, and can be derived by, calculating the work done by gravity as the satellite moves in from infinity., The equations have an interesting relationship: The kinetic energy of the satellite equals, one-half the absolute value of the potential energy. This means that when the radius of, a satellite’s orbit increases, the total energy of the satellite increases. Its kinetic energy, decreases since it is moving more slowly in its higher orbit, but the potential energy, increases twice as much as this decrease in KE., , For a given orbit:, Total energy is constant, , Equation 2 shows the total energy equation for a satellite in an elliptical orbit. This equation uses the value of the semimajor axis a instead of, the radius r., , 236, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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For a given circular orbit, Both PE, KE constant, Total energy increases with radius, , Energy in circular orbits, , G = gravitational constant, M = mass of planet, m = mass of satellite, r = orbit radius, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 237
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Energy in elliptical orbits, , Etot = total energy, a = semimajor axis, 12.18 - Escape speed, , Escape speed: The minimum speed required for, an object to escape a planet’s gravitational, attraction., You know that if you throw a ball up in the air, the gravitational pull of the Earth will, cause it to fall back down. If by some superhuman burst of strength you were able to, hurl the ball up fast enough, it could have enough speed that the force of Earth’s gravity, would never be able to slow it down enough to cause it to return to the Earth., The speed required to accomplish this feat is called the escape speed. Space agencies, frequently fire rockets with sufficient speed to escape the Earth’s gravity as they explore, space., Given enough speed, a rocket can even escape the Sun’s gravitational influence,, allowing it to explore outside our solar system. As an example, the Pioneer 10, spacecraft, launched in 1972, was nearly 8 billion miles away from the Earth by 2003,, and is projected by NASA to continue to coast silently through deep space into the, interstellar reaches indefinitely., , Escape speed, Minimum speed to escape planet’s, gravitational attraction, , At the right is an equation for calculating the escape speed from a planet of mass M. As, the example problem shows, the escape speed for the Earth is about 11,200 m/s, a little, more than 40,000 km/h. The escape speed does not depend on the mass of the object, being launched. However, the energy given to the object to make it escape is a function, of its mass, since the object’s kinetic energy is proportional to its mass., The rotation of the Earth is used to assist in the gaining of escape speed. The Earth’s, rotation means that a rocket will have tangential speed (except at the poles, an unlikely, launch site for other reasons as well). The tangential speed equals the product of the, Earth’s angular velocity and the distance from the Earth’s axis of rotation., , Some escape speeds, , An object will have a greater tangential speed near the equator because there the, distance from the Earth’s axis of rotation is greatest. The United States launches its, rockets from as close to the equator as is convenient: southern Florida. The rotation of, the Earth supplies an initial speed of 1469 km/h (408 m/s) to a rocket fired east from Cape Canaveral, about 4% of the required escape speed., Derivation. We will derive the escape speed equation by considering a rocket launched from a planet of mass M with initial speed v. The, rocket, pulled by the planet’s gravity, slows as it rises. Its launch speed is just large enough that it never starts falling back toward the planet;, instead, its speed approaches zero as it approaches an infinite distance from the planet. If the initial speed is just a little less, the rocket will, eventually fall back toward the planet. If the speed is greater than or equal to the escape speed, the rocket will never return., , 238, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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Variables, We will derive the escape speed by comparing the potential and kinetic energies of the, rocket as it blasts off (subscript 0), and as it approaches infinity (subscript )., , initial, , final, , change, , potential energy, , PE0, , PE, , ǻPE, , kinetic energy, , KE0, , KE, , ǻKE, , gravitational constant, , G, , mass of planet, , M, , mass of rocket, , m, , radius of planet, , r, , initial speed of rocket, , v, , Escape speed equation, , Strategy, 1., , Calculate the change in potential energy of the Earth-rocket system between, launch and an infinite separation of the two., , 2., , Calculate the change in kinetic energy between launch and an infinite separation., , 3., , The conservation of mechanical energy states that the total energy after the, engines have ceased firing equals the total energy at infinity. State this, relationship and solve for v, the initial escape speed., , v = escape speed, G = gravitational constant, M = mass of planet, r = radius of planet, , Physics principles and equations, We use the equation for the potential energy of an object at a distance r from the center, of the planet., , We use the definition of kinetic energy., , Finally, we will use the equation that expresses the conservation of mechanical energy., , ǻPE + ǻKE = 0, Step-by-step derivation, , What is the minimum speed, required to escape the Earth’s, gravity?, , In the first steps we find the potential energy of the rocket at launch and at infinity, and, subtract the two values., , Step, , Reason, , 1., , potential energy at launch, , 2., , potential energy at infinity, v = 11,200 m/s, change in potential energy, , 3., , In the following steps we find the kinetic energy of the rocket at launch and at infinity, and subtract the two values., , Step, , 4., , 5., 6., , Reason, , kinetic energy, , KE = 0, , assumption, change in kinetic energy, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 239
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To complete the derivation, we will substitute both changes in energy into the equation for the conservation of energy, and solve for the critical, speed v., , Step, , 7., , ǻPE + ǻKE = 0, , Reason, , conservation of energy, , 8., , substitute equations 3 and 6 into equation 7, , 9., , solve for v, , 12.19 - Gotchas, The freefall acceleration rate, g, does not depend on the mass that is falling. If you say yes, you are agreeing with Galileo and you are correct., A satellite can move faster and yet stay in the same circular orbit. No, it cannot. The speed is related to the dimensions of the orbit., , 240, , Copyright 2000-2007 Kinetic Books Co. Chapter 12
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12.20 - Summary, Newton's law of gravitation states that the force of gravity between two particles is, proportional to the product of their masses and inversely proportional to the square, of the distance between them., The gravitational constantG appears in Newton's law of gravitation and is the same, everywhere in the universe. It should not be confused with g, the acceleration due, to gravity near the Earth's surface. The value of gvaries slightly according to the, location on the Earth. This is due to local variations in altitude and to the Earth's, bulging shape, nonuniform density and rotation., A circular orbit is the simplest type of orbit. The speed of an object in a circular orbit, is inversely proportional to the square root of the orbital radius. The speed is also, proportional to the square root of the mass of the object being orbited, so orbiting a, more massive object requires a greater speed to maintain the same radius., , Newton’s law of gravitation, , Gravitational acceleration, , Circular orbit, , Johannes Kepler set forth three laws that describe the orbital motion of planets., Kepler's first law says that the planets move in elliptical orbits around the Sun, which, is located at one focus of the ellipse. Most of the planets' orbits in the solar system, are only slightly elliptical., Kepler's second law, the law of areas, says that an orbiting body such as a planet, sweeps out equal areas in equal amounts of time. This means that the planet’s, speed will be greater when it is closer to the Sun., Kepler's third law, the law of periods, states that the square of the period of an orbit, is proportional to the cube of the semimajor axis a, which is equal to one half the, width of the orbit at its widest. For a circular orbit, a equals the radius of the orbit., , Kepler’s second law, , Kepler’s third law, , The orbital energy of a satellite is the sum of its gravitational potential energy (which, is negative) and its kinetic energy. The total energy is constant, though the PE and, KE change continuously if the satellite moves in an elliptical orbit., The escape speed is the minimum speed necessary to escape a planet's, gravitational attraction. It depends on the mass and radius of the planet, but not on, the mass of the escaping object., , Energy in circular orbits, , Energy in elliptical orbits, , Escape speed, , Copyright 2000-2007 Kinetic Books Co. Chapter 12, , 241
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Chapter 12 Problems, , Chapter Assumptions, Unless stated otherwise, use the following values., , rEarth = 6.38×106 m, MEarth = 5.97×1024 kg, , Conceptual Problems, C.1, , Compare planets farther from the Sun to those nearer the Sun. (a) Do the farther planets have greater or lesser orbital speed, than the nearer ones? (b) How does the angular speed of the farther planets compare to that of the nearer ones?, (a), , (b), , i., ii., iii., i., ii., iii., , Outer planets have greater orbital speed., Outer planets have smaller orbital speed., The orbital speed is the same., Outer planets have greater angular speed., Outer planets have smaller angular speed., The angular orbital speed is the same., , C.2, , How much work is done on a satellite as it moves in a circular orbit around the Earth?, , C.3, , According to Kepler's third law, the ratio T2/a3 should be the same for all objects orbiting the Sun, since the factor 4ʌ2/GM is, the same. When this ratio is measured however, it is found to vary slightly. For instance, Jupiter's ratio is higher than Earth's, by about 1%. What are the two main assumptions behind Kepler's third law that are not 100% valid in a real planetary, system?, , C.4, , From your intergalactic survey base, you observe a moon in a circular orbit about a faraway planet. You know the distance to, the planet/moon system, and you determine the maximum angle of separation between the two and the period of the moon's, orbit. Assuming that the moon is much less massive than the planet, explain how you can determine the mass of the planet., , C.5, , In a previous chapter, we used the equation PE = mgh to represent the gravitational potential energy of an object near the, Earth. In this chapter, we use the equation PE = íGMm/r. Explain the reasons for the differences between these two, equations: Why is one expression negative and the other positive? Is r equal to h?, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the first interactive problem in this section to answer the following questions. (a) Given two unchanging, masses, does the force between two masses increase, decrease, or stay the same as the distance between the masses, increases? (b) Given a fixed distance, does the force between two masses increase, decrease or stay the same as the, masses increase?, (a), , (b), , i., ii., iii., i., ii., iii., , Increase, Decrease, Stay the same, Increase, Decrease, Stay the same, , Section 1 - Newton’s law of gravitation, 1.1, , The Hubble Space Telescope orbits the Earth at an approximate altitude of 612 km. Its mass is 11,100 kg and the mass of, the Earth is 5.97×1024 kg. The Earth's average radius is 6.38×106 m. What is the magnitude of the gravitational force that the, Earth exerts on the Hubble?, N, , 242, , Copyright 2007 Kinetic Books Co. Chapter 12 Problems
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1.2, , A neutron star and a black hole are 3.34e+12 m from each other at a certain point in their orbit. The neutron star has a mass, of 2.78×1030 kg and the black hole has a mass of 9.94×1030 kg. What is the magnitude of the gravitational attraction between, the two?, N, , 1.3, , An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of, 3.14×1013 N from the Sun. The mass of the Sun is 1.99×1030 kg. How far is the asteroid from the Sun?, m, , 1.4, , The gravitational pull of the Moon is partially responsible for the tides of the sea. The Moon pulls on you, too, so if you are on, a diet it is better to weigh yourself when this heavenly body is directly overhead! If you have a mass of 85.0 kg, how much, less do you weigh if you factor in the force exerted by the Moon when it is directly overhead (compared to when it is just rising, or setting)? Use the values 7.35×1022 kg for the mass of the moon, and 3.76×108 m for its distance above the surface of the, Earth. (For comparison, the difference in your weight would be about the weight of a small candy wrapper. And speaking of, candy...), N, , 1.5, , Three 8.00 kg spheres are located on three corners of a square. Mass A is at (0, 1.7) meters, mass B is at (1.7, 1.7) meters,, and mass C is at (1.7, 0) meters. Calculate the net gravitational force on A due to the other two spheres. Give the, components of the force., (, , ,, , )N, , Section 2 - G and g, 2.1, , The top of Mt. Everest is 8850 m above sea level. Assume that sea level is at the average Earth radius of 6.38×106 m. What, is the magnitude of the gravitational acceleration at the top of Mt. Everest? The mass of the Earth is 5.97×1024 kg., m/s2, , 2.2, , Geosynchronous satellites orbit the Earth at an altitude of about 3.58×107 m. Given that the Earth's radius is 6.38×106 m and, its mass is 5.97×1024 kg, what is the magnitude of the gravitational acceleration at the altitude of one of these satellites?, m/s2, , 2.3, , Jupiter's mass is 1.90×1027 kg. Find the acceleration due to gravity at the surface of Jupiter, a distance of 7.15×107 m from its, center., m/s2, , 2.4, , A planetoid has a mass of 2.83e+21 kg and a radius of 7.00×105 m. Find the magnitude of the gravitational acceleration at, the planetoid's surface., m/s2, , Section 8 - Interactive problem: Newton’s cannon, 8.1, , Use the simulation in the interactive problem in this section to determine the initial speed required to put the cannonball into, circular orbit., m/s, , Section 9 - Circular orbits, 9.1, , The International Space Station orbits the Earth at an average altitude of 362 km. Assume that its orbit is circular, and, calculate its orbital speed. The Earth's mass is 5.97×1024 kg and its radius is 6.38×106 m., m/s, , 9.2, , An asteroid orbits the Sun at a constant distance of 4.44e+11 meters. The Sun's mass is 1.99×1030 kg. What is the orbital, speed of the asteroid?, m/s, , 9.3, , The Moon's orbit is roughly circular with an orbital radius of 3.84×108 m. The Moon's mass is 7.35×1022 kg and the Earth's, mass is 5.97×1024 kg. Calculate the Moon's orbital speed., , 9.4, , The orbital speed of the moon Ganymede around Jupiter is 1.09×104 m/s. What is its orbital radius? Assume the orbit is, circular. Jupiter's mass is 1.90×1027 kg., , m/s, , m, , Copyright 2007 Kinetic Books Co. Chapter 12 Problems, , 243
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Section 11 - Interactive problem: intercept the orbiting satellite, 11.1 Use the information given in the interactive problem in this section to calculate the speed required to enter the circular orbit, and destroy the rogue satellite. Test your answer using the simulation., m/s, , Section 12 - Interactive problem: dock with an orbiting space station, 12.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) What happens to the speed, of the rocket immediately after firing its rear (Forward thrust) engine? (b) What happens to its speed after a few moments?, (a), , (b), , i., ii., iii., i., ii., iii., , It increases, It decreases, It stays the same, It increases, It decreases, It stays the same, , Section 14 - More on ellipses and orbits, 14.1 A comet's orbit has a perihelion distance of 0.350 AU and an aphelion distance of 45.0 AU. What is the semimajor axis of the, comet's orbit around the sun?, AU, 14.2 The semimajor axis of a comet's orbit is 21.0 AU and the distance between the Sun and the center of the comet's elliptical, orbit is 20.1 AU. What is the eccentricity of the orbit?, , 14.3 When a planet orbits a star other than the Sun, we use the general terms periapsis and apoapsis, rather than perihelion and, aphelion. The orbit of a planet has a periapsis distance of 0.950 AU and an apoapsis distance of 1.05 AU. (a) What is the, semimajor axis of the planet's orbit? (b) What is the eccentricity of the orbit?, (a), , AU, , (b), 14.4 The text states that the eccentricity of an elliptical orbit is equal to the distance from the ellipse's center to a focus, divided by, the semimajor axis. Show that eccentricity is also equal to the positive difference in the perihelion and aphelion distances,, divided by their sum., 14.5 An extrasolar planet's orbit has a semimajor axis of 23.1 AU. The eccentricity of the orbit is 0.010. What is the periapsis, distance (the planet's minimum distance from the star it is orbiting)?, AU, 14.6 The Trans-Neptunian object Sedna was discovered in 2003. By mid-2004, Sedna's orbit was estimated to have a semimajor, axis of 480 AU and an eccentricity of 0.84. (a) What is the perihelion distance of Sedna's orbit? (b) What is the aphelion, distance?, (a), , AU, , (b), , AU, , Section 16 - Kepler’s third law, 16.1 Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in, seconds? (b) What is the period in Earth years? Assume that one Earth year is exactly 365 days, with 24 hours in each day., (a), , s, , (b), , years, , 16.2 Mars orbits the Sun in about 5.94×107 seconds (1.88 Earth years). (a) What is its semimajor axis in meters? (The mass of the, Sun is 1.99×1030 kg.) (b) What is Mars' semimajor axis in AU? 1 AU = 1.50×1011 m., (a), , m, , (b), , AU, , 16.3 A planet orbits a star with mass 2.61e+30 kg. The semimajor axis of the planet's orbit is 2.94×1012 m. What is the period of, the planet's orbit in seconds?, s, , 244, , Copyright 2007 Kinetic Books Co. Chapter 12 Problems
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16.4 An extrasolar planet has a small moon, which orbits the planet in 336 hours. The semimajor axis of the moon's orbit is, 1.94e+9 m. What is the mass of the planet?, kg, 16.5 Jupiter's moon Callisto orbits the planet at a distance of 1.88×109 m from the center of the planet. Jupiter's mass is, 1.90×1027 kg. What is the period of Callisto's orbit, in hours?, hours, 16.6 The Trans-Neptunian object Sedna has an extremely large semimajor axis. In the year 2004, it was estimated to be 480 AU., What is the period of Sedna's orbit, measured in Earth years?, years, , Section 17 - Orbits and energy, 17.1 The Hubble Space Telescope orbits the Earth at an altitude of approximately 612 km. Its mass is 11,100 kg and the mass of, the Earth is 5.97×1024 kg. The Earth's radius is 6.38×106 m. Assume the Hubble's orbit is circular. (a) What is the gravitational, potential energy of the Earth-Hubble system? (Assume that it is zero when their separation is infinite.) (b) What is the, Hubble's KE? (c) What is the Hubble's total energy?, (a), , J, , (b), , J, , (c), , J, , 17.2 What is the least amount of energy it takes to send a spacecraft of mass 3.50×104 kg from Earth's orbit to that of Mars?, (Neglect the gravitational influence of the planets themselves.) Assume that both planetary orbits are circular, the radius of, Earth's orbit is 1.50×1011 meters, and that of Mars' orbit is 2.28×1011 meters. The Sun's mass is 1.99×1030 kg., J, 17.3 How much work is required to send a spacecraft from Earth's orbit to Saturn's orbit around the sun? The semimajor axis of, Earth's orbit is 1.50×1011 meters and that of Saturn's orbit is 1.43×1012 meters. The spacecraft has a mass of 3.71e+4 kg and, the Sun's mass is 1.99×1030 kg., J, 17.4 You wish to boost a 9,550 kg Earth satellite from a circular orbit with an altitude of 359 km to a much higher circular orbit with, an altitude of 35,800 km. What is the difference in energy between the two orbits, that is, how much energy will it take to, accomplish the orbit change? Earth's radius is 6.38×106 m and its mass is 5.97×1024 kg., J, 17.5 A satellite is put in a circular orbit 485 km above the surface of the Earth. After some time, friction with the Earth's, atmosphere causes the satellite to fall to the Earth's surface. The 375 kg satellite hits the Pacific Ocean with a speed of, 2,500 m/s. What is the change in the satellite's mechanical energy? (Watch the sign of your answer.) In this situation,, mechanical energy is transformed into heat and sound. The Earth's mass is 5.97×1024 kg, and its radius is 6.38×106 m., J, 17.6 You launch an engine-less space capsule from the surface of the Earth and it travels into space until it experiences, essentially zero gravitational force from the Earth. The initial speed of the capsule is 18,500 m/s. What is its final speed?, Assume no significant gravitational influence from other solar system bodies. The Earth's mass is 5.97×1024 kg, and its radius, is 6.38×106 m., m/s, , Section 18 - Escape speed, 18.1 Calculate the escape speed from the surface of Venus, whose radius is 6.05×106 m and mass is 4.87×1024 kg. Neglect the, influence of the Sun's gravity., m/s, 18.2 The escape speed of a particular planet with a radius of 7.60e+6 m is 14,500 m/s. What is the mass of the planet?, kg, 18.3 A planet has a mass of 5.69×1026 kg. The planet is not a perfect sphere, instead it is somewhat flattened. At the equator, its, radius is 6.03×107 m, and at the poles, the radius is 5.38×107 m. (a) What is the escape speed from the surface of the planet, at the equator? (b) What is the escape speed at the poles?, (a), , m/s, , (b), , m/s, , Copyright 2007 Kinetic Books Co. Chapter 12 Problems, , 245
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18.4 The average density of Neptune is 1.67×103 kg/m3. What is the escape speed at the surface? Neptune's radius is, 2.43×107 m., m/s, 18.5 The Schwarzschild radius of a black hole can loosely be defined as the radius at which the escape speed equals the speed of, light. Anything closer to the black hole than this radius can never escape, because nothing can travel faster than light in a, vacuum. Not even light itself can escape, hence the name "black hole". (a) Find the Schwarzschild radius of a black hole with, a mass equal to five times that of the Sun. This is a typical value for a "stellar-mass" black hole. The Sun's mass is, 1.99×1030 kg. (b) Find the Schwarzschild radius of a black hole with a mass of one billion Suns. This is the type of black hole, found at the centers of the largest galaxies., , 246, , (a), , m, , (b), , m, , Copyright 2007 Kinetic Books Co. Chapter 12 Problems
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13.0 - Introduction, The study of physics typically begins with the study of, solid objects: You learn how to determine the velocity, of a car as it accelerates down a street, what happens, when two pool balls collide, and so on., This chapter introduces the study of fluids. Liquids, and gases are both fluids. Fluids change shape much, more readily than solids. Pour soda from a can into a, glass and the liquid will change shape to conform to, the shape of the glass. Push on a balloon full of air or, water and you can easily change the shape of the, balloon and the fluid it contains. In contrast to liquids,, The glass and the ice cube in this photo are solid objects. The, gases expand to fill all the space available to them., water is a liquid. The bubbles and the surrounding air are gasses., One reason astronauts wear spacesuits is to keep, their air near them, and not let it expand limitlessly into the near vacuum of space., This chapter focuses on the characteristics exhibited by fluids when their temperature and density remain nearly constant. It covers topics such, as the method of calculating how much pressure water will exert on a submerged submarine, and why a boat floats. Some of the topics apply, to liquids alone, while others apply to both liquids and gases., , 13.1 - Fluid, , Fluid: A substance that can flow and conform, to the shape of a container. Liquids and gases, are fluids., A fluid alters its shape to conform to the shape of the container that surrounds and, holds it. The molecules of a fluid can “flow” because they are not fixed into position as, they would be in a solid. Liquids and gases are fluids, and they are two of the common, forms of matter, with solids being the third. There are other forms of matter as well,, such as plasma (created in fusion reactors) and degenerate matter (found in neutron, stars)., A substance can exist as a solid, a liquid or a gas depending on the surrounding, physical conditions. Factors such as temperature and pressure determine its state. For, the purposes of a physics textbook, we need to be specific about what state of matter, we are discussing at any given time. However, whether a substance is considered a, solid or a fluid may depend on factors such as the time scale under consideration. For, instance, the ice in a glacier can seem quite solid, but glaciers do flow slowly over time,, so treating glacier ice as a fluid is useful to geologists., , Fluids, Can flow, Rate of flow varies, , Fluids, Conform to container, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 247
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13.2 - Density, , Density: Mass per unit volume., Density í to be precise, mass density í equals mass divided by volume. The Greek, letter ȡ (rho, pronounced “roe”) represents density. The SI unit for density is the, kilogram per cubic meter. The gram per cubic centimeter is also a common unit, useful, in part because the density of water is close to one gram per cubic centimeter., Liquids and solids retain a fairly constant density. It requires a great deal of force to, compress water or a piece of clay into a more compact form. A given mass of liquid will, change shape in order to conform to the shape of the container you pour it into, but its, volume will remain nearly constant over a great variety of conditions., In contrast, the volume of a sample of gas changes readily, which means its density, changes easily, as well. For example, when you pump a stroke of air into a bicycle tire,, the volume of air in the pump cylinder is reduced to “squeeze” it into the tire, increasing, the air's density., , Density, Ratio of mass to volume, , Much larger changes can be accomplished using larger increases in pressure., Machinery compresses the air fed into a scuba diving tank, causing its contents to be at, a density on the order of 200 times greater than the density of the atmosphere you, breathe. Before a diver breathes this air, its density (and pressure) are reduced. It, would be lethal at the pressure maintained in the tank., The density of an object can vary at different points based on its composition. A precise, way to state the definition of density is ǻm/ǻV: The mass of a small volume of material, is measured to establish its local density. However, unless otherwise stated, we will, assume that the substances we deal with have uniform density, the same density at all, points. This means that the density can be established by dividing the total mass by the, total volume., The table on the right shows the densities of various materials. Their densities are given, at 0°C and one standard atmosphere of pressure (the density of the air around you can, vary, depending on atmospheric conditions). Exceptions to this are the super-dense, neutron star, which has no temperature in the ordinary sense, and liquid water, whose, density is given at 4°C, the temperature at which it is the greatest. We also supply, some common equivalents for water, for instance relating a liter of water to its mass., The concept of specific gravity provides a useful tool for understanding and comparing, various materials’ densities. Specific gravity divides the density of one material by that, of a reference material, usually water at 4°C. For instance, if a material has a specific, gravity of two, it is twice as dense as water., , ȡ = m/V, ȡ = density, m = mass, V = volume, Units: kg/m3, , Density of various substances, For water at 4º C:, ·1 liter has mass of 1 kg, ·1 cm3 has mass of 1 g, , 248, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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What is the density of the gold, brick?, ȡ = m/V, ȡ = 19.3 kg/0.00100 m3, ȡ = 19,300 kg/m3, 13.3 - Pressure, , Pressure: Final exams,, SATs, free throws in the last, 30 seconds of a tight game,, and driver’s license tests., Pressure: Force divided by, the surface area over which, the force acts., The first definition of pressure above speaks for itself,, but the second deserves further explanation., You experience pressure when you swim. If you dive, deep under the water, you can feel the water pushing, against you with more pressure, more force per, square meter of your body., , Big feet distribute this lynx's weight over a larger surface area,, lowering the pressure they exert on the snow., , For a surface immersed in a fluid, the amount of pressure at a given location in the fluid, is the same for any orientation of the surface. If you place your hand thirty centimeters, underwater, the pressure on your palm is the same no matter how you rotate it. In the, aquarium illustrated on the right, the water exerts a force on the bottom of the tank, but, it exerts force í and pressure í on the sides as well., Pressure equals the amount of force divided by the surface area to which it is applied., As the photograph above shows, some animals, such as the lynx, are able to travel, easily across the surface of snow because their large paws spread the force of their, weight over a large area. This reduces the pressure they exert on the snow, enabling, them to walk on its surface. Animals of similar weight, but with smaller paws, sink into, the snow., People who need to travel across the snow may likewise use snowshoes or skis to, increase surface area and reduce the pressure they exert. In contrast, spike-heeled, shoes concentrate almost all the weight of their wearers over a very small surface area,, and can exert a pressure large enough to damage wood or vinyl floors., , Pressure, Force divided by surface area, , The water in the aquarium exerts force on the walls of the tank as well as its bottom. This is shown in Concept 1. Why does the water exert a, force on the sides of the aquarium? Consider squishing down on a water balloon: The balloon bulges out on its sides. The additional force you, exert on the top is translated into a force on the sides. To return to the aquarium, the downward weight of the water results in a force on the, walls as well as the bottom., The formula in Equation 1 shows how to calculate pressure. It equals the magnitude of the force divided by the area and is a scalar quantity., The SI unit for pressure is the newton per square meter, called a pascal (Pa). One pascal is a very small amount of pressure. The Earth’s, atmosphere exerts about 100,000 pascals of air pressure at the planet’s surface. A bar of pressure equals 100,000 (105) pascals and is, another commonly used unit. Bars are informally called “atmospheres” (atm) of pressure. You may have heard references to “millibars” in, weather reports., In the British system, force is measured in pounds. Pounds per square inch, psi, is a common measure of force. At the Earth’s surface, typical, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 249
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atmospheric pressure is 14.7 psi. Automobile tires are normally inflated to a pressure of, about 24 psi, while road bicycle tires are inflated to pressures as high as 120 psi. The, tire readings reflect the amount of pressure inside the tires over atmospheric pressure., The total (absolute) pressure inside a tire is the sum of atmospheric pressure and the, gauge reading (a total of about 135 psi for the bicycle tire)., , P = F/A, P = pressure, F = force, A = surface area, Units: pascal (Pa), newton/meter2, , Atmospheric pressure at sea, level, Patm = 101,300 Pa § 1 bar, , What is the pressure inside the, bottom of the aquarium caused, solely by the weight of the, water?, P = F/A, P = 927 N / 0.300 m2, P = 3090 Pa, , 250, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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13.4 - Pressure and fluids, As a submarine dives deeper and deeper into the, sea, it encounters increasing pressure. The nightmare, of any submariner is that his craft goes so deep it, collapses under the tremendous pressure of the, water. (Rent the movie U-571 if you want to watch a, Hollywood thriller that deals with pressure at great, depths.), Physicists prefer a little less drama and a little more, measurement in their dealings with pressure. They, have developed an equation to describe the pressure, of a fluid as a function of the fluid’s depth., Navy personnel performing under pressure., , Why does water pressure increase as a submarine, descends? The pressure increases because the amount of water on top of the submarine increases as the vessel does down. The weight of, the water exerts a force over the entire surface area of the submarine’s hull. The water pressure does not depend on the orientation of the, surface. It exists all over the craft: on its top, bottom and sides., The first equation on the right shows how to calculate the pressure at a point underneath the surface of a fluid. The pressure equals the, product of the fluid’s density, the constant acceleration of gravity, and the height of the column of fluid above the point. “Height” means the, distance from the point to the surface of the fluid. The equation states that pressure increases linearly with depth. The water pressure at 200, meters down is twice as great as it is at 100 meters. This equation applies when the fluid is static., The fluid pressure varies solely as a function of the density ȡ and the depth in the fluid, not with the shape of the container holding the fluid. If, you fill a swimming pool and a Coke bottle with water, the water pressure at 0.1 meters below the surface will be the same in both containers., The pressure will also be the same at the bottom of the bottle and on the wall of the pool at the same depth. Whether the surface area is, horizontal or vertical, the pressure is the same., You can add pressures. The total pressure exerted on the exterior of the hull of the submarine equals the sum of atmospheric pressure (the, pressure exerted by the Earth’s air above it) and the pressure of the water above it. The two pressures must be calculated separately and, added as shown in Equation 2. Since the densities of water and air differ, the pressure of the combined column of fluids above the submarine, cannot be calculated as a single product ȡgh., We have implicitly described two types of pressure: absolute and gauge pressure. The total pressure is called the absolute pressure. It is the, sum of the atmospheric pressure and the pressure of the fluid in question, in this case water. The photograph below shows how a Styrofoam®, cup (as shown on the right) was crushed (as shown on the left) by an absolute pressure of 3288 psi when it was submerged to a depth of more, than two kilometers below the ocean’s surface., , The term gauge pressure describes the pressure caused solely by the water (or other fluid), ignoring the atmospheric pressure. The gauge, pressure equals ȡgh, where ȡ, g and h are measured for the fluid alone. To state the same concept in another way: Gauge pressure equals, the absolute pressure minus the atmospheric pressure., Pressures can oppose one other, when they act on opposite sides of the same surface. For example, the “atmospheric” pressure inside an, airplane cabin is allowed to decrease as the plane climbs. The pressure inside your ear can be higher than the pressure of the cabin, since it, may remain at the higher pressure of the atmosphere at the Earth’s surface. In this case, the pressure in parts of your ears is greater than the, pressure outside them, producing a net outward pressure (and force) on your eardrums. The result is that your ear begins to ache., You can reduce the pressure inside your ears by chewing gum or yawning to “pop” them. This stretches and opens the Eustachian tubes,, passages between your ears and throat. Air flows out of your ear into the cabin, balancing the pressures, and your earache disappears., If you look at the large value calculated in the example problem for the absolute pressure on the inner surface of the aquarium’s bottom, you, might wonder why the tank does not burst. Remember that atmospheric pressure is pushing inward on its exterior surfaces as well. The net, pressure on the bottom plate is the gauge pressure due solely to the water, which is not large enough to cause the tank to burst., The density of a liquid does not vary significantly with depth, so using an average density figure in the formula ȡgh provides a good, approximation of the pressure at any depth. The density of gases can vary greatly, so the formula is not as applicable to them, especially if h is, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 251
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large. For instance, the density of the Earth’s atmosphere is about 1.28 kg/m3 at sea, level (and 0°C), but only 0.38 kg/m3 at an altitude of 10,600 m (35,000 ft) above sea, level (at í20°C). The lower layers of the atmosphere are significantly compressed by, the weight of the air above them. Mountain climbers note this difference, remarking that, the air is “thinner” at higher altitudes., , Pressure due to a fluid, Function of:, ·density of fluid, ·acceleration of gravity, ·height of column, , Pressure of a fluid (liquid), P = ȡgh, P = pressure, ȡ = density of liquid, g = acceleration of gravity, h = height of liquid column, , Pressure can be summed, Ptotal = ȈP = P1 + P2 +…+ Pn, Total pressure = sum of pressures, , 252, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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What is the gauge pressure, inside the bottom of the tank due, to the weight of the water alone?, What is the absolute pressure,, including atmospheric pressure,, pressing down inside the, bottom?, PH2O = ȡgh, , PH2O = 2400 Pa, Pabs = 2400 Pa + 101,300 Pa, Pabs = 103,700 Pa, 13.5 - Sample problem: pressure at the bottom of a lake, This lake is at sea level. What is the, absolute pressure at the bottom of the, lake?, , Variables, absolute pressure, , Pabs, , atmospheric pressure, , Patm = 1.013×105 Pa, , gauge pressure due to water, , PH2O, , density of water, , ȡ = 1000 kg/m3, , acceleration of gravity, , g = 9.80 m/s2, , height of water column, , h = 50.0 m, , Strategy, 1., , The atmospheric pressure is given. Find the gauge pressure of the water in the lake., , 2., , Add the atmospheric and gauge pressures to get the absolute pressure., , Physics principles and equations, The gauge pressure of the water at the bottom of the lake is, , PH2O = ȡgh, The absolute pressure at the bottom of the lake is, , Pabs = PH2O + Patm, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 253
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Step-by-step solution, , Patm is known. We find PH2O and add the two quantities together to find the absolute pressure Pabs., , Step, , Reason, , 1., , Patm = 1.013×105 Pa, , standard value, , 2., , PH2O = ȡgh, , gauge pressure equation, substitute values, , 3., 4., , PH2O = 4.90×105 Pa, , evaluate, , 5., , Pabs = PH2O + Patm, , absolute pressure, substitute equations 1 and 4 into equation 5, , 6., , 13.6 - Physics at work: measuring pressure, Atmospheric pressure is not constant, even at a fixed, height like sea level. In Concept 1, you see a simple, barometer. Barometers measure atmospheric, pressure. The terms “barometer” and “barometric, pressure” are sometimes heard in weather reports, because changes in atmospheric pressure often, indicate that a change of weather is on the way., The barometer depicted on the right consists of a, vertical tube, closed at the top. This tube is partially, filled with a liquid, commonly the liquid metal mercury,, and is placed in a container that serves as a, reservoir. The mercury in the tube stands in a column,, with the space at the top of the tube occupied by a, near vacuum, which exerts negligible pressure., , Dial-type barometer, calibrated in millimeters of mercury and millibars., , Increased air pressure pushing down on the surface of the reservoir causes the column, of mercury to rise, until the air pressure and the pressure due to the mercury column, reach equilibrium. The pressure exerted by this liquid column, the product of its density,, the acceleration of gravity, and its height, equals the external air pressure., On a typical day at sea level, air pressure causes the mercury to rise to a height of, about 760 millimeters, or about 30 inches. (The pressure of a one-millimeter column of, mercury is called a torr, after the physicist Evangelista Torricelli.) The design of this, instrument should give you a sense of how strong atmospheric pressure is: It is able to, force a column of mercury, which is denser than lead, to rise more than three-quarters, of a meter., In Concept 2 you see an open-tube manometer, a device for measuring the gauge, pressure of a gas confined in a spherical vessel. The vessel that contains the gas is, connected to a U-shaped tube partially filled with mercury and open to the atmosphere, at its far end. This apparatus allows physicists to accurately determine the gauge, pressure of the gas., , Barometer, Air pressure equals pressure of mercury, ·Height of mercury reflects air pressure, , The pressure inside the spherical vessel on the left-hand side is an absolute pressure. It, presses down on the surface of the left-hand column of mercury, but the higher column, of mercury and the air pressure on the right side push back. When the two columns of, mercury are in equilibrium, the absolute pressure of the gas equals the sum of the, atmospheric pressure and the pressure exerted by the extra mercury on the right-hand, side of the instrument. This equality is shown in Equation 2. In the equation, the product, ȡgh represents the pressure exerted by a column of mercury of height h., The height h of the extra mercury, indicated in the diagram, equals the amount by which, the mercury level on the right is higher than the mercury level on the left. It can be used, with the density of mercury and the acceleration of gravity to calculate the product ȡgh., This product equals the difference between the absolute pressure of the gas in the, vessel and atmospheric pressure. In other words, the product ȡgh gives the gauge, pressure of the gas, in pascals. When the gauge pressure is expressed in “millimeters, of mercury,” or torr, then its numeric value equals the numeric value of the height h,, measured in millimeters., , 254, , Open-tube manometer, Vessel pressure, = atm pressure + mercury pressure, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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Traditional blood pressure gauges (sphygmomanometers) are open-tube manometers., The gauge has an inflatable cuff that is wrapped around your upper arm. Like the, sphere in the diagram, the cuff can be filled with pressurized air. This restricts the flow, of blood to the lower parts of your arm. Air is then released from the cuff until the first, flow of blood can be heard with a stethoscope. At this point, the gauge pressure of the, blood being pumped by your heart equals the cuff’s gauge pressure. This blood, pressure, the systolic pressure, occurs when the heart generates its maximum, pressure., The sphygmomanometer operator (try saying that quickly!) then listens for the part of, the heartbeat cycle when the pressure is the lowest, releasing pressure from the cuff, until its pressure is the same as the lowest blood pressure, and the flow of blood can be, heard continuously. This lower pressure is the diastolic pressure. A young, healthy, human has a systolic blood gauge pressure of about 120 millimeters of mercury and a, diastolic pressure equal to about 80 millimeters of mercury., , P = Patm + ȡgh, P = absolute pressure in vessel, Patm = atmospheric pressure, ȡgh = gauge pressure, , The absolute pressure in the, sphygmomanometer cuff is, 1.177×105 Pa. What is the man’s, blood pressure reading, in torr?, P = Patm + ȡgh, , h = 0.123 m = 123 mm, systolic blood pressure = 123 torr, 13.7 - Archimedes’ principle, , Archimedes' principle: An object in a fluid experiences an upward force equal, to the weight of the fluid it displaces., Archimedes (287-212 BCE) explained why objects float. His principle states that buoyancy, the upward force caused by the displacement of, fluid, equals the weight of the volume of the fluid displaced. For instance, if a boat displaces 300 tons of water, then it experiences an upward, buoyant force of 300 tons. If this buoyant force is greater than the weight of the boat, the boat floats., Archimedes’ principle can be used to explain why a small stone sinks, while a large block of Styrofoam® floats, even if the Styrofoam block is, much heavier than the stone. Stone is denser than water, so the weight of the water it displaces is less than its own weight. This means that, the stone’s weight, directed down, is greater than the upward buoyant force on it. The net force on an underwater stone is downward., In contrast, Styrofoam is less dense than water. A Styrofoam block displaces a weight of water equal to its own weight when it is only partially, submerged. It is in a state of equilibrium since the buoyant force up equals the weight down. In short, it floats., Archimedes’ principle applies at any water depth. The buoyancy of a submerged submarine is the same whether it is 50 meters or 300 meters, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 255
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below the water’s surface, since the density (and weight) of the displaced water, changes only slightly with depth. If this is so, how can a submarine dive or surface? The, submarine changes its weight: It either adds weight by allowing water into its ballast, tanks or reduces its weight by blowing the water out with compressed air., Fish approach the issue in a slightly different way. They change their volume by inflating, or deflating an organ called a swim bladder, filled with gas released from the blood of, the fish. When they increase their volume, they rise because they displace more water, and experience increased upward buoyancy., Archimedes’ principle can also be used to analyze the buoyancy of human beings., People with a high percentage of body fat float more easily than do their slimmer, counterparts. This is because fat is less dense than water, while muscle is denser than, water. In one test for lean body mass, a person is weighed out of water, and then, weighed again while submerged. The difference in the two weights equals the buoyant, force, which allows a calculation of the volume of the displaced water. The average, density of the person, based on his volume and his dry weight, can be used to, determine what percentage of his body is fat., , Archimedes’ principle, Buoyant force:, ·upward force on an object in a fluid, ·equals weight of the displaced fluid, , Triathletes, whose body fat is likely to be very low, demonstrate their appreciation of the, principles of buoyancy by preferring to wear wetsuits during swimming events. Lean, people tend to sink, and a wetsuit helps an athlete float since it is less dense than, water, reducing the energy spent on staying up and allowing more to be spent on, moving forward. Because of this effect, triathlons ban wetsuits in warm water events, where they are not strictly necessary for survival. You wouldn’t want to give those, triathletes any breaks before they bike 180 kilometers and then run over 40 kilometers!, Objects fabricated from materials denser than water can float. A steel boat floats since, its hull encloses air, which means the average density of the volume enclosed by the, boat is less than that of water. Observe what happens when someone steps into a small, boat: It sinks slightly as more water is displaced to balance the person’s weight. If the, boat is overloaded with cargo, or if water enters the hull, its average density will surpass, that of water and the boat will sink (fast-forward to the end of the film A Perfect Storm, for a graphic example of the latter problem)., Often, the concept of buoyancy is applied to water, but it also applies to other fluids,, such as the atmosphere. Blimps and hot air balloons use buoyancy to float in the air. A, blimp contains helium, a gas lighter than air. The weight of the air it displaces is greater, than the weight of the blimp, so it floats upward until it reaches a region of the, atmosphere where the air is less dense and the weight of the blimp equals the weight of, the displaced air., , This chunk of African ironwood, weighs 43 N, and it displaces, 0.0030 m3 of water. What is the, buoyant force on it?, m = ȡV, m = (1000 kg/m3)(0.0030 m3), m = 3.0 kg, mg = (3.0 kg)(9.80 m/s2) = 29 N, F = 29 N, directed up, , 13.8 - Sample problem: buoyancy in water, A stainless steel hook with a worm, dangles underwater at the end of a, fishing line., What is the net downward force that, the bait combination exerts on the, line?, , A stainless steel fishing hook with a worm is in the water at the end of a line, dangled with the intent of attracting the wily fish. For the density of, the hook we use the density of stainless steel, 7900 kg/m3., We will refer to the combination of the hook and worm as “the bait.”, , 256, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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Variables, net force of bait on line, , F, , acceleration of gravity, , g = 9.80 m/s2, , buoyant force on bait combination, , Fb, , hook, , worm, , kg/m3, , ȡw = 1100, , displaced water, , kg/m3, , ȡH2O = 1000 kg/m3, , density, , ȡh = 7900, , volume, , Vh = 2.4×10í8 m3, , Vw = 7.1×10í7 m3, , VH2O, , mass, , mh, , mw, , mH2O, , weight, , mh g, , mw g, , mH2Og, , Strategy, 1., , Calculate the weight of the hook, a downward force., , 2., , Calculate the weight of the worm, another downward force., , 3., , From the combined volumes of the hook and worm, compute the volume and the weight of the displaced water. Use this value for the, upward buoyant force Fb on the bait combination., , 4., , With all the contributing downward and upward forces known, calculate the net force exerted by the bait on the line., , Physics principles and equations, Use the definitions of density and weight,, , Archimedes’ principle states that the upward buoyant force on the bait equals the weight of the water it displaces., Step-by-step solution, Calculate the weight of the hook., , Step, , Reason, , 1., , mh = ȡhVh, , definition of density, , 2., , mhg = ȡhVhg, , multiply by g, evaluate, , 3., , Calculate the weight of the worm., , Step, , 4., , mw g = ȡwVwg, , Reason, , equation 2, for the worm, evaluate, , 5., , Calculate the weight of the displaced water, and from that the buoyancy of the bait combination., , Step, , Reason, , 6., , VH2O = Vh + Vw, , add volumes, , 7., , mH2Og = ȡH2O(Vh + Vw)g, , substitute equation 6 into equation 2, , 8., , evaluate, , 9., , Archimedes’ principle, Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 257
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Finally, the net force is the sum of the buoyancy upward and the weight of the bait downward., , Step, , Reason, , 10. F = Fb +(–mh g) +(–mw g), , net force, , 11., , evaluate, , The negative value of the net force indicates that the bait combination is pulling down on the line., The quantities involved in this problem are rather small: The weight of the bait, its buoyancy and the net downward force all have magnitudes in, the thousandths of newtons. Bait like this will sink, but not very quickly, due to the resistance of water to its motion. For this reason, fishermen, often use lead weights called “sinkers” to cause the bait to sink faster to a depth where fish are feeding., , 13.9 - Sample problem: buoyancy of an iceberg, What fraction of this iceberg is, submerged below the water?, , You may have heard the expression “it’s just the tip of the iceberg”; icebergs are infamous for having nine-tenths of their volume submerged, below the surface of the sea. The composite photograph above shows just how dangerous icebergs can be for navigation. The submerged, portion not only extends downward a great distance, it may also extend sideways to an extent that is not evident from above. A ship might, easily strike the submerged portion without passing especially close to the visible ice., Use the values stated below for the densities of ice and of seawater at í1.8°C, which is the freezing point of seawater in the arctic. The, iceberg is in static equilibrium, moving neither up nor down., Variables, The upward buoyant force on the iceberg is Fb., iceberg, , displaced water, , kg/m3, , ȡH2O= 1030 kg/m3, , density, , ȡice = 917, , volume, , Vice, , VH2O, , mass, , mice, , mH2O, , Strategy, 1., , 2., , Use equilibrium to state that the buoyant force acting on the iceberg equals its weight. Archimedes’ principle allows you to express the, buoyant force in terms of the weight of the displaced water. Replace the equilibrium equation with one stating that the mass of the, iceberg equals the mass of the displaced water., Use the definition of density to replace the masses in the previous equation by products of density and volume. Solve for the ratio, , VH2O/Vice, and evaluate it., Physics principles and equations, Newton’s second law., , ȈF = ma, Archimedes’ principle states that the buoyant force on an object in a fluid equals the weight of the fluid it displaces., The definition of density is, , 258, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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Step-by-step solution, We begin by stating the condition for static equilibrium, and then we apply Archimedes’ principle., , Step, , 1., , Reason, , equilibrium, , 2., , mH2O g = mice g, , Archimedes, , 3., , mH2O = mice, , simplify, , Now we replace the masses in the previous equation by products of densities and volumes, solve for a ratio of volumes, and evaluate., , Step, , 4., , ȡH2OVH2O = ȡiceVice, , Reason, , definition of density, , 5., , rearrange, , 6., , evaluate, , Since the volume VH2O of the displaced water equals the volume of the submerged portion of the iceberg, we have shown that 89.0% of the, iceberg is submerged., For most substances, the solid phase is denser than the liquid phase, meaning a solid sinks when immersed in a liquid composed of the same, substance. Water is very unusual in this respect: the solid phase (ice) floats in liquid water. This proves crucial to life on Earth for a variety of, reasons., , 13.10 - Interactive problem: Eureka!, In this simulation you play the role of the ancient Greek mathematician and physicist, Archimedes. As the story goes, in olden Syracuse the tyrant Hieron suspected that, a wily goldsmith had adulterated one of his kingly crowns during manufacture by, adding some copper and zinc to the precious gold. His Royal Highness asked, Archimedes to discover whether this was indeed the case., Archimedes pondered the problem for days. Then, one day, in the public bath, he, observed how the water level in the pool rose as he eased himself in for a good, soak, and suddenly realized that the answer was right in front of his eyes. Ecstatic,, he supposedly leapt from the bath and ran dripping through the streets of the city, crying “Eureka!” (I have found it!) The adulterated crown was quickly identified, and, the goldsmith roundly punished., Your task is to determine the nature of Archimedes’ insight and apply it in the, simulation to the right. You have two crowns and a bar of pure gold. You have, already observed their masses using a balance scale. You can measure the volume, of a crown or bar in the simulation by dragging it to the bath and noting how much, water it displaces. You see the tube that measures the displaced liquid in the, illustration at the right. Decide which crown has been altered and drag it to the king’s palace. The appropriate consequences will be enacted., , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 259
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13.11 - Pascal’s principle, , Pascal's principle: Pressure in a confined fluid, is transmitted unchanged to all parts of the fluid, and to the containing walls., If you jump down on one side of a waterbed, a person sitting on the other side will get a, jolt up. This illustrates Pascal’s principle: A variation in pressure in the enclosed fluid is, being transmitted unchanged throughout the fluid., In the illustration to the right, you see a five-kilogram mass “balancing” a 50-kilogram, mass. How does something that seems so counterintuitive í a small mass balancing a, large mass í occur?, First, Pascal’s principle asserts that the pressure exerted by the weight of the first, mass on the fluid is transmitted to the second mass unchanged. The two masses, “balance” because the surface area of the fluid under the 50-kilogram mass is 10 times, larger than the surface area under the five-kilogram mass. The pressure is the same, under both masses, but since the surface area is 10 times larger under the more, massive object, the upward force, the pressure times the area, is 10 times greater than, it is under the less massive object, so the system is in equilibrium. Although we focus, on the pressures supporting the weights, according to Pascal’s principle the pressure, acts in all directions at every point in the fluid, so it presses on the walls of the hydraulic, system as well., , Pascal’s principle, An enclosed fluid transmits pressure:, ·unchanged, and, ·in all directions, , In some ways, the equilibrium in such a system is similar to a small mass located at the, end of a long lever arm balancing a large mass located close to the lever’s pivot point., In fact, this similarity has caused systems like those shown to the right to be called, hydraulic levers. You may be familiar with them from the “racks” in car repair shops:, they are the gleaming metal pistons that lift cars., Illustrations of hydraulic levers are shown to the right. In each case the pressure on, both pistons must be the same in order to conform to Pascal’s principle, but the force, will depend on the areas of the pistons. In the example problem you are asked to, calculate the downward force needed on the left piston to lift the small automobile on, the right. It turns out that this force is about 20 newtons (only 4½ pounds)!, Do you get “something for nothing” when you raise a heavy car with such a small force?, No, the amount of work you do equals the work the piston on the right does on the car., You apply less force, but through a greater displacement. If you press your piston down, a meter in the scenario shown in Example 1, the car will rise only about 1.6 millimeters,, as you can verify by considering the incompressible volume of water displaced on each, side., , P1 = P2, F1/A1 = F2/A2, P = pressure, F = force, A = surface area, , How much force must be exerted, on the left-hand piston to lift the, automobile on the right?, P1 = P2, F1/A1 = F2/A2, F1 = mgA1/A2, , 260, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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F1 = 20.4 N, 13.12 - Streamline flow, , Streamline flow: A fluid flow in which the, fluid’s velocity remains constant at any, particular point., Steady, or streamline, flow is one of the characteristics of ideal fluid flow. Streamline, flow is particularly easy to demonstrate with the flow of a gas, although since gasses, are compressible they are not ideal fluids., At the right, you see one streamline traced by smoke in the diagram, flowing around an, automobile. As long as the car does not rotate, the streamline stays the same over time., Any particle of the fluid will follow some streamline, visible or not, as it flows past the, car. As the car is rotated in the video, you get a chance to see the paths followed by, different streamlines of air flowing around various parts of its body., , Streamline flow, At different points, velocities can differ, At any point, velocity constant over time, , Engineers use wind tunnels to photograph streamlines. Powerful fans blow air past an, Image courtesy of Lexus, object like a car or an airplane, and dyes or smoke are injected into the airflow at, several points and carried downstream so that the streamlines are made visible., Engineers analyze the streamlines to investigate the air resistance of a particular car design, or the amount of lift (upward force) generated by, an airplane wing., The velocity of streamline fluid flow can vary from point to point. Air moves past an airplane wing or auto body with different velocities at, different points (for example, it moves faster over the tops of these objects than beneath them). The tangent to a streamline at a point coincides, with the direction of the velocity vectors of the fluid particles passing by the point., In streamline flow, the fluid has a constant velocity at all times at a given point. The velocity of any given air particle in the visible streamline, on the right may change as the flow carries it downstream past the stationary automobile. In particular, a change in the direction of the, streamline reflects a change in velocity. However, all the particles in the streamline pass the same point with the same velocity., How can you conclude that the velocity remains constant at each point? Consider what would happen if the speed of the fluid flow at a point, were to vary over time. If the speed increased, the affected particles would collide with particles ahead of them; if it decreased, particles from, behind would collide. The resulting collisions would cause an erratic flow, changing the streamline, as would changes in the direction of the, particles’ motions. The constancy of the streamlines over time indicates that the velocity at each point does not change., , 13.13 - Fluid continuity, , Fluid equation of continuity: The volume flow, rate of an ideal fluid flowing through a closed, system is the same at every point., Turn on a hose and watch the water flow out, and then cover half the hose end with, your thumb. The water flows faster through the narrower opening. You have just, demonstrated the fluid equation of continuity: How much volume flows per unit time í, the volume rate of flow í stays constant in a closed system. The increased speed of the, flow at the opening balances the decreased cross sectional area there. Of course after, the water leaves the hose with its new speed, it is no longer in the closed system, and, you cannot apply the equation of continuity to the resulting spray of droplets. They, spread out without slowing down., , Fluid continuity, Fluid flows past each point at same rate, , The fluid equation of continuity can be observed in rivers, whose courses approximate, closed systems. River water flows more quickly through narrow or shallow channels, called rapids, and more slowly where riverbeds are wider, and deeper. This relationship is alluded to by the proverb, “Still waters run deep.”, The constancy of fluid flow rate is summarized in the continuity equation that appears as the first line in Equation 1. In this equation, stated for, two arbitrary points P1 and P2, the rate of flow is measured as the mass flow rate. The mass flow rate equals the product of the speed of the, fluid, its density, and the cross sectional area it flows through. It is measured in kilograms per second (kg/s). In a closed system (no leaks, no, inflows) the mass flow rate is the same past all points., Since we assume that an ideal fluid is incompressible, having a density that is constant, we can cancel the density factor from both sides of the, first continuity equation. This enables us to say that the speed of the fluid times its cross sectional area is everywhere the same. This is stated, by the second equation, which expresses continuity in terms of the volume flow rate, represented by R. The volume flow rate is measured in, cubic meters per second (m3/s). If the cross sectional area decreases (as when the pipe illustrated in Equation 1 narrows), the speed of the, fluid flow increases, and R remains the same., , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 261
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13.14 - Bernoulli’s equation, Bernoulli’s equation applies to ideal fluids. It was developed by the Swiss, mathematician and physicist Daniel Bernoulli (1700-1782). The equation is used to, analyze fluid flow at different points in a closed system. It states that the sum of the, pressure, the KE per unit volume, and the PE per unit volume has a constant value., Concept 1 shows an idealized apparatus for determining these three values at various, points in such a system., A simplified form of Bernoulli’s equation is shown in Equation 1. It applies to horizontal, flow, in which the PE of the fluid is everywhere the same. In such a system, the sum of, just the pressure and the kinetic energy per unit volume is constant. Since the, expression for the “KE ” uses the density ȡ of the fluid in place of mass, it describes, energy per unit volume: the kinetic energy density., To illustrate the simplified equation, we use the horizontal-flow configuration shown in, the diagram of Equation 1. The pressure of the fluid is measured where it passes the, gauges. If the speed of the fluid is known at the first gauge, its speed at the second, gauge can be calculated using Bernoulli’s equation., When the sum of the pressure and kinetic energy density equals a constant in a, system, as in the case of horizontal flow, it is often useful to set the sum of these values, at one point equal to the sum of the values at another point. This is stated for points P1, and P2 in Equation 2. An implication of the simplified form of Bernoulli’s equation is the, , Bernoulli’s equation, The sum of:, ·pressure, ·KE / unit volume, ·PE / unit volume, equals a constant in a closed system, , Bernoulli effect: When a fluid flows faster, its pressure decreases., Airplane wings, like the one shown in Equation 2, take advantage of the Bernoulli effect., Air travels faster over the upper surface of the wing than the lower, because it must, traverse a longer path in the same amount of time. A faster fluid is a lower pressure, fluid; the result is that there is more pressure below the wing than above. This causes a, net force up, which is called lift. (The Bernoulli effect is only one way to explain how, wings work. Lift can also be explained using Newton’s third law in conjunction with a, fluid phenomenon called the Coanda effect. The topic of wing lift engenders much, discussion.), In Equation 3 you see a general form of Bernoulli’s equation, which also accounts for, differences in height and the resulting differences in potential energy density. It states, that the sum of the pressure and the kinetic energy density, plus the potential energy, density, is constant in a closed system. At a higher point in such a system, the potential, energy density of the fluid is greater. At that point either the pressure or the kinetic, energy density, or both, must be less than they are at lower points., , For horizontal flow, P + ½ȡv2 = k, P = pressure, ȡ = constant density of fluid, v = speed of flow, k = a constant for the system, , The “Bernoulli effect”, P1 + ½ȡv12 = P2 + ½ȡv22, If v2 > v1, then P1 > P2, ·Net upward force on wing is “lift”, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 263
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General form of the equation, Includes potential energy density, , g = acceleration of gravity, h = height, , What is the speed of the fluid at, point 1?, P1 + ½ȡv12 = P2 + ½ȡv22, , v1 = 0.894 m/s, 13.15 - Physics at work: Bernoulli effect and plumbing, , Drain traps in a plumbing system, Prevent the spread of noxious gases, , Plumbing system with vented sink trap., , The Bernoulli effect can help you understand not only the lofty topic of how airplanes fly, but also the more mundane topic of venting in, household plumbing systems., , 264, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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The illustration above shows a shower, a sink, their wastewater pipe, and a “vent” pipe., Water can flow down the shower drain, through a P-shaped “water trap,” and from there, to a wastewater pipe that connects to a sewer line. The water trap retains a seal of, water that prevents noxious sewer gas from flowing up the waste pipe into the house., The sink in the illustration is likewise connected through a protective P-trap to the waste, pipe., In the illustration, the sink drain is “vented” by an air pipe that extends through the roof, of the house. Because it is called a vent pipe, you might think it allows air to leave, but it, actually allows air in. Several such vents can be seen on the roofs of most houses. The, vents work in conjunction with the water traps that protect the home. What purpose, does a vent serve?, Waste pipes are not pressurized like water supply pipes, and they are always at least, slightly inclined so that they ordinarily stand empty. Imagine that someone turns on the, shower, causing a stream of water to flow through the waste pipe. As the flow, increases, the pressure decreases, in accordance with the Bernoulli effect. If the, pressure in the waste pipe were to decrease enough, it could “suck” the water out of the, sink’s P-trap, thwarting its protective function., , Traps are vented, To counter Bernoulli effect in, wastewater pipes, , This is where the vent comes in: When the pressure decreases at the bottom of the sink’s waste connection due to the shower flow, air flows in, from the exterior (due to atmospheric pressure), restoring the pressure in the connection. The water remains in place in the trap., , 13.16 - The Earth’s atmosphere, The atmosphere is the layer of gas, including the oxygen we need in order to survive,, surrounding the Earth and bound to it by gravity. The atmosphere receives a great deal, of press these days due to environmental concerns such as global warming and ozone, holes. Understanding the nature and dynamics of the atmosphere is proving, increasingly important to human life., In this section we give a brief overview of three topics relating to the Earth’s, atmosphere. The first is the magnitude of atmospheric pressure, and how its existence, was first demonstrated. Next, we discuss briefly why there is no hydrogen in the Earth’s, atmosphere (unlike the atmosphere of Jupiter). Finally, we provide a general sense of, the range of pressures, densities and temperatures of the Earth’s atmosphere at, different altitudes., The very existence of atmospheric pressure was once a topic of debate. After all, you, do not “feel” air pressure (since the fluid inside your body exerts an equal and opposite, pressure) any more than a fish feels water pressure. (At least no fish has ever told us, that it feels this pressure.) In a famous demonstration that showed the existence of, atmospheric pressure, the German scientist Otto von Guericke created a near vacuum, between two copper hemispheres, as shown in Concept 1. Von Guericke, also the, mayor of the town of Magdeburg, where he conducted the demonstration in 1654,, challenged teams of horses to pull the hemispheres apart. The horses failed: The force, of the air pressure was too great for them to overcome., , Air pressure demonstration, Near vacuum inside sphere, Horses pull against air pressure, Air pressure can cause great force!, , This clever demonstration of air pressure may seem incredible until you consider that, the air pressure on the Earth’s surface is about 101 kilopascals, (14.7 pounds per square inch). To get an approximate value for the pressure holding, the hemispheres together, we can simplify matters and assume they acted like halves, of a cube, one meter on a side. We will further assume that a perfect vacuum was, created inside them, so the pressure inside was zero. Each team of horses would then, have had to overcome 101,000 N of force, more than 10 tons. Whoa, Nellie!, Von Guericke demonstrated the surprisingly large magnitude of atmospheric pressure., He might have been startled to know how fast the particles that make up the, At greater altitudes, atmosphere move. Sunlight energizes the molecules in the atmosphere, keeping it in a, Pressure, density, temperature, gaseous state, and causes them to fly energetically about, reaching speeds of up to, decrease, 1600 km/h. The gravitational force of the Earth keeps most of these molecules from, flying off into space, but some molecules í especially the lightest ones í do reach, escape velocity and leave the planet’s atmosphere. This is why there is little or no hydrogen, or any other light gas, in the Earth’s atmosphere., On the average, a lighter molecule moves faster than a heavier molecule of the same energy, meaning the lightest molecules most easily, reach the 11.2 km/s required to escape the Earth’s gravitational pull. The escape velocity is greater on Jupiter, allowing that planet to keep, hydrogen in its atmosphere., The Earth’s atmosphere is a gas, which makes it a fluid. However, since it is a gas, its density changes with its height above the planet’s, surface. The weight of the atmosphere above “compacts” the atoms and molecules below, increasing their concentration (density). We live in, an ocean of air, just as fish dwell in an ocean of water. An important difference between the two is that water is relatively incompressible and, the ocean has essentially the same density, although different temperatures and pressures, at any depth., The diagram in Concept 2 shows the Earth’s atmosphere schematically. The pressure and density of the atmosphere lessen, as does its, temperature, with height. The density of air at sea level and 15°C equals 1.23 kg/m3; at the higher and chillier altitude of 30,000 meters, where, , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 265
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the temperature is í63°C, the density equals 0.092 kg/m3. At the summit of Mount Everest, a human can breathe barely enough oxygen to, stay alive., Conversely, in deep diamond mines, air pressure and density both increase, and the temperature rises rapidly with depth. In the deepest, mines, it would be impossible for miners to work without the introduction of refrigerated air., , 13.17 - Surface tension, , Surface tension: A cohesive, effect at the surface of a, liquid due to the forces, between the liquid’s atoms or, molecules., Above, you see a dewy rose. The drops of water do, not spread and flow to cover the petals on which they, Dewdrops on rose petals form tiny spheres., rest. Instead they contract into tiny spheroids. They, do this because the surface tension of water causes, each drop to try to minimize its own surface area. The bristles of a wet paintbrush, contract into a sleek shape for the same reason. The surface tension of the water on, the bristles causes them to pull in. You see this in Concept 1., Surface tension also enables an insect called a water strider to walk on water. The, creature’s weight, transmitted to the water’s surface through its feet, causes tiny, depressions, but the feet do not break through. The surface tension of the deformed, liquid surface provides an elastic-like restoring force that balances the insect’s weight. A, video of this interesting phenomenon is shown in Concept 2., What causes surface tension? In some liquids like water, the molecules that make up, the liquid attract each other. These molecules are dipoles; they have regions of positive, and negative electrostatic charge. The positive pole of each molecule is attracted to the, negative poles of neighboring molecules, and vice versa., Molecules in the interior of the liquid experience equal attractions in all directions, and, so they experience no net intermolecular force. Molecules at the surface of the liquid,, however, are pulled on only by the molecules below, and by their neighbors in the, surface, so there is a net force pulling them into the interior of the liquid. Because of, this, the surface of the liquid tends to contract and consequently to minimize its own, area. You see the intermolecular forces illustrated in Concept 1., , Surface tension, Contraction of surface of liquid, Due to mutual attraction of molecules, , In the case of liquid dewdrops, the minimum surface area is roughly spherical. In the, case of the water strider, the elastic-like upward force of surface tension on its feet, results from the water’s tendency to flatten, and so minimize the area, of its surface., Water has a strong surface tension, but that tension can be reduced. For example,, when you heat water, you reduce its surface tension because the faster moving, molecules do not attract each other as much as they would at cooler temperatures., Diminished surface tension allows other substances that might be in the water í say,, butter í to rise to the surface. Cooks know this (at least implicitly), and they serve soups, hot because they will be more flavorful. Adding soap to water also reduces its surface, tension í and it can cause a water strider to sink!, , Water strider, Surface tension lets insect walk on, water, Restoring force balances weight, , 13.18 - Gotchas, Pressure increases with depth in a fluid. Yes, it does. The farther below the surface of a fluid an object is, the more fluid above it and the, greater the pressure on it., Buoyant force increases with an object’s depth in water. Only if more of the object is getting submerged, in which case the buoyant force does, increase. But a soda can five meters below the surface, and one 500 meters below the surface, both experience the same buoyant force. This, force is equal in magnitude to the weight of the displaced water., Two surfaces have the same pressure on them, so the pressure must exert the same force on each. No, pressure is force divided by area, so if, one surface has a greater area, it experiences a greater force., Pressure acts in every direction. Yes, this is stated by Pascal’s principle. For instance, water pressure pushes both down on the top and up on, the bottom of a scuba diver exploring under the sea., , 266, , Copyright 2000-2007 Kinetic Books Co. Chapter 13
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13.19 - Summary, Fluids are substances, liquids and gases, that can flow and conform to the shape of, the container that holds them., , Definition of density, A material’s density is the amount of mass it contains, per unit volume. Density is, represented by the Greek letter ȡ and has units of kg/m3. Unless otherwise stated,, substances are assumed to be of uniform density, which means they have the same, density at all points., , ȡ = m/V, Definition of pressure, , Pressure is the amount of force on a surface per unit area. The unit of pressure is, the pascal (Pa), which is equal to 1 N/m2., , P = F/A, , Fluids can exert forces, and therefore pressure, just as solid objects can. For an, object immersed in a liquid, the pressure is the product of the liquid’s density, the, acceleration of gravity g, and the object’s depth. Gauge pressure is the pressure, due solely to the liquid, while absolute pressure is the pressure of the liquid plus, atmospheric pressure., , Pressure of a liquid, , Buoyancy is the upward force that results when an object is placed in a fluid, the, force that causes a ship to float. Archimedes’ principle states that the magnitude of, the buoyant force is equal to the weight of the fluid displaced by the object., Pascal’s principle applies to confined fluids. It states that an enclosed fluid will, transmit pressure unchanged in all directions., To simplify the study of fluid flow, we often assume an ideal fluid flow. This means, that the flow is streamline flow í it has a constant velocity at every fixed point í and, that it is irrotational. The fluid is also assumed to be incompressible and nonviscous., One property of ideal fluid flow is stated by the fluid equation of continuity. The, amount of fluid flowing past every point in a closed system is the same. In other, words, the volume flow rate is constant regardless of the size of the area through, which the fluid flows., Another property of ideal fluid flow is described by Bernoulli’s equation. The sum of, the pressure, kinetic energy density, and potential energy density is constant in a, closed system. For horizontal flow, the faster a fluid flows, the lower its pressure., This is called the Bernoulli effect., The atmosphere exerts pressure because it is a fluid. But since it is a gas, air, density and pressure decrease noticeably with altitude. Despite often being ignored, in day-to-day life, air pressure is actually (and demonstrably) quite large., , P = ȡgh, Pascal’s principle, , P1 = P2, F1/A1 = F2/A2, Fluid equation of continuity, , v1ȡA1 = v2ȡA2, v1A1 = v2A2 = R, Bernoulli’s equation, , For horizontal flow: P + ½ ȡv2 = k, P1 + ½ ȡv12 = P2 + ½ ȡv22, General form:, P1 + ½ ȡv12 +ȡgh1, = P2 + ½ ȡv22 +ȡgh2, , Surface tension is an effect seen with certain liquids such as water. Polar molecules, on the surface of the liquid are attracted toward the interior by unbalanced, intermolecular forces, causing the surface to contract, and consequently to minimize its own area and exhibit some elastic-like properties., , Copyright 2000-2007 Kinetic Books Co. Chapter 13, , 267
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Chapter 13 Problems, , Chapter Assumptions, Unless stated otherwise, use the following values:, Atmospheric pressure at the Earth's surface: Patm, , = 1.013×105 Pa, , Density of pure water = 1000 kg/m3, Density of seawater = 1030 kg/m3, “Standard temperature and pressure” means 0°C and the atmospheric pressure stated above., , Conceptual Problems, C.1, , You are standing in your driveway. You measure the pressure inside a bicycle tire with your tire gauge and get a reading of, 60 psi. You then don your space suit, take the tire into outer space and repeat the measurement, this time getting a reading of, 75 psi. Air was neither added to nor removed from the tire, and its temperature did not change. What explains this, discrepancy?, , C.2, , Scuba divers are instructed to exhale slowly but continuously as they rise to the surface in an emergency situation (such as, losing a tank). How is it possible for them to do this?, , C.3, , If an astronaut took a full bottle of water and ejected it from the pressurized interior of the International Space Station out into, space, what would happen? Your friend claims that there would be a sudden overpressure of almost 15 psi from inside the, bottle, and that it would expand and explode violently. Do you agree? Explain your answer., Yes. The bottle will explode., No. The bottle will not explode., , C.4, , Lurid science fiction stories sometimes dramatize a deep-space event known as "explosive decompression": The villain, ejects an innocent victim, without a spacesuit, from a spaceship's airlock, and the victim's eyes bug out and then he or she, explodes. Such an event is enacted on the nearly airless surface of Mars in the Schwarzenegger film Total Recall. These, scenes are inaccurate, but if you are suddenly ejected into space, you should be concerned about the danger of a sudden, expansion of a substance in your body. What is this substance?, , C.5, , A barometer is constructed by inverting a closed-end tube full of mercury and submerging its open end in a mercury reservoir, that is open to the air. The mercury in the tube will sink, leaving a vacuum at the closed end, until the system reaches, equilibrium with the atmospheric pressure outside the reservoir. For a given atmospheric pressure, does the height of the, column depend on the diameter of the tube? Explain., , C.6, , A barometer is constructed using a closed-end tube containing a vacuum above a column of mercury, as described in the, , Yes, , No, , textbook. On a certain day, the pressure exerted solely by this column of mercury at the bottom of the tube is 1.024×105 Pa., Check each of the following quantities that are equal to this measurement., Absolute pressure at bottom, Atmospheric pressure, C.7, , In a legendary and probably apocryphal story, a physics professor poses a question on a test, "How would you use a, barometer to determine the height of a tall building?" In the story, a brilliant but rebellious physics student artfully avoids giving, the "correct" answer but gives instead a long list of plausible alternative answers, including the following..., The kinematic answer: I would drop the barometer from the top of the building and time its fall. The equation, , ǻy = vit + (1/2)at2 would then tell me the building's height., The pendulum answer: I would tie the barometer to a long string, lift it slightly above the ground, and swing it from the top of, the building. The equation, , would then tell me the building's height., , The geometric answer: On a sunny day, I would measure the height of the barometer, the length of its shadow, and the, length of the building's shadow. I would then use similar triangles to compute the building's height., The human-engineering answer: I would go to the building manager and say, "I have here a fine scientific instrument that I, will give to you if you tell me the building's height!", What answer was the professor really looking for?, , 268, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems
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C.8, , An empty boat is placed in a freshwater lake and a mark is painted on the hull at the waterline, a line corresponding to the, surface of the water when the vessel is floating upright. The same boat is then transported to Jupiter, and placed into a pool, of fresh water that has been prepared just for this comparison experiment. The acceleration due to gravity on Jupiter is 2.6, times what it is on Earth. The new waterline is noted on Jupiter. Compared to the waterline mark on Earth, where is the new, waterline mark located on the hull of the boat? Ignore atmospheric effects., The new waterline mark is, , C.9, , i. higher, ii. lower, iii. at the same place, , on the hull., , An empty boat is placed in a freshwater lake and a mark is painted on the hull at the waterline, a line corresponding to the, surface of the water when the vessel is floating upright. The boat is then transported to the Dead Sea, where the liquid density, is about 1.2 times that of fresh water due to the high concentration of salts. A waterline mark is noted in the Dead Sea., Compared to the first waterline mark, where is the new waterline mark located on the hull of the boat?, The new waterline mark is, , i. higher, ii. lower, iii. at the same place, , on the hull., , C.10 You hold a ping-pong ball and a steel ball bearing of the same diameter so that they are submerged underwater. Which one, experiences the greater buoyant force? Explain your answer., i. The ping-pong ball, ii. The ball bearing, iii. The buoyant forces are the same, , ., , C.11 A ping-pong ball and a steel ball bearing of the same diameter are thrown into a swimming pool. The ball floats, while the, bearing sinks. Which one experiences the greater buoyant force? Explain your answer., i. The ping-pong ball, ii. The ball bearing, iii. The buoyant forces are the same, C.12 A boat carrying a load of bricks is floating in a canal lock. One of the crew members throws a brick overboard, and it sinks., Does the level of the water in the lock rise or fall? Explain your answer., It rises, , It falls, , C.13 Explain how you would float a battleship in a thimbleful of water. This is a real question, not a trick question, except that you, have to make one idealizing assumption: Water does not consist of discrete molecules, but is a "fluid" at every scale of, magnitude., C.14 Susie is out fishing on a calm, sunny day. She hooks an old tire that is resting on the lake bottom, and hauls it into her boat., In principle, how does the water level of the lake change?, The lake's water level, , i. drops very slightly, ii. does not change, iii. rises very slightly, , C.15 Susie is out fishing again on another calm, sunny day. She sees a chunk of firewood floating in the water, and hauls it into, her boat. In principle, how does the water level of the lake change?, The lake's water level, , i. drops, ii. does not change, iii. rises, , C.16 Here is a trick that is often performed as a physics demonstration. A vacuum cleaner hose and wand are attached to the, "back end" of a canister-type vacuum cleaner so that the wand directs a fountain of air straight upward. Then a light ball is, placed on top of the fountain where it bobs around and even tumbles to the side, but always returns to the top of the fountain, without actually falling off. Explain how this trick works., C.17 People are warned not to stand near open doors on airplanes in flight, because they can get "sucked" out of the door. Explain, how this might happen., , Section Problems, Section 2 - Density, 2.1, , Calculate the average population density in people/km2 for each of the following geopolitical entities. (a) The United States,, whose population is 293×106 people, and land area is 9.37×106 km2. (b) The world, whose population is 6356×106 people,, and land area is 149×106 km2. (c) Siberia, whose population is 25.1×106 people, and land area is 13.5×106 km2. (d) Hong, Kong, whose population is 6.70×106 people, and land area is 1098 km2., , Copyright 2007 Kinetic Books Co. Chapter 13 Problems, , 269
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2.2, , (a), , people/km2, , (b), , people/km2, , (c), , people/km2, , (d), , people/km2, , The density of air at standard atmospheric pressure and 15°C is 1.23 kg/m3. (a) What is the total mass of the air in a, rectangular room that measures 5.25 m × 4.20 m × 2.15 m? (b) What is the weight of the air in the room? (c) What is the, weight of the air in the Pentagon office building in Washington, DC, which has a volume of 2.2×106 m3?, kg, , (a), , 2.3, , (b), , N, , (c), , N, , If you have ever toured a facility (such as a cyclotron laboratory) where people have to be protected against radiation, there, may have been lead bricks lying around, and you may have been given the opportunity to heft one. It is a surprising, experience. The dimensions of a standard brick are 9.2 cm × 5.7 cm × 20 cm, and the density of lead is 11,300 kg/m3. (a), What is the mass of a (standard) lead brick? (b) What is the weight of the brick, in pounds?, , 2.4, , (a), , kg, , (b), , lb, , In the opening sequence of the movie, Raiders of the Lost Ark, the intrepid explorer Indiana Jones deftly swipes a golden idol, from a Mayan temple, instantly replacing it with a bag of sand of the same size to neutralize the ancient weight-based booby, trap protecting it. (a) The density of gold is 19,300 kg/m3. If the idol's mass is 11.3 kg and it is solid gold, calculate its volume., (b) The density of silica sand is 1220 kg/m3. What is the mass of a sandbag of equal volume? (c) Is the scene realistic?, m3, , (a), (b), (c), , kg, Yes, , No, , Section 3 - Pressure, 3.1, , 3.2, , (a) A fashionable spike heel has an area of 0.878 cm2. When a 61.4 kg woman walking in this shoe sets her full weight down, on the heel, what is the pressure it exerts on the floor? (b) The heel of a "sensible" shoe has an area of 38.3 cm2. When the, same woman sets her weight on this heel, what is the pressure?, (a), , Pa, , (b), , Pa, , During the first half of the twentieth century, research into the behavior of materials at ultrahigh pressures was conducted, using huge hydraulic presses. In 1958 scientists had a sudden insight: instead of building ever larger presses to exert ever, larger forces on material samples of a given area, why not use relatively modest forces and an extremely small area? They, invented the diamond anvil pressure cell. This device incorporates two opposed diamonds (the "anvils") whose sharp points, have been slightly filed off to produce roughly circular flat surfaces with a diameter of 250 μm. A minute sample of material is, placed between these surfaces and the diamonds are pressed together using leverage. (a) If the diamonds are 2.15 cm from, the pivot point of the lever and a scientist bears down with a force of 255 N at a distance of 14.2 cm from the pivot point, what, force presses the diamonds together? (b) What is the pressure exerted on the sample?, , 3.3, , (a), , N, , (b), , Pa, , A 106 kg man is traveling through the snow. He finds that when he wears only his boots, the sole of each of which has an, area of 136 cm2, he sinks into the snow up to his calves. Figuring that he will never get to Grandma's house this way, he dons, his snowshoes, made of a lightweight flat material with no holes. Each snowshoe can be modeled as a rectangle, 37.0 cm, wide and 115 cm long, with a half-circle added to the front and the back as a "toe" and a "heel." (a) What pressure does the, man exert on the frozen crust of the snow when he puts his entire weight on one booted foot? (b) What pressure does the, man exert when he puts his weight on one snowshoe?, , 3.4, , (a), , Pa, , (b), , Pa, , An automobile has four tires, each one inflated to a gauge pressure of 24.0 psi, or 1.66×105 Pa. Each tire is slightly flattened, by its contact with the ground, so that the area of contact is 17.5 cm by 12.0 cm. What is the weight of the automobile?, N, , 3.5, , An advertisement for a certain portable vacuum cleaner shows off its power with a photograph of the vacuum-cleaner wand, suspending a bowling ball by the strength of its suction. The vacuum cleaner can maintain a moderate vacuum inside the, apparatus at an absolute pressure of 3.53×104 Pa (against an outside atmospheric pressure of 1.01×105 Pa) when the intake, wand is closed. The wand is a hollow metal cylinder with an inside diameter of 3.19 cm. What is the weight of the heaviest ball, , 270, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems
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the vacuum cleaner can lift?, N, 3.6, , Water is generally said to be nearly incompressible. The deepest part of the ocean abyss lies at the bottom of the Marianas, trench off the Philippines, at a depth of nearly eleven kilometers. At a depth of 10.0 km, the measured water pressure is an, incredible 103 MPa (that's megapascals). (a) If the density of seawater is 1030 kg/m3 at the surface of the ocean, and its bulk, modulus is B = 2.34×109 N/m2, what is its density at a depth of 10.0 km? (Hint: Use the volume stress equation from the study, of elasticity, ǻP = íB (ǻV/Vi), where the object undergoing the stress has an initial volume Vi, and experiences a change in, volume, ǻV, when the pressure changes by ǻP.) (b) By what factor does the density increase?, kg/m3, , (a), (b), , Section 4 - Pressure and fluids, 4.1, , Seawater has a density of 1030 kg/m3. The Marianas Trench is a deep undersea canyon in the Pacific Ocean off the, Philippines. Assuming the seawater is incompressible, and ignoring the contribution of atmospheric pressure, what is the, pressure in this trench (a) at a depth of 1.00 km? (b) at a depth of 5.00 km? (c) at a depth of 10.0 km? (Empirically measured, pressures are a little larger than those given by these calculations because seawater compresses slightly at great depths.), (a), , 4.2, , Pa, , (b), , Pa, , (c), , Pa, , A photograph in the text shows how a Styrofoam® cup gets crushed by great pressure deep under the surface of the sea., Before the cup was crushed, experimenters used colored pens to write data on it, including the absolute pressure (3288 psi), at the depth to which they planned to submerge the cup. (You can inspect this data by right-clicking at a spot on the, photograph and selecting Zoom In from the popup menu that appears.) At what depth below the ocean's surface is the, pressure equal to 3288 psi? Use the value 1030 kg/m3 for the density of seawater., m, , 4.3, , (a) The gauge pressure at the bottom of a particular column of water, open at the top, is equal to 1.013×105 Pa, which is, atmospheric pressure. How tall is this column? (b) The gauge pressure at the bottom of a particular column of mercury, open, at the top, is also equal to Patm. How tall is the mercury column? (c) Convert your answer to part b from meters into, millimeters. (d) Convert your answer to part b from meters into inches. (Meteorologists often express variations in atmospheric, pressure in terms of "inches" or "millimeters" of mercury.), m, , (a), , 4.4, , (b), , m, , (c), , mm, , (d), , in, , A creamy salad dressing is made up of heavy cream, corn oil, and vinegar (as well as a pinch of dry mustard, salt and, pepper). You put the cream, oil, and vinegar in a glass jar. The density of the cream is 994 kg/m3, the oil 880 kg/m3, and the, vinegar 1000 kg/m3. The salad dressing liquids separate into three layers. (a) What is the order of the layers from top to, bottom? (b) If the cream layer is 2.20×10í2 m tall, the oil layer 2.80×10í2 m tall, and the vinegar layer 1.60eí2 m tall, what is, the gauge pressure at the bottom of the jar?, (a), , (b), 4.5, , i., ii., iii., iv., v., , Cream/oil/vinegar, Oil/cream/vinegar, Oil/vinegar/cream, Vinegar/cream/oil, Vinegar/oil/cream, Pa, , To supply the plumbing system of a New York, office building, water needs to be pumped to a, tank on the roof, where its height will provide a, "head" of pressure for all the floors. The, vertical height between the basement pump, and the level of the water in the tank is 95.3 m., What gauge pressure does the pump have to, apply to the water to get it up to the tank?, Pa, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems, , 271
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Section 5 - Sample problem: pressure at the bottom of a lake, 5.1, , In the movie Creature from the Black Lagoon, the depth of the freshwater lagoon at its muddy and inscrutable bottom where, the Creature lurks is 15.2 m. (a) What is the gauge pressure at the bottom of the lagoon? (b) What is the absolute pressure at, the bottom of the lagoon? (c) Who played the Creature in this classic 1954 horror film?, (a), , Pa, Pa, , (b), (c), , 5.2, , i., ii., iii., iv., , Ben Chapman, Johnny Depp, Groucho Marx, Fay Wray, , Saturn's moon Titan is the largest moon in the solar system. Its mostly-nitrogen atmosphere exerts a pressure of 1.60×105 Pa, at the moon's surface (a pressure about 60% greater than Patm at the surface of the Earth). Scientists speculate that it may, have lakes and seas of "gasoline," a mixture of liquid methane and ethane. Suppose that there is a gasoline lagoon on Titan,, 43.7 m deep, and consisting of 63.7% methane and 36.3% ethane, which have densities of 465 kg/m3 and 570 kg/m3, respectively. (a) Assuming that the acceleration due to gravity on Titan is 1.35 m/s2, what is the gauge pressure at the bottom, of the lagoon? (b) What is the absolute pressure at the bottom of the lagoon?, (a), , Pa, , (b), , Pa, , Section 6 - Physics at work: measuring pressure, 6.1, , 6.2, , The gauge pressure at the bottom of a pint of beer, at a depth of 14.6 cm, is 1450 Pa. (a) What is the density of the beer? (b), What is the absolute pressure at the bottom of the pint?, (a), , kg/m3, , (b), , Pa, , (a) Convert 1.05 bar to torr, using the relationships 1 bar = 105 Pa and 1 atm = 1.013×105 Pa = 760 torr. Express your answer, to the nearest torr. (b) There is a photograph of a dial-type barometer in the textbook, calibrated in both millibars and, millimeters of mercury. On this barometer's scales, 1000 millibars (1 bar) appears to be approximately equal to 750 mm (750, torr) of mercury. Is this approximation good to the nearest torr?, (a), (b), , torr, Yes, , No, , 6.3, , The mercury column of an open tube manometer has a height of 0.0690 m. What is the absolute pressure inside the, manometer vessel?, , 6.4, , The pressure inside a manometer vessel is 1.42e+5 Pa. How high is the column of mercury in the open tube of the, manometer?, , Pa, , m, , Section 7 - Archimedes’ principle, 7.1, , The sizes of ships are commonly expressed in "tons displaced". If a naval vessel displaces 7.67e+3 tons, this means it, displaces that weight of water. What is the volume of the water displaced by this ship? Use the fact that 1 ton is equal to, 8.90×103 N, and take the density of seawater to be 1030 kg/m3., m3, , 7.2, , A cylindrical metal can with a radius of 4.25 cm is floating upright in water. A rock is placed in the can, causing it to sink, 0.0232 m deeper into the water. What is the weight of the rock?, , 7.3, , The dry weight of a man is 845 N. When he is submerged underwater, the reading on the scale is only 43.2 N. (a) What is the, , N, volume of the man? (b) What is the density of the man? (c) The density of fat is 209 kg/m3, and the density of the "lean" parts, of the human body (muscle, bones, and so on) is 2670 kg/m3. What percentage of the man's body mass is "lean body mass"?, (a), , m3, , (b), , kg/m3, , (c), , %, , Section 8 - Sample problem: buoyancy in water, 8.1, , 272, , You are fishing off a bridge. Suddenly you feel a tug on the vertical fishing line. Elated, you begin hauling in your catch at, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems
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constant speed. The creature rears its head above the water and it is...a rubber tire! (a) If the tire is made entirely of hard, rubber, with volume 6800 cm3, and density 1190 kg/m3, then what is the tension on your fishing line after you pull the tire out, of the water? Assume that the tire is made entirely of rubber, it is a tire (not an inner tube), and it is punctured so you are not, pulling up any water. (b) What is the tension on your fishing line before you pull the tire out of the water? Ignore any drag, forces from the water., , 8.2, , (a), , N, , (b), , N, , You are fishing off a bridge and feel a tug on the vertical line. This time, your lucky catch is an old boot. (a) Assume that the, boot is not punctured, so that as you lift it out of the water at constant speed, you haul up one bootful, or 7500 cm3, of water, along with the boot. If the neoprene rubber making up the boot has volume 435 cm3 and density 1240 kg/m3, then what is the, tension on your fishing line after you pull the boot out of the water? (b) What is the tension in your fishing line before you pull, the boot out of the water? Ignore any drag forces., (a), , N, , (b), , N, , Section 9 - Sample problem: buoyancy of an iceberg, 9.1, , A shipwrecked mariner is stranded on a desert island. He seals a plea for rescue in a 1.00 liter bottle, corks it up, and throws, it into the sea. If the mass of the bottle, plus the message and the air inside, is 0.451 kg, what percentage of the volume of the, bottle is submerged as it bobs away? Take the density of seawater to be 1030 kg/m3. For simplicity, assume the bottle and its, contents have a uniform density., %, , 9.2, , (a) A block of balsa wood is placed in water. The density of the wood is 125 kg/m3. What percentage of the block is, submerged? (b) A block of maple wood is placed in water. The density of the wood is 683 kg/m3. What percentage of the, block is submerged? (c) A block of ebony wood is placed in water. The density of the wood is 1200 kg/m3. What percentage, of the block is submerged?, %, , (a), (b), , %, , (c), , %, , Section 10 - Interactive problem: Eureka!, 10.1 Use the information given in the interactive problem in this section to determine which crown is not made of pure gold., Confirm your answer by using the simulation., The 3.20 kg crown, The 2.70 kg crown, , Section 11 - Pascal’s principle, 11.1 An automobile having a mass of 1750 kg is placed on a hydraulic lift in a garage. The piston lifting the car is 0.246 m in, diameter. A mechanic attaches a pumping mechanism to a much smaller piston, 1.50 cm in diameter, which is connected by, hydraulic lines to the lift. She pumps the handle up and down, slowly lifting the car. What is the force exerted on the small, piston during each downward stroke?, N, , Section 13 - Fluid continuity, 13.1 An incompressible fluid flows through a circular pipe at a speed of 15.0 m/s. The radius of the pipe is 5.00 cm. There is a, constriction of the pipe where the radius is only 3.20 cm. How fast must the fluid flow through the constricted region?, m/s, 13.2 The open end of a garden hose is directed horizontally, at a height of 1.25 m above the ground. Water issues from the hose, and follows a falling parabolic trajectory to strike the ground 2.41 m away. A gardener holding the hose wishes to water some, plants that are 5.12 m distant. What fraction of the hose end should she cover with her thumb? Assume that she continues to, hold the hose end horizontally at the same height, and be careful to tell the fraction covered, not the fraction left open., , Section 14 - Bernoulli’s equation, 14.1 A stream of water is flowing through the horizontal configuration shown. The speeds v 1 and v 2 are 2.95 m/s and 5.35 m/s,, respectively. The pressure P2 is 7.36×104 Pa. What is P1? (Hint: the numbers on the pressure dials are not correct - that, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems, , 273
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would be too easy!), Pa, , 14.2 A stream of alcohol (density 790 kg/m3) is, flowing through the pipeline shown. The, speeds v 1 and v 2 are 11.7 m/s and 15.8 m/s., The gauge pressures P1 and P2 are, 2.77×105 Pa and 1.15×105 Pa, and the height, h1 is 3.60 m. What is h2?, m, , 14.3 An open can is completely filled with water, to a depth of 20.6 cm. A hole is punched in the can at a height of 1.7 cm from the, bottom of the can. Bernoulli's equation can be used to derive the following formula for the speed of the water flowing from the, hole., , In this formula, h represents the depth of the submerged hole below the surface of the water. (a) How fast does the water, initially flow out of the hole? (b) How fast does the water flow when the can is half empty?, (a), , m/s, , (b), , m/s, , Section 16 - The Earth’s atmosphere, 16.1 The relationship between pressure and altitude in the Earth’s atmosphere is complicated by the fact that the density of air is, not constant, but decreases with height according to an equation called the law of atmospheres, which we do not show here., This law assumes that the atmosphere is at a constant overall Kelvin temperature T., Because of its nonconstant density and its ill-defined upper limit, the pressure of the atmosphere cannot be described in, terms of some atmospheric depth h below the “top” of the atmosphere by a simple equation like P = ȡgh. Instead, the, atmospheric pressure P is a function of the altitude y above the ground, as given by the following equation that can be, derived as an immediate consequence of the law of atmospheres:, , In this pressure equation, Patm is atmospheric pressure at sea level (y, , = 0), m is the average mass in kilograms of an, , atmospheric molecule, g is the acceleration due to gravity, and k is Boltzmann’s constant (k = R/NA = 1.38×10–23 J/K)., (a) If the dry atmosphere is composed of 78.1% nitrogen molecules (mN = 28.0 u), 21.0% oxygen molecules, (mO = 32.0 u), and 0.930% argon molecules (mA = 39.9 u), plus trace amounts of other substances, what is the average, mass of an atmospheric molecule in atomic mass units u?, (b) Convert the average atmospheric molecular mass from atomic mass units to kilograms, using the relationship, 1 u = 1.66×10–27 kg., (c) Now use the pressure equation given above, together with an overall atmospheric temperature of T = 264 K (which is, about –9 °C), to find the theoretical pressure of the atmosphere at 1000 m, 5000 m, and 10,000 m (a common cruising, altitude for commercial jetliners)., (a), , 274, , u, , (b), , kg, , (c), , Pa at 1000 m, , Pa at 5000 m, , Pa at 10,000 m, , Copyright 2007 Kinetic Books Co. Chapter 13 Problems
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14.0 - Introduction, This chapter will give you a new take on the saying, “What goes around comes, around.”, An oscillation is a motion that repeats itself. There are a myriad of examples of, oscillations: a child playing on a swing, the motion of the Earth in an earthquake, a, car bouncing up and down on its shock absorber, the rapid vibration of a tuning fork,, the diaphragm of a loudspeaker, a quartz in a digital watch, the amount of electric, current flowing in certain electric circuits, etc.!, Motion that repeats itself at regular intervals is called periodic motion. A traditional, metronome provides an excellent example of periodic motion: Its periodic nature is, used by musicians for timing purposes. Simple harmonic motion (SHM) describes a, specific type of periodic motion. SHM provides an essential starting point for, analyzing many types of motion you often see, such as the ones mentioned above., SHM has several interesting properties. For instance, the time it takes for an object, to return to an endpoint in its motion is independent of how far the object moves., Galileo Galilei is said to have noted this phenomenon during an apparently lessthan-engrossing church service. He sat in the church, watching a chandelier swing back and forth during the service, and noticed that the, distance the chandelier moved in its oscillations decreased over time as friction and air resistance took their toll. According to the story, he, timed its period í how long it took to complete a cycle of motion í using his pulse. To his surprise, the period remained constant even as the, chandelier moved less and less. (Although this is a well-known anecdote, apparently the chandelier was actually installed too late for the story, to be true.), To begin your study of simple harmonic motion, you can try the simulation to the right. A mass (an air hockey puck) is attached to a spring, and, glides without friction or air resistance over an air hockey table, which you are viewing from overhead. When the puck is pushed or pulled from, its rest position and released, it will oscillate in simple harmonic motion., A pen is attached to the puck, and paper underneath it scrolls to the left over time. This enables the system to produce a graph of displacement, versus time. A sample graph is shown in the illustration to the right. A mass attached to a spring is a classic configuration used to explain SHM,, and the graph of the mass's displacement over time is an important element in analyzing this form of motion., Using the controls, you can change the amplitude and period of the puck’s motion. The amplitude is the maximum displacement of the puck, from its rest position. The period is the time it takes the puck to complete one full cycle of motion., As you play with the controls, make a few observations. First, does changing the amplitude change the period, or are these quantities, independent? Second, does the shape of the curve look familiar to you? To answer this question, think back to the graphs of some of the, functions you studied in mathematics courses., , 14.1 - Simple harmonic motion, , Simple harmonic motion: Motion that follows a, repetitive pattern, caused by a restoring force, that is proportional to displacement from the, equilibrium position., At the right, you see an overhead view of an air hockey table with a puck attached to a, spring. Friction is minimal and we ignore it. The only force we concern ourselves with is, the force of the spring on the puck., Initially, the puck is stationary and the spring is relaxed, neither stretched nor, compressed. This means the puck is at its equilibrium (rest) position. Imagine that you, reach out and pull the puck toward you. You see this situation in Concept 1 to the right., The spring is pulling the puck back toward its equilibrium position but the puck is, stationary since you are holding onto it., , Simple harmonic motion, Repeated, consistent back and forth, motion, Caused by a restoring force, , Now, you release the puck. The spring pulls on the puck until it reaches the equilibrium, point. At this point, the spring exerts no force on the puck, since the spring is neither stretched nor compressed. As it reaches the equilibrium, point, the puck’s speed will be at its maximum. You see this in Concept 2., The puck’s momentum means it will continue to move to the left beyond the equilibrium point. This compresses the spring, and the force of the, spring now slows the puck until it stops moving. You see this in Concept 3. At this point, the puck’s velocity is zero., Both the displacement of the puck from the equilibrium position and the force on it are now the opposites of their starting values. The puck is as, far from the equilibrium point as it was when you released it, but on the opposite side. The spring exerts an equal amount of force on the puck, as it initially did, but in the opposite direction. The force will start to accelerate the puck back to the right., , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 275
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The motion continues. The spring expands, pushing the puck to the equilibrium point., The puck passes this point and continues on, stretching the spring. It will return to the, position from which you released it. There, the force of the stretched spring causes the, puck to accelerate to the left again. Without any friction or air resistance, the puck would, oscillate back and forth forever., As you may have noted, “equilibrium” means there is no net force present. It does not, mean “at rest” since the puck is moving as it passes through the equilibrium position. It, is where the spring is neither stretched nor compressed., The motion of the puck is called simple harmonic motion (SHM). The force of the spring, plays an essential role in this motion. Two aspects of this force are required for SHM to, occur. First, the spring exerts a restoring force. This force always points toward the, equilibrium point, opposing any displacement of the puck. This is shown in the diagrams, to the right: The force vectors point toward the equilibrium position., Second, for SHM to occur, the amount of the restoring force must increase linearly with, the puck’s displacement from the equilibrium point. Why can a spring cause SHM?, Springs obey Hooke’s law, which states that F = íkx. The factor k is the spring, constant and it does not vary for a given spring. As x (the displacement from, equilibrium) increases in absolute value, so does the force. For instance, as the puck, moves from x = 0.25 m to x = 0.50 m, the amount of force doubles. In sum, since a, spring causes a restoring force that increases linearly with displacement, it can cause, SHM., , At equilibrium, Force is zero, Speed is at maximum, , We have extensively used the example of a puck on an air hockey table here, but this is, just one configuration that generates SHM. For example, we could also hang the puck, from a vertical spring and allow the puck’s weight to stretch the spring until an, equilibrium position was reached. If the puck were then pulled down from this position, it, would oscillate in SHM, since the net force on the puck would be proportional to its, displacement from equilibrium but opposite in sign., , Far position, Force is equal/opposite initial force, Speed is zero, , Restoring force, Proportional to displacement from, equilibrium, Opposite in direction, , 14.2 - Simple harmonic motion: graph and equation, At the right, the puck is again moving in SHM, and a graph of its motion is shown. In this case, we have changed our view of the air hockey, table so the puck moves vertically instead of horizontally. This puts the graph in the usual orientation. We continue to measure the, displacement of the puck with the variable x, which is plotted on the vertical axis. The horizontal axis is the time t., Unrolling the graph paper underneath the puck as it moves up and down would create the graph you see, the blue line on the white paper. The, graph traces out the displacement from equilibrium of the puck over time as it moves from “peaks” where its displacement is most positive, to, “troughs” where it is most negative. It starts at a peak, passes through equilibrium, moves to a trough, and so on. After four seconds, it has, returned to its initial position for the second time., The graph might look familiar to you. If you have correctly recognized the graph of a cosine function, congratulations! A cosine function, describes the displacement of the puck over time. You see this function in Equation 1., This graph represents the puck starting at its maximum displacement. When t = 0 seconds, the argument of the cosine function is zero radians, and the cosine is one, its maximum value. (In describing SHM, the units of the argument of the cosine must be specified as radians.) Because, the function used for this graph multiplies the cosine function by an amplitude of three meters, the maximum displacement of the puck (this is, always measured from equilibrium) is also three meters., Equation 2 shows the general form of the equation for SHM. The parameters A, Ȧ, and ij are called the amplitude, angular frequency, and, , 276, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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phase constant, respectively. The argument of the cosine function, Ȧt + ij, is called the, phase. In sections that follow we will explain how these parameters are used to, describe SHM., , Graphing simple harmonic, motion, Cosine (or sine) function describes, displacement, , For this graph:, x(t) = 3 cos (ʌt) meters, , Simple harmonic motion, equation, x(t) = A cos (Ȧt + ij), 14.3 - Period and frequency, , Period: Time to complete one full cycle of motion., Frequency: Number of cycles of motion per second., The period specifies how long it takes an object to complete one full cycle of motion. The letter T represents period, which is measured in, seconds. A convenient way to calculate the period is to measure the time interval between two adjacent peaks, as we do in Equation 1. In the, example shown there, it takes two seconds for the puck to complete one full cycle of motion., The frequency, represented by f, specifies how many cycles are completed each second. It is the reciprocal of the period. The graph in, Equation 2 is the same as in Equation 1. Its frequency is 0.5 cycles per second., Frequency has its own units. One cycle per second equals one hertz (Hz). This unit is named after the German physicist Heinrich Hertz, (1857í1894). You may be familiar with the hertz units from computer terminology: The speed of computer microprocessors used to be, specified in megahertz (one million internal clock cycles per second) but microprocessors now operate at over one gigahertz (one billion cycles, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 277
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per second). Radio stations are also known by their frequencies. If you tune into an AM, station shown on the dial at 950, the frequency of the radio waves transmitted by the, station is 950 kHz., , Period, Time of one complete cycle of motion, T represents period, Units: seconds (s), , Frequency, f = 1/T, Cycles per second, Units: hertz (Hz), , What is the period? What is the, frequency?, T = 4.0 s, f = 1/T = 0.25 Hz, , 278, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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14.4 - Angular frequency, , Angular frequency: Frequency measured in, radians/second., In the equation for SHM shown in Equation 1, the parameter Ȧ is the angular frequency, and it is the coefficient of time in the equation for SHM. Its units are radians per second., The angular frequency equals 2ʌtimes the frequency. The relationship between, frequency and period can be used to restate this equation in terms of the period. Both, these equations are shown in Equation 2., You may have noticed that Ȧalso stands for the angular speed of an object moving in a, circle, which is measured in radians per second, as well. If an object makes a complete, loop around a circle in one second, its angular speed will be 2ʌ radians per second., Similarly, an object in SHM that completes a cycle of motion in one second has an, angular frequency of 2ʌ radians per second. This is indicative of a relationship between, circular motion and SHM that can be productively explored elsewhere., , Angular frequency, x(t) = A cos (Ȧt + ij), Angular frequency is Ȧ, Units: radians per second (rad/s), , Angular frequency and period, Ȧ = 2ʌf, Ȧ = 2ʌ/T, T = period, f = frequency, , The function x(t) = A cos(Ȧt), describes this graph. What is the, angular frequency, Ȧ?, Ȧ = 2ʌ/T, T = 3.5 seconds, Ȧ = 2ʌ/3.5, Ȧ = 1.8 rad/s, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 279
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14.5 - Amplitude, , Amplitude: Maximum displacement from, equilibrium., The amplitude describes the greatest displacement of an object in simple harmonic, motion from its equilibrium position., In Concept 1, you see the now familiar air hockey puck and spring, as well as a graph of, its motion. The amplitude is indicated. It is the farthest distance of the puck from the, equilibrium point., The equation for SHM is shown again in Equation 1, with the amplitude term, highlighted. The amplitude is the absolute value of the coefficient of the cosine function., The letter A stands for amplitude. Since the amplitude represents a displacement, it is, measured in meters., , Amplitude, Maximum displacement from equilibrium, , Why does the amplitude equal the factor outside the cosine function? The values of the, cosine range from +1 to í1. Multiplying the maximum value of the cosine by the, amplitude (for example, four meters for the function shown in Example 1) yields the, maximum displacement., , x(t) = A cos (Ȧt + ij), Amplitude is |A|, Units: meters (m), , What is the amplitude?, Amplitude = |A| = 4 m, 14.6 - Interactive problem: match the curve, In the simulation on the right, you control the amplitude and period for a puck on a, spring moving in simple harmonic motion. With the right settings, the motion of the, puck will create a graph that matches the one shown on the paper., Determine what the values for the amplitude and period should be by examining the, graph. Assume you can read the graph to the nearest 0.1 m of displacement and, the nearest 0.1 s of time, and set the values accordingly. Press GO to start the, action and see if your motion matches the target graph. If it does not, press RESET, to try again., Review the sections on amplitude and period if you have difficulty solving this, problem., , 280, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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14.7 - Velocity, In Concept 1, you see a graph of an object in simple harmonic motion. The graph, shows the displacement of the object versus time. At any point the slope of the graph is, the object’s instantaneous velocity. The slope equals ǻy/ǻx. In this graph, this is the, change in displacement per unit time, which is velocity., You can consider the relationship of velocity and displacement by reviewing the role of, force in SHM. Consider an object attached to a spring, where the spring is stretched, and then the object is released. The spring force pulls the object until it reaches the, equilibrium point, increasing the object’s speed., Once the object passes through the equilibrium point, the spring is compressed and its, force resists the object’s motion, slowing it down. Because the object speeds up as it, approaches the equilibrium point and slows down as it moves away from equilibrium, its, greatest speed is at the equilibrium point., When the spring reaches its maximum compression, the object stops for an instant. At, this point, its speed equals zero. The spring then expands until the object returns to its, initial position, with the spring fully extended. Again, the object stops for an instant, and, its speed is zero., , Velocity in SHM, Velocity constantly changes, ·Extreme velocities at equilibrium, ·Zero velocity at endpoints, , In the paragraphs above, we discussed the motion in terms of speed, not velocity, so, we could ignore the sign and focus on how fast the object moves. The object’s velocity, will be both positive and negative as it moves back and forth. You see this alternating, pattern of positive and negative velocities in the graph in Equation 1., When the displacement is at an extreme, the velocity is zero, and vice-versa. One way, to state the relationship between the displacement and velocity functions is to say they, are ʌ/2 radians (90°) out of phase. An equivalent way to express this without a phase, constant is to use a cosine function for displacement and a sine function for velocity,, and this is what we do. This relationship can also be derived using calculus. In, Equation 1, you see both a velocity graph and the function for velocity., The second equation shown in Equation 1 states that the maximum speed vmax is the, amplitude of the displacement function times the angular frequency. To understand the, source of this equation, recall that the maximum magnitude of the sine function is one., When the sine has a value of í1 in the velocity equation, the velocity reaches its, maximum value of AȦ., , v(t) = íAȦ sin (Ȧt + ij), vmax = AȦ, v = velocity, A = amplitude, Ȧ = angular frequency, t = time, ij = phase constant, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 281
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What is the velocity at t = 4.0, seconds?, v(t) = íAȦ sin (Ȧt + ij), , v(4.0 s) = (í2ʌ) sin (8ʌ/3) m/s, v(4.0 s) = (í6.28)(0.866) m/s, v(4.0 s) = í5.4 m/s, 14.8 - Acceleration, For SHM to occur, the net force on an object has to be proportional and opposite in sign, to its displacement. Again, we use the example of a mass attached to a spring on a, friction-free surface, like an air hockey table., With a spring like the one shown in Concept 1 to the right, Hooke’s law (F = íkx), states the relationship between net force and displacement from equilibrium. This, equation for force enables you to determine where the acceleration is the greatest, and, where it equals zero. The magnitudes of the force and the acceleration are greatest at, the extremes of the motion, where x itself is the greatest. This is the point where the, object is changing direction. Conversely, x = 0 at the equilibrium point, so F = 0 and, the object is not accelerating there., The first equation shown in Equation 1 enables you to calculate the acceleration of an, object in SHM as a function of time. This equation can be simplified by noting that the, amplitude times the cosine function, the rightmost term in the equation, is the function, for the object’s displacement, x(t). We replace the terms A cos Ȧt by x(t) to derive the, second equation, which relates the acceleration directly to the object’s displacement., This equation says that the acceleration at a particular time equals the negative of the, angular frequency squared times the object’s displacement at that time., , Acceleration in SHM, Proportional to force, ·Zero at equilibrium, ·Maximum at extremes, , Finally, the third equation reveals that the maximum acceleration of the object is the amplitude times the square of the angular frequency. This, equation is a consequence of the first equation., , 282, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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a(t) = íAȦ2 cos (Ȧt + ij), a(t) = íȦ2x(t), amax = AȦ2, a = acceleration, A = amplitude, Ȧ = angular velocity, x = displacement, t = time, ij = phase constant, , What is the acceleration at t = 1.6, seconds?, a(t) = íAȦ2 cos (Ȧt + ij), , a(1.6 s) = í7.6 m/s2, 14.9 - A torsional pendulum, The torsional pendulum shown in Concept 1 is another device that exhibits simple, harmonic motion. A torsional pendulum consists of a mass suspended at the end of a, stiff rod, wire or spring. It does not swing back and forth. Instead, the mass at the, bottom is initially rotated by an external torque away from its equilibrium position. The, elasticity of the rod supplies a restoring torque, causing the mass to rotate back to the, equilibrium position and beyond. The mass rotates in an angular version of simple, harmonic motion., Earlier, we stated that for SHM to occur, the force must be proportional to displacement., Since torsional pendulums rotate, we must use angular concepts to analyze them. With, a torsional pendulum, a restoring torque, not a force, acts to return the system to its, equilibrium position. The restoring torque is proportional to angular displacement, just, as a restoring force is proportional to (linear) displacement. The moment of inertia of the, system takes on the role that mass plays in linear SHM., The same analysis that applies to linear displacement, velocity and acceleration applies, equally well to angular displacement, angular velocity and angular acceleration. In, Equation 1, you see an equation that states the nature of the restoring torque. It equals, the negative of the product of the torsion constant and the angular displacement., , Torsional pendulum, Exhibits simple harmonic motion, Use rotational concepts to analyze, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 283
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The formula in Equation 2 calculates the period of the pendulum. When the period and, the torsion constant are known, the moment of inertia can be calculated, as shown in, Example 1. This makes torsional pendulums useful tools for experimentally determining, the moments of inertia of complex objects., , Restoring torque, IJ = íțș, IJ = torque, ț = torsion constant, ș = angular displacement, Units for ț: N·m/rad, , Period, , T = period, I = moment of inertia, ț = torsion constant, , The torsional pendulum has a, period of 3.0 s. What is its, moment of inertia?, , I = (0.088 N·m/rad)(3.0 s)2/4ʌ2, I = 0.020 kg·m2, , 284, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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14.10 - A simple pendulum, Old-fashioned “grandfather” clocks, like the one you see in Concept 1, rely on the, regular motion of their pendulums to keep time. A typical pendulum is constructed with, a heavy weight called a “bob” attached to a long, thin rod. The bob swings back and, forth at the end of the rod in a regular motion., We approximate such a system as a simple pendulum. In a simple pendulum, the bob is, assumed to be concentrated at a single point located at the very end of a cable, and the, cable itself is treated as having no mass. The system is assumed to have no friction and, to experience no air resistance. When such a pendulum swings back and forth with a, small amplitude, its angular displacement closely approximates simple harmonic, motion. This means the period does not vary much with the pendulum’s amplitude. This, regularity of period is what makes pendulums useful in clocks., For SHM to occur, the restoring force or torque needs to vary linearly with, displacement. In the case of a pendulum, the motion is rotational, so the torque must be, linearly proportional to the angular displacement., , Simple pendulum, Point mass at end of massless rod, Approximates simple harmonic motion, , In Equation 1, you see a free-body diagram of the forces on the pendulum bob. The, tension in the cable exerts no torque on the pendulum since it passes through its center, of rotation, so the weight mg of the bob exerts the only torque. The lever arm of this, weight equals the length L of the cable times sin ș. For small angles, the angle, expressed in radians is a very close approximation of the sine of the angle. (The error is, less than 1% for angles less than 14°.) This means that the resulting torque is roughly, proportional to the angular displacement, and the condition for SHM is approximated,, with a torsion constant of mgL., In Equation 2, you see the equation for the period of a simple pendulum. When the, angular amplitude is small and the approximation mentioned above is used, the period, depends solely on the length of the cable and the acceleration of gravity., A pendulum can be an effective tool for measuring the acceleration caused by gravity, using the equation just mentioned. The length L of the cable is measured and the, pendulum is set swinging with a small amplitude. The period T is then measured. The, value of g can be calculated using the rearranged equation g = 4L(ʌ/T)2., , Restoring torque, IJ = ímgL sin ș § ímgLș, IJ = torque, m = mass, g = acceleration of gravity, L = length of pendulum, ș = angular displacement, , Period, , T = period, L = length of pendulum, g = acceleration of gravity, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 285
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What is the period of this, pendulum?, , T = 2.3 s, 14.11 - Interactive problem: a pendulum, On the right is a simulation of a simple pendulum: a bob at the end of a string. You, can control the length of the string, and in doing so change the period of the, pendulum. Your goal is to set the length so that the period is 2.20 seconds. As the, pendulum swings, you will see a graph reflecting the angular displacement of the, bob., Calculate and set the value for the string length to the nearest 0.05 m using the dial,, then use your mouse to drag the bob to one side and release it to start the, pendulum swinging. There may not be enough room in the window to show the, entire length of the string, but we will show the motion of the bob and the resulting, period. If you do not set the length correctly, press RESET to try again. Refer to the, section on simple pendulums if you do not remember the equation for the period., You may want to double-check your work by creating an actual pendulum with a, string of the correct length. You can time it: Ten cycles of its motion should take, about 22 seconds., For small angles, the angular displacement of a pendulum approximates simple, harmonic motion and the graph looks sinusoidal. Try smaller and larger angles and observe the graphs. How sinusoidal do they look to you?, (In the simulation, decreasing the string length makes it easier to create large angular displacements.) You can check the box labeled "SHM" to, draw a sinusoidal graph of SHM motion in black underneath your red graph. The black graph shows simple harmonic motion for the amplitude, you choose and the period calculated by the pendulum equation. If the amplitude is small, you might not see the black graph, because the two, graphs match so closely., , 14.12 - Period of a physical pendulum, Not all pendulums are simple. A physical pendulum is a rigid extended object (not a, point mass) pivoting around a point. In Concept 1, you see a violin acting as a physical, pendulum., The equation for the period of a physical pendulum is shown in Equation 1. The, distance h is the distance from the pivot point to the center of mass of the object. As, always, the moment of inertia must be calculated about the pivot point., The example problem shows how to calculate the period of a meter stick used as a, pendulum in the Earth’s gravitational field. The period is 1.6 seconds. With the use of a, meter stick, this is a result you can verify for yourself. If the stick has a hole close to one, end, put an unbent paper clip through the hole (otherwise, pinch the end very loosely, between your fingers), and set the stick swinging. Ten swings should take, approximately 16 seconds., , Physical pendulum, Mass is distributed, Motion approximates SHM, , 286, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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Physical pendulum period, , T = period, I = moment of inertia, m = mass, g = acceleration of gravity, h = distance from pivot to center of, mass, , What is the period of the, swinging meter stick?, , I = 1/3 × mass × length2, , 14.13 - Damped oscillations, We have considered many types of oscillations, and up until now assumed the periodic motion continued without change. But most real-world, oscillations are damped, which means they are subject to forces like friction that cause the amplitude of the motion to decrease over time., Mountain bike shock absorbers provide an excellent demonstration of damped oscillations. A shock absorber often combines a spring with a, sealed container of fluid. Shock absorbers lessen the jolts of a bumpy trail., To explain this in more detail, let’s consider what happens when a bike equipped with such a shock absorber hits a bump. The force from the, bump compresses the spring, with the result that less of the force from the bump passes to the rest of the bike (and the rider)., The spring then supplies a restoring force. In the absence of any other force, the rider and bike would in principle then move forever in simple, harmonic motion. However, inside a shock absorber, the spring moves a piston in a sealed cylinder of fluid. The fluid supplies what is called a, damping force., In Concepts 1 and 2, you see a diagram of this system. The fluid (typically oil) provides a force that opposes the motion of the piston. The, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 287
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damping force always opposes (resists) the motion of an object, which means, sometimes it acts in the same direction as the restoring force (when the object moves, away from equilibrium), and sometimes in the opposite direction (when the object, moves toward equilibrium). At all times, however, it is opposing the motion., Instead of moving in SHM, the system moves back to its equilibrium point and stops, or, it may oscillate a few times with smaller and smaller amplitude before resting at its, equilibrium point. The fluid “dampens” the motion, reducing the amplitude of the, oscillations. The result is a relatively fast yet smooth return to the equilibrium position., The resistive force of the fluid in a system like this is often proportional to the velocity,, and opposite in direction. In Equation 1, you see the equation for the damping force. It, equals the negative of b (the damping coefficient) times the velocity. (You may note that, this is similar to the formula for air resistance, where the drag force depends on the, square of the velocity.) The negative sign indicates that the damping force opposes the, motion that causes it., , Damped oscillations, Damping causes oscillations to diminish, , In Equation 2, you also see the equation for the net force FN. The net force is the sum, of the restoring forces and the damping force. (If you look at the equation, it may seem, that two negatives combine to make a larger number, but the sign of the velocity is the, opposite of the displacement as the system moves toward equilibrium.), The graph in Equation 3 illustrates three types of damping. The blue line represents a, critically damped system. The damping force is such that the system returns to, equilibrium as quickly as possible and stops at that point., The green line represents a system that is overdamped. The damping force is greater, than the minimum needed to prevent oscillations. The system returns to equilibrium, without oscillating, but it takes longer to do so than a critically damped system., The red line is a system that is underdamped. It oscillates about the equilibrium point,, with ever diminishing amplitude., With certain shock absorbers, the system can be adjusted, which means that the, damping coefficient can be tuned based on rider preferences. Beginners often prefer an, underdamped system. The bike bounces a bit but there is less of a “jolt” because the, shock absorber acts more slowly. Advanced riders sometimes prefer a critically damped, or overdamped “harder” ride, trading off a less smooth ride in exchange for regaining, control of the bicycle more quickly., , Damping force, Opposes motion, Often proportional to velocity, , Damping force, Fd = íbv, Fd = damping force, b = damping coefficient, v = velocity, , Net force, ȈF = íkx í bv, ȈF = net force, k = spring constant, x = displacement, b = damping coefficient, v = velocity, , 288, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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Types of damped harmonic, motion, Critically damped: blue line, Overdamped: green line, Underdamped: red line, 14.14 - Forced oscillations and resonance, , Forced oscillation: A periodic external force, acts on an object, increasing the amplitude of its, motion., External forces can dampen, or reduce, the amplitude of harmonic motion. For instance,, in a mass-spring system, friction reduces the amplitude of the mass’s motion over time., External forces can also maintain or increase the amplitude of an oscillation,, counteracting damping forces., Consider a child on a swing. Friction and air resistance are damping forces that reduce, the amplitude of the motion. On the other hand, an external force like a person pushing,, as you see in Concept 1, can increase the amplitude. When an external force increases, the amplitude, forced oscillation occurs., , Forced oscillations, External force in direction of motion, Amplitude increases, , An external force that acts to increase the amplitude of oscillations is called a driving, force. The driving force oscillates at a frequency called the driving frequency., The natural frequency of a system is the frequency at which it will oscillate in the, absence of any external force. Systems have natural frequencies based on their, structure. The closer the driving frequency is to the natural frequency, the more, efficiently the driving force transfers energy to the system, and the greater the resulting, amplitude. This is why you push a child on a swing “in sync” with the swing’s motion., The resulting phenomenon is called resonance. When the driving and natural, frequencies are the same, the result is called perfect resonance., There are several famous/infamous cases of forced oscillations and resonance. In, Equation 1, you see a movie of the Tacoma Narrows Bridge. A few months after it was, built in 1940, strong winds caused the bridge to oscillate at its natural frequency, and, the amplitude of the oscillations increased over time until the bridge collapsed. The, precise cause of the collapse is a matter of some debate, but the resonant oscillations, played a large part., The Bay of Fundy in Nova Scotia provides another famous example. The tides vary, greatly in the bay with the water level changing by as much as 16 meters. One reason, for the dramatic tides is that the natural frequency of the bay, the time it takes for a, wave to go from one end to the other, is close to the driving frequency of the tide cycle,, which is about 12.5 hours., , Resonance, Ȧ § Ȧn, Ȧ = driving frequency, Ȧn = natural frequency, , As a third example, the natural frequency of one- to three-story buildings is close to the, driving frequency supplied by some earthquakes, which is why these buildings (very common in San Francisco) often sustain the heaviest, damage during quakes., In Equation 2, you see a graph called a resonance curve. It is a graph of amplitude versus frequency for a system that has both a damping, force and an external driving force. We call the natural (angular) frequency Ȧn and use Ȧ to indicate the driving frequency. As the driving, frequency Ȧ approaches the natural frequency Ȧn, the amplitude increases dramatically., Natural frequencies can be “natural,” but in some cases they can also be controlled. Electric circuits, such as those used to tune radios to, stations of different frequencies, are designed so that humans can change the natural frequency of the circuit. As you turn the radio dial, you, are changing the natural frequency of the circuit. It then “tunes in” a driving frequency from a radio station that matches the natural frequency of, the circuit. These concepts have entered everyday language. People say that “an idea resonates with me.” Such everyday speech is good, , Copyright Kinetic Books Co. 2000-2007 Chapter 14, , 289
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physics; they mean the “driving frequency” of the idea is close to the “natural frequency”, of their own beliefs., , Frequency and amplitude, Amplitude increases as Ȧ approaches, Ȧn, 14.15 - Gotchas, To calculate the amplitude of an object moving in SHM, measure the difference between two successive peaks of its graph. No, that is the, period you just measured. The amplitude is the height of a peak of the graph above the horizontal (time) axis., The slope at any point on the displacement graph of an object in SHM is its velocity. Yes, you are correct. This is a point that is true of any, displacement graph, not just an SHM graph., , 14.16 - Summary, Simple harmonic motion (SHM) is a kind of repeated, consistent back and forth, motion, like the swinging of a pendulum. It is caused by a restoring force that varies, linearly with displacement., The displacement associated with such motion can be described with a sinusoidal, function, typically a cosine. The displacement is zero at equilibrium and maximum at, the extreme positions., Just as with other types of repetitive motion, the period of SHM is the amount of, time required to complete one cycle of motion. The frequency is the number of, cycles completed per second. It is the reciprocal of the period. The unit of frequency, is the hertz (Hz), equal to one inverse second., Angular frequency is the frequency measured in radians per second. It is, represented by the Greek letter Ȧ and is seen in the function for harmonic motion., The amplitude of harmonic motion is the maximum displacement from equilibrium. It, is represented by A and appears as the coefficient of the cosine in the displacement, function for SHM., , x(t) = A cos (Ȧt + ij), f = 1/T, Ȧ = 2ʌf, v(t) = –AȦ sin (Ȧt + ij), a(t) = –AȦ2 cos (Ȧt + ij), , The velocity and acceleration functions for SHM are also sinusoidal. The maximum velocity occurs at equilibrium, and it is zero at the, extremes. Acceleration is the opposite: zero at equilibrium and maximum at the extremes. These relationships follow from the general nature of, velocity as the instantaneous slope of the displacement graph, and acceleration as the slope of velocity., A simple pendulum displays simple harmonic motion in its angular displacement, provided that the amplitude of the motion is small. Instead of, a restoring force, there is a restoring torque due to gravity. The period of a pendulum depends upon the length of the pendulum and the, acceleration of gravity., The simple pendulum is a special case of the more complicated physical pendulum. In general, the period of a physical pendulum depends, upon its moment of inertia, mass, and the distance from the pivot point to its center of mass, as well as the acceleration of gravity., Sometimes a damping force opposes oscillatory motion. A typical damping force is proportional to the velocity of the object, which changes, with time., A force that acts with the restoring force can maintain or increase the amplitude of oscillations. Forced oscillations occur when such a driving, force is present. The natural frequency of a system is the frequency at which it will oscillate in the absence of external force. As the frequency, of the driving force approaches the natural frequency, energy is transferred more efficiently and the system’s oscillation amplitude increases., When these frequencies are approximately equal, resonance occurs., , 290, , Copyright Kinetic Books Co. 2000-2007 Chapter 14
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Chapter 14 Problems, , Chapter Assumptions, The general form of the equation of motion for an object in SHM is x(t), , = A cos (Ȧt + ij)., , Conceptual Problems, C.1, , A bouncing ball returns to the same height each time. Is this an example of simple harmonic motion? Explain your answer., , C.2, , In old pocket watches, a balance wheel acts as a torsional pendulum, rotating with a fixed period. If a pocket watch is running, slow, the period of the balance wheel is too long. Would you add or remove mass from the outer edge of the balance wheel to, correct it?, , Yes, , No, , Remove mass, Add mass, C.3, , What are the units for the damping coefficient constant?, N/m, kg/s, kg/s2, , Section Problems, Section 0 - Introduction, 0.1, , Using the simulation in the interactive problem in this section, answer the following questions. (a) If you increase the, amplitude, does the period increase, decrease, or stay the same? (b) What does the shape of the curve look like?, (a), , (b), , i., ii., iii., i., ii., iii., iv., , Increase, Decrease, Stay the same, Line, Parabola, Sinusoidal function, Circle, , Section 3 - Period and frequency, 3.1, , 3.2, , Consider the minute hand on a clock. (a) Compute the frequency of its motion in cycles per second. State your answer to, three significant digits. (b) Do the same for the hour hand., (a), , Hz, , (b), , Hz, , A graph of the displacement of an object, moving in SHM is shown. Determine the, frequency of the object's motion. (Assume you, can read the graph points to two significant, figures.), Hz, , Section 4 - Angular frequency, 4.1, , What is the angular frequency of the second hand on a clock? (State your answer using three significant figures.), rad/s, , 4.2, , A potter's wheel rotates with an angular frequency of 1.54 rad/s. What is its period?, s, , Copyright 2007 Kinetic Books Co. Chapter 14 Problems, , 291
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Section 5 - Amplitude, 5.1, , What is the amplitude of an object moving in SHM if its displacement in meters is described by:, (a) x(t) = 5 cos(t í ʌ/2), (b) x(t) = 4 cos(2ʌt) í cos(2ʌt), (c) x(t) = 4 cos2(ʌt) í 4 sin2(ʌt), (a), , 5.2, , m, , (b), , m, , (c), , m, , The displacement of an object moving in SHM, is graphed as shown. What is its amplitude of, motion? (Assume you can read the graph, points to two significant figures.), m, , 5.3, , An object moving in SHM has an amplitude of 3.5 m and a period of 4.0 s. Which of the following equations could describe its, displacement over time?, x(t) = 2.0 cos (3.5t), x(t) = 3.5 cos (t + 2.0), x(t) = 3.5 cos ((ʌ/2)t + 2.0), x(t) = 3.5 cos (2.0ʌt), , 5.4, , The displacement of an object in meters is described by the function x(t) = 4.4 cos (7.4ʌt, seconds. What are the (a) amplitude, (b) frequency, (c) period of the object's motion?, , + ʌ/2) where t is measured in, , m, , (a), (b), , Hz, , (c), , s, , Section 6 - Interactive problem: match the curve, 6.1, , Using the simulation in the interactive problem in this section, what (a) amplitude and (b) period should be used to match the, graph?, (a), , m, , (b), , s, , Section 7 - Velocity, 7.1, , In a car engine, a piston moves in SHM with an amplitude of 0.410 m. The engine is running at 2400 rpm, which is an angular, frequency of 251 rad/s. What is the maximum speed of the piston?, m/s, , 7.2, , The equation for the displacement in meters of an object moving in SHM is x(t) = 1.50 cos (4.20t) where t is in seconds., (a) What is the maximum speed of the object? (b) At what time does it first reach the maximum speed?, (a), , m/s, , (b), , s, , Section 8 - Acceleration, 8.1, , A ball on a spring moves in SHM. At time t = 0 s, its displacement is 0.50 m and its acceleration is í0.72 m/s2. The phase, constant for its motion is 0.84 rad. What is the ball's displacement at t = 3.4 s?, m, , 8.2, , A platform moves up and down in SHM, with amplitude 0.050 m. Resting on top of the platform is a block of wood. What is, the shortest period of motion for the platform so that the block will remain in constant contact with it?, s, , 292, , Copyright 2007 Kinetic Books Co. Chapter 14 Problems
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Section 9 - A torsional pendulum, 9.1, , An irregularly-shaped 1.4 kg object is suspended from a wire with a known torsion constant of 0.49 N·m/rad. The object's, period is 1.2 seconds. What is the object's moment of inertia for rotations about this axis?, kg · m2, , 9.2, , A thin square slab of material is suspended, "on edge" at the end of a torsion pendulum, so, that the axis of rotation passes through the, center of the square, parallel to an edge. The, mass of the slab is 0.78 kg and the length of, an edge is 0.28 m. The torsion constant of the, wire is 6.2 N·m/rad. What is the period of, motion when the slab oscillates?, s, , Section 10 - A simple pendulum, 10.1 You need to know the height of a room, but you have no tape measure. You fasten one end of a string to the ceiling of the, room, and tie a small rock at the other end so it almost touches the floor. You start this simple pendulum swinging slightly, and, measure its period, which is 3.56 seconds. How tall is the room?, m, 10.2 On the moon of a distant planet, an astronaut measures the period of a simple pendulum, 0.85 m long, and finds it is 4.7, seconds. Back on Earth, she could throw a rock 13 m straight up (while wearing her spacesuit). With the same effort, how far, up can she throw the same rock at her present location? Ignore the effects of air resistance., m, 10.3 It is the year 2305 and the tallest structure in the world has an insane height of 3.19×106 m above the surface of the Earth. A, pendulum clock that keeps perfect time on the surface of the Earth is placed at the top of the tower. How long does the clock, take to register one elapsed hour? The radius of the Earth is 6.38×106 m and its mass is 5.97×1024 kg., minutes, , Section 11 - Interactive problem: a pendulum, 11.1 Using the simulation in the interactive problem in this section, what is the length of string needed to achieve the desired, period for the pendulum?, m, , Copyright 2007 Kinetic Books Co. Chapter 14 Problems, , 293
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15.0 - Introduction, Waves can be as plain to see as the ripples in a pond or as invisible as the, electromagnetic waves emanating from a cellular phone. Mechanical waves, like, those in a pond, require a medium in order to propagate. Electromagnetic waves í, including radio waves and light í require no medium and can travel in the near, vacuum of space. Electromagnetic waves rely on the interaction of electric and, magnetic fields to propagate through space., In this chapter, we focus on mechanical waves. These are waves in which a, vibration causes a disturbance to travel through a medium. You are familiar with a, variety of mechanical waves: water waves in the ocean, sound waves in the air, or, waves along a string if you shake an end up and down. These waves exist due to, the movement of particles that make up a medium, such as water molecules in the, ocean or gas molecules in the air., Waves carry energy from place to place: a relatively small amount with a sound, wave, a much larger amount with a tsunami wave., Although many mechanical waves travel, sometimes across great distances, there, is no net movement of the medium through which they propagate. The 15th century Italian scientist and artist Leonardo da Vinci described this, key attribute when he said: “It often happens that the wave flees the place of its creation, while the water does not.”, Use the simulation to the right to begin your exploration of waves. It consists of a string stretched across the screen. A hand on the left holds, the string. By shaking the hand up and down, you can generate a variety of waves in the string., When you open the simulation, press GO to send a wave down the string. You will see the hand begin to shake the string, causing a wave to, travel from left to right., The control panel has two input gauges that allow you to vary the amplitude and frequency of the wave. As you may remember from your study, of simple harmonic motion, amplitude is the maximum displacement of a wave from equilibrium. Frequency is the number of cycles per second., You can vary these parameters and observe changes in the shape of the wave. Also in the control panel is an output gauge that displays the, wavelength, the distance between successive peaks of the wave., The string’s tension and other properties remain constant., When you run the simulation, make sure you observe the differences between a wave with higher frequency and one with lower frequency., This is an important fundamental characteristic of a wave., Then try three quick experiments. First, does changing the frequency of a wave also change its wavelength? Change the frequency and, observe what happens to the wavelength., Second, does changing the frequency result in any change in amplitude? Again, you can vary the frequency and note any change., Finally, as you change the frequency and amplitude, does the wave travel down the string any faster or slower? For example, does a wave with, a very large amplitude travel noticeably faster than one with a very small amplitude?, The simulation is intended to let you conduct a preliminary exploration of topics that will be presented in this chapter. Answer what questions, you can above and then proceed to the rest of the chapter, which covers the topics in more depth., , 15.1 - Mechanical waves, , Mechanical waves: Vibrations in a medium., Ocean breakers, the rolling wave of a crowd in a sports stadium, the back and forth, vibrations in a Slinky®: These are a few of the many kinds of waves you can see. Some, mechanical waves are invisible to the eye but detectable by the ear, such as the sound, waves generated by musical instruments., Mechanical waves are vibrations in a medium, traveling from place to place without, causing any net movement of the medium. You may be familiar with “the wave” in a, football or baseball stadium. The wave travels around the stadium, the result of, spectators standing and then sitting in a rolling succession. As the fans oscillate up and, down, they create what is called a disturbance or waveform. The location of the, disturbance changes as the wave moves through the stadium, but the wave’s medium,, the crowd, stays put., , Mechanical waves, Disturbances in a medium, , A wave in a stadium is a useful example, but it is not a true mechanical wave. Mechanical waves, such as a wave in a string, result from an, initial force (a vibration up and down or to and fro) followed by a continuing sequence of interactions between particles in the medium. In a, stadium wave, the particles of the medium (the people) do not typically exert physical forces on one another to propagate the wave (since peer, pressure is not a physical force)., , 294, , Copyright 2007 Kinetic Books Co. Chapter 15
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Mechanical waves have some common properties. First, they require a physical, medium, such as air, a string or a body of water. Mechanical waves cannot move, through a vacuum., Second, mechanical waves require a driving excitation to get the wave started. The, vibrations then propagate, via interactions between particles, through the medium., In this fashion, waves transfer energy from place to place. When you hear a sound, you, are hearing energy that has been transferred by a wave through air or water (or even, if, you are listening for buffalo, the ground)., All of the waves in this chapter are traveling waves in which the disturbance moves from, one point to another. Concept 2 shows a wave moving down a string, caused by a hand, shaking the string. The illustration shows three successive moments in time. You can, track the position of the first peak as it moves down the string over time. It moves with a, constant speed v. The other peaks also move down the string with the same speed., , Traveling waves, Vibrations that travel through a medium, , The peaks do not move through the medium in all waves. In what are called standing, waves, the locations in the medium where peaks and troughs (the "low" parts of the wave) occur are fixed. These waves, caused by the, reflection or interaction of traveling waves, are discussed in a later chapter., , 15.2 - Transverse and longitudinal waves, , Transverse wave: Particles in a medium, vibrating perpendicular to the direction the, wave is traveling., Longitudinal wave: Particles in a medium, vibrating parallel to the direction the wave is, traveling., Waves can be classified by the relationship between their direction of travel, and the, direction of the motion of the particles in the medium., Imagine that two people stretch a Slinky between them, and one shakes the Slinky up, and down. This causes a wave to move along the Slinky, as shown in Concept 1. The, wave moves to the right with a velocity called vwave in the diagram., , Transverse waves, Particles vibrate perpendicular to, direction of wave, , Although the wave moves to the right, the particles that make up the medium move up, and down. An individual particle of the Slinky is highlighted in red in the diagram to the, right, and its movement is shown with the vertically directed arrows. The direction of the, wave is perpendicular to the motion of the particles of the medium. This type of wave, is called a transverse wave., Many types of mechanical waves are transverse waves, including those caused by, shaking a Slinky up and down, the vibrations of a violin string, and certain types of, earthquake waves., Now imagine that instead of shaking the Slinky up and down, a person pulls the Slinky, to the left and then pushes it to the right, as shown in Concept 2. This causes the spring, to be stretched and then compressed., , Longitudinal waves, Particles vibrate parallel to direction of, , This disturbance again travels horizontally along the Slinky, and again we show its, wave, velocity as vwave. The wave consists of regions in which the coils of the spring are tightly, packed, followed by regions in which the coils are widely spaced. A particle of the, Slinky, again marked with a red dot, oscillates horizontally, parallel to the direction the wave is traveling. This type of wave is called a, longitudinal wave., , Sound is a longitudinal wave that consists of alternate compressions and rarefactions of air. Individual air particles oscillate back and forth, and, a sound wave travels through the air, where it can be detected by a sophisticated instrument: the human ear., In both transverse and longitudinal waves, the particles do move, but there is no net motion of the particles after each cycle. A particle moves, up and down, or back and forth, but it returns to its initial position. It oscillates like a mass attached to a spring., A single source of vibration, such as an earthquake, can create both transverse and longitudinal waves. In an earthquake, the longitudinal, waves (P waves, for primary waves) travel at about 8 km/s, while the transverse waves (S waves, for secondary waves) are slower, moving at, about 5 km/s. By noting when each type of wave arrives at a given seismographic station, a seismologist can determine the distance of the, earthquake from that station. Using data from several stations, the seismologist can triangulate the location of the earthquake’s epicenter., The motion of the particles that make up a wave can be complex, with ocean waves serving as one example. You can see this in Concept 3,, where the wave moves to the left and the water molecules near the surface move in circles. This means the molecules’ motion involves vertical, and horizontal components. Their motion is both perpendicular and parallel to the direction of the wave. An ocean wave displays both, , Copyright 2007 Kinetic Books Co. Chapter 15, , 295
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longitudinal and transverse properties., , Some waves both transverse and, longitudinal, Water waves, ·Wave travels horizontally, ·Particle motion perpendicular and, parallel, , 15.3 - Periodic waves, , Wave pulse: A single disturbance caused by a, one-time excitation., Periodic wave: A continuing wave caused by a, repeated vibration., Two types of transverse waves are shown to the right. Concept 1 shows a single wave, pulse in a string. The hand shakes up and down once and the wave pulse moves from, left to right along the string., Concept 2 shows a periodic wave. The hand moves continuously up and down. In a, periodic wave, each particle in the string moves through a repeated cycle of rising to a, peak, falling to a trough, and then returning again to a peak. The procession of, wavefronts moving down the string is called a wave train., , Wave pulse, Caused by a single up and down motion, , If you observe a particular crest in the periodic wave, it will move horizontally along the, string over time. This is more apparent in an animation. In a static diagram, the wave, can appear to be stationary, though it is moving down the string as the velocity vector, indicates. Click on the illustration to see an animation., In this chapter, we focus on waves in which the particles are vibrating in simple, harmonic motion. This vibration will be caused by something (in this case, a hand), moving or vibrating in simple harmonic motion. The result is the type of sinusoidal wave, you see to the right., , Periodic wave, Continuing wave caused by a repetitive, vibration, , 296, , Copyright 2007 Kinetic Books Co. Chapter 15
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15.4 - Amplitude, , Amplitude: The maximum displacement of a, particle in a wave from its equilibrium position., Several terms discussed in earlier topics such as simple harmonic motion also apply to, waves, including amplitude., At the right, you see a transverse wave caused by a hand shaking a string. The, amplitude of the wave is the distance between a particle at its maximum displacement í, a peak or trough í and the particle at its rest or equilibrium position. The horizontal line, in the diagram is the equilibrium position for the particles in the string. The amplitude by, convention is positive. Since amplitude is a distance, it is measured in meters., A wave’s amplitude is related to the energy it carries. Waves with greater amplitude, carry more energy. You can experience this relationship at the beach; you may barely, notice a small-amplitude wave crashing into you, while a large-amplitude wave may, knock you off your feet!, , Amplitude, Distance between rest point and, maximum displacement, ·Height of peak, , What is the amplitude of this, wave?, A = 0.20 m, 15.5 - Wavelength, , Wavelength: The distance between adjacent, peaks., The wavelength of a wave is the distance between adjacent peaks in the wave. This will, be the same distance as that between adjacent troughs, or any two successive points, on the wave with the same vertical displacement and direction of particle motion., The Greek letter lamda (Ȝ) represents wavelength. At the right, you see the wavelength, measured for a transverse periodic wave in a string. The unit for wavelength is the, meter., Also on the right is a table that shows the wavelengths of a variety of waves., , Wavelength, Distance between adjacent wave peaks, , A variety of wavelengths, , Copyright 2007 Kinetic Books Co. Chapter 15, , 297
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What is the wavelength of this, wave?, Ȝ = 0.60 m, 15.6 - Period and frequency, , Period T of a wave, Time a particle takes to complete a, cycle of motion, , Period: Amount of time for a particle in a wave, to complete a cycle of motion., Frequency: Number of wave cycles per second., The definitions of period and frequency may look familiar from your study of simple, harmonic motion., The period of a wave equals the amount of time required for a particle of the medium to, move through a complete cycle of motion., At the top of this section are four time-lapse “snapshots” of a transverse wave moving, through a string. The particle marked in red moves vertically up and down. The amount, of time it takes to rise to a peak, fall to a trough and return to its initial position is the, period. Because the period is an interval of time, its unit is the second., As the particle oscillates up and down through a full cycle of motion, the wave travels to, the right a distance of one wavelength. Frequency is the number of full cycles of motion, per second. Frequency (cycles/second) equals the reciprocal of the period, (seconds/cycle). The unit for frequency is the hertz (Hz), equal to one cycle per second., , Period and frequency, Period: time to complete a cycle, ·Units: seconds (s), Frequency: number of cycles per, second, ·Reciprocal of period, ·Units: hertz (Hz), , In Concept 2, we show a graph related to the transverse wave, which is not a depiction of the wave itself, but a graph of the motion of a particle, over time. The scale of its horizontal axis is time, not position. The particle oscillates up and down in SHM, like the red particle used in the, wave illustration at the top of this section. This graph could be generated in a fashion akin to the graphs you saw in the chapter on simple, harmonic motion, where we rolled graph paper below a mass that had a “pen” attached to it. In this case, we would roll the paper under the red, circle to record its location over time., The period of the wave itself can be measured as the difference in time between two equivalent points (such as adjacent peaks) on the graph., The frequency of the wave is the reciprocal of the period. The particle in Concept 2 has a period of 2.0 seconds, so its frequency is 0.5 cycles, per second, and that is the frequency of the associated wave., The example problem asks you to determine the frequency and period of another transverse wave. The transverse motion of a single particle in, this wave is graphed with time on the horizontal axis., , 298, , Copyright 2007 Kinetic Books Co. Chapter 15
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What are the frequency f and the, period T of the wave?, f = 3.0 cycles/2.0 s, f = 1.5 Hz, T = 1/ f = 1 /1.5 cycles, T = 0.67 s, 15.7 - Wave speed, How fast a wave moves through a medium is called its wave speed. Different types of, waves have vastly different speeds, from 300,000,000 m/s for light to 343 m/s for sound, in air to less than 1 m/s for a typical ocean wave. The wave in the string to the right, might be moving at, say, 15 m/s., The speed of a mechanical wave depends solely on the properties of the medium, through which it travels. For example, the speed of a wave in a string depends on the, linear mass density and tension of the string. This relationship is explored in another, section., For periodic waves there is an algebraic relationship between wave speed, wavelength, and period. This relationship is shown in Equation 1., This relationship can be derived by considering some of the essential properties of a, wave. Wavelength is the distance between two adjacent wave peaks. The period is the, time that elapses when the wave travels a distance of one wavelength. If you divide, wavelength by period, you are dividing displacement by elapsed time. This is the, definition of speed., , Wave speed, How fast a wave travels, , Because frequency is the reciprocal of period, the speed of a wave also equals the, wavelength times its frequency. Both of these formulations are shown to the right., Since the speed of a wave is dictated by the physical characteristics of its medium, its, speed must be constant in that medium. In the example of the string mentioned above,, the constant speed is determined by the linear density and tension of the string., Because the speed of a wave in a medium is constant, the product of its wavelength, and frequency is a constant. This means that for a wave in a given medium the, wavelength is inversely proportional to the frequency. Increase the frequency of the, wave and the wavelength decreases. Decrease the frequency and the wavelength, increases., Consider sound waves. Different sounds can have different frequencies. If this were not, the case, there would be no music. For example, consider the first four notes of, Beethoven’s Fifth Symphony (the famous “dut dut dut daaah”). As played by the violins,, the first three identical notes are the G above middle C and have a frequency of, 784.3 Hz and a wavelength of 0.434 m, while the fourth note (E flat) has a frequency of, 622.4 Hz and wavelength of 0.547 m., , v = wave speed, Ȝ = wavelength, T = period, f = frequency, , Copyright 2007 Kinetic Books Co. Chapter 15, , 299
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The light’s wavelength is, 6.0×10í7 m. The light completes, 5.0 cycles in 10í14 seconds. What, is the light’s wave speed?, v = Ȝf, , v = (6.0×10í7 m)(5.0×1014 Hz), v = 3.0×108 m/s, 15.8 - Wave speed in a string, This section examines in detail the physical factors that determine the speed of a, transverse wave in a string. The factors are the force on the string (the string’s tension), and the string’s linear density. Linear density is the mass per unit length, m/L. It is, represented with the Greek letter μ (pronounced “mew”)., The relationship of wave speed to the string’s tension and linear density is expressed in, Equation 1. The equation states that the wave speed equals the square root of the, string tension divided by the linear density of the string., Your physics intuition may help you understand why wave speed increases with string, tension and decreases with string density., Consider Newton’s second law, F= ma. If the mass of a particle is fixed, a larger force, on the particle will result in a greater acceleration. When a string under tension is, shaken up and down, the tension acts as a restoring force on the string, pulling its, particles back toward their rest positions. The greater the tension, the greater this, restoring force and the faster the string will return to equilibrium. This means the string, will oscillate faster (its frequency increases). Because wave speed is proportional to, frequency, the speed will increase with the tension., , Wave speed in a string, Increases with string’s tension, Decreases with string’s linear density, , Now let’s assume that the tension is fixed, and compare wave speeds in strings that, have differing linear densities (mass per unit length). Newton’s second law says that for, a given restoring force (tension), the particles in the more massive string will have less, acceleration and move back to their rest positions more slowly. The wave frequency, and wave speed will be less., The equation on the right is a good approximation when the amplitude of the wave is, significantly smaller than the overall length of the medium though which the wave, moves. The force that causes the wave must also be significantly less than the tension, for this equation to be accurate. The equation can be derived using the principles, discussed in this section., , v = wave speed, F = string tension, m = string mass, L = string length, , 300, , Copyright 2007 Kinetic Books Co. Chapter 15
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The green parrot is trying to, dislodge the other bird. How fast, will the wave he creates travel?, , 15.9 - Interactive problem: wave speed in a string, In this interactive problem, two strings are tied together with a knot and stretched, between two hooks. String 1, on the left, is twice as long as string 2, on the right., Both strings have the same tension., In the simulation, each string is plucked at its hook at the same instant. The, resulting wave pulses travel inward toward the knot. The wave pulse in string 1, starts at twice the distance from the knot as the wave pulse in string 2., You want the wave pulses to meet at the knot at the same instant. To accomplish, this, set the linear density of each segment of string. When you increase a string, segment’s linear density in this simulation, the string gets thicker., The minimum linear densities you can set are 0.010 kg/m. Set a convenient linear, density for string 1; your choice for this linear density determines the appropriate, linear density for string 2, which you must calculate., Enter these values using the dials in the control panel and press GO to start the, wave pulses. If they do not meet at the knot at the same instant, the simulation will, pause when either of the pulses reaches the knot. Press RESET and enter different, values for the linear densities to try again., , 15.10 - Mathematical description of a wave, When a mechanical wave travels through a medium, the particles in the medium, oscillate. Consider the diagram in Concept 1 showing a transverse periodic wave. The, particles of the string oscillate vertically and the wave moves horizontally., The vertical displacement of the highlighted particle will change over time as it, oscillates. We show its displacement in Concept 1 at an instant in time., In this section, we analyze a wave in which the particles oscillate in simple harmonic, motion. An equation that includes the sine function is used to describe a particle’s, displacement. The equation relates the vertical displacement of the particle to various, factors: the horizontal position of the particle, the elapsed time, and the wave’s, amplitude, frequency and wavelength. When all these factors are known, the vertical, position of a point can be determined at any time t., , Particles, Equation 1 describes a wave moving from left to right. The variable y in the equation is, the vertical displacement of a particle at a given horizontal position away from its, Oscillate in, equilibrium position at a particular time. To use the equation, you must assume the, wave has traveled the length of the string, and the time t is some time after this has occurred., , simple harmonic motion, , In the equation, the variable x is the particle’s position along the x axis, which does not change for a given particle. The variable A is the, wave’s amplitude; the variable Ȝ is the wavelength; and the variable ƒ is the wave’s frequency., The argument of the sine function is called the phase. As a wave sweeps past a particle located at a horizontal position x, the phase changes, linearly with respect to the elapsed time t. The phase is an angle measured in radians. The angle in the wave equation must be expressed in, , Copyright 2007 Kinetic Books Co. Chapter 15, , 301
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radians., The equation in Equation 1 describes a transverse wave moving from left to right. For a, wave moving from right to left, the minus sign inside the phase is switched to a plus, sign, reversing the sign of the coefficient of time., Equation 1 assumes that a particle at position x = 0 at time t = 0 is at the equilibrium, position y = 0. You can add what is called a phase constant to the equation to create a, new equation describing a wave with a different initial state. For example, suppose a, constant angle such as ʌ/2 radians were added to the argument of the sine function., Then, the particle at x = 0 at time t = 0 would be at its maximum positive displacement,, because the sine of ʌ/2 equals one. A phase constant does not change the shape of a, wave, but rather shifts it back or forward along the horizontal axis by the same amount, at all times. Note that as the phase is increased by an integer multiple of 2ʌ radians, the, sine function describing the wave behaves as if there were no change at all., If you contrast the equation here to the equation for simple harmonic motion, you will, note that the equation for a traveling wave requires two inputs to determine the vertical, displacement of a particle. Both equations include time, but the equation in this section, also requires knowing the x position of a particle in the medium. With a wave, the, vertical displacement is a function not only of time, but also of position in the medium,, while the position is not a factor in SHM., , Equation for traveling wave, Function relates particle’s vertical, displacement y to:, ·particle’s horizontal position x, ·elapsed time t, ·wave’s amplitude, wavelength,, frequency, , The equations to the right can be used with either transverse or longitudinal waves., When applied to longitudinal waves, the oscillation of the particles occurs parallel to the, direction of travel of the wave, and then we would use the variable s instead of y to, represent the horizontal displacement of a particle away from its equilibrium position., , y = A sin (2ʌx/Ȝ í 2ʌft), y = particle’s vertical displacement, A = amplitude of wave, x = particle’s horizontal position, Ȝ = wavelength, f = frequency, t = elapsed time, For wave motion toward íx:, y = A sin (2ʌx/Ȝ + 2ʌft), 15.11 - Gotchas, A mechanical wave can travel with or without a medium. No. Mechanical waves must have a medium. This is why, in the vacuum of space,, there is total silence. Sound, a mechanical wave, cannot travel without a medium such as air., The medium carrying a wave does not move along with the wave. That is correct. The medium oscillates, but it does not travel with the wave., This differentiates wind, which consists of moving air, from a sound wave. With a sound wave, the air remains in place after the sound wave, has passed through., The amplitude of a wave has no effect on the speed of the wave. That is correct. The speed of a wave is determined by the properties of the, medium. This means that if you are in a hurry, it is no use yelling at people!, Wave speed in a string is a function of frequency, so if I increase the wave frequency, the wave speed will increase, too. No. The speed of a, wave in a string is fixed by the tension and linear density of the string. Increasing wave frequency will cause a decrease in wavelength, but no, change in wave speed., Amplitude is the same as the vertical displacement y of a particle in a wave. No, the amplitude A is the maximum positive vertical displacement, of a particle, while at a time t the instantaneous vertical displacement y can be anywhere between +A and íA., , 302, , Copyright 2007 Kinetic Books Co. Chapter 15
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15.12 - Summary, Mechanical waves are oscillations in a medium. This chapter discussed traveling, waves: disturbances that move through a medium., There are two basic wave types. In a transverse wave, the particles in the medium, oscillate perpendicularly to the direction that the wave travels. In a longitudinal, wave, the particles oscillate parallel to the direction the wave travels. Some waves,, such as water waves, exhibit both transverse and longitudinal oscillation., , T = 1/f, Wave speed, , A wave can come as a single pulse, or as a periodic (repeating) wave. In this, chapter, we analyze periodic waves whose particles oscillate in simple harmonic, motion., , Wave speed in a string, The amplitude of a wave is the distance from its equilibrium position to its peak. The, wavelength is the distance between two adjacent peaks. The period is the time it, takes for a particle to complete one cycle of motion, for example, moving from peak, to peak. Frequency is the number of cycles that are completed per second., Wave speed is the speed with which a wave moves through a medium. It is equal to, the wavelength of the wave times its frequency., The wave speed in a string increases with the tension of the string and decreases with the string’s linear density., , Copyright 2007 Kinetic Books Co. Chapter 15, , 303
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Chapter 15 Problems, , Conceptual Problems, C.1, , Many children are lined up at an ice cream stand. If the child at the back pushes the child in front of him, and she in turn, pushes the child in front of her, and so on, will they create a transverse or longitudinal wave in the line? Explain., Transverse, Longitudinal, , C.2, , When sports spectators do "the wave," are they making a transverse or longitudinal wave? Explain., Transverse, Longitudinal, , C.3, , Suppose a wave is moving along a chain at 10 m/s. Does that mean each link of the chain moves at 10 m/s? Explain., Yes, , C.4, , A wave is traveling through a particular medium. The wave source is then modified so that it now emits waves at a higher, frequency. Does the wavelength increase, decrease, stay the same, or does it depend on other factors? Explain., i., ii., iii., iv., , C.5, , No, , Increases, Decreases, Stays the same, Depends on other factors, , Radio waves all travel through space at the same speed. If an AM station broadcasts at a frequency 7×105 Hz and an FM, station broadcasts at 90.3×106 Hz, which station's radio waves have the longer wavelength? Explain., i. AM station, ii. FM station, iii. Both the same, , C.6, , You have two strings of different linear densities tied together and stretched end-to-end between a pair of supports. What, happens to the speed of a wave that passes from the more dense string to the less dense one?, i. The speed decreases, ii. The speed increases, iii. No change, , C.7, , Waves of identical frequency and amplitude are simultaneously launched down two identical wires of equal mass and length,, although one wire is tauter than the other. Which wave will arrive at the other end first, or will they arrive at the same time?, i. Wave on tauter string arrives first., ii. Wave on looser string arrives first., iii. They arrive at the same time., , Section Problems, Section 0 - Introduction, 0.1, , 304, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Does changing the, frequency of a wave in the simulation change its wavelength? (b) Does changing the frequency of a wave in the simulation, change its amplitude? (c) Does changing the amplitude in the simulation cause the the wave to move noticeably faster or, slower down the string?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , Copyright 2007 Kinetic Books Co. Chapter 15 Problems
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Section 4 - Amplitude, 4.1, , What is the amplitude of the wave that is, shown? Enter the answer using two significant, figures., m, , 4.2, , A water strider sits on a the surface of a still lake. A rock thrown into the lake creates a series of sinusoidal ripples that pass, through the location where the strider sits, so that it rises and falls. The distance between its highest and lowest locations is, 0.008 m. What is the amplitude of the wave?, i. 0.002 m, ii. 0.004 m, iii. 0.008 m, , Section 5 - Wavelength, 5.1, , What is the wavelength of the wave that is, shown? Enter the answer using two significant, figures., m, , 5.2, , In a foggy harbor, a tugboat sounds its foghorn. Bobbie stands on shore, 7.5×102 m away. The foghorn's sound wave, completes 1.1×103 cycles on its way to Bobbie. What is the wavelength of the sound wave?, m, , Section 6 - Period and frequency, 6.1, , A wave has a frequency of 575 Hz. What is its period?, s, , 6.2, , If 12 wave crests pass you in 26 seconds as you bob in the ocean, what is the frequency of the waves?, Hz, , Section 7 - Wave speed, 7.1, , An important wavelength of radiation used in radio astronomy is 21.1 cm. (This wavelength of radiation is emitted by excited, neutral hydrogen atoms.) This radiation travels at the speed of light, 3.00×108 m/s. Compute the frequency of this radio wave., Hz, , 7.2, , A wave has a speed of 351 m/s and a wavelength of 2.4000023468268647 meters. What is its period?, s, , 7.3, , Suppose you are standing on a boat in the ocean looking out at a person swimming in the water. The waves swell under your, boat at a frequency of 0.80 Hz, while traveling with a wave speed of 1.5 meters per second. At any given moment, you count, that there are exactly 6 complete waves between you and the swimmer. How far away is the swimmer?, , 7.4, , The radio wave from a station broadcasting at 1.3×106 Hz has a wavelength of 230 m. What is the wave's speed?, , meters, , m/s, , Copyright 2007 Kinetic Books Co. Chapter 15 Problems, , 305
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7.5, , Two waves travel in the same medium at the same speed. One wave has frequency 5.72×105 Hz and wavelength 0.533 m., The other wave has frequency 6.13×105 Hz. What is the second wave's wavelength?, m, , 7.6, , Two waves travel in the same medium at the same speed. One has wavelength 0.0382 m and frequency 9.67×106 Hz. The, other has wavelength 0.04460000398122414 m. What is the period of the second wave?, s, , Section 8 - Wave speed in a string, 8.1, , A wave is traveling through a 35.0-meter-long cable strung with a tension of 35,000 newtons. The mass of this length of cable, is 10.2 kilograms. What is the speed of a wave that is traveling in the cable?, m/s, , 8.2, , A 0.340 gram wire is stretched between 2 points that are 75.0 cm apart. The tension in the wire is 620 N. When the string is, plucked, a wave is created with wavelength equal to twice the length of the string. This wave's frequency is called the, "fundamental frequency" of the string. What is this fundamental frequency?, Hz, , 8.3, , The speed of longitudinal waves in a fluid (sound waves, for example) is given by, , where B is a constant property of the fluid called its bulk modulus and ȡ is the density of the fluid. For water at 4ºC,, B = 2.15×109 N/m2 and ȡ = 1000 kg/m3. What is the speed of longitudinal waves in water at this temperature?, m/s, 8.4, , When a certain string has a tension of 34 N, a wave travels along it at 68 m/s. What is the linear density of the string?, kg/m, , Section 9 - Interactive problem: wave speed in a string, 9.1, , Use the information given in the interactive problem in this section to answer the following question. If the linear density of, string 1 is 0.020 kg/m, what linear density for string 2 will cause the wave pulses will reach the knot at the same instant? Test, your answser using the simulation., kg/m, , Section 10 - Mathematical description of a wave, 10.1 A wave is described by the equation y = (4.5×10í2) sin (1.9x í 3.2t), where lengths are measured in meters and time in, seconds. What is the displacement of a particle at position x = 2.7 m at time t = 5.8 s? Note: Remember that the argument of, the sine function in the wave equation is expressed in radians., m, 10.2 A wave is defined by the equation y = 2.1 sin (4.2ʌx + 6.4ʌt). Time is measured in seconds and lengths are in meters. What, are the wave's (a) direction of motion; (b) amplitude; (c) wavelength; (d) frequency?, (a), , 306, , i., ii., iii., iv., , (b), , Left to Right, Right to Left, Up to down, Down to Up, m, , (c), , m, , (d), , Hz, , Copyright 2007 Kinetic Books Co. Chapter 15 Problems
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10.3 A taut string is agitated by a mechanical oscillator, producing a transverse wave that is described by the equation, , where time is measured in seconds, and lengths in meters. Find (a) the direction of travel of the wave, (b) the wave's, amplitude, (c) the wavelength, and (d) the frequency., (a), , (b), , i., ii., iii., iv., , Left to right, Right to left, Up to down, Down to up, m, , (c), , m, , (d), , Hz, , 10.4 A wave is described by the equation y = A sin (2.3x í 1.8t), where lengths are measured in meters, and time in seconds., What is the earliest time t (> 0) when the displacement of a particle at position x = 1.6 m is at a maximum?, s, , Copyright 2007 Kinetic Books Co. Chapter 15 Problems, , 307
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16.0 - Introduction, Sounds are so commonplace that it is easy to take them for granted, but they are a, central part of the human experience. When you think of sound, you may think of, your favorite song or an alarm clock that goes off early and loud. To a physicist,, though, both a pleasant song and a shrill alarm are mechanical longitudinal waves, consisting of regions of high and low pressure. The physics of sound waves is the, topic of this chapter., Sounds can be classified as audible, infrasonic or ultrasonic. Audible sounds are in, the frequency range that can be heard by humans. Infrasonic sounds are at, frequencies too low to be heard by humans, but animals such as elephants and, whales use them to communicate over great distances. Ultrasonic sounds are at, frequencies too high to be perceived by humans. They are used by bats for sonar, and by doctors to see inside the human body., You may have used the speed of sound in air to estimate the distance to a, thunderstorm. The flash from the lightning reaches you almost instantaneously,, while the sound from the thunder takes more time. Sound travels at approximately, 343 m/s in air at 20°C, so for every third of a kilometer of distance to the lightning,, the sound of the thunder lags the flash of light by about one second., Sound travels slowly enough in air that manmade objects such as airplanes can catch and pass their own sound waves. You may have heard, the result when a plane is flying faster than the speed of sound: a sonic boom. A small sonic boom is also the cause of the "crack" of a whip, as, the tip of the whip travels faster than the speed of sound., You may begin your study of sound with the simulation to the right, which allows you to experiment with a loudspeaker that causes sound, waves to travel through a tube filled with air particles. One set of particles is colored red to emphasize that all the particles just oscillate back, and forth; they do not travel along with the wave., Sound waves can be described with the same parameters that are used to describe transverse mechanical waves: amplitude, frequency and, wavelength. Recognizing these parameters in a longitudinal wave may require some practice., When you open the simulation, press GO to send a sound wave through the air. You will see the loudspeaker's diaphragm vibrate horizontally., This causes the nearby air particles to vibrate and a longitudinal wave to travel from left to right along the length of the tube., Observe the differences between this wave and the transverse waves you saw in strings. You should be able to see how the particles of the, medium (air) oscillate parallel to the direction the wave travels in a longitudinal wave, as opposed to the perpendicular motion of the particles, in a transverse wave., The simulation lets you control the loudspeaker to determine the amplitude and frequency of the wave. As with any wave, the amplitude is the, maximum displacement of a particle from its rest position, and the frequency is the number of cycles per second. You can vary these, parameters and observe changes in the motion of the loudspeaker and in the properties of the sound wave. You can also observe how the, wavelength changes when you alter the frequency., Humans can identify different sound waves by pitch, which is related to frequency. If you have audio on your computer, turn it on and listen to, the pitch created by a particular wave. Then increase the frequency and hear how the pitch changes. The loudness of a sound wave is related, to its amplitude. Increase the amplitude of the wave in the simulation and note what you hear., , 16.1 - Sound waves, Sound waves are longitudinal mechanical waves in a medium like air generated by, vibrations such as the plucking of a guitar string or the oscillations of a loudspeaker., Sound waves are caused by alternating compression and decompression of a medium., In Concept 1, a loudspeaker is shown. As the loudspeaker diaphragm moves forward, it, compresses the air in front of it, causing the air particles there to be closer together., This region of compressed air is called a condensation. The pressure and density of, particles is greater in a region of condensation. This compressed region travels away, from the loudspeaker at the speed of sound in air., The diaphragm then pulls back, creating a region in which there are fewer particles., This region is a rarefaction, and the pressure there is lower. The rarefaction also travels, away from the loudspeaker at the speed of sound. The velocity of the wave is indicated, with the orange vector v in the diagram. The back-and-forth motion of an individual, particle is indicated with the black arrows., , Sound waves, Are longitudinal, , The illustration for Concept 2 also shows the alternating regions of condensation and, rarefaction as the loudspeaker oscillates back and forth. The wavelength is the distance between two successive areas of maximum, condensation or rarefaction. As with transverse waves, the wavelength is measured along the direction of travel. The wavelength can be, readily visualized as the distance between the midpoints of the two regions of condensation shown in the diagram., , 308, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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With a wave in a string, a complete cycle of motion occurs when a particle in the string, starts at a point (perhaps a peak), moves to a trough, and returns to a peak. With a, sound wave, an analogous cycle passes from compression to rarefaction and then, returns to compression., To measure the period, you can note how long it takes for an air particle to pass, through a complete cycle at a given location. As with transverse waves, the frequency, of a sound wave is the number of cycles completed per second., Concept 3 illustrates the motion of an individual particle in a sound wave. Refresh the, browser page to see an animation of the particle's motion. The particle oscillates back, and forth horizontally as regions of high and low pressure pass by, also horizontally., The particle is first pushed to the right as an area of higher pressure passes, and then, pulled to the left by a region of lower pressure. As high and low pressure regions pass, by, the individual particles oscillate in simple harmonic motion., The harmonic oscillation of the particles distinguishes a sound wave from wind. Wind, causes particles to have net displacement; it moves them from one location to another., There is no net movement of air particles over time due to sound waves. The particles, oscillate back and forth around their original locations., , Structure of a sound wave, Condensation, followed by rarefaction, , Two aspects of the behavior of gas molecules will help you understand sound waves., First, we have used increased density and increased pressure to define condensation,, and decreased density and decreased pressure to define rarefaction. Density and, pressure are correlated. The ideal gas law (which you may study in a later chapter), states that, everything else being equal, pressure increases with the number of gas, molecules in a system. This means that the pressure is greater in regions of, condensation than in regions of rarefaction. Other factors (such as temperature) also, influence pressure, but in this discussion we treat them as constant., Second, the speed of sound can be understood in terms of the behavior of air, molecules. The speed v of a sound wave does not equal the speed of the loudspeaker’s, motion. That may seem odd. How could the waves move faster or slower than the, object that causes them?, , Particles in a sound wave, Move in simple harmonic motion, , The diagram we use simplifies the nature of the motion of air molecules in a wave. Air molecules at room temperature move at high speeds, (hundreds of meters per second) and frequently collide. On average, their location is stationary since their motion is random, so we can draw, them as stationary to reflect their average position. On the other hand, they are always moving, and when the loudspeaker moves, it changes, the velocity of molecules that are already in motion. The change in velocity caused by the loudspeaker is transmitted through the air by multiple, collisions as a function of the random speeds of the molecules. The faster the molecules are moving, the more frequently they will collide. As, the air becomes warmer, for example, the average thermal speed of the molecules increases, as does the speed of sound. Properties of the, medium itself, not the speed of the loudspeaker, determine the speed of sound., , 16.2 - Human perception of sound frequency, When a sound wave composed of alternating high and low pressure regions reaches a, human ear, the wave vibrates the eardrum, a thin membrane in the outer ear. The, vibrations are then carried through a series of structures to generate signals that are, transmitted by the auditory nerve to the brain, which interprets them as sound., There is a subjective relationship between the frequency of a sound wave and the pitch, of the sound you hear. Pitch is the distinctive quality of the sound that determines, whether it sounds relatively high or low within a range of musical notes. A home smoke, alarm issues a high-pitched beep, while a foghorn emits a low-pitched rumble. The, human ear is extremely sensitive to differences in the frequency of sound waves., A pure tone sound consists of a sound of a single frequency. The tones produced by, musical instruments combine waves with several frequencies, but each note has a, fundamental frequency that predominates. The note middle C on a piano has a, fundamental frequency of 262 cycles per second (Hz); the lowest and highest notes of a, piano have frequencies of 27.5 and 4186 Hz, respectively. Orchestras tune to a note of, 440 Hz (the A above middle C). To experiment with frequency and pitch yourself, try the, interactive simulation in the next section., , Audible frequencies, , A young person can hear sounds that range from 20 to 20,000 Hz. With age, the ability of humans to perceive higher frequency sounds, diminishes. Middle-aged people can hear sounds with a maximum frequency of about 14,000 Hz., The human ear is sensitive to a wide range of frequencies, but other animals can perceive frequencies that humans cannot. Whales emit and, hear sounds with a frequency as low as 15 Hz. Bats emit sounds in a frequency range from 20,000 Hz up to 100,000 Hz and then listen to the, reflected sound to locate their prey. Dogs (and cats) can detect frequencies more than twice as high as humans can hear, and some dog, whistles operate at frequencies that the animals can hear but humans cannot., Interestingly, when it comes to certain sounds, the ear is not the most sensitive part of the human body. Sometimes you can feel sounds even, when you cannot hear them. Some contrabass musical instruments are designed to play notes below the lowest limit of human hearing, (20 Hz). For example, the organ in the Sydney (Australia) Town Hall can play a low, low C that vibrates at only 8 Hz, a rumbling that can only, , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 309
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be felt by the audience members. Low frequency sounds are also used in movies and some arcade games to increase tension and suspense., , 16.3 - Interactive problem: sound frequency, In this interactive simulation, you can experience the relationship between sound, wave frequency and pitch., The simulation includes a virtual keyboard. (As you might expect, your computer, must have a sound card and speakers or a headphone for you to be able to hear, the musical notes.), Above the keyboard is an oscilloscope, used to display the sound wave. The, oscilloscope graphs the waves with time on the horizontal axis. Each division on the, horizontal axis represents one millisecond., On the vertical axis, the oscilloscope graphs an air particle’s displacement from, equilibrium as a function of time, as the sound wave passes by. The wave is, longitudinal, and peaks and troughs on the graph correspond to the particle’s, maximum displacement, which occurs along the direction of the wave’s motion., Although we do not provide units on the vertical axis, displacements of the particles, in audible sound waves are generally in the micrometer range., The frequency of ordinary musical notes is high enough that the oscilloscope graphs, time in milliseconds. The middle C on the keyboard (a white key near its midpoint) has a frequency of about 262 Hz (cycles per second), or, 0.262 cycles per millisecond. A full cycle of this sound wave would span slightly less than four squares on the horizontal axis of the, oscilloscope., When you start the simulation, the instrument is set to "synthesizer." When you press a key, you will hear a tone that plays at the same, intensity for as long as you hold down the key. Even this simple synthesizer tone consists of several frequencies, but one fundamental, frequency predominates, and it is displayed on the oscilloscope., You can use the oscilloscope to compare the frequencies of various sound waves, as well as using your ears to compare various pitches. Do, they relate? Specifically, do higher pitched musical notes have a higher or lower frequency than lower pitched musical notes? If you know how, to play notes on the piano that are an octave apart, compare the frequencies and wavelengths of these sounds. What are the relationships?, You can also set the simulation to hear notes from a grand piano. These tones are even more complex than the synthesizer and again we, display just the fundamental frequency. To simulate a piano's sound, the notes will fade away even if you hold down the key, but the, oscilloscope will continue to display the initial sound wave., This simulation is designed to give you an intuitive sense of the frequencies of different musical notes. If you know how to play the piano, even, something as simple as "Chopsticks," play a song and observe the waves that make up that tune. You can also play a note and see how close, a friend can come to guessing it. Some people have a capability called perfect pitch, and can tell the note or frequency correctly every time., , 16.4 - Sound intensity, , Sound intensity: The sound power per unit area., Sound carries energy. It may be a small amount, as when someone whispers in your, ear, or it may be much more, as when the sonic boom of an airplane rattles windows, or, when a guitar amplifier goes to 11., Sound intensity is used to characterize the power of sound. It is defined as the power of, the sound passing perpendicularly through a surface area. Watts per square meter are, typical units for sound intensity., The definition of sound intensity is shown in Equation 1. An intensity of approximately, 1×10í12 W/m2 is the minimum perceptible by the human ear. An intensity greater than, 1 W/m2 can damage the ear., As a sound wave travels, it typically spreads out. You perceive the loudness of the, sound of a loudspeaker at an outdoor concert differently at a distance of one meter than, you do at 100 meters. The intensity of the sound diminishes with distance., , Sound intensity, Sound power that passes, perpendicularly through a surface area, Diminishes with square of distance from, sound source, , Equation 2 is used to calculate the intensity of sound when it spreads freely from a, single source. The intensity diminishes with the square of the distance from the source., It does so because the sound energy in this case is treated as being distributed over the surface of a sphere whose radius increases with time., The denominator of the expression for intensity is 4ʌr2, the expression for the surface area of a sphere., The example problem to the right asks you to find the relative sound intensities experienced by two listeners, one twice as far as the other from, the fireworks. In this scenario, the sound is four times as intense for the closer listener, but this does not mean he hears the sound as four, times louder. The loudness of sounds as perceived by human beings has a logarithmic relationship to sound intensity. This topic is explored in, another section., Here we have focused on sound that freely expands in all directions. However, sound can also reflect off surfaces such as walls. Concert halls, are designed to take advantage of this reflection to deliver a full, rich sound to the audience., , 310, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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I = P/A, I = sound intensity, P = power perpendicular to surface, A = surface area, Units: watts/meter2, , Sound spreading radially, , P = power of sound source, r = distance from sound source, , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 311
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How many times more intense is, the sound for the man than for, the woman?, , Sound for man is 4 times as intense, 16.5 - Sound level in decibels, , Sound level: A scale for measuring the, perceived intensity of sounds., To compare the intensity of two sounds, you could directly calculate the ratio of their, intensities. However, humans do not perceive a sound with twice the intensity as being, twice as loud, so a different system may be used. Loudness is subjective, and having, an objective measurement that corresponds to human perception is convenient. This, measurement is called the sound level (or sometimes, more confusingly, the intensity, level)., The human ear is sensitive to an extraordinary range of sound intensities; at the, extremes, humans can perceive sounds whose intensities differ by a factor of, 1,000,000,000,000. Although they can hear a broad range of intensities, people do not, distinguish between them finely. For instance, the human ear cannot very well, distinguish a sound that has an intensity of 1.0 W/m2 from one with an intensity of, 0.50 W/m2., , Sound level, Used to measure perceived loudness, ·Units: decibels (dB), ·Logarithmic scale (20 dB is ten times, more intense than 10 dB), , To reflect humans’ perception of differences in sound intensity, scientists use a, logarithmic scale. The common unit for the sound level is the decibel (dB), or one-tenth of a “bel,” a unit named after Alexander Graham Bell. A, logarithmic scale provides an appropriate tool for describing the human perception of sounds., In calculating a sound level ȕ, you start by dividing the intensity of the sound being measured by a reference sound intensity that approximates, the lowest intensity humans can hear. This reference intensity is 1×10í12 W/m2. Then you calculate the common logarithm (to the base 10) of, this ratio, which gives the sound level in bels, and finally you multiply that value by 10 to express the level in decibels. This is the first equation, shown to the right., To practice calculating sound levels in decibels, consider a sound with an intensity of 1.5×10í11 W/m2. It has 15 times the sound intensity of, the reference intensity of 1×10í12 W/m2. The base-10 logarithm of 15 is 1.2. Multiplying this value by 10 decibels yields a sound level of 12, decibels. A sound level increase of 10 dB means the intensity increases by a factor of 10. In calculations of sound levels, both the numbers, , 1×10í12 W/m2 and 10 (decibels) are considered exact., You may note that the reference intensity of 1×10í12 W/m2 corresponds to a decibel reading of zero. At zero decibels, a human ear can still, barely hear sound. The pressure exerted by this sound is very, very slight: It displaces particles of air by about one hundred-billionth of a meter., It is possible to have negative decibel sounds, perturbations in air pressure so slight the human ear cannot detect them., Tests indicate that a one to three decibel change in sound level is about the smallest change most humans can perceive. A general rule of, thumb is that a human will perceive a tenfold increase in intensity as sounding twice as loud. A 50 dB sound is 10 times more intense than a, 40 dB sound í remember, it is a logarithmic scale í and a typical human would say it sounds twice as loud., , 312, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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The sound level equation can be recast in terms of sound power, as shown in the, second equation to the right. If you know the relative power of two sound sources, you, can use this equation to compare their relative loudness to the human ear, provided the, listener is equidistant from the two sound sources. The reference power P0 in this, equation is the power at the source that results in the reference intensity at the location, of interest. For instance, you might determine this value for a loudspeaker for an, audience member 75 meters away., , ȕ = sound level, I = intensity of sound, reference intensity I0 = 1×10í12 W/m2, , P = sound power of source, P0 = reference sound power, Units: decibels (dB), 16.6 - Sample problem: sound level, Sitting on a sofa, your roommate, hears 100 dB from your stereo, which, supplies 10 W to each speaker. He, says this is lame and requests a, system with a maximum 120 dB. How, many watts of power should the new, stereo supply to each speaker?, , Assume the sound power of the loudspeakers equals the power supplied by the stereo., Variables, current power per speaker, , P1 = 10 W, , reference power level, , P0, , current sound level, , ȕ1 = 100 dB, , proposed power per speaker, , P2, , proposed sound level, , ȕ2 = 120 dB, , What is the strategy?, 1., , State two equations that relate sound level to power, for both the current system and the proposed system., , 2., , Subtract the two equations and solve the resulting equation to determine the power of the new stereo., , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 313
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Physics principles and equations, The equation for sound level with respect to the power of the sound source is, , Mathematics principles, , log (a) í log (b) = log (a/b), Step-by-step solution, , Step, , Reason, , 1., , sound level power equation, , 2., , current system, , 3., , proposed system, , 4., , subtract, , 5., , difference of logarithms, , 6., , divide and simplify, , 7., , take antilogarithm, , 8., , P2 = 1000 W, , solve, , To increase the sound level by 10 decibels, the sound power (and sound intensity) must increase tenfold. Raising the maximum sound intensity, of the system by 20 decibels requires a 100-fold increase in sound power and stereo power. The result will be loud enough to ensure the entire, dormitory hears your music!, We hope they appreciate your taste., , 16.7 - Doppler effect: moving sound source, , Doppler effect: A change in the frequency of a, wave due to motion of the source and/or the, listener., You experience the Doppler effect when a train races past you while sounding its, whistle. As the train is approaching, you perceive the whistle as emitting sound of one, frequency, and as it moves away, the perceived frequency of the whistle drops to a, lower pitch., This effect is named for the Austrian physicist who first analyzed it, Christian Doppler, (1803-1853). Doppler’s research concerned light from stars, but his principles apply to, sound also., , Sound source stationary, Wavelength, frequency constant, , In the example described above, the frequency of the sound emitted by the train is, constant. The Doppler effect occurs because of the motion of the source of the sound, the train whistle. It is moving first toward and then away, from you, and you are standing still. (What is moving and what is still is relative to sound’s medium, the air.), You see this situation illustrated on the right. In Concept 1, the train and listener are both stationary. The diagram shows the peaks of the, sound waves as they emanate from the train and radiate in all directions, including toward the listener. They are equally spaced, which means, their wavelength is constant, as is their frequency., In Concept 2, the train is moving to the right, toward the listener, at a source velocity vs. As you see, the peaks of the sound waves at the, listener are closer together than in Concept 1, which means the wavelength is shorter and the frequency at the listener is higher., , 314, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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The motion of the train causes these changes in wavelength and frequency. To, understand this, consider two successive regions of condensation generated by the, train’s horn. The first moves toward the listener. The train continues to move forward,, and the next time the horn creates a region of condensation, it will be closer to the prior, one than if the train were stationary. The regions arrive more frequently because of the, motion of the train toward the listener., Concept 3 shows the effect perceived by a listener for whom the train is moving away., The sound waves reach this listener less frequently, and he hears a lower pitched, sound., The Doppler effect is quantified using the two equations shown to the right. (They apply, when the sound source is moving; a different set of equations is used when the listener, is moving.) The first equation shows how to calculate the frequency when the source of, the sound moves directly toward a stationary listener; the second is used when the, source moves directly away. If the source is moving in some other direction, the, component of its velocity directly toward or away from the listener must be used in the, formulas. The speed of sound changes with temperature, air density and so on; a value, of 343 m/s is often used., , Source moves toward listener, Sound wavefronts arrive closer together, Listener hears higher frequency, , Source moves away from, listener, Sound wavefronts arrive farther apart, Listener hears lower frequency, , Source moves toward listener, , fL = frequency perceived by listener, fs = frequency emitted by source, vs = speed of the source, v = speed of sound in air § 343 m/s, , Source moves away from, listener, , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 315
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16.8 - Sample problem: Doppler effect, The frequency of the train whistle, when the train is not moving is 495, Hz. What sound frequency does each, person hear?, , Variables, speed of train, , vs = 50.0 m/s, , speed of sound, , v = 343 m/s, , frequency of train whistle when, train is motionless, , fs= 495 Hz, , frequency heard by front listener, , fL1, , frequency heard by rear listener, , fL2, , What is the strategy?, 1., , Use the equation for determining the Doppler effect when a sound-emitting object is moving toward a listener., , 2., , Then use the equation for determining the Doppler effect when a sound-emitting object is moving away from a listener., , Physics principles and equations, The equation for a sound source approaching a listener, , The equation for a sound source moving away from a listener, , Step-by-step solution, We start by calculating the frequency perceived by the front listener. The train is approaching that listener., , Step, , Reason, , 1., , Doppler equation, , 2., , substitute values, , 3., , fL1 = 579 Hz, , evaluate, , Now we calculate the frequency perceived by the rear listener, for whom the train is moving away., , Step, , Reason, , 4., , Doppler equation, , 5., , substitute values, , 6., , fL2 = 432 Hz, , evaluate, , This is quite a noticeable Doppler effect. The listener on the right hears a frequency roughly one-third higher than the listener on the left. If the, train were playing a musical note, the listener on the right would hear a pitch about five semitones (piano keys) higher than the listener on the, left., , 316, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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16.9 - Supersonic speed and shock waves, After studying the Doppler effect equations, you might, wonder: What happens if the speed of the sound, source equals the speed of sound? The relevant, Doppler equation shows that you would have to divide, by zero, yielding infinite frequency, a troubling result., Equally troubling was the result when aircraft first, tried to break the sound barrier (the speed of sound)., In the first half of the 20th century, the planes tended, to fall apart as much as the equation does. It was not, until 1947 that a plane was able to fly faster than the, speed of sound, as excitingly shown in the movie The, Right Stuff, proving that sound was not a barrier after, all., , This boat is traveling faster than the speed of waves in water., Its wake forms a two-dimensional "Mach cone" on the water's surface., , The Doppler effect equations do not apply if the sound source is moving at or above the, speed of sound. Concept 1 shows the result when a sound source moves at the speed, of sound. The leading edges of the sound waves bunch up at the tip of the aircraft as, the plane travels as fast as its own sound waves., Concept 2 shows the result when an aircraft exceeds the speed of sound. Aircraft, capable of flying that fast are called supersonic. The plane travels faster than its own, sound waves, and the waves spool out behind the plane creating a Mach cone., The surface of the Mach cone is called a shock wave. Supersonic jets produce shock, waves, which in turn create sounds called sonic booms. As long as a jet exceeds the, speed of sound, it will create this sound. A rapid change in air pressure causes the, sonic boom. Shock waves may be visible to the human eye because a rapid pressure, decrease lowers temperature and causes water molecules to condense, resulting in fog., You may have heard other sonic booms, such as the report of a rifle or the crack of a, well-snapped whip. The boom indicates that the bullet or the tip of the whip has moved, faster than the speed of sound. The example of the whip shows that the moving object, can be silent and still create a shock wave., , When object is at speed of, sound, It travels as fast as its own sound waves, , Equation 1 shows how the sine of the angle of a Mach cone can be calculated as a ratio, of speeds. The inverse ratio, of the speed of the object to the speed of sound, is known, as the Mach number. A fighter jet described as a “Mach 1.6 plane” can move as fast as, 1.6 times the speed of sound, or about 550 m/s (more than 1200 mph). You see this, described mathematically in Equation 2. Because the speed of sound varies with factors, like temperature, the exact speed of a Mach 1.6 plane depends on its environment., , Supersonic speed, Object exceeds speed of its sound, waves, ·Sound waves form Mach cone, ·Surface of cone is shock wave, ·Angle ș is called Mach angle, , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 317
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Mach angle, , ș = Mach angle, vsound = speed of sound, vobject = speed of object, , Mach number, , 16.10 - Gotchas, You hear sound because air molecules move from the sound’s source to your ears.No. Air molecules oscillate back and forth as a sound wave, passes by, but there need be no net motion of the particles that make up the medium., The speed of sound can vary significantly depending on the medium it travels through. Yes. The speed of sound varies widely. Sound travels, nearly five times faster through water than through air, for instance, and faster still through solids, including many metals., The greater the amplitude of a sound wave, the faster the wave moves. No. The speed of a sound wave is dependent solely on the properties, of the medium through which it moves., A 20 dB sound is twice as intense as a 10 dB sound. No, the decibel measurement system is logarithmic. The 20 dB sound has ten times the, intensity: ten times as many watts per square meter. A typical person perceives it as twice as loud., The degree of Doppler shift experienced by a listener is independent of how far the listener is from a moving sound source. This is true:, Distance is not a factor in determining the Doppler shift., , 318, , Copyright Kinetic Books Co. 2000-2007 Chapter 16
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16.11 - Summary, Sound waves are longitudinal mechanical waves. Instead of the peaks and troughs, of transverse waves, sound waves are composed of condensations and rarefactions, of the medium through which they travel. Particles in a sound wave move in simple, harmonic motion along the direction that the wave travels., The intensity of a sound is the sound power passing perpendicularly through a unit, area. For a sound that spreads radially, such as from a fireworks explosion in, midair, the sound intensity is inversely proportional to the square of the distance, from the source., Do not confuse sound intensity with the sound level. The sound level takes into, account the logarithmic perception of sound by the human ear., The Doppler effect is the change in frequency of a wave due to the relative motion, of the source and/or the listener. A common example is the change in frequency, heard as the siren on a fire engine races by you. As the fire engine moves toward, you, the sound waves “pile up” and their wavelength decreases. Since the speed of, sound remains the same, this results in you hearing a higher frequency., , Speed of sound in air, , Sound intensity, , I = P/A, Sound spreading radially, , Sound level, , Doppler effect, source moves toward, listener, , Doppler effect, source moves away, from listener, , Mach angle, , Copyright Kinetic Books Co. 2000-2007 Chapter 16, , 319
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Chapter 16 Problems, , Chapter Assumptions, Unless stated otherwise, use 343 m/s for the speed of sound., , Conceptual Problems, C.1, , Why do people at public events use cone-shaped bullhorns to address the crowd? (We are referring to the non-batterypowered type!), , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Does the pitch get higher,, lower or does it stay the same as the frequency increases? (b) Does the pitch get higher, lower or does it stay the same as, the amplitude increases?, (a), , (b), , i., ii., iii., i., ii., iii., , Higher, Lower, Stays the same, Higher, Lower, Stays the same, , Section 3 - Interactive problem: sound frequency, 3.1, , Use the simulation in the interactive in this section to determine whether higher pitched notes have higher or lower, frequencies than lower pitched notes., i. Higher, ii. Lower, iii. The same, , Section 4 - Sound intensity, 4.1, , You are standing 7.0 meters from a sound source that radiates equally in all directions, but it is too loud for you. How far, away from the source should you stand to experience one third the intensity that you did at 7.0 meters?, m, , 4.2, , 15.0 meters from a sound source that radiates freely in all directions, the intensity is 0.00460 W/m2. What is the rate at which, the source is emitting sound energy?, W, , 4.3, , A longitudinal wave spreads radially from a source with power 345 W. What is the intensity 40.0 meters away?, W/m2, , 4.4, , Sound spreads radially in all directions from a source with power 15.3 W. If the intensity you experience is 3.00×10í6 W/m2,, how far away are you from the source?, m, , Section 5 - Sound level in decibels, 5.1, , The sound of heavy automobile traffic, heard at a certain distance, has a sound level of about 75 dB. What is the intensity of, the noise there?, W/m2, , 5.2, , Jane and Sam alternately pound a railroad spike into a tie with their hammers. The crew chief has a migraine, and notes that, Jane's hammer blows cause a sound with intensity 5.0 times greater than the sound that Sam makes when he swings his, hammer. What is the difference in sound level between the two sounds?, dB, , 5.3, , 320, , A grade-school teacher insists that the the overall sound level in his classroom not exceed 64 dB at his location. There are 25, students in his class. If each student talks at the same intensity level, and they all talk at once, what is the highest sound level, , Copyright 2007 Kinetic Books Co. Chapter 16 Problems
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at which each one can talk without exceeding the limit?, dB, 5.4, , A cafe is known for its tasty blended juice drinks. The owner worries that if too many blenders are running in the preparation, area at once, the noise will drive customers away. She determines that the maximum allowable sound level at the cash, register should be 84 dB. If each blender in the cafe emits sound with a level of 79 dB at the register, how many blenders can, be running at once before the maximum is exceeded?, blenders, , 5.5, , The human eardrum has an area of 5.0×10–5 m2. (a) What is the sound power received when the incident intensity is the, minimum perceivable, 1.0×10–12 W/m2? (b) What is the sound power at the eardrum for conversational speech, which has a, sound level of 65 dB? (c) What is the sound power at the threshold of pain, 140 dB?, (a), , 5.6, , W, , (b), , W, , (c), , W, , Pete's family is watching a movie at home, with the TV emitting sound at a level of 71 dB as heard by the family. His dad, turns on the popcorn popper, which emits sound that has a level of 68 dB as heard by the family. (a) What is the total sound, intensity perceived by Pete's family? (b) What is the total sound level?, (a), , W/m2, , (b), , dB, , 5.7, , Suzy revs her monster truck, producing a sound with a sound level of 84 dB as heard by Mrs. DiNapoli across the street., Next door to Suzy, Mike guns his station wagon which produces a sound with a sound level of 71 dB at Mrs. DiNapoli's. At, this poor neighbor's house across the street, what is the unbearable ratio of the intensity of the truck's sound to the intensity of, the wagon's sound?, , 5.8, , Ju Yeon sits in a park where people come to fly small radio-controlled airplanes. She can hear down to a sound level of, 20 dB. If a plane emits sound with a power of 1.6eí6 W, how far away will the plane be when she is just able to hear it?, m, , Section 7 - Doppler effect: moving sound source, 7.1, , Passengers on a train hear its whistle at a frequency of 735 Hz. What frequency does someone standing by the train tracks, hear as it moves directly toward her at a speed of 22.5 m/s?, , 7.2, , A particular ambulance siren alternates between two frequencies. The higher frequency is 617 Hz as heard by the ambulance, driver. If the ambulance drives directly away from you at a speed of 16.8 m/s, what frequency do you hear as the higher, frequency?, , Hz, , Hz, 7.3, , A small plane is taxiing directly away from you down a runway. The noise of the engine, as the pilot hears it, has a frequency, 1.13 times the frequency that you hear. What is the speed of the plane?, m/s, , 7.4, , A train moves away from you at 29.0 m/s, sounding its horn. You perceive the frequency of the horn's sound as being, 415 Hz. What is the frequency heard by the train's passengers?, Hz, , 7.5, , A stuntwoman is preparing to take a punch, crash through a "candy glass" window, and fall a long distance. The script calls, for her to emit a piercing scream just before she hits the "ground." In reality, she will land on a waiting airbag. Lights! Camera!, Action! The primary camera crew, filming from her starting height, hears her last-instant scream at a frequency 3.77 kHz. Her, scream has a frequency of 4.05 kHz when she is at rest. How far did she fall? Report this as a positive distance., m, , Section 9 - Supersonic speed and shock waves, 9.1, , The Mach angle of a jet traveling at supersonic speed is 19.4°. What is the Mach number?, , 9.2, , What is the Mach angle of a jet moving at Mach 2.2?, °, , Copyright 2007 Kinetic Books Co. Chapter 16 Problems, , 321
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17.0 - Introduction, In this chapter, we will discuss what happens when, two or more traveling waves combine with each other,, as when waves meet in a pond, a pool or even a, bathtub. The result is higher peaks and lower troughs., These waves can “pass through” each other and then, regain their original shape and direction, in contrast to, the collision of two particles, such as tennis balls,, which alters the motion of the balls., A stringed musical instrument like a violin or a guitar, uses the reflection and recombination of waves from, the end of a bowed or plucked string to create its, Intersecting ripples from two different wave sources in a pond., magical sound. You also hear the result of waves, combining when you listen to music on loudspeakers, or as a live performance of a band or a symphony. Music halls are designed to take, advantage of the reflection and combination of sound waves to produce an effect, that audience members will find particularly pleasing., You can begin your study of the result of combining waves with the simulations to, the right. In Interactive 1, you control two wave pulses traveling on the same string., One pulse starts on the left and moves right, and the other starts on the right and, moves left. You determine the amplitude and width of each pulse, as well as, whether it is a peak or a trough. Set the parameters for each pulse and press GO to, see what happens when the pulses meet. If you want to see the pulses combining in, slow motion, press the “time step” arrow to advance or reverse time in small, increments., A challenge for you: Can you set the pulses so that when they meet they exactly, cancel each other out, causing the string to be completely flat for an instant?, In Interactive 2, two traveling waves combine as they move toward one another., You determine the amplitudes and wavelengths of these waves. Again, set the, parameters of the waves and press GO to see the result of combining the waves., Can you create a single combined wave that does not move either to the left or to, the right? If you do, you will have created what is called a standing wave. You will, find that by making the settings of the two waves identical, except for travel, direction, you can create such a wave., , 17.1 - Combining waves: the principle of superposition, In much of this chapter, we discuss what happens, when traveling waves combine. In this section, we, consider the less complicated case of what occurs, when two wave pulses combine in a string, as you, see above. This will help us illustrate the principle of, superposition, which is more readily viewed with, pulses than with traveling waves., , Two traveling wave pulses on a string., , 322, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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In Concept 1 below, we show what occurs when two peaks like the ones in the illustration above combine. We show the result at four, successive times. A blue pulse is traveling from left to right and a red pulse is traveling in the opposite direction. You can see that at each, instant the combined pulse is determined by adding the vertical displacements of the two pulses at every point along the string., , Superposition of waves, Combined wave = sum of waves, Add wave displacements at each point, , In the description above, we are applying the principle of superposition: The wave that results when two or more waves combine can be, determined by adding the displacements of the individual waves at every point in the medium. This is sometimes more tersely stated as: The, resulting wave is the algebraic sum of the displacements of the waves that cause it. (“Algebraic” means that you add positive and negative, displacements as you would any signed numbers; no trigonometry is required.), You see this principle in play in Concept 1. For instance, when the two peaks meet at time t = 2.5 s, the resulting peak’s displacement equals, the sum of the displacements of the two separate peaks. This is an example of constructive interference, which occurs when the amplitude of a, combined pulse or wave is greater than the amplitude of any individual pulse or wave., In Concept 2 below, a peak meets a trough. Except for the different directions of displacement, the pulses are identical. The two pulses cancel, out completely when they occupy the same location on the string, and the string momentarily has zero displacement at each point. Positive and, negative displacements of the same magnitude sum to zero. This is destructive interference: The amplitude of the combined pulse or wave is, less than the amplitude of either individual pulse or wave., , Peak meets trough, Displacement is reduced, , You may have experimented with this in the introduction simulation, but if you did not, you can go back and see what happens when peak, meets peak, when peak meets trough, and finally, when two troughs meet. The result in each case is that the combined displacement is the, sum of the displacements of all the pulses., When the string is “flat” in Concept 2, it may seem there is no motion because you are looking at a static diagram. In fact, some of the string, particles are moving up and some are moving down. The particles that were part of a peak are moving down and will be part of a trough. This, is readily witnessed in the introductory simulation., In this section, we used transverse wave pulses on a string to illustrate superposition, but the principle of superposition can also be applied to, longitudinal waves (for instance, the combined sound wave produced by two stereo speakers). Acoustical (sound) engineers rely on, constructive interference to create louder sounds and destructive interference to mask noises. For instance, noise-reducing headphones, contain a microphone that detects unwanted noise from the environment. A circuit then creates a sound wave that is an inverted version of the, noise wave, with peaks where the noise has troughs, and vice-versa. When this wave is played through the headphones, it destructively, interferes with the unwanted ambient noise. The same technique is used to reduce the noise from fans in commercial heating and ventilation, systems., , Copyright 2000-2007 Kinetic Books Co. Chapter 17, , 323
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17.2 - Standing waves, , Standing wave, Created by identical waves moving in, opposite directions, , Standing wave: Oscillations with a stationary, outline produced by the combination of two, identical waves moving in opposite directions., When two identical traveling waves move in opposite directions through a medium, the, resulting combined wave stays in the same location. The resulting wave is called a, standing wave. Individual particles oscillate up and down, but unlike the case of a, traveling wave, the locations of the peaks and troughs of a standing wave stay at fixed, positions., Consider the illustration above, showing waves on a string. The standing wave is, formed by the combination of two traveling waves. We show three “snapshots” in time, of two identical waves traveling in opposite directions, and the combined wave they, create. The traveling waves are shown in colors. These waves have the same, frequency and amplitude, but are traveling in opposite directions. The result is a, standing wave that does not move along the string., , Standing wave, Nodes: no displacement, Antinodes: maximum displacement, , In the first snapshot above, the two traveling waves are out of phase, and they destructively interfere. The combined standing wave has a, constant zero displacement. In the second snapshot, the traveling waves have moved slightly, and the standing wave exhibits some peaks and, troughs. In the third snapshot, the traveling waves exactly coincide, in phase, to constructively interfere, and the standing wave has its largest, peaks and troughs. As the traveling waves continue to move, the standing wave’s peaks and troughs will diminish until the traveling waves are, again out of phase and the standing wave is flat., If you find this difficult to visualize, you can see a simulation of the creation of a standing wave by clicking on the whiteboard above. You can, have the simulation move slowly, step by step, by pressing the arrow buttons. You can also press “show components” if you want to see the, component traveling waves. Press REPLAY to restart the simulation., Along a standing wave, there are some fixed positions where there is no displacement and others where there is the maximum displacement., The positions with no displacement are called nodes. The positions where the wave has the greatest displacement (the peaks and troughs) are, called antinodes. Adjacent nodes (and antinodes) are separated by a constant distance. Looking at the illustration for Concept 2, you can see, that two adjacent nodes are separated by one half the wavelength of the wave. The same is true for two adjacent antinodes., The standing wave depicted above is created by the combination of transverse waves. Longitudinal waves, such as sound waves, can combine, to form standing waves as well., , 324, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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17.3 - Reflected waves and resonance, , Pulse on string, Reflection is initial pulse inverted, , We have considered standing waves formed by two waves generated by separate, sources, but they can also be formed by a single wave reflecting off a fixed point. This is, the basis behind the sound production of many musical instruments. We will discuss, this topic using the example of a piece of string, with one end connected to a wavemaking machine that vibrates sinusoidally, moving the string up and down, and the, other end fixed to a wall., To start, consider what happens when the wave machine generates a single pulse, as, illustrated above. We show the pulse at three positions over time. When the pulse, reaches the wall, it “yanks” on the wall. Newton’s third law dictates that there will be an, equal “yank” in the opposite direction, which sends a reflected pulse back down the, string. The reflected pulse is inverted from the original, so when a peak reaches a wall,, a trough returns in the opposite direction., , Wave on string, , A wave is a continuous series of pulses. When the wave machine vibrates continuously,, Reflects at wall, each pulse will reflect off the wall, resulting in an inverted wave moving at the same, Creates standing wave, speed in the direction opposite to the original wave. The reflected wave travels down, the string toward the wave machine. We also consider the wave machine as fixed,, which is reasonable if its vibrations are small in amplitude. The wave then reflects off that piece of machinery just as it reflected off the wall., The wave machine continues to vibrate, sending wave pulses down the string. If it vibrates in synchronization with the reflected wave, the result, will be two traveling waves of equal amplitude, frequency and speed, moving in opposite directions on the same string. This system has, created the condition for a standing wave: two identical traveling waves moving in opposite directions. You see this illustrated in Concept 2., On the other hand, if the wave machine is not in sync with the reflected wave, it will continue to add “new” waves to the string that will combine, in more complicated ways, and there may be no obvious pattern of movement on the string., If the wave machine works in synchronization with the reflected waves to create a standing wave, we say that it is working in resonance. Its, motion reinforces the waves, and the amplitude of the resulting standing wave will be greater than the amplitude of the vibrations of the wave, machine. This is akin to you pushing a friend on a swing. If you time the frequency of your “pushes” correctly, you will send your friend higher., We will discuss next how this frequency is determined., , 17.4 - Harmonics, , Fundamental frequency: The, frequency of a standing wave, in a vibrating string that has, two nodes., Harmonic: A frequency of a, standing wave in a string that, has more than two nodes., , A tuning peg is used to change the frequency of a guitar string., , We have considered vibrating strings fairly abstractly. However, strings connected to two fixed points are the basis for musical instruments, such as violins, cellos and so forth. To put it another way: Orchestras have “string” sections., In this section, we want to put your knowledge of standing waves into practice. To do so, we ask a question: When a musician plays a note,, what determines the frequency at which the string will vibrate? To put the question another way, what are the possible frequencies of a, standing wave on a string fixed at both ends?, , Copyright 2000-2007 Kinetic Books Co. Chapter 17, , 325
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To answer this question, we begin by considering the number of nodes that must be, present on the vibrating string. The minimum number of nodes is two: There must be a, node at each end of the string because the string is attached to two fixed points. These, nodes might be the only two, but there may be intermediate nodes as well. Drawings of, standing waves with zero, one and two intermediate nodes are shown in Concept 1, below., , Harmonics, , for n = 1, 2, 3, …, fn = nth harmonic, v = wave speed, Ȝn = wavelength of nth harmonic, L = string length, , A tension of 417 N is applied to a, 1.56 m string of mass 0.00133 kg., What is the fundamental, frequency?, , v = 699 m/s, , ƒ1 = (1)(699 m/s)/2(1.56 m), ƒ1 = 224 Hz, , 326, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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Harmonics, Standing waves at specific wavelengths, Corresponding frequencies are, harmonics, First harmonic is fundamental frequency, , The fundamental frequency of the string occurs when the only nodes are at the ends of the string. The fundamental frequency is also called the, first harmonic. The second harmonic has one additional node in the middle of the string, the third harmonic two such nodes, the fourth, harmonic three such nodes, and so on. Each harmonic is created by a particular mode of vibration of the string., The fundamental frequency occurs when the wavelength of the standing wave is twice the length of the string, because two adjacent nodes, represent half a wave. In general, the wavelength of the nth harmonic is twice the length of the string divided by n, with n being a positive, integer. That is, the wavelength of the nth harmonic on a string of length L is Ȝn = 2L/n., The frequency of a wave is its speed divided by the wavelength, which lets us restate the equation above in terms of frequency. The equation, to the right is the basic equation for harmonic frequencies. The harmonic frequencies are positive integer multiples of the fundamental, frequency v/2L. Let’s say the fundamental frequency ƒ1 of a string is 30 hertz (Hz). The second harmonic ƒ2 will be 60 Hz, the third harmonic, ƒ3 90 Hz, and so forth., You see these principles in play in musical instruments like the piano. Its strings are of different lengths, which is one factor in determining their, fundamental frequency. Other factors that you see in the equations are also employed in musical instruments to determine a string’s, fundamental frequency. Pianos have thicker and thinner strings. A string’s mass per unit length (its linear density) will partly determine its, fundamental frequency, by changing the wave speed on the string., In addition, string instruments are “tuned” by changing the tension in a string. You will see a guitarist frequently adjusting her instrument by, turning a tuning peg, as you see at the top of this page, which determines the tension in a string. Along with linear density, this is the other, factor that determines wave speed. A guitarist will also press her finger on a single string to temporarily create a string with a specific length, and fundamental frequency., A harmonic is sometimes called a resonance frequency or a natural frequency. Musicians also use other terms dealing with frequencies and, harmonics. An overtone or a partial is any frequency produced by a musical device that is higher than the fundamental frequency. Unlike a, harmonic, the frequency of an overtone does not necessarily bear any simple numerical relationship to the fundamental., Some overtones are harmonics í that is, whole-number multiples of the fundamental í but some are not. Musical instruments such as drums,, whose modes of vibration can be very complex, create non-harmonic overtones. Although harmonic overtones are “simple” integer multiples of, the fundamental, they are often referred to by numbers that are, confusingly, “one off” from the numbers for harmonics: The first overtone is the, second harmonic, and so on. Each musical instrument has a characteristic set of overtones that creates its distinctive timbre., , 17.5 - Interactive problem: tune the string, Let’s say you play an unusual instrument in the orchestra, the alto pluck, an, instrument specifically designed for physics students. The concert is underway and, your big moment is coming up, when you get to play a particular note on the pluck., You are supposed to play the G above middle C, a note that has a frequency of, 392 Hz. You produce the correct note by setting the string length and the harmonic, number. Remember that harmonic numbers are multiples of the fundamental, frequency of the string. For this instrument, you are limited to harmonic numbers in, the range of one to four., When you set a harmonic number higher than one, a finger will touch the string at a, position that will cause the string to vibrate with the harmonic number you selected., It does so by creating a standing wave node at the appropriate location. For, instance, if you choose a harmonic number of four, there will be three nodes, between the two ends of the string, and the finger will be one-fourth of the way, along the string. If you see a musician such as a cellist performing, you will see that, he sometimes lightly places his fingers at locations along a string to create, harmonics in just this way. He also frequently presses a string firmly against the, fingerboard to create a different, shorter, string length., The string length of the alto pluck can range from 1.00 to 2.00 meters, but within that range, you are restricted to values that will produce, frequencies found on the musical scale. You will find that as you click on the arrows for length, the values will move between appropriate string, lengths. The harmonic values are easy to set: Just choose a number from one to four!, The description above is reasonably complicated; it is impressive that musicians with some instruments must make similar determinations as, they play. In terms of approaching this problem, you will want to start with the equation, , ƒn = nv/2L, that enables you to calculate the frequency of the nth harmonic. The wave speed in your stringed instrument is 588 m/s., , Copyright 2000-2007 Kinetic Books Co. Chapter 17, , 327
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There is only one solution to the problem we pose, given the range of string lengths and harmonics together with the wave speed we provide., If you are not sure how to proceed, try solving the equation above for the string length and entering the known values. You will still have, another variable, n, left in the equation. However, since you know the range of string lengths, and that n must be an integer from one to four,, you will be able to solve the problem., To test your answer, set a string length and harmonic number and press GO to see a hand come down and pluck the string so you can hear, the resulting note and see its frequency. The resulting musical note is displayed on the sounding board. Press RESET to start over., , 17.6 - Sample problem: string tension, The frequency of the standing wave, on the string is 329 Hz. What is the, tension on the string?, , The values shown in the problem are representative of the lowest frequency string on a guitar. To answer the question, you must first, determine the harmonic of the standing wave. You can do so by inspecting the diagram above. We exaggerated the amplitude of the wave to, make the nodes and antinodes more visible., Variables, string length, , L = 0.640 m, , string mass, , m = 0.00435 kg, , frequency, , ƒ = 329 Hz, , harmonic number, , n, , wave speed, , v, , tension, , T, , What is the strategy?, 1., , Determine the harmonic by counting the number of nodes shown in the string above., , 2., , Calculate the wave speed from the frequency, string length and harmonic number., , 3., , Calculate the tension using the equation for wave speed on a string., , Physics principles and equations, Nodes are locations where there is no displacement. The harmonic number is one less than the number of nodes, including the nodes at the, ends of the string., The equation for the nth harmonic, , The wave speed on a stretched string, , 328, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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Step-by-step solution, First, we determine the harmonic number by looking at the wave above. Then, we calculate the wave speed using the equation for the, frequency of the nth harmonic., , Step, , 1., , n=4, , Reason, , harmonic is one less than number of nodes, , 2., , frequency of nth harmonic, , 3., , solve for wave speed, , 4., , substitute values, , 5., , v = 105 m/s, , evaluate, , Now we use the equation that relates wave speed to tension, mass, and string length., , Step, , Reason, , 6., , wave speed on string, , 7., , solve for tension, , 8., , substitute values, , 9., , T = 74.9 N, , evaluate, , 17.7 - Wave interference and path length, We described wave interference resulting from a phase difference, or differing initial, conditions for two waves moving in the same direction. Interference also results when, two waves travel from different starting points and meet. To illustrate this, we use the, example of two longitudinal traveling waves produced by two loudspeakers., To the right, we show a person listening to the loudspeakers. The vibrating speakers, create regions of pressure that are greater than atmospheric pressure (condensation), and less (rarefaction). We show this pattern of oscillation emanating from each speaker, in Concept 1. We assume the speakers create waves with the same amplitude and, wavelength, and that there is no phase difference between them. We focus on the point, where the two waves combine just as they reach the listener’s ear., In Concept 1, we position the two speakers so that the listener is equidistant from them., The distance from a speaker to the ear is called the path length. Since the waves travel, the same distance, they will be in phase when they arrive. This means peaks and, troughs exactly coincide with each other., , Path lengths the same, Constructive interference, , When two peaks combine, they double the pressure increase. When two troughs combine, they also add, and the pressure decrease is, doubled. The listener hears a louder sound. In short, the waves constructively interfere., The waves would also constructively interfere if one speaker were placed one wavelength farther away. Peaks would still meet peaks and, troughs would still meet troughs. In fact, if the difference in the distances between the loudspeakers and the listener is any integer multiple of, the wavelength, the waves constructively interfere. (This does assume that the loudspeakers vibrate at a constant frequency, an unusual, assumption for most music.) You see this condition for constructive interference stated in Equation 1., Can we arrange the speakers so that the waves destructively interfere? Yes, by moving one speaker one-half wavelength away from the, listener. This is shown in Concept 2. When the waves combine, peaks will meet troughs and vice versa. This will also occur if the, speaker/listener distances differ by half a wavelength, or 1.5 wavelengths, or any half-integer multiple of the wavelength. This condition for, destructive interference is stated as an equation on the right. If the waves have the same amplitude at the ear and destructively interfere, completely, the result is silence at the listener’s position., , Copyright 2000-2007 Kinetic Books Co. Chapter 17, , 329
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The principle of conservation of energy applies to sound waves. When two waves, destructively interfere at some position, the energy does not disappear. Rather, there, must be another area where the waves are constructively interfering, as well. For any, area of silence, there must be a louder area. Positioning loudspeakers or designing, concert halls to minimize the severity of these “dead spots” and “hot spots” is a topic of, interest to audiophiles and sound engineers alike., , Path lengths differ by one-half, wavelength, Destructive interference, , Constructive interference:, ǻp = nȜ, Destructive interference:, , ǻp = difference in path lengths, n = 0, ±1, ±2, …, Ȝ = wavelength, 17.8 - Beats, , Beats, Periodic variations in amplitude, produced by two waves of different, frequencies, , Beats: The pattern of loud/soft sounds caused by two sound sources with, similar but not identical frequencies., So far in this chapter, we have studied combinations of traveling waves with the same amplitude, frequency and wavelength. You may wonder, , 330, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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what happens if waves that have different properties are combined, and in this section,, we consider such a situation., To be specific, we consider two traveling waves with the same amplitude but slightly, different frequencies, using sound waves as our example. We examine the waves and, their combination at a fixed x position. You see the graphs above of two such waves, over time at a particular position. Note that in this case the graphs that you see are, displaying the displacement over time of a particle at a fixed position in a medium, carrying longitudinal waves., The waves were created by tuning forks, and the combined wave is shown below them., When the waves combine, they produce a wave whose amplitude is not constant, but, instead varies in a repetitive pattern. For sound waves, these “beats” are heard as a, repeating pattern of variation in loudness in a wave of constant frequency., Musicians sometimes use beats to tune their instruments. Sounds that are close in, frequency produce audible beats, but the beats disappear when the frequencies are the, same. A guitar player, for example, might tune the “A” string by playing an “A” on, another string at the same time and adjusting the tension of the “A” string until there are, no longer any audible beats. This occurs when the frequencies match exactly, or are, close enough that the beats are so far apart in time they can no longer be heard., We hear beats because the waves constructively and destructively interfere over time., When they constructively interfere, there is greater condensation and rarefaction of the, air at our ears and we hear louder sounds. Destructive interference means a smaller, change in pressure and a softer sound., , Beat frequency, ƒbeat = ƒ1 í ƒ2, ƒbeat = beat frequency, ƒ1 = frequency of sound one, ƒ2 = frequency of sound two, (ƒ1 > ƒ2), , The beat frequency equals the number of times per second we hear a cycle of loud and, soft. This is computed as shown in Equation 1, as the difference of the original, frequencies. When the frequencies are the same, the beat frequency is zero, as when, two strings are perfectly in tune. Humans can hear beats in sound waves at frequencies, up to around 20 beats per second. Above that frequency, the beats are not, distinguishable., , What is the beat frequency when, these two waves combine?, ƒ = Ȧ/2ʌ, ƒ1 = (3380 rad/s)/2ʌ = 538 Hz, ƒ2 = (3330 rad/s)/2ʌ = 530 Hz, ƒbeat = ƒ1 íƒ2, ƒbeat = 538 Hz í 530 Hz, ƒbeat = 8 Hz, 17.9 - Gotchas, A standing wave has its maximum displacement at an antinode. Yes, that is correct. An antinode is the opposite of a node, where no motion, occurs., The locations where peaks and troughs occur are constant in a standing wave. That is correct, and this is the distinguishing point between a, standing wave and a traveling wave., I see a standing wave on a string with two fixed ends and a single antinode. The wave has two nodes. Yes. The two fixed ends are nodes., Two waves traveling in the same direction can cause a standing wave. No. Waves traveling in opposite directions can cause a standing wave., However, a wave from a single source when it is reflected can cause a standing wave, because the reflected wave is traveling in the opposite, direction., , Copyright 2000-2007 Kinetic Books Co. Chapter 17, , 331
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17.10 - Summary, The principle of linear superposition says that whenever waves travel through a, medium, the net displacement of the medium at any point in space and at any time, is the sum of the individual wave displacements. The superposition of waves, explains the phenomenon of interference., Destructive interference occurs when two waves in the same medium cancel each, other, either partially or fully. If two sinusoidal waves have the same wavelength,, destructive interference will happen when the waves are close to being completely, out of phase, meaning that their phase constants differ by ʌ radians (or 180°)., Constructive interference occurs when two waves in the same medium reinforce, each other. This happens when the waves are close to being in phase for waves, with the same wavelength., , Harmonics, , Interference, , Constructive: ǻp = nȜ, Destructive:, , Intermediate interference is a general term for a situation where the difference, between the phases of two interfering waves, called the phase shift, is somewhere, between 0 and ʌ radians., When two identical waves traveling in opposite directions in the same medium, interfere, they produce a standing wave. In standing waves, there are points called, nodes that experience no displacement at all. The points that experience the, maximum displacement are called antinodes. Standing waves can be either, transverse, as with the oscillations of a piano string, or longitudinal, as with the, sound waves in an organ pipe., , Beat frequency, , fbeat = f1 – f2, , Identical waves from different sources can interfere with each other at a point in space based on the distances they travel to that point, called, their path lengths. Provided they start out in phase, if the path lengths of two waves to a certain point differ by an integer number of, wavelengths, they will constructively interfere at that point. If the path lengths differ by a half-integer number of wavelengths, they will exhibit, completely destructive interference., Another type of interference occurs when two waves have different frequencies. In this case, beats are produced. The waves alternate, between constructive and destructive interference., , 332, , Copyright 2000-2007 Kinetic Books Co. Chapter 17
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Chapter 17 Problems, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the first interactive problem in this section to create a situation where the pulses will exactly cancel each, other out. (a) If the left pulse is a peak, what should the right pulse be? (b) If the amplitude of the left pulse is 2.00 m, what, should the amplitude of the right pulse be? (c) If the width of the left pulse is 1.00 m, what should the width of the right pulse, be?, (a), , Peak, , Trough, , (b), , m, , (c), , m, , Section 4 - Harmonics, 4.1, , What is the fundamental frequency of a 4.65×10í3 kg, 2.50-meter string under 438 newtons of tension?, Hz, , 4.2, , A string that is under a tension of 389 N has a linear mass density of 0.0220 kg/m. Its fundamental frequency is 440 Hz (an A, note). How long is the string?, m, , 4.3, , A string is fixed to two eye hooks 0.940 m apart. The tension in the string is 505 N and its mass is 2.30×10í4 kg. What is the, frequency of the 3rd harmonic?, , 4.4, , You drape a string over a pulley and hang a mass of 35.0 kg from one end. You tie the other end to a point a distance L from, the pulley so that the string is horizontal between this point and the pulley. The string has a linear mass density of, 0.0470 kg/m. If you want the difference in frequency between any two consecutive harmonics to be 35.0 Hz, what will the, distance L have to be?, , 4.5, , A musical instrument is constructed using a 0.550-meter wire with a mass of 6.90×10í4 kg. If the fundamental frequency is to, be 523 Hz, what is the required tension in the wire?, , Hz, , m, , N, 4.6, , A harp string plays the A above middle C, vibrating at its fundamental frequency of 440 Hz. The tension in the string is, 57.8 N, and its mass density is 5.72×10í4 kg/m. What is the distance between the nodes of the standing wave pattern?, m, , Section 5 - Interactive problem: tune the string, 5.1, , Use the information given in the interactive problem in this section to calculate (a) the string length and (b) the harmonic, required to play a G note. Test your answer using the simulation., (a), (b), , m, i., ii., iii., iv., v., , 1, 2, 3, 4, 5, , Copyright 2007 Kinetic Books Co. Chapter 17 Problems, , 333
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Section 7 - Wave interference and path length, 7.1, , The apparatus shown demonstrates, interference of sound waves. A pure tone from, the speaker takes both paths to the listener's, ear, and the resulting difference in path, lengths then creates interference. The length, of one path can be changed by sliding the, adjustable U-shaped tube at the top. You, adjust this tube to find a position that results in, complete destructive interference. From this, point you then measure how far you must pull, the end of the tube up to first hear destructive, interference again. If this distance is 9.20 cm, what is the frequency of the sound that is being used? (Use 343 m/s for the, speed of sound.), Hz, , 7.2, , Two speakers emit sound of identical frequency and amplitude, in phase with each other. The frequency is 415 Hz. If the, speakers are 9.00 meters apart, and you stand on the line directly between them, 3.47 m from one speaker, are you standing, closer to a point of constructive or destructive interference? (Use 343 m/s for the speed of sound.), Constructive, Destructive, , 7.3, , Two speakers are mounted 5.00 meters apart on the ceiling, at a height of 3.05 meters above your ears. You plug up one ear, with a cotton ball, and stand with your other ear directly under the midpoint of a line connecting the speakers. The speakers, emit identical sound waves, which are in phase when they reach your ear. You start walking, remaining directly under the, connecting line, until you first encounter a point of complete destructive interference. You have walked 0.250 meters. What is, the frequency of the sound coming from the speakers? (Use 343 m/s for the speed of sound.), Hz, , Section 8 - Beats, 8.1, , Two violinists are playing their "A" strings. Each is perfectly tuned at 440 Hz and under 245 N of tension. If one violinist turns, her peg to tighten her A string to 251 N of tension, what beat frequency will result? Express your answer to the nearest 100th, of a Hz., Hz, , 8.2, , Two identical strings are sounding the same fundamental tone of frequency 156 Hz. Each string is under 233 N of tension., The peg holding one string suddenly slips, reducing its tension slightly, and the two tones now create a beat frequency of, three beats per second. What is the new tension in the string that slipped?, , 8.3, , Two cellists play their C strings at their fundamental frequency of 65.4 Hz. They are identical strings at the same tension. One, of the cellists plays a glissando (slides her finger down the string) until she has shortened it to an effective length of 15/16 the, length of the other cellist's C string. What beat frequency will result? Express your answer to the nearest tenth of a Hz., , 8.4, , Consider the musical note "A above middle C", known as "concert pitch" or "A440." The frequency of this note is 440 Hz by, international agreement. In the chromatic scale, the frequency of a sharp is a factor of 25/24 higher than the note, and the, frequency of a flat is a factor of 24/25 lower. If the "A sharp" and the "A flat" notes corresponding to A440 are played together,, what will be the resulting beat frequency? (State your answer to the nearest Hz.), , 8.5, , You hold two tuning forks oscillating at 294 Hz. You give one of the forks to your friend who walks away at 1.50 m/s. What, , N, , Hz, , Hz, beat frequency do you hear? Use 343 m/s as the speed of sound and give your answer to the nearest 100th of a Hz., Hz, , 334, , Copyright 2007 Kinetic Books Co. Chapter 17 Problems
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18.0 - Introduction, Thermodynamics is the study of heat (“thermo”) and, the movement of that heat (“dynamics”) between, objects. A kitchen provides an informal laboratory for, the study of thermodynamics. Manufacturers offer, numerous kitchen devices designed to facilitate the, flow of heat: stovetops and ovens, convection ovens,, toasters, refrigerators, and more. Heat flow changes, the temperature of what is being cooked or cooled,, and that can be monitored with thermometers., This chapter starts with a few basic thermodynamics, concepts, namely how temperature is measured,, what temperature scales are, and what is meant by, heat. It then begins the discussion of the relationship, between heat and temperature., , Kitchen appliances are engineered using the principles of thermodynamics., , 18.1 - Temperature and thermometers, Although temperature is an everyday word, like energy it is surprisingly hard to define., For now, we ask that you continue to think of temperature simply as something, measured by a thermometer. Warmer objects have higher temperatures than cooler, objects., Traditional thermometers rely on the important principle that any two objects placed in, contact with each other will reach a common temperature. For instance, when a, traditional fever thermometer is placed under your tongue, after a few minutes the flow, of heat causes it to reach the same temperature as your body., While there are many different types of thermometers available, they all rely on some, physical property of materials in order to measure temperature. In the “old” days, body, temperature was measured with a glass thermometer filled with mercury, a material that, expands significantly with temperature and whose expansion is proportional to the, change in temperature., , Thermometers, Measure temperature based on physical, , Today, a wide variety of physical properties are used to determine temperature. Some, properties, medical clinics use thermometers that measure temperature with plastic sheets, containing a chemical that changes color with temperature. Battery-powered digital thermometers rely on the fact that a resistor’s resistance, changes with temperature. Ear thermometers use thermopiles that are sensitive to subtle changes in the infrared radiation emitted by your, body; this radiation changes with your temperature., , 18.2 - Temperature scales, In the United States, the Fahrenheit system is the most common measurement system, for temperature. The units in this system are called degrees. In most of the rest of the, world, however, temperatures are measured in degrees Celsius. Physicists use the, Celsius scale or, quite often, another scale called the Kelvin scale. All three scales are, shown on the right., There are two things required to construct a temperature scale. One is a reference, point, such as the temperature at which water freezes at standard atmospheric, pressure. As shown to the right, the three scales have different values at this reference, point. Water freezes at 273.15 kelvins (273.15 K), 0° Celsius (0°C) and 32°, Fahrenheit (32°F). Notice that the standard terminology for the Kelvin scale avoids the, use of “degrees.” Water freezes at 273.15 kelvins, not 273.15 degrees Kelvin., The other requirement is to pick another reference point, such as the temperature at, which water boils at standard atmospheric pressure, and establish the number of, degrees between these two points. This determines the magnitude of the units of the, scale. The Celsius and Kelvin scales both have 100 units between the freezing and, boiling points of water. This means that their units are equal: a change of 1 C° equals a, change of 1 K. (Changes in Celsius temperatures are indicated with C° instead of °C.), In contrast, there are 180 degrees between these temperatures in the Fahrenheit, system., , Temperature scales, Kelvin, Celsius, and Fahrenheit, Water freezes at 0°C, Absolute zero is 0 K, Unit of Celsius = unit of Kelvin, , Copyright 2007 Kinetic Books Co. Chapter 18, , 335
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Another important concept is shown in the illustration to the right: absolute zero. At this temperature, molecules (in essence) cease moving., Reaching this temperature is not theoretically possible, but temperatures quite close to this are being achieved. Absolute zero is 0 K, or, í273.15°C., To standardize temperatures, scientists have agreed on a common reference point called the triple point. The triple point is the sole, combination of pressure and temperature at which solid water (ice), liquid water, and gaseous water (water vapor) can coexist. It equals, 273.16 K at a pressure of 611.73 Pa. The triple point is used to define the kelvin as an SI unit. One kelvin equals 1/273.16 of the difference, between absolute zero and the triple point., If you are a sharp-eyed reader, you may have noticed the references to both 273.16 and 273.15 in this section. The freezing point of water is, typically stated as 273.15 K (0°C) because this is its value at standard atmospheric pressure, but at the triple point pressure, water freezes at, 273.16 K (0.01°C)., , 18.3 - Temperature scale conversions, Since the Celsius and Kelvin scales have the same number of units between the, freezing and boiling points of water, it takes just one step to convert between the two, systems, as you see in the first conversion formula in Equation 1. To convert from, degrees Celsius to kelvins, add 273.15. To convert from kelvins to degrees Celsius,, subtract 273.15., Since water freezes at 32° and boils at 212° in the Fahrenheit system, there are 180, degrees Fahrenheit between these points, compared to the 100 units in the Celsius and, Kelvin systems. To convert from degrees Fahrenheit to degrees Celsius, first subtract, 32 degrees (to establish how far the temperature is from the freezing point of water) and, then multiply by 100/180, or 5/9, the ratio of the number of degrees between freezing, and boiling on the two systems. That conversion is shown as the second equation in, Equation 1. If you further needed to convert to kelvins, you would add 273.15., To switch from Celsius to Fahrenheit, you first multiply the number of degrees Celsius, by 9/5 (the reciprocal of the ratio mentioned above) and then add 32., In Example 1, you see the conventionally normal human body temperature, 98.6°F,, converted to degrees Celsius and kelvins., , Temperature scales:, conversions, TK = TC + 273.15, TC = (5/9)(TF í 32), TK = Kelvin temperature, TC = Celsius temperature, TF = Fahrenheit temperature, , Convert 98.6°F to Celsius and, Kelvin., TC = (5/9)(TF í 32.0), TC = (5/9)(98.6°F í 32.0), TC = (5/9)(66.6) = 37.0ºC, TK = TC + 273.15, TK = 37.0 + 273.15 = 310 K, , 336, , Copyright 2007 Kinetic Books Co. Chapter 18
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18.4 - Absolute zero, , Absolute zero, It cannot get colder, Molecules at minimum energy state, 0 K, í273.15°C, , Absolute zero: As cold as it can get., Absolute zero is a reference point at which molecules are in their minimum energy state (quantum theory dictates they still have some energy)., It does not get colder than absolute zero; nothing with a temperature less than this minimum energy state can exist., Physics theory says it is impossible for a material to be chilled to absolute zero. Instead, it is a limit that scientists strive to get closer and closer, to achieving. Above, you see a photograph of scientists who chilled atoms to less than a hundred-billionth of a degree above absolute zero. At, that temperature, the atoms changed into a state of matter called a Bose-Einstein condensate., Although we exist around the relatively toasty 293 K thanks to the Sun and our atmosphere, the background temperature of the universe í the, temperature far from any star í is only about 3 K. Brrrr., , 18.5 - Heat, , Heat: Thermal energy transferred between, objects because of a difference in their, temperatures., If you hold a cold can of soda in your hand, the soda will warm up and your hand will, chill. Energy flows from the object with the higher temperature í your hand í to the, object with the lower temperature í the soda. This energy that moves is called heat., Ovens are designed to facilitate the flow of heat. In the diagram to the right, you see, heat flowing from the oven coils through the air to the loaf of bread. As with other forms, of energy, heat can be measured in joules. It is represented by the capital letter Q., Physicists do not say an object has heat. Heat refers solely to the flow of energy due to, temperature differences. Heat transfers thermal energy that is internal to objects,, related to the random motion of the atoms making up the objects., , Heat, Energy flow due to temperature, difference, Not a property of an object, , Heat is like work: It changes the energy of an object or system. It does not make sense, to say “how much work a system has”, nor does it make sense to say “how much heat, the system has”. Just as work is done by a system or on a system, heat as thermal, energy can enter a system or leave a system., Having said that heat is measured in joules, we will backtrack a little in order to explain, some other commonly used units. These units measure heat by its ability to raise the, temperature of water. For example, a calorie raises one gram of water from 14.5°C to, 15.5°C. The British Thermal Unit (BTU) measures the heat that would raise a pound of, water 1° on the Fahrenheit scale., The unit we use to measure one property of food í the calories you see labeled on the, back of food packages í is actually a kilocalorie (good marketing!). This is sometimes, spelled with a capital C, as in Calories. Food calories measure how much heat will be, released when an object is burned. A Big Mac® hamburger contains 590 Calories, or, 590,000 calories. This amount of energy equals about 2,500,000 J. If your body could, capture all this energy, if it were 100% efficient and solely focused on the task, the, energy from a Big Mac would be enough to allow you to lift a 50 kg weight one meter, about 5000 times. We will skip the calculation for the french fries., , Q represents heat, Units: joules (J), , Copyright 2007 Kinetic Books Co. Chapter 18, , 337
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18.6 - Internal energy, , Internal energy: The energy associated with the, molecules and atoms that make up a system., In the study of mechanics, energy is an overall property of an object or system. The, energy is a function of factors like how fast a car is moving, how high an object is off the, ground, how fast a wheel is rotating, and so forth., In thermodynamics, the properties of the molecules and/or atoms that make up the, object or system are now the focus. They also have energy, a form of energy called, internal energy. The internal energy includes the rotational, translational and vibrational, energy of individual molecules and atoms. It also includes the potential energy within, and between molecules., , Internal energy, Energy of system’s atoms, molecules, , To contrast the two forms of energy: If you lift a pot up from a stovetop, you will, increase its gravitational potential energy. But in terms of internal energy, nothing has, changed. The potential energy of the pot’s molecules based on their relationship to each other has not changed., , However, if you turn on the burner under the cooking pot, the flow of heat will increase the kinetic energy of its molecules. The molecules will, move faster as heat flows to the pot, which means the internal energy of the molecules of the pot increases., , 18.7 - Thermal expansion, , Thermal expansion: The, increase in the length or, volume of a material due to a, change in its temperature., You buy a jar of jelly at the grocery store and store it, on a pantry shelf. When it comes time to open the jar,, the lid refuses to budge. Fortunately, you know that, placing the jar under hot water will increase your odds, of being able to twist open the lid., , Expansion joints allow bridge sections to expand without breaking., , By using hot water to coax a lid to turn, you are implicitly using two physics principles., First, most materials expand as their temperature increases. Second, different materials, expand more or less for a given increase in temperature. The metal lid of the jar, expands more than the glass container as you increase their temperatures, effectively, lessening the “grip” of the lid on the glass. (The temperature of the lid is also likely to, increase faster, another factor that accounts for the success of the process.), The expansion of materials due to a temperature change can be useful at times, as the, jar-opening example demonstrates. Sometimes, it poses challenging engineering, problems. For example, when nuclear waste is stored in a rock mass, heat can flow, from the waste to the rock, raising the rock’s temperature and causing it to expand and, crack. This could allow the dangerous waste to leak out. Knowing the exact rate of, expansion can help engineers design storage intended to prevent cracking., Good engineering takes expansion into account. For instance, bridges are built with, expansion joints, like the one shown at the top of this page, that accommodate, expansion as the temperature increases., , Thermal expansion, Most materials expand with increased, temperature, Different materials expand at different, rates, , 18.8 - Thermal expansion: linear, , Thermal linear expansion:, Change in the length of a, material due to a change in, temperature., Most objects expand with increased temperature; how, much they expand varies by material. In this section,, we discuss how much they expand in one dimension,, , 338, , Copyright 2007 Kinetic Books Co. Chapter 18
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along a line. Their expansion is measured as a, fraction of their initial length., In Equation 1, you see the equation for linear expansion. The change in length equals, the initial length, times a constant Į (Greek letter alpha), times the change in, temperature. The constant Į is called the coefficient of linear expansion and depends, on the material. A table of coefficients of linear expansion for some materials is shown, above. These coefficients are valid for temperatures around 25°C., Differing coefficients of linear expansion can be taken advantage of to build useful, mechanisms. A bimetallic strip, shown to the right, consists of two metals with different, coefficients of linear expansion. As the strip increases in temperature, the two materials, expand at different rates, causing the strip to bend. Since the amount of bending is a, function of the temperature, such a strip can be used in a thermometer to indicate, temperature. It can also be used as a thermostat to control appliances, such as coffee, pots and toasters. In these appliances, the bending of the strip interrupts a circuit and, turns off the power when the appliance has reached a specified temperature., , Thermal expansion: linear, Measured along one dimension, Constant Į depends on material, , Significant changes in temperature cause fairly minor changes in length. For instance,, in Example 1, we calculate the expansion of a 0.50 meter copper rod when its, temperature is increased 80 C°. The increase in length is just 6.6×10í4 meters, less, than a millimeter., Some materials, like carbon fiber, have negative coefficients of expansion, which, means they shrink when their temperature increases. By blending materials with both, positive and negative coefficients, engineers design systems that change shape very, little with changes in temperature. The Boeing Company pioneered the use of negative, coefficient materials in airplanes and satellites., , Rods of same material, Expansion proportional to initial length, , Linear expansion, ǻL = LiĮǻT, L = length, Į = coefficient of linear expansion, ǻT = change in temperature, Coefficient calibrated for K or °C, , Copyright 2007 Kinetic Books Co. Chapter 18, , 339
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The copper rod is heated from, 15°C to 95°C. What will its, increase in length be?, ǻL = LiĮǻT, ǻT = 95°C í 15°C = 80 C°, ǻL = (0.5 m)(1.65×10í5 1/C°)(80 C°), ǻL = 6.6×10í4 m, 18.9 - Sample problem: thermal expansion and stress, What stress does the aluminum rod, exert when its temperature rises, 20 K?, , Above, you see an aluminum rod heated by the Sun and held in place with concrete blocks. Since the rod increases in temperature, its length, also increases. This exerts a force on the concrete blocks. Stress is force per unit area, and an equation for tensile stress was presented in, another chapter. Young’s modulus for aluminum is given; it relates the fractional increase in length (the strain) to stress. You are asked to find, the stress that results from the increase in temperature., Variables, thermal expansion coefficient, , Į = 2.31×10í5 1/C°, , Young’s modulus, , Y = 70×109 N/m2, , temperature change, , ǻT = 20 K, , initial length, , Li, , change in length, , ǻL, , tensile stress, , F/A, , You may notice that the initial length of the rod is not known. It is not needed to answer the question., What is the strategy?, 1., , Combine the equations for thermal expansion and tensile stress to write an equation to calculate tensile stress from the temperature, change., , 2., , Use the equation to compute the stress in this case., , Physics principles and equations, We will use the equations for thermal expansion and tensile stress. Tensile stress is measured as force per unit area, or, , ǻL = LiĮǻT, F/A = Y(ǻL/Li), , 340, , Copyright 2007 Kinetic Books Co. Chapter 18, , F/A.
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Step-by-step solution, We start by substituting the expression for the change in length from the thermal expansion equation into the tensile stress equation, and then, do some algebraic simplification., , Step, , Reason, , 1., , F/A = Y(ǻL/Li), , tensile stress equation, , 2., , ǻL = LiĮǻT, , thermal expansion equation, Substitute equation 2 into equation 1, , 3., , 4., , F/A = YĮǻT, , Licancels out, , The equation we just found does not depend on the initial length. We know all the values needed to calculate the tensile stress in the aluminum, rod., , Step, , Reason, , 5., 6., , enter values, F/A = 3.2×107 N/m2, , multiply, , An aluminum rod with a radius of 0.025 meters (about one inch) exerts more than 60,000 newtons of force (equivalent to the weight of a large, elephant) against perfectly rigid supports when its temperature increases 20 C°! In this problem, we have ignored the expansion of the, concrete in which the aluminum is embedded., , 18.10 - Thermal expansion of water, Water exhibits particularly interesting expansion properties when it nears its freezing, point. Above 4°C, water expands with temperature, as most liquids do. However, water, also expands as it cools from 4°C to 0°C, a significant and unusual phenomenon., Below 0° water once again contracts as it cools. The consequence is that liquid water is, most dense at a temperature of around 4°C., This pattern of expansion means that lakes and other bodies of water freeze from the, top down. Why? In colder climates, as the autumn or winter seasons approach and the, air temperature drops to near freezing or below, the water in a lake cools. When the, water is cold but still above 4°C, it contracts when it chills. Since it becomes denser, it, sinks. This brings warmer water to the surface, which cools in turn. Eventually, the, entire lake reaches 4°C. But when the top layer then becomes colder still, it no longer, sinks. Below 4°C, the water expands, becoming less dense. It floats atop the warmer, water., As the cold water at the top of the lake further cools and freezes, it forms a floating layer, of ice that insulates the water below. Water is also atypical in that its solid form, ice, is, less dense than its liquid form and floats on top of it. Fish and other aquatic life can live, in the relatively warm (and liquid) water below, protected by a shield of ice., , Thermal expansion of water, Water contracts and sinks as it cools,, until 4°C, From 4°C to 0°C, water expands and, stays on top, Then ice forms on top and floats, , If water always expanded with increasing temperature for all temperatures above 0°C,, and contracted with decreasing temperature, the coldest water would sink to the bottom, where it might never warm up. Water’s negative coefficient of expansion in the temperature range from 0°C to 4°C is crucial to life on Earth. If, ice did not float, oceans and lakes would freeze from the bottom to the top. This would increase the likelihood that they would freeze entirely,, since they would not have a top layer of ice to insulate the liquid water below and their frozen depths would not be exposed to warm air during, the spring and summer., , Copyright 2007 Kinetic Books Co. Chapter 18, , 341
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18.11 - Thermal expansion: volume, , Thermal volume expansion:, Change in volume due to a, change in temperature., The equation for thermal linear expansion is used to, calculate the thermally induced change in the size of, an object in just one dimension. Thermal expansion, or contraction also changes the volume of a material,, and for liquids (and many solids) it is more useful to, determine the change in volume rather than expansion along one dimension. The, expansion in volume can be significant. Automobile cooling systems have tanks that, capture excess coolant when the heated fluid expands so much it exceeds the, radiator’s capacity. A radiator and its overflow tank are shown in Concept 1 on the right., The formula in Equation 1 resembles that for linear expansion: The increase is, proportional to the initial volume, a constant, and the change in temperature. The, constant ȕ is called the coefficient of volume expansion., Above, you see a table of coefficients of volume expansion for some liquids and solids., The coefficients for liquids are valid for temperatures at which these substances remain, liquid., For solid materials like copper and lead, the coefficient of volume expansion ȕ is about, three times the coefficient of linear expansion Į, because the solid expands linearly in, three dimensions., , Thermal expansion: volume, Volume increases with temperature, Constant ȕ varies by material, Increase is proportional to initial volume, , ǻV = Vi ȕǻT, V = volume, ȕ = coefficient of volume expansion, ǻT = change in temperature, Coefficient calibrated for K or °C, , For solids, ȕ § 3Į, ȕ = coefficient of volume expansion, Į = coefficient of linear expansion, , 342, , Copyright 2007 Kinetic Books Co. Chapter 18
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The temperature of 2.0 L of water, increases from 5.0° C to 25° C., How much does its volume, increase?, ǻV = Vi ȕǻT, ǻT = 25°C í 5.0°C = 20 C°, ǻV = (2.0 L)(2.07×10í4 1/C°)(20 C°), ǻV = 0.0083 L, 18.12 - Specific heat, , Specific heat: A, proportionality constant that, relates the amount of heat, flow per kilogram to a, material’s change in, temperature., Specific heat is a property of a material; it is a, proportionality constant that states a relationship between the heat flow per kilogram of, a material and its change in temperature., A material’s specific heat is determined by how much heat is required to increase the, temperature of one kilogram of the material by one kelvin. A material with a greater, specific heat requires more heat per kilogram to increase its temperature a given, amount than one with a lesser specific heat. In spite of its name, specific heat is not an, amount of heat, but a constant relating heat, mass, and temperature change., The specific heat of a material is often used in the equation shown in Equation 1. The, heat flow equals the product of a material’s specific heat c, the mass of an object, consisting of that material, and its change in temperature. The illustration in Equation 1, shows how specific heat relates heat flow to change in temperature. As you can see, from the graph, lead increases in temperature quite readily when heat flows into it,, because of its low specific heat., , Specific heat of a material, Relates heat and temperature change,, , In contrast, water, with a high specific heat, can absorb a lot of energy without changing per kilogram, much in temperature. Temperatures in locations at the seaside, or having humid, atmospheres, tend to change very slowly because it takes a lot of heat flow into or out of the water to accomplish a small change in, temperature. Summer in the desert southwest of the United States is famous for its blazing hot days and chilly nights, while on the east coast, of the country the sweltering heat of the day persists long into the night. Materials with large specific heats are sometimes informally called, “heat sinks” because of their ability to store large amounts of internal energy without much temperature change., Above, you see a table of some specific heats, measured in joules per kilogram·kelvin. The specific heat of a material varies as its temperature, and pressure change. The table lists specific heats for materials at 25°C to 30°C (except for ice) and 105 Pa pressure, about one atmosphere., Specific heats vary somewhat with temperature, but you can use these values over a range of temperatures you might encounter in a physics, lab (or a kitchen)., , Copyright 2007 Kinetic Books Co. Chapter 18, , 343
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Q = cmǻT, Q = heat, c = specific heat (J/kg·K), m = mass, ǻT = temperature change in C° or K, , How much heat is required to, increase the coffee's, temperature 68 K?, Q = cmǻT, Q = (4178 J/kg·K)(0.74 kg)(68 K), Q = 210,000 J, 18.13 - Sample problem: a calorimeter, A calorimeter is used to measure the, specific heat of an object. The water, bath has an initial temperature of, 23.2°C. An object with a temperature, of 67.8°C is placed in the beaker., After thermal equilibrium is, reestablished, the water bath's, temperature is 25.6°C. What is the, specific heat of the object?, , In a calorimeter, a water bath is placed in a well-insulated container. The temperature of the water bath is recorded, and an object of known, mass and temperature placed in it. After thermal equilibrium is reestablished, the temperature is measured again. From this information, the, specific heat of the object can be calculated. (We ignore the air in this calculation.), The use of a calorimeter depends on the conservation of energy. In the calorimeter, heat flows from the object to the water bath (or vice-versa, if the object is colder than the water). Because the calorimeter is well insulated, negligible heat flows in or out of it. The conservation of energy, allows us to say that the heat lost by the object equals the heat gained by the water bath., The water bath consists of the water and the beaker containing it. If the mass of the water is much greater that the mass of the beaker,, relatively little heat will be transferred to the beaker and it can be ignored in the calculation. We do that here., , 344, , Copyright 2007 Kinetic Books Co. Chapter 18
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Variables, water, , object, , mass, , mw = 0.744 kg, , mo = 0.197 kg, , initial temperature, , Tw = 23.2°C, , To = 67.8°C, , specific heat, , cw = 4178 J/kg·K, , co, , heat transferred to water, , Qw, , heat transferred from object, , Qo, , final temperature, , Tf = 25.6°C, , What is the strategy?, 1., , Use conservation of energy to state that the heat lost by the object equals the heat gained by the water bath. Apply the specific heat, equation to write the conservation of energy equation in terms of the masses, temperatures, and specific heats., , 2., , Solve for the unknown specific heat, substitute the known values, and evaluate., , Physics principles and equations, By the conservation of energy, the heat gained by the water bath (beaker plus water) equals the heat lost by the object. The sum of the heat, transfers is zero., , Qw + Qo = 0, We use the equation below to relate the heat flow to specific heat., , Q = cmǻT, Step-by-step solution, We start with an equation stating the conservation of energy for this experiment. Then, the heat values in the equation are written with, expressions involving specific heat., , Step, , 1., , Qw + Qo = 0, , Reason, , conservation of energy, , Qw = íQo, 2., , Qo = comoǻT, , specific heat equation, , 3., , Qo = como(Tf – To), , initial and final temperatures for object, , 4., , Qw = cwmw(Tf – Tw), , specific heat equation, , 5., , cwmw(Tf – Tw) = – como(Tf – To) substitute equations 3 and 4 into equation 1, , Now we solve the equation for the unknown specific heat of the object and evaluate., , Step, , Reason, , 6., , solve for specific heat of object, , 7., , substitute, , 8., , co = 897 J/kg·K, , evaluate, , Based on the values in the table of specific heats, it appears that the material may consist of aluminum., , Copyright 2007 Kinetic Books Co. Chapter 18, , 345
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18.14 - Phase changes, , Phase change: Transformation between solid, and liquid, liquid and gas, or solid and gas., When you pop some ice cubes into a drink, they will melt. Heat flows from the warmer, drink to the cooler ice cubes. Let’s say the ice cubes start at í10°C, cooler than the, freezing point of water, and they are dropped into a pot of hot coffee. Initially, heat, flowing to the ice cubes raises their temperature. But at 0°C, heat will flow to the ice, cubes from the still warm coffee without the cubes changing temperature. That is, it, takes energy to liberate the water molecules from the crystal structure of the ice and, allow them to move freely at the same temperature through the coffee. This occurs as, the ice melts, changing phase from a solid to a liquid. Phase changes between solid,, liquid and gas do not change an object’s temperature, but they do require heat transfer., Phase changes occur as heat flows into or out of a substance. An ice cube melts in hot, coffee, but the icemaker in a freezer causes water to change from a liquid to ice. In a, freezer, heat is transferred from the liquid water to the cooler freezer., , Phase change, Transformation between states, Consumes energy or releases energy, Temperature stays constant, , In days of yore, refrigerators were called “iceboxes” because ice was used to cool the contents of the box. Heat would flow from the warmer air, to the cooler ice, cooling the air. As the ice warmed and then melted, or changed phase, it would be replaced with a new block. Modern, refrigerators continue to use phase changes (between liquid and gas), but they employ substances other than water., The temperature at which a substance changes phase depends on the substance. For instance, water melts at 0°C at atmospheric pressure,, but iron melts at 1538°C., Some substances can “skip” the liquid state by transforming directly from a solid to a gas or vice-versa. This is called sublimation. Mothballs, sublimate: They transform from a solid directly into a gas. “Dry” ice (solid carbon dioxide) is another solid that sublimates directly into gas at, atmospheric pressure. Frost in your freezer is an example of sublimation in the reverse direction. In this case, gaseous water vapor changes, directly into solid ice., , 18.15 - Latent heat, , Latent heat: Energy required, per kilogram to cause a phase, change in a given material., Heat flow can cause a substance to change phases, by converting it between a solid and a liquid, or a, liquid and a gas., Latent heat describes how much energy per kilogram, is required for a given substance to change phase. It, Latent heats of fusion and vaporization., is a proportionality constant, expressing the, relationship between heat and mass as shown in Equation 1. The constant depends on, the material and on the phase change. Different amounts of energy are required to, transform a material between its liquid and solid states than between its liquid and, gaseous states., The latent heat of vaporization is the amount of heat per kilogram consumed when a, given substance transforms from a liquid into a gas, or released when the substance, transforms from a gas back to a liquid. The latent heat of fusion is the heat flow per, kilogram during a change in phase between a solid and a liquid., The table above shows the latent heats of fusion and vaporization for various, substances. For instance, you need 3.34×105 J of energy to convert a kilogram of ice, (at 0°C) to liquid water. Continued flow of heat into the water will raise its temperature, until it reaches 100°C. At this temperature, it will take 2.26×106 joules of heat to turn it, into a gas, about seven times as much as it took to convert it to a liquid., Salt causes ice to melt, a phenomenon called “freezing point depression.” When you, add rock salt to the crushed ice in a hand-cranked ice cream freezer, you force the ice, to melt. Heat flows from the resulting saltwater solution into the ice as it changes phase, from solid to liquid, resulting in a slurry having a temperature far colder than 0°C. Heat, then flows from the ice cream solution into this mixture, and the ice cream freezes., , 346, , Latent heat, Energy required per kg to change state, Latent heat of fusion: solid to liquid, Latent heat of vaporization: liquid to gas, Amount same in either “direction”, , Copyright 2007 Kinetic Books Co. Chapter 18
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Heat required for phase changes, Q = Lf m, Q = Lv m, Q = heat, m = mass, Lf = latent heat of fusion (J/kg), Lv = latent heat of vaporization (J/kg), , A glass contains 0.0370 kg of ice, at 0°C. How much heat transfers, to the ice as it melts without, changing temperature?, Q = Lf m, Q = (3.34×105 J/kg)(0.0370 kg), Q = 1.23×104 J, 18.16 - Sample problem: watching ice melt, The glass contains 0.160 kg of water, at 30.0°C and 0.0370 kg of ice at, 0.00°C. What is the resulting, temperature of the water at thermal, equilibrium after the ice melts?, , In the insulated container, the only source of the heat to melt the ice is the surrounding water. The water’s temperature will decrease as the ice, melts. The ice will melt while staying at 0°C. There is a tricky part to solving this problem: When the ice melts it turns into water, and this extra, water must be accounted for when calculating the final temperature., Variables, It is important to distinguish between the two masses of water in the final mixture: the mass that was initially liquid, and the mass that was, initially solid ice. We use the subscripts L and S to distinguish these masses of water., , Copyright 2007 Kinetic Books Co. Chapter 18, , 347
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In the section on latent heat, we calculated the heat transferred to the same amount of ice as it melted to be 1.23×104 J. Here, we need the, heat transferred from the water, not to the ice. Since the water loses heat, we state this as í1.23×104 J, with a negative sign., mass of liquid water, , mL = 0.160 kg, , mass of solid ice, , mS = 0.0370 kg, , heat transferred from water to melt ice, , Q = í1.23×104 J, , initial temperature of liquid water, , TLi = 30.0°C, , initial temperature of solid ice, , TSi = 0.00°C, , temperature of liquid water after ice melts, , TLf, , temperature of ice-melt, , TSf = 0.00°C, , final temperature of liquid water plus melted ice, , T, , heat transferred from liquid water for thermal equilibrium, , QL, , heat transferred to melted ice for thermal equilibrium, , QS, , specific heat of water, , c = 4178 J/kg·K, , What is the strategy?, 1., , Calculate the temperature of the water after it loses heat to melt the ice, using the specific heat of water., , 2., , As the water reaches thermal equilibrium, heat transfers from the originally liquid water, to the melted ice. These are the only heat, transfers in the system, so they sum to zero. Use this fact and the specific heat equation to calculate the temperature of the total mass, of water at the end., , Physics principles and equations, The specific heat equation, , Q = cmǻT, As the two masses of water reach thermal equilibrium, the heat transferred from the originally liquid water plus the heat transferred to the, melted ice must sum to zero., , QL + QS = 0, Step-by-step solution, First we calculate the temperature of the liquid water after it gives up heat to melt the ice. We use the specific heat of water, 4178 J/kg·K., , Step, , 1., , Q = cmǻT, , Reason, , specific heat equation, substitute values, , 2., 3., , ǻT = –18.4 K = –18.4 C°, , solve, , 4., , TLf = TLi + ǻT, , calculate water temperature, , TLf = 30.0°C + (–18.4°C), TLf = 11.6°C, , 348, , Copyright 2007 Kinetic Books Co. Chapter 18
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Now we use the fact that the heat transfers sum to zero as the two masses of water reach thermal equilibrium to calculate the final temperature, of the total mass of water., , Step, , Reason, , 5., , QL + QS = 0, , equation above, , 6., , cmL(T – TLf) + cmS(T – TSf) = 0, , specific heat equation, , 7., , solve for T, , 8., , substitute values, , 9., , T = 9.42°C, , evaluate, , 18.17 - Conduction, , Conduction: The flow of thermal energy, directly through a material without motion of, the material itself., When a frying pan is placed on a burner, heat flows from the burner to the pan. The, heat then spreads through the pan, soon reaching the handle even though the handle is, not in direct contact with the burner. This process illustrates the flow of thermal energy, via conduction., Conduction is the direct flow of thermal energy without a net motion of the materials, involved. The heat flows from the bottom of the pan to its handle. Heat also flows by, conduction where the burner and the bottom of the pan are in direct contact., , Conduction, Heat flow within object, , Conduction results from interactions at the atomic level of particles in an object:, ·Does not involve “bulk” motion, molecules, atoms and, in metals, electrons. Particles with greater average energy, collide with nearby lower-average-energy particles, increasing their energy. Heat spreads as particles collide with their neighbors, and those, neighbors collide with their neighbors, and so on. In this way, heat flows throughout the object. Although motion is involved at this level, the pan, itself does not move, and this is part of the definition of conduction: The flow of thermal energy within an object that does not involve motion., Different materials conduct heat at different rates. This is the topic of the next section., , 18.18 - Thermal conduction quantified, For appliances that heat or chill (such as ovens or, refrigerators), and indeed for an entire house or, apartment, thermal conduction is the dominant form, of unintended heat transfer. Since this transfer of, heat is costly and wasteful, a good amount of effort is, spent reducing it. In this section, we focus on building, materials that are designed to minimize this heat, transfer, keeping the interior of a house cool during, the summer and warm during the winter., The thermal conductivity of a material specifies how, quickly heat transfers through it. Above, you see a, Values for k and R., table of thermal conductivity values for materials such, as wood, insulating foam, glass, and some metals., The letter k represents the thermal conductivity of a material. Materials with small values for k are used as insulators to reduce the transfer of, heat., The units for thermal conductivity are joules per second-meter-Kelvin. Since watts are joules per second, this is the same as watts per meterKelvin, and these are the units used in the table above., , Pc represents the rate of heat transfer through conduction. As shown in Equation 1, Pc is defined as the heat transferred divided by the, elapsed time. It is calculated for a slab of material like a glass window pane using the equation shown in Equation 2. The rate of heat transfer, increases with the thermal conductivity k of the material that makes up the slab, the area of the slab, and the temperature difference between, its two sides. It decreases with the thickness of the material. The equation also can be used to explain why the units for k have meters in the, denominator, not “meters squared.” Because there is a length term (thickness) in the denominator of the heat-transfer equation, a length term, cancels out of the units., , Copyright 2007 Kinetic Books Co. Chapter 18, , 349
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The insulating effectiveness of building materials is usually specified by an R-value. A, material with a high R-value is a good insulator. For a slab of material of a given, thickness, the thermal resistance R is defined, as you see in Equation 3, as the, thickness of the slab divided by its thermal conductivity. This is often called the R-value, in the building trade. The SI units for R are square meters·kelvin per watt, and the, building material ratings are then often called RSI-values. We show metric values with, that name in the table above., In North America, R-values based on the British measurement system are commonly, used for building materials. These R-values have units of square feet·Fahrenheit, degrees·hours per British thermal unit. We list them also in the table above. The, thickness of the material for both RSI- and R-values is one inch, which is typical for, building materials., The term k in the equation in Equation 2 can be replaced with an expression involving, R. This yields another equation, also shown in Equation 3, for the rate of heat transfer., Considering these equations and studying the table above can help you understand, why certain materials are chosen in construction. First, increasing the thickness of a, material increases its R-value, so you see thicker insulating material in a colder, environment where the temperature difference between inside and outside is greater., Second, some materials have high k (low R) values, making them unlikely choices for, insulators. For instance, building a house out of copper would lead to high heating costs, (not to mention building costs). Building materials such as polyurethane foam are, effective insulators and can be combined with other reasonably good insulators such as, wood for even greater energy efficiency., Third, materials can be combined. Double-paned windows trap a quantity of an inert, gas like argon between two layers of glass. Argon has a high R value and considerably, reduces the rate of heat transfer through the window., , Heat conduction, Rate of heat transfer depends on:, ·thermal conductivity, ·area, ·temperature difference, ·thickness, , Rate of heat transfer, definition, Pc = Q/t, Pc = rate of heat transfer (J/s), Q = heat transferred, t = time, , Rate of heat transfer, calculated, , k = thermal conductivity, A = area, ǻT = temperature difference, L = thickness, k units: J/s·m·K = W/m·K, , Thermal resistance, R = L/k, , R = thermal resistance, R units (SI): m2·K/W, R units (British): ft2·F°·h/Btu, , 350, , Copyright 2007 Kinetic Books Co. Chapter 18
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Heat transfers through the, window at a rate of 1700 J/s., What is its thermal conductivity, constant?, , k = 0.93 W/m·K, 18.19 - Convection, , Convection: Heat transfer through a gas or, liquid caused by movement of the fluid., Gases and liquids usually decrease in density when they are heated (liquid water near, 0°C is a notable exception). When part of a body of liquid or gas is heated, the warmed, component rises because of its decreased density, while the cooler part sinks. This, occurs in homes, where heat sources near the floor heat the nearby air, which rises and, moves throughout the room. The warmer air displaces cooler air near the ceiling,, causing it to move near the heat source, where it is heated in turn. This transfer of heat, by the movement of a gas or liquid is called convection., All kitchen ovens, like the one shown in Concept 1, rely largely on convection for, baking. The heating element at the bottom of the oven warms the air next to it, causing, it to rise. The heated air then reaches the food in the oven to warm it, while the cooler, air sinks to the bottom of the oven. So-called “convection ovens” speed this process, with fans that cause the air to circulate more quickly., , Convection, Heat transfer due to movement in gases, and liquids, , Convection occurs in liquids as well as in gases. If you stir spaghetti sauce as it heats, you are accelerating the process of convection. Again,, your goal is to uniformly distribute the thermal energy., If you see a hawk soaring upward without flapping its wings, it may be riding what is called a “thermal.” As the Sun warms the ground, the, nearby air also becomes warmer. In the process, it becomes less dense, and is forced upward by air that is cooler and denser. A bird can ride, this upward draft., , Copyright 2007 Kinetic Books Co. Chapter 18, , 351
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18.20 - Radiation, , Radiation: Heat transfer by electromagnetic, waves., If you place your hand near a red-hot heating element and feel your hand warm up, you, are experiencing thermal radiation: the transfer of energy by electromagnetic waves., You correctly think of objects like the heating element as radiating heat; in fact, every, object with a temperature above absolute zero radiates energy., Radiation consists of electromagnetic waves, which are made up of electric and, magnetic fields. Radiation needs no medium in which to travel; it can move through a, vacuum. The wavelength of radiation varies. For instance, red light has a wavelength of, about 700 nm, and blue light a wavelength of about 500 nm. Infrared and ultraviolet, radiation are two forms of radiation whose wavelengths are, respectively, longer and, shorter than those of visible light., , Radiation, Heat transfer by electromagnetic, radiation, , All objects radiate electromagnetic radiation of different wavelengths. For instance, you, see the red-hot stove coil because it emits some visible light. The coil also emits, infrared radiation that you cannot see but do feel as heat flowing to your hand, and it, emits a minimal amount of ultraviolet radiation too., Although any particular object radiates a range of wavelengths, there is a peak in that, range, a wavelength at which the power output is the greatest. This peak moves to, shorter wavelengths as the temperature of the object increases. Understanding the, exact form of the spectrum of thermal radiation wavelengths requires concepts from, quantum physics, and its derivation was one of the early triumphs and verifications of, quantum theory., Bodies with temperatures near the temperature of the surface of the Earth emit mostly, infrared radiation. In the photograph in Concept 2, called an infrared thermograph, you, see the radiation emitted by a horse. Since areas of inflammation in the body are, unusually warm, and emit extra thermal radiation, veterinarians can use photographs, like this to diagnose an animal’s ailments. They are created by a digital or film camera, that assigns different (visible) colors to different intensities of (invisible) infrared, radiation in a process called false color reproduction., , Radiation, All objects emit radiant energy, , Sunlight is a form of radiation and is crucial to life on Earth. The Sun emits massive, amounts of energy in the form of radiation: 3.9×1026 joules every second. Some of that, strikes the Earth, where it warms the planet and supplies the energy that plants use in, photosynthesis., The amount of power radiated by a body is proportional to the fourth power of its, absolute temperature, its surface area, and a factor called its emissivity. The Sun emits, tremendous amounts of radiation energy because it is quite hot (about 6000 K) and vast, (with a surface area of about 6×1018 m2).Only a small portion of the total power emitted, by the Sun reaches the Earth. Even this fraction is an enormous amount: 1.8×1017, watts, about 100 times what human civilization consumes. The average solar power, striking the Earth’s atmosphere in regions directly facing the Sun is about 1370 watts, per square meter. This value is called the solar constant., Not all parts of the Earth directly face the Sun, and some radiation is reflected or, scattered by the Earth’s atmosphere before it reaches the surface. Different regions on, Earth receive different average amounts of power per square meter: Measurements, show that on average about 240 W/m2 (watts per square meter) reaches the Earth's, surface. North America is estimated to receive radiation of 150 W/m2, on the average., , Sun and Earth, Sun emits enormous amounts of radiant, energy, Some strikes the Earth’s atmosphere, , Given the dimensions of an average American house, this amount of radiation supplies, about four times as much power as the household consumes. Using this clean, “renewable resource” constructively for human purposes challenges both engineers and, physicists. It also provides a significant opportunity to conserve non-renewable energy, resources that would otherwise be used to heat, cool, and light our houses., Efforts have been underway to take advantage of this form of energy for years., Photovoltaic cells convert solar energy directly into electricity., The energy in sunlight can also be used without transforming it into electricity. The heat, of sunlight can be used to heat both water and the interior of a house. Devices called, solar collectors are becoming increasingly common. They are typically black; you see, one in Concept 4. Solar collectors heat water for uses ranging from hot showers to, warm swimming pools. Typically, the collectors contain tubes or rods that hold water,, and after it is heated the water is circulated to the desired locations in the building., , Solar collector, Uses solar energy that reaches the, Earth, , Current solar collectors come in a variety of shapes and sizes. The engineers designing them have two main challenges. One is to maximize, , 352, , Copyright 2007 Kinetic Books Co. Chapter 18
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the amount of solar radiation that strikes the collector. Some collectors, particularly those designed for industrial or research purposes, use, mirrors to focus sunlight onto the water tubes, increasing the amount of solar radiation that reaches the tubes., The second engineering challenge is to efficiently transfer the heat energy of sunlight to the water inside the collector tubes and then circulate, the heat out of the collector. There are many designs intended to maximize the efficiency of this process., Residential systems can capture usable amounts of energy. One current system is capable of capturing 1000 Btu (British thermal units) per, square foot of collector per day. A typical residential gas or oil furnace can supply on the order of 1.2 million Btu a day, which means a 10-by20 foot solar panel supplies about one-sixth of the energy of a typical home furnace., If you are interested in solar collectors, you can view a web site with information on them. You will study the physics of electromagnetic, radiation in detail in a later chapter., , 18.21 - Interactive summary problem: pop the cork, You just bought a bottle of Pierrot, the water from ancient limestone caves deep in, the French Alps, filtered by pure quartz crystals. But you did not realize the bottle, came with a cork, and you have no corkscrew. Fortunately, your knowledge of, thermal physics comes in handy., You remember that the density of ice is 9% less than the density of water. This, means that water expands quite a bit when it freezes into ice. If you let the water in, the bottle freeze, the expansion of the ice will push the cork out., If 89.0% of the water freezes, the expanding ice will just push the cork out. But if, more than 89.0% of the water freezes, the ice will expand too much and the bottle, will break. You want to remove just enough heat from the water so that exactly, 89.0% of it turns to ice., In the interactive simulation on the right, you control the amount of heat removed, from the water. The bottle contains 0.750 kg of water and its temperature is now, 15.0°C. You need to remove enough heat to reduce the temperature of all the, water to 0°C, and then remove enough additional heat to freeze 89.0% of it. To do, these calculations, you will need to use the specific heat of water, 4178 J/kg·K, and the latent heat of fusion of water, 3.34×105 J/kg. We ignore, the glass bottle itself in these calculations. Heat is removed from the bottle, but much more heat (about 50 times more) is removed from the, water. Also, while the volume of the glass bottle decreases slightly as it cools, the expansion of the ice is much greater. Similarly, the small air, space at the top of the bottle has little effect., Set the amount of heat to be removed from the water, then press GO. If you are right, the ice will push the cork out. Press RESET to try again., If you have trouble calculating the correct amount of heat transfer, review the sections in this chapter on specific heat and latent heat, and the, sample problem that combines the two concepts., , 18.22 - Gotchas, Heat is the same as temperature. No, heat is a flow of energy. Temperature is a property of an object. The flow of heat will change the, temperature of an object, and a thermometer measures the object’s temperature., The Fahrenheit temperature system is the wave of the future. If you think so, can I interest you in buying a record player?, Two rods of the same material experience the same increase in temperature, which means they must have expanded by the same amount of, length. Only if they were the same initial length. Their percentage increase would be the same in any case., You throw a football upward. You have not increased the internal energy of the air within the football. Correct: You have not increased the, internal energy of the molecules inside the football. (You have increased its translational kinetic energy and its rotational kinetic energy and its, gravitational PE by throwing it upward, but not its internal energy.), , Copyright 2007 Kinetic Books Co. Chapter 18, , 353
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18.23 - Summary, Temperature can be thought of simply as what thermometers measure., Thermometers rely on physical properties that vary with temperature in a reliable,, reproducible way., Different scales are used to measure temperature. The Fahrenheit scale is, commonly used to measure temperature in the United States, but the Celsius scale, is used in most of the rest of the world, and the Kelvin scale is used in science., Reference points such as the freezing and boiling points of water define a, temperature scale and how to convert between scales. Another important (though, ultimately unreachable) temperature point is absolute zero, where molecules are at, their minimum energy state. This is 0 K on the Kelvin scale, or í273.15°C., Heat is the flow of energy between objects resulting from a difference in, temperature. Like work, heat changes the energy of an object, but we cannot say, that an object has a certain amount of heat, any more than we could say it contains, a certain amount of work. Heat can change the internal energy of a system. Heat is, commonly measured in joules, or sometimes in calories or British Thermal Units. A, food Calorie is actually a kilocalorie, or 1000 calories., , TK = TC + 273.15, TC = (5/9)(TF – 32), Thermal expansion, , ǻL = LiĮǻT, ǻV = ViȕǻT, Specific heat, , Q = cmǻT, Latent heat, , Objects change in length and volume when their temperature changes, in a process, called thermal expansion. The change in length of an object depends on the, coefficient of linear expansion of the material from which it is made, which is, represented by the symbol Į. Its change in volume depends on the coefficient of, volume expansion, represented by ȕ. Water has the unusual property that it, expands with decreasing temperature from 4°C to 0°C; this is why ice floats., , Q = Lvm, , Specific heat is a property of a material, not an object, that relates the amount of, heat to the change in temperature for a unit of mass of the material., , Pc = Q/t, , The transformation of an object from one state (solid, liquid, or gas) to another is a, phase change. Phase changes require heat to be either added or taken away. The, latent heat of a material is the energy consumed or released per kilogram during a, particular phase change. The latent heat of fusion is the heat flow during a change, between solid and liquid; the latent heat of vaporization is the heat flow during a, change between liquid and gas., , Q = L fm, , Thermal conduction, , R = L/k, , Ways in which heat is transferred include conduction, convection, and radiation., Conduction is the flow of thermal energy directly through a material without motion, of the material itself. Thermal resistance, sometimes called the R-value, determines, the rate of heat transfer by conduction. Convection transfers heat by the bulk movement of molecules in a gas or liquid. Radiation transfers, energy by means of electromagnetic waves. Every object emits electromagnetic radiation., , 354, , Copyright 2007 Kinetic Books Co. Chapter 18
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Chapter 18 Problems, , Conceptual Problems, C.1, , The standard value for a healthy human's body temperature is 37°C, to the nearest Celsius degree, although this value is, only an approximation. In fact, the temperature of a healthy person varies by as much as 1.5°C each day, being lowest in the, morning and peaking in the evening. On North American fever thermometers, a "normal" body temperature is usually, indicated as 98.6°F. (a) Explain how this "normal" value in the Fahrenheit system is obtained from the "normal" value in the, Celsius system. (b) Explain why stating the value for a "normal" human temperature with three significant digits, as 98.6°F, is, misleading. (c) What would be a better "normal" value in the Fahrenheit system?, (a), (b), (c), , °F, , C.2, , A new website comes out promoting the Cold Water Diet. It proclaims: "On the CWD, you eat all the carbs and fat you want,, and just drink cold water. One liter of cold water has a mass of 1000 g, and it takes 1.0 calorie to raise the temperature of 1.0, g of water by 1 Celsius degree. If you ingest a liter of water at 0°C, it will take 37,000 calories to raise the temperature up to, 37°C, which is body temperature. That energy has to come from your body, so you can watch those pounds melt away!" What, is wrong with the website's claim?, , C.3, , A steel block is heated so that the length of each side increases 1%. What happens to its mass?, i., ii., iii., iv., , It increases 1%, It increases 3%, It increases 3.0301%, It does not change, , C.4, , You want to use a calorimeter to measure the specific heat of a very small object that you suspect has a low specific heat, value. Assume that you can use any of three liquids to fill the calorimeter: water, ethyl alcohol (which has a specific heat of, 2549 J/kg·K), or mercury (which has a specific heat of 139 J/kg·K). Which liquid would you choose, and why?, , C.5, , Why does a tiled floor feel colder to your bare feet than a carpeted floor, even though they are both at room temperature?, , C.6, , In coastal towns, breezes tend to blow offshore (from the land to the ocean) during the night and early morning, and onshore, (from the ocean to the land) during the afternoon and evening. What factors primarily contribute to this effect?, , Water, , Ethyl alcohol, , Mercury, , Heat conduction, Heat convection, Specific heat of land vs water, Salinity of the water, Thermal expansion, C.7, , In outer space environments, such as on the International Space Station, why must every piece of machinery or electronic, equipment have its own cooling fan? This makes for a noisy environment. Hint: What is different compared to an Earth-based, laboratory, and why should this matter?, , C.8, , When people are extremely cold and are not prepared with warm clothing, they will sometimes crouch and curl up into a tight, roll, hugging their knees to their chests. How does such a position conserve body heat?, , Section Problems, Section 2 - Temperature scales, 2.1, , Daniel Fahrenheit first proposed the temperature scale that bears his name in 1724. Originally, his reference points were a, well-mixed slurry melted from equal weights of ice and salt (similar to what you use in a hand-cranked ice cream freezer) for 0, degrees, and the temperature of the healthy human body for 12 degrees. Later he subdivided each one of his original, "degrees" into eight equal parts to define degrees Fahrenheit. In modern degrees Fahrenheit, what is the temperature of (a), the ice-salt mixture? (b) Fahrenheit's "healthy human"?, (a), , °F, , (b), , °F, , Copyright 2007 Kinetic Books Co. Chapter 18 Problems, , 355
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Section 3 - Temperature scale conversions, 3.1, , Convert each of the following temperatures to the indicated scale: (a) 100.00 K to the Celsius scale; (b) 100.00 K to the, Fahrenheit scale; (c) 72°F to the Celsius scale; and (d) í273.15°C to the Fahrenheit scale., (a), , °C, , (b), , °F, , (c), , °C, , (d), , °F, , 3.2, , What Celsius temperature is the same when converted to Fahrenheit? (There is only one such temperature.), , 3.3, , An iron rod is heated from 66.0°F to 280°F. Express the increase in temperature in Celsius., , °, , °C, 3.4, , In 1731 the French scientist Rene Reaumur proposed the Reaumur temperature scale. The reference points for his scale, were the freezing and boiling points of water (0°R and 80.0°R respectively). Like the Fahrenheit scale in northern Europe, his, scale was widely used in France until the 1960's, when the Celsius scale was officially adopted. (a) Convert 37.0°C to, Reaumur. (b) Convert 56.5°F to Reaumur. (c) Convert 256 K to Reaumur., (a), , °R, , (b), , °R, , (c), , °R, , Section 8 - Thermal expansion: linear, 8.1, , The plumbing in an old house uses lead pipes (which are now considered hazardous). Hot water is run through a section of, pipe, increasing its temperature from 19.4°C to 36.8°C. If the pipe is initially 3.90 m in length, what is the change in the length, of the pipe?, , 8.2, , A square copper plate has a circular hole in its center. Each side of the plate is 23.52 cm, and the radius of the circular hole is, 7.827 cm. If the plate's temperature increases 174.3 K, what are the new lengths of (a) the side of the plate and (b) the radius, of the hole? Express your answers to four significant digits., , m, , 8.3, , (a), , cm, , (b), , cm, , A metal rod is 2.673 m long at 23.25°C. When its temperature is increased to 168.4°C, the length of the rod is 2.681 m. What, is the metal's coefficient of linear expansion? Express your answer to four significant digits., 1/C °, , 8.4, , The aptly-named Steel Bridge over the Willamette River in Portland, Oregon, has a central span steel truss that is 64.1 m, long. What is its change in length over the year from the average minimum annual temperature of 0.90°C to the average, maximum of 26.3°C? Use a coefficient of linear expansion of 1.17×10í5 (1/C°)., m, , 8.5, , A ring of silver has an inside diameter of 2.355 cm at a temperature of, 23.48°C. A steel rod has a diameter of 2.361 cm. To what temperature should you raise the ring so it will just slip onto the, rod? Assume the rod does not change temperature. Express your answer to four significant digits., , 8.6, , An aluminum ring has an inside diameter of 1.793 cm, and an iron rod has a diameter of 1.798 cm, both at 18.73°C. If you, increase the temperature of both the ring and the rod, at what common temperature will the ring just slip onto the rod?, , °C, , °C, 8.7, , An aluminum beam shows a length of L meters as measured with a steel measuring tape, with both the beam and the tape at, a temperature Ti. The temperature (of both the beam and the measuring tape) then increases by ǻT. Develop an expression, for the value the tape now shows for the length of the beam. Hint: Find out how long a meter becomes on the steel measuring, tape, find the new length of the aluminum beam, then think about how to combine the two results., , 8.8, , A clock is designed with a simple pendulum, consisting of a mass at the end of an aluminum rod, whose period is supposed, to be equal to one second. The clock is designed to keep accurate time at 20.0°C, but is operated at a constant temperature, of í80.0°C. (a) Will the clock run fast or slow? Explain. (b) How many periods will the pendulum go through before the clock is, off by 1.00 s? (The coefficient of linear expansion of aluminum is 2.31×10í5 1/C°.), (a), (b), , 356, , Fast, , Slow, periods, , Copyright 2007 Kinetic Books Co. Chapter 18 Problems
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Section 9 - Sample problem: thermal expansion and stress, 9.1, , A copper rod at a temperature of 22.82°C is bolted in place between two rigid walls. What stress does the rod exert (the force, on each of the perfectly rigid walls, divided by the area of contact) when it is heated to a temperature of 274.2°C? (The, coefficient of linear expansion for copper is 1.65×10í5 1/C°, and Young's modulus for the material is 1.08×1011 N/m2.), N/m2, , 9.2, , A rod has a cross-sectional area of 0.001615 m2. When the rod's temperature is 12.23°C, it is placed between two rigid walls, that are exactly as far apart as the length of the rod. The rod's temperature then increases to 27.23°C, and the rod exerts a, force of 5.334×104 N on each wall. If Young's modulus for the material of which the rod is made is 1.05×1011 N/m2, what is the, coefficient of linear expansion of the rod?, 1/C °, , Section 11 - Thermal expansion: volume, 11.1 A lead ball has a volume of 94.3 cm3 at 19.3°C. What is the change in volume when its temperature changes to 32.3°C?, cm3, 11.2 A solid copper ball with radius 1.35 cm increases in temperature from 15°C to 86°C. What is the change in its volume?, cm3, 11.3 A hot water tank holds exactly 80.0 gallons of water at 115°F. What is the volume of the water at 40.0°F?, gallons, 11.4 The density of mercury at 0°C is 13,595 kg/m3. What is its density at, 100.0°C?, kg/m3, 11.5 Fever thermometers are often made of a glass tube filled with a liquid. Older thermometers were often filled with mercury., Newer ones are filled with red-tinted ethyl alcohol. You have two identical glass tube thermometers, one filled with mercury, and the other filled with ethyl alcohol. The degree marks on the mercury thermometer are 1.32×10í3 m apart. How far apart, are they on the alcohol thermometer? (Ignore the expansion of the glass in your calculations. The coefficient of volume, expansion is 1.12×10í3 1/C° for ethyl alcohol.), m, 11.6 A cup is made from a thin sheet of aluminum, shaped into a cylinder open at one end. At 13.4°C, the radius of the cylinder is, 3.24 cm and its height is 7.56 cm. (a) What is the change in volume of the cup when it is raised to a temperature of 78.2°C?, (The coefficient of linear expansion for aluminum is 2.31×10í5 1/C°.) (b) If the cup is initially exactly filled to the brim with, water, and the water is raised to the same temperature as the cup, what volume of water bulges out over the top of the cup?, (The coefficient of volume expansion for water is 20.7×10í5 1/C°.), (a), , cm3, , (b), , cm3, , 11.7 The temperature of a block of lead is raised from 0°C to 100°C. What is the percentage change in its density? (The density, of lead at 0°C is 11,300, , kg/m3.), , %, , Section 12 - Specific heat, 12.1 A block of iron at 35.0°C has mass 2.30 kg. If 3.50×104 J of heat are transferred to the block, what is its resulting, temperature?, °C, 12.2 You want to take a bath with the water temperature at 35.0°C. The water temperature is 38.0°C from the hot water tap and, 11.0°C from the cold water tap. You fill the tub with a total of 191 kg of water. How many kilograms of water from the hot, water tap do you use?, kg, 12.3 An orchid grower hears that the temperature is expected to fall below freezing overnight, so she puts a large barrel in her, orchid greenhouse and fills it with 315 kg of water at 21.0°C. In the morning, the water is at 1.5°C. How much heat left the, water during the night? (Express the answer as a positive amount.), J, , Copyright 2007 Kinetic Books Co. Chapter 18 Problems, , 357
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12.4 A 1.32 kg sample of some material increases in temperature from 13.5°C to 39.2°C when 1.98×104 J of heat energy is, transferred to it. What is the specific heat of the material?, J/kg · K, 12.5 A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and 2.5% of its kinetic energy at the time it, reaches the ground is transformed into internal energy in the ball itself. (The rest of the energy is transmitted to the ground,, converted into sound energy, and so on.) What is the ball's increase in temperature?, K, 12.6 A 0.00340 kg lead bullet is traveling at 473 m/s when it hits the bull's-eye of a practice target. Half of the bullet's kinetic, energy is transferred to the bullet in the form of heat. What is the resulting temperature increase in the bullet? Assume that, the specific heat of the lead does not vary significantly over this temperature., K, 12.7 A lead block and a silver block are both at the temperature of 41°C. They then each absorb the same amount of heat, and,, without any exchange of heat between them or with the environment, their new final temperatures are again the same. (a) If, the mass of the lead block is 5.0 kg, what is the mass of the silver block? (b) If they each absorbed 6.9×103 J, what is their, final temperature?, (a), , kg, , (b), , °C, , 12.8 As a body of water gives up heat energy to the air above it, the air temperature increases. Assume 1.0 kg of water (about one, liter) decreases 1.0 K in temperature, giving up its energy to a nearby body of air. (a) What mass of air could the energy from, the water increase by 1.0 K in temperature? (b) Assume the density of the air is 1.2 kg/m3. What is the volume of this mass of, air?, (a), , kg, , (b), , m3, , 12.9 A samovar is a Russian urn, often made of silver, for serving tea. A large samovar made from 4.62 kg of silver is filled with, 11.3 kg of tea, at the same temperature. How much heat is required to raise the temperature of both the samovar and tea by, 34.7 C°? (Assume the specific heat of tea is the same as that of water, 4178 J/kg·K.), J, , Section 13 - Sample problem: a calorimeter, 13.1 A large handful of identical gemstones, totaling 195 carats (one carat has a mass of 200 mg) at temperature 73.2°C, is, placed in a calorimeter filled with 0.594 kg of water at 13.6°C. After thermal equilibrium is established, the water temperature, is 14.1°C. What is the specific heat of the gemstone material?, J/kg · K, 13.2 An Italian chef, who is very scientific in her approach to cooking, measures the specific heat of a porcini mushroom by placing, it in a calorimeter filled with 0.678 kg of olive oil at 13.2°C. The mushroom has mass 0.0802 kg and temperature 24.3°C, before being placed into the calorimeter. The olive oil's temperature is 14.7°C after thermal equilibrium is reached. The, specific heat of olive oil is 1970 J/kg·K. What is the specific heat of the mushroom?, J/kg · K, , Section 15 - Latent heat, 15.1 An ice cube with a mass of 0.0410 kg at 0°C melts to water with no change in temperature. How much heat does the ice, absorb while melting?, J, 15.2, A large pot is placed on a stove and 1.2 kg of water at 14°C is added to the pot. The temperature of the water is raised evenly, to 100°C just before it starts to boil. (a) What amount of heat is absorbed by the water in reaching 100°C? (b) The water then, boils until all of it has evaporated, turning to water vapor at 100°C. How much heat does the water absorb in this process?, (a), , J, , (b), , J, , 15.3 How much heat is required to change 1.83 kg of solid lead at 135°C to liquid lead at 327°C?, J, , 358, , Copyright 2007 Kinetic Books Co. Chapter 18 Problems
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15.4 Steam at 100°C is injected into 1.8 kg of water at 22°C in a well-insulated container, where it condenses and mixes with the, existing water, reaching thermal equilibrium. If the final temperature of the well-mixed water is 25°C, what is the mass of the, injected steam?, kg, , Section 18 - Thermal conduction quantified, 18.1 A window opening is 1.1 m by 1.3 m, and the wall thickness is 0.12 m. The temperature difference between the inside and, outside is 28°C. What is the rate of heat transfer through the opening if it is filled with (a) concrete, or (b) wood (fir)?, (a), , W, , (b), , W, , 18.2 A material has thermal conductivity of 0.380 W/m·K. What is its thermal resistance for a one-inch thick layer, in SI units of, m2·K/W?, m2 · K/W, 18.3 You want to add insulation under the roof of your house with a thermal resistance of 31.0 ft2·F°⋅h/Btu. (a) What thickness of, fiberglass should you add, in inches? (b) What thickness of plywood would do the job?, (a), , in, , (b), , in, , 18.4 A circular pond of radius 12 m has a layer of ice 0.25 m thick. On a spring day, the temperature of the water below the ice is, 4.0°C and the temperature of the air above is 11°C. What is the rate of heat transfer through the ice, in kW?, kW, 18.5 A cube of wood and a cube of concrete, each, 0.17 m on a side, are placed side by side. One, of the long faces of the rectangular prism, formed by the two cubes is held at 17°C, and, the opposite long face is held at 32°C. What is, the total rate of heat transfer through the, cubes?, W, , 18.6 A copper rod and an aluminum rod with identical cross-sections are joined end to end. The free end of the copper rod is held, at a constant temperature of 94°C, and the free end of the aluminum rod is held constantly at 13°C. Where the rods meet, the, temperature is 45°C. If the copper rod is 2.4 m long, how long is the aluminum rod? (For the temperatures in the range, discussed here, the thermal conductivity of copper is 401 W/m·K and the thermal conductivity of aluminum is 237 W/m·K., Ignore the heat loss through the sides of the rods.), m, , Section 21 - Interactive summary problem: pop the cork, 21.1 Use the information given in the interactive problem in this section to calculate the heat flow required to pop the cork. Test, your answer using the simulation., J, , Copyright 2007 Kinetic Books Co. Chapter 18 Problems, , 359
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19.0 - Introduction, We live in a vast ocean of gas called the atmosphere. Deeper than the seas, it, provides the oxygen we breathe, guards us against harmful radiation from space,, helps to regulate the temperature, and makes flight possible., In addition to supporting life, gases are used in many engines, and this is an, important topic of study in physics. Transferring heat to a gas can cause it to, expand. As the gas expands, it is able to do useful work. To understand how, engines work, it is necessary to understand the behavior of gases., In this chapter and elsewhere in this textbook, we focus on ideal gases. An ideal, gas can be modeled as a number of small, hard spheres (atoms or molecules), moving rapidly around a container and colliding with each other and the container, walls in perfectly elastic collisions. This aspect of the study of gases, which focuses, on the motion of gas particles, is called the kinetic theory of gases., To begin your study of gases, try the simulation to the right. In the container, each, particle represents a gas atom or molecule. A gauge in the simulation provides a, readout of the gas pressure inside the container. The pressure is a function of, various factors including the volume of the container and the speed and the number of the particles., The pressure measured in the simulation is caused by the collisions of the particles with the walls of the container. Each time a particle strikes, a wall of the container, it exerts a force on the wall and the wall exerts an equal amount of force on the particle. The greater the speed with, which the particle strikes the wall, the greater the force the particle exerts on the wall., As detailed below, you can change three properties of the gas. The changes you make will be reflected solely in changes in the pressure of the, gas. Other properties of the gas could also change, but in this simulation we hold those other properties constant., Clicking the up and down arrows for volume causes an external mechanism to raise or lower the container’s lid. As the volume changes, the, gas particles continue to move at the same average speed. How do you think gas pressure will relate to the volume of the container? Will, decreasing the volume increase or decrease the pressure? Try it. Does the simulation help you see why there is a relationship between volume, and pressure? Consider how it alters the frequency of the collisions., You can also vary the temperature of the gas. The volume will not change as you do this, since the lid’s position remains fixed. If you increase, the temperature of the gas, you increase the average speed of the particles in the gas, and if you decrease the temperature, you decrease their, average speed. Considering how the force and frequency of the collisions affects pressure, what do you think is the relationship between, temperature and pressure when the volume is constant? Again, run the simulation to confirm your hypothesis., The final controller allows you to add or subtract particles from the chamber. The particles you add will move at the same average speed as the, gas particles already there. Do you think adding particles will result in more or less pressure? To test your conclusion, run the simulation to add, particles and observe the resulting pressure., When you have finished answering the questions above, you have experimented with some of the essential properties of a gas. The simulation, should enable you to see relationships between gas pressure and the volume, temperature, and quantity of a gas., , 19.1 - Ideal gas, , Ideal gas: A gas that can be modeled as a set of, particles having random elastic collisions. The, collisions are the only interactions between the, particles., The concept of an ideal gas is used frequently in physics, both in the study of gases, and later in the field of thermodynamics, where ideal gases play an important role. Real, gases tend to behave like ideal gases at low enough densities (where quantum, mechanical effects are not important). This makes ideal gases a good model for, analyzing gas behavior. Here, we summarize the major properties defining an ideal gas., Some gases, like carbon dioxide (CO2) and oxygen (O2), are composed of atoms, , Ideal gas, Molecules widely separated, collisions, , bonded together in molecules. Other gases, like neon (Ne) and helium (He), are made, Interact only in elastic, up of particles that are individual atoms. For the sake of brevity, we will simply say that, gases contain molecules (or particles), rather than repeatedly writing “atoms or, molecules.” When we analyze a gas, it is one made up of a single substance, such as oxygen or carbon dioxide., , An ideal gas consists of a large number of molecules. They are separated on average by distances that are large relative to the size of a, molecule. They move at high speeds and collide frequently., , 360, , Copyright 2000-2007 Kinetic Books Co. Chapter 19
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Using descriptions like “large” and “frequently” does not convey the nature of gases as well as specific numbers. Consider nitrogen molecules, (N2), the most common molecule in our atmosphere. In one cubic meter of nitrogen molecules at 0°C and one atmosphere of pressure, there, are 2.69×1025 molecules. They occupy just 0.07% of the space and move at an average speed of 454 m/s. The molecules run into one another, a lot: There are 9.9×1034 collisions each second between molecules in that space. The condition of 0°C (273 K) at one atmosphere is called, standard temperature and pressure and is often used in calculations involving gases., The molecules collide elastically with both each other and the container walls. The molecules only exert forces on one another when they, collide. Forces that they exert on one another at other distances are negligible., Each molecule can be considered as a small, hard sphere. In elastic collisions, both kinetic energy and momentum are conserved. The, velocities of the molecules change after a collision, but the total momentum and kinetic energy of the molecules remain the same., The walls of the container are rigid. When a molecule collides with a wall, its speed does not change although its velocity does (because its, direction changes). Newtonian physics can be used to analyze the collisions of the molecules with each other and with the walls of the, container., , 19.2 - Gas pressure, Gas molecules not only collide with each other, but also with the walls of any container., During these collisions, the molecules exert a force on the walls. Dividing the total, amount of force caused by these collisions by the surface area of the walls yields the, absolute pressure exerted by the gas., In this section, we consider three factors that qualitatively influence gas pressure. If you, tried the simulation in the introduction section, some of this discussion should be, familiar to you. In this section, we show two containers of gas. For visual clarity, we, show only a small fraction of the actual number of particles in the containers. Each, particle in the simulation represents about 3.4×1018 molecules. We have done the, calculations assuming that these are nitrogen molecules in a container about 10í2 m, wide. In the container on the left in each diagram, the pressure is one atmosphere and, the temperature is 0°C. We vary one property at a time in each illustration., Consider two containers of equal volume containing different numbers of gas particles, moving at the same speed. This is shown in Concept 1. The pressure will be greater in, the container on the right (the one that contains more molecules). Why? With everything, else equal, the collisions are more frequent in the container with the greater number of, molecules, which means the molecules collectively exert more outward force on the, walls of the container. Greater average force on the same amount of surface area, means the pressure is greater., The relationship between molecular speed and pressure is shown in Concept 2. Now, each container contains the same number of molecules, but the container on the right, has faster moving molecules (at a greater temperature) and, thus, greater pressure., Why does pressure increase with the speed of the molecules? There are two reasons, why this is the case. First, when molecules move faster, they collide more frequently, with the walls of the container. Second, when a molecule strikes the wall and rebounds,, the molecule exerts more force at greater speeds than at slower speeds. (Imagine the, force involved in a tennis ball rebounding off you when it is thrown at 1 m/s versus, 20 m/s.) Since the average speed of molecules increases with temperature, this means, pressure also increases with temperature., , Pressure increases with, Number of molecules, , Pressure increases with, Speed of molecules (temperature), , In Concept 3 the temperature and the number of molecules are the same for each, container. We change the volume of the container on the right. The pressure, decreases when the container volume increases; pressure is inversely proportional to, volume. In a larger container, the molecules collide less frequently with the walls. The, surface area of the container also increases, so pressure (force divided by area), decreases for that reason, too., , Pressure decreases with, Volume of container, , Copyright 2000-2007 Kinetic Books Co. Chapter 19, , 361
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19.3 - Boyle’s and Charles’ gas laws, In this section we take a macroscopic view of ideal gases to discuss their overall, properties. Several laws state the relationship among macroscopic properties such as, pressure, volume, temperature, and the amount of gas (number of molecules). We start, here with Boyle’s and Charles’ laws., Boyle’s law is named for Irish scientist Robert Boyle, who in 1662 discovered that the, absolute pressure of a fixed amount of gas at a constant temperature is inversely, proportional to its volume. To put it another way, if you decrease the volume of a gas,, its pressure increases proportionally. Boyle’s law is stated in Equation 1: The product of, the initial pressure and volume of a gas equals the product of its final pressure and, volume., Jacques Charles, a Frenchman who worked a century later than Boyle, also studied, gases. One way to state Charles’ law is that for a fixed amount of gas at a constant, pressure, the ratio of its volume to its absolute temperature remains constant. You see, this in Equation 2 using the initial and final volumes and temperatures of a gas., , Boyle’s law, PiVi = PfVf, , These gas laws are empirical; that is, they are determined to be true through, experiment and measurement rather than being derived from other laws., , P = absolute pressure, V = volume, Temperature, quantity of gas constant, , Charles' law, , V = volume, T = Kelvin temperature, Pressure, quantity of gas constant, , 362, , Copyright 2000-2007 Kinetic Books Co. Chapter 19
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0.020 m3 of gas expands to, 0.040 m3. Its initial pressure was, 5.0×105 Pa. Assume its, temperature is constant. What is, its final pressure?, PiVi = PfVf, Pf = PiVi/Vf, Pf = (5.0×105 Pa)(0.020 m3)/0.040 m3, Pf = 2.5 × 105 Pa, 19.4 - Avogadro’s number and moles, , Avogadro’s number: 6.022×1023., Mole: The amount of a substance that contains, Avogadro’s number of particles., The concept of moles is used in the study of gases. You might find that this section, reviews what you learned in a chemistry class, where this concept also figures, prominently., Physicists often need to deal with large quantities of molecules when they analyze, gases. For instance, one cubic meter of air at sea level contains about 3×1025, molecules. In addition, scientists may need to quantify a given amount of gas by its, number of molecules rather than its mass. To do so, they use moles and Avogadro’s, number., , Avogadro’s number, Number of atoms in 12 g of carbon-12, NA = 6.022×1023 particles, , One mole is defined as the amount of a substance containing the same number of, particles as there are atoms in 12 grams of carbon-12. Carbon-12 is one form, or, isotope, of carbon, having six protons and six neutrons in its nucleus. Scientists have, determined that there are, to six significant figures, 6.022 14×1023 atoms in 12 grams of, carbon-12, and have named this value Avogadro’s number (after the 19th century Italian, physicist Amedeo Avogadro). Avogadro’s number is indicated by NA., A mole contains about 6.022×1023 particles of a substance. The abbreviation for mole is, “mol”. The concept of the mole is important enough that the mole is one of the seven, fundamental units of the Système Internationale., Avogadro’s number specifies the number of particles per mole of a substance. It can be, applied to any kind of object. For example, if you had Avogadro’s number of golf balls,, or one mole, you would have 6.022×1023 balls (and their mass would be 2.8×1022 kg)., , Mole, Of course, moles are more typically used to specify the number of molecules in a, quantity of gas. Moles are used to focus on the number of molecules, not their mass., Contains Avogadro’s number of, For example, a mole of hydrogen atoms contains the same number of atoms as a mole, particles, of lead, even though one mole of lead atoms weighs 207 times more than the hydrogen, atoms. If you have studied chemistry, you may be familiar with combining substances in terms of their molar amounts to achieve the correct, ratio of molecules., The following discussion will prove useful if you need to use atomic mass units. The atomic massscale relates moles and the mass of individual, molecules. The atomic mass unit, symbolized by u, is based on carbon-12. By definition, one atom of carbon-12 has a mass of exactly 12, atomic mass units, or 12 u. One atomic mass unit equals 1.660 54×10í27 kg. Since Carbon-12 contains six protons and six neutrons, which, have about the same mass and make up almost all the mass of the atom, the mass of a proton or neutron is approximately one atomic mass, unit., A particle with an atomic mass of 24 u is twice as massive as a carbon-12 atom. This means a mole of the 24 u substance will have twice as, , Copyright 2000-2007 Kinetic Books Co. Chapter 19, , 363
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much mass as a mole of Carbon-12., The mass per mole of a substance is called its molar mass. When measured in grams, (or, to be precise, g/mol) the molar mass of a particular substance has the same, numerical value as the atomic mass of the substance. Why? The atomic mass of an, atom of carbon-12 is 12 u by definition, and a mole of carbon-12 atoms is exactly 12, grams, by the definition of Avogadro’s number., Consider a substance with an atomic mass of 36 u. Each particle will be three times as, massive as an atom of carbon-12. One mole of the substance has the same number of, particles as one mole of carbon-12. This means a mole of this substance would have a, mass that is three times the mass of a mole of carbon-12, or 36 grams. In physics, the, molar mass is usually stated in kg/mol instead of g/mol., , How many helium atoms are in, the balloon?, (0.43 moles)(6.022×1023 atoms/mol), 2.6×1023 helium atoms, 19.5 - Ideal gas law, , Ideal gas law: The product of a gas’s pressure, and volume is proportional to the amount of gas, and its temperature., The ideal gas law shown in Equation 1 is an empirical law that describes the, relationship of pressure, volume, amount of gas, and temperature in an ideal gas. It, states that the pressure of an amount of an ideal gas times its volume is proportional to, the amount of gas times its temperature. The pressure is the absolute, not gauge,, pressure. (Absolute pressure equals gauge pressure plus atmospheric pressure.) The, small n in the equation is the number of molecules, measured in moles. The, temperature is measured in kelvins. R is a constant called the universal gas constant., Its value is 8.314 51 J/mol·K., Everyday observations confirm the main precepts of the ideal gas law. When you inflate, a bicycle tire, you increase the amount of gas in the tire, which increases both its, pressure and volume. Once it is inflated and its volume is held nearly constant by the, tire, an increase in temperature increases the pressure of the gas within. A hot day or, the rolling friction from riding increases its temperature. You are told to check the, pressure of a tire when it is “cold” since the correct pressure for a tire is established for, 20°C or so. Checking tire pressure when the tire is hot may cause you to deflate the, tire unnecessarily., Sometimes, it is useful to be able to apply the ideal gas law when the number of, particles N is known, rather than the number of moles. Since the number of particles is, the number of moles times Avogadro’s number, we can rewrite the ideal gas law as, shown in Equation 2. The constant k here is called Boltzmann’s constant, which equals, the gas constant R divided by Avogadro’s number NA. Its value is 1.380 66×10í23 J/K., While the first statement of the ideal gas law relates P, V and T to the macroscopic, properties R and n of a gas, the second relates them to the microscopic properties k, and N., If this is the ideal gas law, are there non-ideal gases? Yes, there are. For instance,, water vapor (steam) is a non-ideal gas. The ideal gas law provides accurate results, when gas molecules bounce into each other with perfectly elastic collisions. If the, molecules also interact in other ways, as they do with steam, the law becomes a less, accurate predictor of the relationship between pressure, volume, temperature, and the, amount of gas. In the case of water vapor, steam tables are used to relate these values., In Example 1, the volume of one mole of gas at standard pressure and temperature is, calculated. The volume is 2.25×10í2 m3 (22.5 liters, or about six gallons). One mole of, any ideal gas occupies this same volume under these conditions. In 1811, Avogadro, first proposed that equal volumes of all gases at the same pressure and temperature, contained the same number of molecules., , Ideal gas law, Pressure times volume proportional to:, ·molar quantity of gas, ·Kelvin temperature of gas, , Ideal gas law, number of moles, PV = nRT, P = absolute pressure (Pa), V = volume, n = moles, R = universal gas constant, R = 8.31 J/mol·K, T = Kelvin temperature, , Ideal gas law, number of, particles, PV = NkT, N = number of particles, k = R/NA = Boltzmann's constant, k = 1.38×10í23 J/K, , 364, , Copyright 2000-2007 Kinetic Books Co. Chapter 19
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What is the volume of one mole, of gas at standard temperature, and pressure? Standard, temperature is 273 K and, standard pressure is 1.01×105, Pa., PV = nRT, , V = 2.25×10í2 m3, 19.6 - Sample problem: tank of air, The emergency scuba tank is filled, with 8.50×10í2 m3 of air at, atmospheric pressure and 21.0°C., Immediately after the tank is filled, the, absolute pressure is 1.53×107 Pa and, the volume is 8.37×10í4 m3. What is, the temperature of the air in the tank, at that time?, , Scuba divers may carry small tanks like the one shown above to give them air in case of an emergency. Air that is at atmospheric pressure,, 1.01×105 Pa, is pumped into the tank. The process of filling the tank increases the temperature and pressure of the air. The temperature will, decrease as heat flows from the warmer tank to its cooler surroundings, but we assume the compression process occurs quickly enough for, heat flow to be negligible. A value for atmospheric pressure is stated below., Scuba diving equipment manufacturers use an estimated breath size of 1.6 liters, so this tank would hold a little over 50 typical breaths., Assuming a diver might breathe about 15 times a minute, an average under normal conditions, the emergency tank would give her around, three minutes of air. However, she might be breathing a little more rapidly if she needs to use the tank!, Variables, initial volume of air, , Vi = 8.50×10í2 m3, , initial pressure, , Pi = 1.01×105 Pa, , initial temperature, , Ti = 21.0°C, , final volume of air, , Vf = 8.37×10í4 m3, , final pressure, , Pf = 1.53×107 Pa, , final temperature, , Tf, , moles of air, , n, , What is the strategy?, 1., , Model the air as an ideal gas. State the ideal gas law, and isolate the parameters that are constant, R and the amount n of gas, on one, side of the equation., , 2., , Since the values mentioned above are constant, for both the initial and final states of the gas the “other” sides of the equations are equal, , Copyright 2000-2007 Kinetic Books Co. Chapter 19, , 365
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to one another. Set them equal to each other in an equation., 3., , Solve for the final temperature and evaluate., , Physics principles and equations, The ideal gas law, , PV = nRT, The temperature is given in Celsius and will need to be converted to Kelvin to apply the ideal gas law., Step-by-step solution, We start by using the ideal gas law to write an equation relating the initial and final conditions of the gas., , Step, , 1., , PiVi = nRTi, , Reason, , ideal gas law, initial condition, , 2., , divide by temperature, , 3., , ideal gas law, final condition, , 4., , from steps 2 and 3, , We can solve for the final temperature because all the other values are known. The initial temperature was stated in Celsius, but the ideal gas, law requires that kelvins be used, so we add 273 to the Celsius temperature to convert., , Step, , Reason, , 5., , solve for Tf, , 6., , substitute values, , 7., , Tf = 439 K, , evaluate, , 8., , Tf = (439 – 273) = 166°C, , convert to Celsius, , This is very hot! When tanks like this are filled, they are usually held under cold water to increase the heat flow out of the tank and reduce their, temperature increase., , 19.7 - Interactive problem: pressure of an ideal gas, In this simulation, a small chamber contains an ideal gas consisting of 26 gas, particles held at a constant temperature of 400 K. The gas has a pressure of, 897 Pa. The chamber volume is currently set at 1.6×10í22 m3., You can change the volume of the chamber, within a range of 0.5×10í22 m3 to, 2.5×10í22 m3. You also can vary the number of gas particles, from 10 to 30. Your, task is to set the number of particles and the volume of gas so that the gas pressure, is reduced to 552 Pa. The temperature of the gas in this simulation remains fixed at, 400 K., When you launch the simulation, you will see the gas particles moving around as, determined by the settings stated above. After running for a few moments, the, simulation will pause so that you can enter your values. Determine how to alter the, volume and the number of molecules to achieve the target pressure of 552 Pa., Specify these values using the up and down arrows in the control panel and press, GO to apply the new settings to the gas., A text message will tell you whether the gas reached the correct pressure. You can, also note the pressure in the pressure gauge in the simulation., If you get the wrong answer, you can try again. Press RESET to return the gas to its initial conditions, specify new values, and press GO again., If you have trouble with this simulation, review the section on the ideal gas law., , 366, , Copyright 2000-2007 Kinetic Books Co. Chapter 19
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19.8 - Interactive checkpoint: volume of an ideal gas, A quantity of an ideal gas occupies a, volume of 2.18 m3 at 288 K, with an, absolute pressure of 1.45×105 Pa. If, its temperature increases to 298 K, and its pressure increases to, 2.15×105 Pa, what is the volume?, , Answer:, , Vf =, , m3, , 19.9 - Kinetic energy and temperature, In this chapter, we have shown that by varying the speed of its molecules, we can affect, the pressure of a gas. We also have mentioned that the molecules’ speed increases, with temperature. Now we will get more specific about this relationship., Rather than showing a relationship between speed and temperature, it proves more, productive to show a relationship between translational molecular kinetic energy and, temperature. To put it another way, we find it more useful to discuss the relationship, between the molecular speed squared and temperature., The molecules that compose a gas move at a variety of speeds. As the temperature of, a gas increases, the average speed of its molecules increases, as does the molecules’, average kinetic energy. The kinetic theory of gases links the temperature of a gas to the, average kinetic energy of the molecules that make up the gas., The relationship is quantified in Equation 1. The average translational kinetic energy of, a particle in an ideal gas equals (3/2)kT, with k being Boltzmann’s constant and T the, Kelvin temperature., The equation for the kinetic energy of a particle has an interesting implication: The, kinetic energy of gas molecules is solely a function of temperature. It is independent of, other factors, such as pressure or volume. As the graph in Equation 1 illustrates, when, the temperature of a gas increases, the kinetic energy of the molecules that make it up, increases linearly., , Kinetic energy of ideal gases, As temperature of gas increases:, ·average molecular speed increases, ·average translational KE of molecule, increases, , Now we move from considering the kinetic energy of a particle to the kinetic energy of, an amount of an ideal monatomic gas. A monatomic gas is one whose particles are, single atoms. The internal kinetic energy of a monatomic gas is the sum of the, individual kinetic energies of all the atoms., You see the equation for the internal energy of an ideal monatomic gas in Equation 2. It, can be derived from the first equation by multiplying by the number of atoms. The, kinetic energy of one molecule is (3/2)kT, as described above. For n moles of gas,, there are nNA atoms, so the total energy is (nNA)(3/2) kT. Since k equals R/NA, the, internal energy of n moles of monatomic gas equals (3/2)nRT., This energy makes up all the internal energy of an ideal monatomic gas, because the, only form of energy of this type of gas is translational (linear) kinetic energy. There is no, potential energy in any ideal gas. Since there are no forces between the molecules,, there can be no energy due to position or configuration. The nature of monatomic, molecules (single atoms) means they have neither rotational nor vibrational energy., , Average KE of molecule, ideal, gas, , Although we have focused on ideal monatomic gases, the internal energy and, temperature of ideal diatomic and polyatomic gases are also linearly related., , k = Boltzmann’s constant, T = Kelvin temperature, , Copyright 2000-2007 Kinetic Books Co. Chapter 19, , 367
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Internal energy, ideal monatomic, gas, , Eint = internal energy of gas, n = amount of gas in moles, R = gas constant, , Two moles of an ideal, monatomic gas are at 320 K., What is the average KE of an, atom?, , What is the total internal energy, of the gas?, , Eint = 8.0×103 J, , 368, , Copyright 2000-2007 Kinetic Books Co. Chapter 19
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19.10 - Gotchas, Using the wrong temperature scale. Make sure you use Kelvin, not Celsius (and certainly not Fahrenheit!), when applying the equations in this, chapter., , 19.11 - Summary, In this chapter, you studied ideal gases, ones where the particles (atoms or, molecules) can be modeled as interacting with each other and the walls of their, container only in elastic collisions., Avogadro’s number is 6.022×1023. It is a dimensionless number used to specify a, quantity of matter. A mole contains Avogadro’s number of particles., The pressure of a gas increases with the number of molecules of the gas and its, temperature, and decreases as its volume increases. The ideal gas law expresses, this relationship in an equation. The ideal gas law can be stated in terms of the, number of moles of a gas, or the number of particles., The average kinetic energy of a gas molecule and the internal energy of a given, quantity of gas can be calculated from its absolute temperature., , Boyle’s law, , PiVi = PfVf, Charles’ law, , Ideal gas law, , PV = nRT = NkT, Energy and temperature, , Copyright 2000-2007 Kinetic Books Co. Chapter 19, , 369
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Chapter 19 Problems, , Chapter Assumptions, Use the following values for constants:, , NA = 6.02×1023, R = 8.31 J/mol·K, k = 1.38×10–23 J/K., In problems which require you to know the atomic weights of atoms or molecules, use the following:, 12.0 u for a carbon atom (C), 4.00 u for a helium atom (He), 1.00 u for a hydrogen atom (H), 14.0 for a nitrogen atom (N), 20.2 u for a neon atom (Ne), 16.0 u for an oxygen atom (O), 44.0 u for a carbon dioxide molecule (CO2), 18.0 u for a water molecule (H2O), , Section Problems, Section 3 - Boyle’s and Charles’ gas laws, 3.1, , To prepare for a magic trick, a magician squeezes an inflated balloon into a small box of volume 0.0760 m3. If the balloon, initially had a volume of 0.130 m3 at an absolute pressure of 1.01×105 Pa, and the temperature of the gas inside it does not, change, what is the pressure in the balloon after it is squeezed into the box?, Pa, , 3.2, , A scuba tank contains 2.83×10í2 m3 of, compressed air at an absolute pressure of, 2.07×107 Pa. If all this air is released with no, change in temperature, so that it has a, pressure of 1.01×105 Pa (standard, atmospheric pressure), what volume does it, occupy?, m3, , 3.3, , A balloon is filled with air outside on a very cold day, at an absolute pressure of 1.01×105 Pa and temperature í12.0°C. The, balloon has a volume of 0.133 m3. It is then brought indoors and its temperature increases to 22.0°C at the same pressure., What is the balloon's volume indoors?, m3, , 3.4, , An experimental air compressor is designed to work by cooling the air to reduce its volume, without any change in pressure., To reduce the volume of a quantity of air at 34.0°C to 87.0% of its original volume, what temperature does the air have to, reach?, K, , 370, , Copyright 2007 Kinetic Books Co. Chapter 19 Problems
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Section 4 - Avogadro’s number and moles, 4.1, , How many molecules are there in 0.017 moles of carbon dioxide gas, CO2?, molecules, , 4.2, , The molar mass of silver is 108 g/mol. What is, the mass in kilograms of the 2.3×1024 atoms, of silver that make up the chalice in the, illustration?, kg, , 4.3, , The atomic mass of carbon is 12.0 u and of oxygen is 16.0 u, rounded to 3 significant figures. (a) What is the molecular mass, of a carbon dioxide molecule, CO2, in atomic mass units? (b) What is the molar mass of carbon dioxide in kg/mol? (c) How, many molecules are in 0.0123 kg of carbon dioxide?, u, , (a), , 4.4, , (b), , kg/mol, , (c), , molecules, , The molecular mass of water is 18 u. The mass of a teaspoon of water is 4.9×10í3 kg. If one teaspoon of water were spread, evenly over the surface of the Earth, what would be the surface number density, that is, the number of water molecules per, square meter? (Assume the Earth is a smooth sphere with radius 6.4×106 m.), molecules/m2, , Section 5 - Ideal gas law, 5.1, , 5.2, , An ideal gas occupies 0.14 m3 at an absolute pressure of 1.2×105 Pa and temperature 280 K. (a) How many moles of gas, are there? (b) How many molecules?, (a), , moles, , (b), , molecules, , At 265 K, the pressure of 0.0453 mol of an ideal gas is 1.42×105 Pa. What volume does the gas occupy?, m3, , 5.3, , A balloon holds 0.0140 m3 of an ideal gas at a temperature of 35.0°C. If the amount of gas is 0.680 mol, what is the pressure, of the gas?, Pa, , 5.4, , At a pressure of 2.32×105 Pa, 6.41×1025 molecules of an ideal gas occupy 0.752 m3. What is the temperature of the gas?, K, , 5.5, , A room with dimensions 2.52 × 4.57 × 5.61 m is filled with air at pressure 1.01×105 Pa. Air can move into or out of the room, through a crack under the door. The pressure stays constant as the temperature increases from 18.0°C to 24.0°C. What is the, change in the number of moles of air in the room? Treat the air as an ideal gas. (Be careful with the sign of your answer.), , 5.6, , The atmosphere on Mars is almost all carbon, dioxide (CO2), and has a pressure of about, , mol, , 6.9×102 Pa, much less than that of Earth. At, í22°C (a warm day on Mars), what is the, mass of one cubic meter of the atmosphere?, (Assume the atmosphere is completely CO2.), kg, , Copyright 2007 Kinetic Books Co. Chapter 19 Problems, , 371
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5.7, , A quantity of an ideal gas is held in a cylinder with a movable piston. At a temperature of 245°C, the gas has a volume of, 0.0158 m3 and pressure 2.12×105 Pa. The piston then moves so the volume of the gas is 0.0287 m3 and the pressure is, 1.67×105 Pa. What is the resulting temperature of the gas, in kelvins?, K, , Section 7 - Interactive problem: pressure of an ideal gas, 7.1, , Using the simulation in the interactive problem in this section, if there are 20 particles in the chamber, what volume will result, in a pressure of 552 Pa? Calculate your answer first, then check it with the simulation., m3, , Section 9 - Kinetic energy and temperature, 9.1, , If 8.1 mol of an ideal gas occupy 0.26 m3 at a pressure of 1.4×105 Pa, what is the average translational kinetic energy of a, molecule of the gas?, J, , 9.2, , An ideal monatomic gas has an internal energy of 1.10×104 J. If there are 2.91 moles of the gas, what is its temperature?, , 9.3, , The average translational kinetic energy of a molecule of a quantity of an ideal gas is 1.10eí20 J. What is the temperature of, the gas?, , K, , K, 9.4, , Helium, the second most abundant element in the universe, can exist as a monatomic gas. At a pressure of 1.1×105 Pa, a, quantity of helium gas occupies a volume of 0.16 m3. What is the internal energy of the gas? Treat helium as an ideal gas., J, , 372, , Copyright 2007 Kinetic Books Co. Chapter 19 Problems
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20.0 - Introduction, The desire to build more powerful and efficient engines led engineers and scientists, to embark on pioneering research into the relationship of heat, energy and work., The engines that powered the Industrial Revolution primarily used energy provided, by burning coal., Much has changed since the first engines were used to drain marshes and pump, water from the coalmines of Great Britain. New sources of energy, from gasoline for, automobiles to nuclear fuel for electric power generators, are used to power, engines. Engines are now used in applications undreamt of by their first designers,, for whom the horseless carriage and nuclear fission would have been sciencefictional fantasies., Despite the changes, much of the fundamental science developed in the 19th, century is still used to analyze engines. The goal of an engine, to get useful work, out of a heat source, has not changed. Although the technology has advanced and, made engines more efficient, the physics of heat engines as established by 19th, century engineers and scientists applies equally well to the steam locomotives of, their era and to modern electrical power plants powered by nuclear fission., This chapter focuses on two topics, using the engine as the basis of much of the discussion. One topic is the first law of thermodynamics, the, relationship between the energy supplied to an engine and how much work it does. The other topic is the role of gases in the functioning of an, engine. Many engines use a gas to function; applying some basic principles of how gases behave proves very useful in understanding the, functioning of engines., The simulation on the right will get you started on your study of engines. Here you see a heat engine, a device that uses heat as its energy, source to do work. There is gas (represented by some bouncing molecules) in the central chamber of the engine, a piston on top, and a “hot, reservoir” on the left from which heat can be allowed to transfer into the gas. In this case, we consider the cylinder, the gas and the piston to be, the system. The reservoirs on the sides allow heat to flow in and out of the system. When heat flows into the engine’s gas, its temperature will, rise (reflecting an increase in its internal energy), and if the piston is free to move, the gas will expand, pushing it up., In this simulation, you will see what occurs during two distinct engine processes. In the first process, the piston is locked in place while heat is, transferred to the gas. You control the amount of heat that is transferred. The internal energy of the gas is displayed in an output gauge. In the, first step, you should ask yourself: What is the relationship between the amount of heat transferred to the gas and the change in the internal, energy of the gas? Consider the principle of conservation of energy when you ponder your response to this question, and then test your, hypothesis., In the second process, no heat is allowed to flow in or out of the engine, but when you press GO, the piston will be unlocked so it can move. A, gauge will show you how much work the gas does as it expands or contracts, changing the piston’s position. At the end of this process, note, how much work the gas did and again its change in internal energy. How do the values for the work and change in internal energy relate during, this process?, One last calculation (there are a few here, but you will have taught yourself the first law of thermodynamics when you complete this exercise):, How does the amount of heat you initially transferred to the gas relate to its change in internal energy and the work it does? Write down these, three values and see if you note a fundamental mathematical relationship. You can also apply the physics you learned earlier; consider the, principle of conservation of energy and the work-energy theorem. (The work-energy theorem in its most general form says that the work done, by an external force on a system equals its change in total energy, which includes thermal energy as well as mechanical energy.), If these are too many questions, read on. The first law of thermodynamics begins this chapter., , 20.1 - First law of thermodynamics, , First law of thermodynamics: The net heat transferred to a system equals the, change in internal energy of the system plus the work done by the system., The first law states that the net heat transferred into or out of a system equals the change in the internal energy of the system plus any work, the system does. It is written as an equation in Equation 1, with Q representing the net heat transferred to the system. You saw this law at work, in the simulation in the introduction to this chapter. First, heat was transferred to the gas in the engine, increasing the internal energy of the, system. The increase in the temperature of the gas in the engine reflected the increase in internal energy. When you pressed GO again, the, piston rose. The engine did work, and its internal energy decreased. The amount of work done by the engine equaled the magnitude of its, change in internal energy as the piston rose., We use the simple engine in Concept 1 to discuss the first law in more depth. Heat flows from the flame into the container, causing the internal, energy of the gas it contains to increase. As the gas’s internal energy increases, its temperature and the average speed of its molecules, increases. As the gas’s temperature increases, its pressure increases and it may expand. The gas does positive work when it expands,, applying a force to the moving lid in the direction of its displacement., , Copyright 2007 Kinetic Books Co. Chapter 20, , 373
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It is important to define what we mean by “system.” Here we define the system as the, engine that includes the container, with its moveable lid, and the gas. When heat flows, into the engine, the gas’s temperature changes, but the container’s temperature is, assumed to stay the same. A rod is attached to the lid. The pressure of the gas pushes, against the lid, and this can force both the rod and the lid upwards. The rod might be, attached to the rest of the engine, such as the drive assembly of an automobile. When, the gas expands, it does work on the rod., The first law is mandated by the principle of conservation of energy. Conservation of, energy applies to isolated systems, and the system of the engine is not isolated. But the, engine is part of some larger system that is isolated. Heat energy flows into the engine., This energy must be conserved: Energy can be neither created nor destroyed in the, process. That means the heat flow increases the internal energy of the gas. This, energy, however, can do work, increasing the energy of some object outside of the, engine. Work done by the gas reduces its internal energy., The work done by the gas equals the force it exerts times how far the lid moves. That, force equals the pressure times the surface area of the lid. Calculating the work done by, the gas is easier when the pressure is constant than when the gas pressure varies., Although it is not the typical way to analyze engines, you could think of the process, entirely in terms of energy. Let’s further simplify the engine, ignoring the rod and any, forces acting on the lid other than gravity and the gas pressure from inside the engine., Heat flows in, raising the gas’s internal energy and, if the lid rises, increasing the, potential energy of the disk on top (by mgǻy). The sum of the increase in the gas’s, internal energy and the increase in potential energy of the disk equals the amount of net, heat flow., The first law applies to any engine process, for any initial and final state of the gas plus, any work done. For instance, in the simulation in the introduction, heat was added and, then the piston was allowed to move, but this could have happened simultaneously, or it, could have happened in a series of steps., , First law of thermodynamics, Heat transferred to system:, ·increases internal energy and/or, ·causes system to do work, , Q = ǻEint + W, Q = net heat transferred to system, ǻEint = change in internal energy, W = work done by system, , Mathematical signs are important here. A net flow of heat into the engine means Q is, positive; a net flow of heat out of the engine means Q is negative. Work done by the, gas raising the lid is positive, since the force is in the direction of the displacement., When the lid moves down, the work done by the gas is negative., The relationship between heat, work and internal energy is illustrated in the example, problem to the right. If 130 J of heat is transferred to an engine and its internal energy, increases by 75 J, the first law of thermodynamics dictates that it must do 55 J of work., The change in internal energy and the work done by the gas must add up to equal the, net heat transferred., , The internal energy of the gas, increases 75 J. How much work, is done by the gas?, Q = ǻEint + W, W = Q í ǻEint, W = 130 J í 75 J, W = 55 J, 20.2 - James Joule and the first law, Although it is accepted today as one of the fundamental tenets of science, the proponents of the first law of thermodynamics faced a skeptical, scientific community. Most scientists did not view heat as another form of energy. The prevalent theory in the early 1800s was the caloric, theory, which proposed that a caloric fluid was added to matter as it was heated. This addition of fluid explained why a metal rod expanded, when it was heated: It expanded because it contained more caloric fluid. One legacy of this belief is that heat still has its own units, the calorie, and the British thermal unit (Btu)., Between 1843 and 1850, the English scientist James Joule performed a series of experiments that showed that heat was another form of, energy. He used equipment similar to the apparatus shown in Concept 1. Rather than using heat to raise the temperature, Joule used, mechanical energy. As the diagram illustrates, a falling weight causes the paddles to rotate, and the work of the paddles on the water causes, its temperature to increase., Joule applied the work-energy theorem to conclude that the work done by the paddles equaled the amount of change in the potential energy of, the weight. By measuring the change in height and the mass of the weight, he could quantify the change in its PE and so the work done by the, paddles. He also measured the increase in the temperature of the water., Joule performed his experiment repeatedly. He showed beyond doubt that the change in the potential energy of the weight, which equaled the, , 374, , Copyright 2007 Kinetic Books Co. Chapter 20
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amount of work done by the paddles, was always proportional to the increase in the, temperature of the water. This proved a relationship between mechanical energy and, temperature., Joule used British units. In his experiments, he calculated that 772.5 foot-pounds of, work would increase the temperature of one pound of water by one degree Fahrenheit., Today, that value has been refined to 778 foot-pounds. Joule’s measurement was, impressively accurate., Joule also knew how to use the specific heat of water to relate an amount of heat, transferred to a quantity of water to its increase in temperature. He knew how much, heat it would take to raise the temperature of a given amount of water by one degree,, and with his apparatus, he could also calculate how much mechanical energy it would, take to do the same. He showed that one calorie of heat equals an amount of, mechanical energy that today we would describe as 4.19 J. (Since scientists today, know heat is just one form of energy, they do not use the calorie unit as often, but, instead use the joule as the unit for all forms of energy, as well as for work.), , Heat as energy, , Joule showed heat is a form of energy, Change of PE proportional to:, Modern scientists still differentiate between heat (the transfer of energy) and the internal ·water temperature increase, ·heat required for same increase, energy of an object. Heat changes the internal energy of an object or system, just as, work does. Work and heat are two ways to change some form or forms of an object’s, energy. They reflect a process. It does not make sense to refer to “the work of an object” or “the heat of an object.” Rather, it is correct to state, how much work is done on or by an object, or how much heat is transferred into or out of the object. The result is a change in the object’s, energy., , 20.3 - Heat engines, A heat engine uses the energy of heat to do work. Many engines have been designed, to take advantage of heat energy. To cite two: A steam engine in an old locomotive and, the internal combustion engine in a modern automobile both use heat as the source of, energy for the work they do., A heat engine is shown on the right. It is the container with a lid and piston on top,, between a hot reservoir on the left and a cold reservoir on the right. In the engines we, will consider, the container encloses a gas. The gas is cooler than the hot reservoir but, warmer than the cold reservoir. Heat flows spontaneously from the hot reservoir into the, gas in the engine, where the energy can be used to do work. Heat will also flow, spontaneously from the gas to the cold reservoir. Otherwise, the container is insulated, and no heat flows through any other mechanism into or out of the engine., The reservoirs are large enough that they can supply or absorb as much heat as we, like. We control the flow of heat between a reservoir and the container by opening a, hole in the insulating wall to allow heat to flow, and closing it to stop the flow. The, reservoirs are not part of the system we consider when applying the first law of, thermodynamics., The amount of gas in the engine stays constant. The temperature and pressure are, assumed to be the same everywhere in the gas at any moment. When we analyze a, heat engine, we will assume it is ideal: that there is no friction to reduce its efficiency, and that heat flows uniformly and instantly through the system., , Heat engines, Heat flows into engine, Engine uses heat to do work, Heat flows out of engine, Engine cycle: system returns to initial, state, , The engine goes through a sequence of processes that collectively are called an engine, cycle. (One process or step in an engine cycle is sometimes called a stroke.) At the end, of each cycle, the engine returns to its initial state. For instance, during one process,, heat might flow into the engine from the hot reservoir so the piston rises. During another, process, heat might flow out to the cold reservoir as the piston falls. No matter what the, processes, at the end of an engine cycle, the engine returns to its initial configuration:, The piston is in its initial position and the gas is back to its initial volume, pressure and, temperature., Since the gas returns to its initial state, its internal energy does not change during a, complete engine cycle. This means we can concentrate on the two other quantities in, the first law, namely heat and work. To apply the first law and other equations, we, consider the work done by the gas. The gas does positive work as it lifts the piston. If, the piston moves down and compresses the gas during another part of the cycle, the, gas does negative work., The distance the piston moves up or down might be the same. However, the amount of, work that occurs will typically differ, since the purpose of the engine is to do net positive, work on the piston during a complete engine cycle. The gas applies more force when it, expands (to move a crankshaft, for example) than when it contracts, so the net work, done by the engine in a cycle is positive. The net transfer of heat into the engine during, a complete engine cycle equals the net work it does. The net transfer of heat equals the, heat in from the hot reservoir minus the heat out to the cold reservoir., , For a complete engine cycle, W = Qh í Qc, W = net work done by engine, Qh = heat in, Qc = heat out, Internal energy returns to initial value, , Copyright 2007 Kinetic Books Co. Chapter 20, , 375
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You see this stated as an equation to the right. The net work done by the engine during, a cycle equals the heat transferred into the engine from the hot reservoir minus the heat, that flows out to the cold reservoir. Heat that flows into the engine from the hot reservoir, is called Qh, and heat that flows out to the cold reservoir is called Qc. We treat both Qh, and Qc as positive quantities. This means the net flow of heat equals Qh í Qc. If heat, flows only out of the engine during a process in an engine cycle, then the net heat flow, for that process is negative., This equation is a special case of the first law of thermodynamics. The law states that, the net heat flow equals the work done plus any change in internal energy. Since the, internal energy is not changed after a complete engine cycle (the engine returns to its, initial state), the net work done by the engine equals the net flow of heat., , During an engine cycle, the heat, transfers are as shown. What is, the net work done by the, engine?, W = Qh íQc, W = 3200 J í 1800 J, W = 1400 J, 20.4 - The ideal gas law and heat engines, Many engines use a gas as their working substance. Heat is transferred to the working, substance, which expands and does work. Water (both in liquid and gaseous form) is, another common working substance. However, water vapor is far from an ideal gas,, and in this textbook, we focus on an ideal gas as the working substance., In addition to the first law of thermodynamics, the ideal gas law is very useful in, analyzing the processes in an engine cycle. Here, we want to briefly review this law,, and show how it is used in analyzing heat engines. The ideal gas law relates pressure,, volume, temperature and the amount of gas. You see the law stated in Equation 1., In the heat engines we consider, the quantity of gas enclosed by the container is, constant. This leaves three variable properties of a gas in the ideal gas law: the gas’s, pressure, volume and temperature. The product of the pressure and volume is, proportional to the temperature. If the temperature of the gas increases, for instance,, then the product of its pressure and volume must increase, as well., The first law of thermodynamics states that when heat flows into an engine, the heat, energy increases the gas’s internal energy and/or causes it to do work. A change in, internal energy will be reflected in the gas’s temperature; greater internal energy means, a higher temperature. This means the volume or the pressure of the gas will increase,, or both., To apply these two principles, consider Example 1. The piston is locked in place so that, the gas can do no work. This means that the heat transferred to the engine solely, increases the internal energy of the gas. That increase in internal energy is reflected by, an increase in temperature. Since the piston is locked, the volume of the gas is constant, and the ideal gas law enables us to conclude that its pressure must increase, proportionally to the temperature increase., , Ideal gas law, PV = nRT, When n is constant:, , P = pressure, V = volume, n = number of moles of gas, R = gas constant, T = temperature (K), , 376, , Copyright 2007 Kinetic Books Co. Chapter 20
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The engine piston is locked in, position and heat flows into the, gas. What happens to the, pressure, volume, and, temperature?, Volume constant (piston locked), Temperature increases, Pressure increases proportionally, 20.5 - Engine processes, In this section, we will review an engine cycle. It, involves the three processes described below. To, understand these processes, watch the gauges that, display the pressure, volume, molar quantity and, temperature of the gas. We assume that the pressure, and temperature are uniform throughout the gas., We start with one mole of gas with a volume of, 2.44×10í2 m3 at 292 K and a pressure of, 1.00×105 Pa. The piston is locked and no heat is, allowed to flow in or out. This state of the engine is, depicted in the illustration at the top of the page., Initial state of heat engine., , First process, Volume held constant, Heat flows in, Pressure and temperature increase, , First process. Now, 1350 joules of heat are allowed to flow into the gas. The energy inflow increases the internal energy of the gas, which is, reflected by its increased temperature. Since the piston is locked, the gas’s volume cannot change. This means its pressure must increase., You see this directly above. Its pressure has increased from 1.00 to 1.36×105 Pa and its temperature has increased from 292 to 400 K., , Copyright 2007 Kinetic Books Co. Chapter 20, , 377
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Second process, Gas does work on piston, Pressure and temperature decrease, Heat flows out, , Second process. Now, we release the lock on the piston. This allows the gas to expand, doing work by moving the piston. The internal energy, of the gas and its temperature decrease. A hole in the insulating wall opens to the cold reservoir, and 1074 J of heat also flow out of the, engine. As you can see above, the gas’s temperature and pressure are now less than they were when the cycle started., , Third process, Piston does work on gas, Engine returns to initial state, Cycle complete, , Third process. The insulating walls to the reservoirs are again closed, so no heat flows. The piston pushes down, returning the gas to its initial, volume. This is a lesser amount of work than was done when the piston moved up, because the gas pressure is always less during this, process than during the expansion process. This work done on the gas increases its temperature to its initial value. The pressure returns to its, initial state., This is one of many possible engine cycles. We will examine some specific engine cycles in more detail later. For now, we focus on the, thermodynamic processes that can make up part of an engine cycle., , 20.6 - Pressure-volume graphs and heat engines, , Pressure-volume graphs, Multiple paths between same pressurevolume points, , In an engine, the pressure and volume of the gas change as heat is transferred to the engine and the engine does work. These changes can, be tracked with what is called a pressure-volume graph. In a pressure-volume graph, pressure is plotted on the vertical axis and volume on the, horizontal axis. You see two engine processes diagramed above, one in blue and the other in red. They illustrate how two different processes, can cause a gas to move between the same initial and final states., We will now describe these processes step-by-step. The gas starts at 2.43×10 5 Pa of pressure, 1.00×10í2 cubic meters of volume, and a, temperature of 293 K., , 378, , Copyright 2007 Kinetic Books Co. Chapter 20
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Process 1, Volume increases as pressure, decreases, Temperature does not change, , Process 1. The gas’s pressure is then reduced as its volume is increased in the process labeled 1, which is immediately above. The gauges, and piston position in Concept 2 show the state of the engine at the end of the process. The temperature does not change during this process., During the first process, the amount of heat added to the engine equals the work done on the piston because the internal energy and, temperature of the gas do not change. The ideal gas law must be obeyed, so as the volume increases, the pressure decreases; their product, remains constant at each point along the path. Path 1 reflects one path by which the gas can change from its initial to its final state., , Process 2, a. Volume increases at constant, pressure; temperature increases, , Process 2, step a. Now we show a different process by which the gas can be caused to move between the same two states. To make the, comparison clearer, we continue to show path 1 on the graph. The second process starts with the same pressure, volume and temperature as, before. This process has two distinct steps. In the first step, labeled 2a in the graph above, heat energy flows into the gas from the hot reservoir, as before. This increases the internal energy of the gas (as indicated by the increase in temperature). The graph is horizontal, indicating a, constant pressure. We could maintain the pressure by having the gas press against a constant weight., , Process 2, b. Pressure decreases at constant, volume; temperature returns to initial, , Process 2, step b. Next, since we have the same volume that was reached by the path 1 process, we lock the piston into place, keeping the, volume constant. The pressure is too high, so we allow heat to flow out to the cold reservoir. This decreases the temperature of the gas, and at, this fixed volume, this means the pressure must decrease proportionally. You see this step labeled 2b in the graph above., This two-step process arrives at the same final pressure-volume point as the first process, but in a different way. In the first, the gas’s, temperature remained constant while its pressure and volume constantly changed. In the second, first the volume was changed at constant, pressure, and then the pressure was changed at constant volume. The temperature changed during both of the steps of process 2., The difference in the paths reflects an important point: Gases can change from one combination of pressure and volume to another by, experiencing different histories., Both the processes sketched above obey the first law of thermodynamics and the ideal gas law. The first law is obeyed because there is the, same net flow of heat into the engine in both cases. In process 1, all the heat is used for work, and the gas’s internal energy stays the same., (To make the engine realistic, some heat should flow to the cold reservoir, since no engine is 100% efficient.) In process 2, more heat is added, during step 2a than in process 1, since in this step the gas does the same amount of work as in all of process 1 and its internal energy, increases. Heat flows out of the engine during step 2b, making the net heat into the engine the same for processes 1 and 2., The ideal gas law is also obeyed. The product of the pressure and volume is always proportional to its temperature. For instance, on path 2b,, as the gas’s temperature decreases, its pressure decreases proportionally., , Copyright 2007 Kinetic Books Co. Chapter 20, , 379
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20.7 - Classifying thermal processes, , Gas cycles, Four types of gas processes:, ·Constant volume: isochoric, ·Constant pressure: isobaric, ·Constant temperature: isothermal, ·No heat transfer: adiabatic, , As a heat engine operates, the gas expands or contracts and its pressure changes. A gas can move from one pressure-volume state to, another by different thermal processes. Four thermal processes are commonly studied and used in engines. These processes are, characterized as follows., 1., , Does the gas’s volume remain constant? When it stays the same, the process is called a constant-volume process. This is also called, an isochoric process., , 2., , Is the gas’s pressure constant? If so, the process is called isobaric., , 3., , Does any heat flow in or out of the engine during the process? If there is no heat flow, the process is called adiabatic., , 4., , Does the temperature of the gas stay the same? If it does, the process is called isothermal., , Above, we show each of these processes on a pressure-volume graph. Although we use an arrow to show each process occurring in a, particular direction, the arrows could be reversed without changing the types of the processes. Any of these processes can occur in either, direction., , 20.8 - Gotchas, A gas is hotter after an engine cycle is complete. No, the definition of an engine cycle is that the engine returns to its initial state., Adding heat to an engine immediately causes it to do work. Although that is likely the purpose of the heat transfer, the heat can also increase, the internal energy of the engine. It may be later in the cycle that the engine uses the energy to do work., Doing work on an engine can cause the temperature of its gas to increase. This is true. It can also cause heat to flow out of the engine as well., The first law of thermodynamics is the same as the principle of conservation of energy. Essentially, this is true. The first law explicitly factors in, work, a way to add or subtract energy from a system. However, both physics principles are statements of conservation. Energy cannot, magically appear or disappear; it can always be accounted for., James Joule demonstrated a fundamental relationship between heat, temperature and work. Yes. He showed how both work and heat could, increase the temperature (and energy) of a substance., , 20.9 - Summary, The first law of thermodynamics relates heat, work, and internal energy, and is a restatement of the law of conservation of energy. It states that the net heat transferred, to a system equals its change in internal energy plus the work done by the system., , First law of thermodynamics, , Q = ǻEint + W, A heat engine uses thermal energy to do useful work. An engine contains a working, substance, usually a gas, that goes through a series of thermal processes. During a, thermal process, heat can be transferred to the gas from a hot reservoir, or can flow, Engine cycle, out to a cold reservoir, or it could be that no heat exchange takes place. In an, W = Qh – Qc, analogous way, during the process, energy transfers can also take place via work:, work can be done on the system, work can be done by the system on its, surroundings, or no work might be done by the system. An engine cycle is a series, of processes that returns the gas to its original temperature, pressure, and volume. During a cycle, because the gas has returned to its original, state and the change in its internal energy is zero, the net work done equals the difference between the heat transferred into the engine and, the heat transferred out., When the amount of gas is constant, the ideal gas law states that the product of pressure and volume is proportional to the temperature of the, gas., Four thermal processes that occur frequently in engines are constant-volume, isobaric, adiabatic and isothermal., , 380, , Copyright 2007 Kinetic Books Co. Chapter 20
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Chapter 20 Problems, , Conceptual Problems, C.1, , A gas is enclosed in a container whose walls are made of a material that is flexible and which acts as a perfect insulator for, all practical purposes. A thermal process takes place on the gas. What would be the best model for the type of process?, i., ii., iii., iv., , C.2, , Constant-volume, Isobaric, Adiabatic, Isothermal, , A gas is sealed in a container with rigid walls that allow heat to flow through them. A thermal process takes place on the gas., What would be the best model for the type of process?, i., ii., iii., iv., , Constant-volume, Isobaric, Adiabatic, Isothermal, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) What is the relationship, between the amount of heat transferred to the gas and the change in the internal energy of the gas during the first process, when the volume is fixed? (b) How do the values for the work and change in internal energy relate during the second process, when no heat flows in or out of the gas? (c) In general, how does the amount of heat transferred to the gas relate to the, change in internal energy and the work it does?, (a) The heat is, , i. greater than the change in internal energy., ii. equal to, iii. less than, (b) The work done is, i. equal to, the change in internal energy., ii. the negative of, (c) The net heat transferred is, i. greater than the sum of the change in internal energy and the work done., ii. equal to, iii. less than, , Section 1 - First law of thermodynamics, 1.1, , In the gym, you do a workout routine that causes you to lose 3.4×105 J of heat while performing 1.7×105 J of work. With, yourself as the system, state the following quantities for the first law of thermodynamics, with the appropriate signs: (a) Q, (b), ǻEint, (c) W., (a), , J, , (b), , J, , (c), , J, , 1.2, , A system performs 356 J of work on its surroundings while losing 133 J of internal energy. Determine the heat transferred to, the system during this process., , 1.3, , A sealed container of gas with a piston is well-insulated, so that no heat is transferred to it or from it. The gas performs, 1700 J of work on its surroundings. What is its change in internal energy?, , J, , J, 1.4, , An ideal monatomic gas is held in a well-insulated container with a piston, so that no heat is transferred to it or from it. There, are 1.75 moles of gas in the container. During a process, the temperature of the gas increases from 297 K to 353 K. (a) What, is the work associated with the process? Make sure to include the correct sign. (Hint: there is an equation in the chapter on, the Kinetic Theory of Gases for the internal energy of an ideal monatomic gas that will help you solve this problem.) (b) Is, work done on the gas, or does the gas do work on its suroundings?, (a), (b), , J, i. Work is done on the gas, ii. Work is done by the gas, , Copyright 2007 Kinetic Books Co. Chapter 20 Problems, , 381
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Section 3 - Heat engines, 3.1, , During a cycle, a heat engine receives 2.78×104 J of heat from the hot reservoir and gives up 1.83×104 J of heat to the cold, reservoir. (a) What is the net work performed during one cycle? Make sure to include the correct sign. (b) Is work done on the, system, or does the system do work on its suroundings?, (a), (b), , 3.2, , J, i. Work is done on the system, ii. Work is done by the system, , A heat engine does 2.4×105 J of work during each cycle. If the engine loses 1.8×105 J of heat to the cold reservoir in each, cycle, how much heat is transferred from the hot reservoir during the cycle?, J, , 3.3, , A mountain climber gains 897 m in elevation on a particular hike. Consider the climber's body as a heat engine, with heat for, work provided by burning food (the "hot reservoir") and excess heat radiated into the mountain air (the "cold reservoir"). (a) If, her mass is 62.3 kg, how much work does she do changing her gravitational potential energy? (b) If she radiates 1.67×105 J, of heat during the climb, and her internal energy doesn't change, how much heat does the food supply?, , 3.4, , 382, , (a), , J, , (b), , J, , The horsepower rating commonly used for engines indicates the rate at which they can do work. A one-horsepower engine, can do 746 J of work each second. A lawn mower has a 6.50 horsepower engine that runs at 3800 rpm. Each revolution of, the engine represents one engine cycle. (a) What amount of work does the engine do on each cycle? (b) If the hot reservoir of, the engine supplies 142 J of heat each cycle, hot much heat must the engine expel to the cold reservoir?, (a), , J, , (b), , J, , Copyright 2007 Kinetic Books Co. Chapter 20 Problems
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21.0 - Introduction, Physics often seems to be about possibilities, but it is also about limits. You can, think of the first law of thermodynamics as stating a limit: An engine cannot do more, work during an engine cycle than the heat added to it. If it did, it would defy the, principle of conservation of energy. If such an engine existed, it would be a source, of “free energy.” Alas, no such engine exists., If an engine cannot perform an amount of work greater than the energy added to it,, can an engine just “break even”? This is not an idle question: The efficiency of an, engine, how much work it does per amount of energy added, is a crucial measure of, an engine’s utility., Automobiles, for instance, can take advantage of about one-fourth of the energy, produced by the combustion of gasoline to do useful work, making them about 25%, efficient., The second law of thermodynamics establishes the theoretical limit of efficiency: It, states that no heat engine can be 100% efficient; that is, no engine can convert all, of the heat supplied to it into an equal amount of work. In theory, engines could be, built with efficiencies of 99.9%, but never 100%. This limit cannot be reached with any conceivable improvements in engineering; it is a, theoretical limit that in principle can never be realized. The second law of thermodynamics and the topic of efficiency are two areas of focus for, this chapter., The concept of entropy provides another way to study thermal processes. In general terms, entropy is a measure of how ordered a system is,, and is another useful tool for understanding the efficiency and limits of engines., The simulation on the right lets you explore the relationship between heat, work and efficiency in an engine cycle. During the engine cycle, the, engine will do work and return to its initial condition at the end of the cycle. The heat engine will perform one cycle when you press GO. You, can set the amount of heat transferred from the hot reservoir to the engine during the cycle, and the amount of heat the engine expels to the, cold reservoir. At the end of the cycle, the work done by the engine and the engine's efficiency are calculated and displayed., You can add from 50 to 500 joules of heat. These are small amounts of heat, appropriate for a toy or model engine. More than 500 joules will, exceed the safety limits of the engine., We want you to observe two principles at work in this process. First, apply what you learned about the first law of thermodynamics to this, engine. What relationship do you expect between the net heat transferred to the engine and the net work done by the engine? (The engine, does positive work on the piston as the system expands and raises the piston, and a smaller amount of negative work as the piston falls.) Make, a hypothesis and then test it with the simulation., Second, you will encounter the second law of thermodynamics: No engine is 100% efficient. The energy that flows out of the engine in the form, of heat to the cold reservoir cannot be used to do useful work. The greater the heat that flows out to the cold reservoir, the less efficient the, engine. In the interest of realism, the simulation requires a realistic amount of heat to flow to the cold reservoir., Even the most efficient practical engines, like those in electric generation plants, run at less than 60% efficiency. See how efficient you can, make the simulation engine; it can be much more efficient than the average engine. But, try as you might, you will find that reaching 100%, efficiency is a goal you cannot achieve., , Copyright 2007 Kinetic Books Co. Chapter 21, , 383
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21.1 - Efficiency, , Efficiency: The ratio of the net work done by an, engine during a cycle to the heat energy, supplied to the engine., Engine design has been refined for hundreds of years, as engineers have sought to, increase the power and reliability of engines while decreasing their size. Equally, importantly, they have sought to increase the efficiency of engines, the amount of useful, work an engine does divided by the amount of thermal energy supplied to it., The first formula in Equation 1 states the definition of the efficiency of a heat engine., The efficiency equals the net work W done by the engine divided by the heat Qh, transferred to the system from the hot reservoir. This ratio is often stated as a, percentage. A typical internal combustion automobile engine has an efficiency of about, 25%, while the diesel and coal powered engines in electrical plants have efficiencies, ranging from 40% to 60%. These numbers are for the engines alone. The overall, systems í the entire car, the whole electrical plant í run at lower total efficiencies due, to inefficiencies outside the engines., , Engine efficiency, e = W/Qh, , The second formula for engine efficiency shown in Equation 1 provides a way to, calculate engine efficiency from the heat Qh added to the engine, and the heat Qc that, flows out., , e = engine efficiency, , The second equation for engine efficiency is derived below from the definition (the first, equation). The derivation uses the first law of thermodynamics: The net heat flow, equals the work done by the engine plus its change in internal energy. Since the, internal energy of an engine is the same at the beginning and end of a cycle, the net, heat flow equals the work done by the engine., , Qc = heat out during cycle, , Step, , W = net work during engine cycle, Qh = heat in during cycle, , Reason, , 1., , e = W/Qh, , definition of efficiency, , 2., , W = Qh – Qc, , first law of thermodynamics, , 3., , substitute equation 2 into equation 1, , 4., , divide, What is the efficiency of this, engine?, , e = 1 í (4700 J)/(8400 J), e = 0.44 = 44%, 21.2 - Second law of thermodynamics, , Second law of thermodynamics: No heat engine, can transform 100% of the energy supplied to it, into work during a cycle., There are several equivalent ways to express the second law of thermodynamics. The, definition above states the law in a form that is one of its important consequences:, There is a limit to the amount of work that can be done by a heat engine supplied with a, certain amount of energy during a cycle. This is called the Kelvin-Planck statement of, the second law., To take a step back for a moment: The first law of thermodynamics states that the net, heat transferred to a system equals the net work done by the system plus the change in, its internal energy. That provides one limit to how much work an engine can do: no, more than the net heat transferred to it. In essence, the first law is about energy, , 384, , Second law of thermodynamics, Engines cannot transform 100% of heat, into work during a cycle, , Copyright 2007 Kinetic Books Co. Chapter 21
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conservation. Energy cannot be created by a process; it must stay constant., The second law is a little more dire: It says that during an engine cycle, engines do less, work than the energy transferred into them. In principle, the work could equal 99.9% of, the energy that flows into an engine, but it can never equal 100%., , Heat flows spontaneously only from hot, to cold, , This law is not about designing an efficient engine. It is instead a physical law that limits, the efficiency of any engine. No engine can be 100% efficient., Another way to state the second law, called the Clausius statement, is: There can be no, process whose sole final result is the transfer of heat from a cooler object to a warmer, object. Heat flows spontaneously from an object at a higher temperature to an object at, a lower temperature. Heat will not flow in the other direction unless compelled to do so., This direction of heat flow agrees with your everyday experience. If you place a quart of, ice cream in a hot car, you expect the ice cream to get warmer, not colder. Energy flows, from the hotter air to the cooler ice cream., Stating the law in this fashion may help you to understand why no engine can be 100%, efficient. The illustrations at the right show a conceptual diagram of a heat engine. Heat, is allowed to flow into the engine from the hot reservoir. Some heat also flows out to the, cold reservoir. The maximum amount of work the engine can do during a cycle is the, net flow of the heat, the heat in minus the heat out., , Non-spontaneous heat flow, Work required to force heat to flow from, cold to hot, , An enterprising engineer might think she could “recycle” the heat absorbed from the, engine by the cold reservoir, say by connecting the cold reservoir to the hot reservoir. The Clausius statement of the second law says that heat, will not flow spontaneously from the cold reservoir to the hot., To solve this problem, she might attach another engine to force heat to flow from the cold reservoir to the hot. This is indeed possible: Air, conditioners and refrigerators cause heat to flow from a cooler region to a warmer one. However, this heat flow is not spontaneous or free; it, requires work and comes at a price, as the electrical bills for air conditioners and refrigerators indicate., , 21.3 - Reversible and irreversible processes, , Reversible process: A process in which a, system can be returned to its initial state without, the addition of energy., Irreversible process: Energy must be added to, a system to return it to its initial state after an, irreversible process occurs., Some of the crucial underpinnings of the theories concerning engine efficiency rely on, the concepts of reversible and irreversible processes. In general, a process is a series, of steps that move a system from one state to another. When a system undergoes a, reversible process, it can be returned to its initial state without the addition of energy., , Reversible process, System can be returned to initial state, without adding energy, , Every real process is irreversible, although some are close to being reversible., Consider, for example, a puck attached to a spring on an air hockey table, as shown in, Concept 1. Imagine the spring is initially compressed and the puck is held in place., When the puck is released, the compressed spring will push the puck to the right until it, pauses momentarily. This is a process: The spring pushing the puck until it pauses. In, an ideal system (no friction, no air resistance), the puck will return to its initial position, as the spring contracts with no additional energy required. The process of spring, expansion can be thought of as reversible., In contrast, if you drop an egg and it breaks, several changes take place that cannot be, easily reversed (that is quite an understatement). For example, the breaking of the egg, creates sound energy through vibrations in the air. This reduces the energy of the egg., You do not expect to be able to put it back together again without extraordinary effort., This is an irreversible process. It would take an extreme amount of time and effort, (Humpty Dumpty inevitably comes to mind) to return the egg to its initial state., , Irreversible process, Cannot be reversed without adding, energy, Video runs only "one way", , One way to think about reversibility is to imagine videotaping a process. If you can, easily decide whether the tape is being played forward or backward, the process is, irreversible. If you watched a videotape of an egg shattering, you would know the, direction the tape is being played. On the other hand, with the puck and the spring, you would not be able to discern easily if the tape were, being played forward or backwards. (If you watched it for several cycles of motion, however, you would note that the extreme positions of the, puck were less far from equilibrium as the system lost energy to dissipative forces like air resistance and friction.), , Copyright 2007 Kinetic Books Co. Chapter 21, , 385
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These principles can be stated in terms of a system and its environment. With an irreversible process, a system can be returned to its initial, state but its environment must change. This is particularly applicable to heat engines, where the “environment” may be modeled as hot and, cold reservoirs. The engine mechanism, the gas and the piston, can be returned to their initial state, but their environment changes. The hot, reservoir becomes a little less hot and the cold reservoir becomes warmer each time the engine completes a cycle., , 21.4 - Entropy, , Entropy: A property of a system. When heat is, transferred to a system, its entropy increases., Entropy is a concept used to describe the state of a system, and it is a property of a, system, just like pressure, volume, temperature or internal energy. Some properties, may be directly measured or read from a gauge. A thermometer will tell you an object’s, temperature and a meter stick will tell you its length. For other properties, such as, kinetic energy, there are no direct measures; there are no “KE gauges.” A property like, kinetic energy must be computed from factors that can be measured, mass and speed., Similarly, there are no direct “entropy gauges” available to scientists or students. In this, section, we focus on the relationship of entropy, temperature and heat without, concerning ourselves too much about exactly what is meant by “entropy.” In other, words, we will start our discussion of entropy by considering some properties that can, be observed and are familiar to you, and focus on entropy itself once this groundwork, has been laid. In fact, the concept of entropy was historically developed in a process, like this. It arose out of the quest to understand the relationship between temperature, and heat., , Entropy, Property of objects, systems, Increases as heat is transferred to, object, , When heat flows into a system, its entropy increases. When heat flows out of a system,, its entropy decreases. The change in entropy equals the heat divided by the, temperature at which the heat flow occurs., We express this as a formula in Equation 1. To apply this equation, the process must be, reversible and the temperature measured in kelvins. Since absorbing or expelling heat, will change the temperature of the system, this change in entropy must be measured for, a small amount of heat, or the system must be large enough that it can expel or absorb, a fair amount of heat with only a negligible change in temperature. The units for entropy, are joules/kelvin., This equation is useful for two reasons. First, it provides the tool for computing a, change in entropy. Second, it (finally!) provides a definition of temperature more formal, than just “something measured by a thermometer”: Temperature is the slope of an, entropy-heat curve. (Solving the equation on the right for temperature shows why this is, the case.), The entropy of a system, like any property, depends only on the system's state, not how, the system arrived there. There are many processes that change a system from a, particular initial state to a particular final state. The system’s resulting change in entropy, is the same for any of these processes. (This is analogous to gravitational potential, energy, where only the initial and final positions of an object matter, not the path it took, between them.), At the right, you see a graph of temperature plotted against entropy. As heat flows into, an object, its temperature increases, as does its entropy. The area under the curve,, positive or negative, equals the amount of heat transferred., , ǻS = change in entropy, Qrev = net heat transferred in reversible, process, T = temperature (K), Units: joules/kelvin (J/K), , The curve is not a straight line. The horizontal section occurs at a first-order phase transition, where heat added causes the substance to, change from solid to liquid, or liquid to gas. At phase transitions, adding heat increases the entropy, but not the temperature., , 386, , Copyright 2007 Kinetic Books Co. Chapter 21
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Heat flows to a large block in a, reversible process. What is the, change in the entropy of the, block? Assume its temperature, is constant., , ǻS = (45.0 J)/(285 K), ǻS = 0.158 J/K, 21.5 - Second law of thermodynamics: entropy, , Second law of thermodynamics: Entropy, increases, or at best remains constant, in any, isolated system., If nature abhors a vacuum, as the saying goes, then it revels in messes. The universe, tends toward increasing disorder. So if your life feels like it is becoming increasingly, chaotic, well, you are just going with the flow., A common non-physicist’s statement of the second law is “systems tend toward, disorder,” a paraphrase of the definition above, which is the second law stated in terms, of entropy. Entropy is considered as a measure of a system’s disorder. The disorder of, a system either stays constant, or increases. Your room never spontaneously becomes, more orderly., You might think that the second law is violated because you can reduce the entropy of, an object by cooling it. For instance, you can place a hot drink inside a refrigerator and, cool the drink., However, the second law applies to isolated systems and the refrigerator does not, function as an isolated system. If you place your hand by the back of a refrigerator, you, will realize that refrigerators emit heat. This heat increases the entropy of the, surrounding atmosphere. When we correctly apply the second law to the isolated, system of the refrigerator and surrounding atmosphere, we find that entropy stays, constant, or increases., A series of reversible processes can, in theory, leave the entropy of a system, unchanged. In practice, however, no process is perfectly reversible, and the entropy of, an isolated system increases as processes occur., , Entropy and the second law, The entropy of an isolated system can, only increase, The total entropy of the universe is, increasing, Carpe diem, , In an isolated system:, , S = entropy, , Above, we asserted that the heat flow out of a refrigerator would increase the entropy of, a system. We will show how this is true, using a particular instance., We rely on the formulation of the second law that states that heat flows spontaneously only from a hotter object to a colder one. Also recall that, the entropy change equals the heat flow divided by the temperature at which the heat flow occurs. The colder the temperature, the larger the, increase in entropy., In Example 1 on the right, 1100 J of heat flows from an object at 250 K to an object at 130 K. To calculate the system’s entropy change, we, consider the two blocks separately. For the warmer object, we divide the heat by the temperature of the warmer object. Since heat flows out of, the warmer object, that heat flow is negative. The change in entropy of the hotter object is í4.4 J/K. Since heat flows into the cooler object, that, heat flow is positive. Doing similar calculations for the cooler object, but dividing by the cooler object’s lower temperature, we determine that its, entropy change is +8.5 J/K. The net change in entropy for the system is 4.1 J/K. The positive sign for the change in entropy tells us that the, entropy increases and the second law is obeyed., This example helps to illustrate why entropy increases with spontaneous heat flow. Heat flows from hot to cold, and as the example illustrates,, , Copyright 2007 Kinetic Books Co. Chapter 21, , 387
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the reduction of entropy in the hotter object is more than matched by the increase in, entropy of the cooler object. This must always be the case since the change in entropy, equals the heat divided by the temperature, so a hotter object “expels” less entropy than, the colder object “absorbs.” In our example, we treated the two objects as being at, constant temperatures, but even if their temperatures changed, the spontaneous flow of, heat would still cause the entropy of the system to increase., , What is the change in entropy of, the system during this, spontaneous heat transfer?, ǻS = ǻS1 + ǻS2, , ǻS = í4.4 + 8.5 = 4.1 J/K, 21.6 - Entropy and disroder (oops: disorder), , Entropy: Sometimes defined as a measure of, the disorder of a system., The concept of entropy as disorder has entered popular culture thoroughly enough that, this section both uses, and critiques, this popular definition. Entropy often is described, as measuring the order, or really the disorder of a system. A system that has more, disorder has more entropy. For instance, when a new deck of playing cards is shuffled,, it goes from being well ordered (the cards are arranged by suit and number) to, disordered (the cards can be in many possible arrangements). The deck’s entropy has, increased., In thermodynamics, entropy is more often discussed in terms of heat and temperature., The entropy of an object increases as its temperature increases. For instance, when ice, is heated so that it melts, and then the liquid is further heated until it turns into steam,, the water’s entropy increases. On the other hand, in ice the water molecules are, “ordered” because of their crystalline structure. In a gas, the molecules have no defined, positions, they move randomly, and a particular water molecule may wander far from its, initial position. Transferring thermal energy into the system increases its temperature,, and its “disorder” increases as its temperature increases., , Entropy and disorder, “Popular” definition of entropy: measure, of disorder, As temperature increases, so does, disorder, , In some respects these examples work well, providing a visual sense or metaphor to the abstraction of entropy. On the other hand, they are not, particularly rigorous, and they do not necessarily link the underlying reality of entropy í its relation to heat í to what we actually see., For instance, what we perceive as more “ordered” might in fact have greater entropy. Consider the example of ice cubes and liquid water. What, looks more “orderly”: a glass containing jagged irregular chunks of ice, or the same glass filled with an equivalent amount of liquid water? The, glass of water “looks” more orderly, but considering water at the molecular level, the ice is more orderly., Something is also missing if entropy is only considered to be the “disorder” of a system. Consider an unusual trick deck of cards in which every, card is the two of spades. You can shuffle all you like without creating a more “disordered” deck. Entropy as disorder depends on the number, of different ways the elements of a system can be arranged. For instance, you can create more disorder by shuffling a standard 52-card deck, than one that consists of, say, only 13 cards since the 52-card deck can be arranged in many more ways (8.07×1067) than can the 13-card one, (only 6.23×109)., In summary, considering entropy as disorder, and increasing entropy as increasing disorder, can be a useful metaphor at times, but it also has, its limits and potential traps., , 388, , Copyright 2007 Kinetic Books Co. Chapter 21
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21.7 - Maximum engine efficiency and reservoir temperatures, Heat engines are everywhere in the modern world. They convert heat into useful work., These engines always consist of a “hot” source reservoir from which thermal energy is, taken í for example, the coal burning firebox in an old-fashioned train engine í and a, “cold” exhaust reservoir where thermal energy is expelled. In the case of a train’s steam, engine, that cold reservoir is the atmosphere., Higher efficiency í turning more of the heat into work í is what heat engine designers, strive for. This results in less fuel consumption, which saves money, extends natural, resources and reduces pollution., We can state an inequality, shown on the right, expressing the maximum efficiency of, any engine as determined by the temperatures of its hot and cold reservoirs. In this, section, we derive this inequality. To do so, we consider the engine, including its, reservoirs, as an isolated system, and assume the reservoirs are large enough that the, heat flow does not appreciably change their temperatures., , Maximum efficiency of heat, engine, , During an engine cycle, the entropy of the hot reservoir decreases as heat flows out of it, and the entropy of the cold reservoir increases as heat flows into it. The entropy of the, gas does not change after the completion of a cycle because the engine returns to its, initial state., The second law of thermodynamics dictates that the entropy of an isolated system must, increase or stay the same during any process, including a complete engine cycle. This, means the decrease in entropy of the hot reservoir must be matched or exceeded by, the increase in entropy of the cold reservoir. The net entropy increases., , e = efficiency, Tc = temperature of cold reservoir (K), , Variables, , Th = temperature of hot reservoir (K), , engine efficiency, , e, , net change in entropy, , ǻS, hot reservoir, , cold reservoir, , temperature, , Th, , Tc, , heat transferred, , Qh, , Qc, , change in entropy, , ǻSh, , ǻSc, , What is the strategy?, 1., , Calculate the change in entropy for the hot and cold reservoirs. The net change in entropy, which cannot be negative, is the sum of, these two quantities. Write this as an inequality, and then rearrange the inequality so the ratio of reservoir temperatures is on one side., , 2., , Apply the definition of engine efficiency to the inequality., , Physics principles and equations, The definition of entropy, , The entropy statement of the second law, , ǻS 0, An equation for engine efficiency, , Copyright 2007 Kinetic Books Co. Chapter 21, , 389
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Step-by-step derivation, We start with the entropy inequality, and use that to write an inequality with the ratio of reservoir temperatures on one side., , Step, , Reason, , 1., , entropy never decreases, , 2., , definition of entropy, cold reservoir, , 3., , definition of entropy, hot reservoir, , 4., , substitute equations 2 and 3 into inequality 1, , 5., , multiply by Tc/Qh, , The inequality in step 4 is a concise statement of the entropy changes during an engine cycle: The increase in entropy of the cold reservoir, outweighs the decrease in entropy of the hot reservoir., Now we use the equation for engine efficiency to prove the inequality that describes the maximum engine efficiency., , Step, , Reason, , 6., , equation stated above, , 7., , add inequality 5 to equation 6, , 8., , Simplify, , The inequality derived in step 8 shows the maximum possible efficiency for an engine with given hot and cold reservoir temperatures. The, efficiency limit 1 í Tc/Th is maximized when the cold reservoir is as cold as possible, and the hot reservoir is as hot as possible. For instance,, an engine that has a cold reservoir of 300 K and a hot reservoir of 1000 K has an efficiency limit of 1 í 0.3, or 70%. One appeal of diesel, engines is that they run “hotter” than gasoline engines, which is a reason why diesel engines are more fuel-efficient., There are practical limits to engine efficiency. Reservoirs at 10 K and 10,000 K could make for an extremely efficient engine in principle, but it, is not possible to build a cost-efficient engine that would maintain reservoirs at these temperatures. Factors other than efficiency, such as cost, and the ability to supply power, also affect the design of engines., , 21.8 - Carnot cycle and efficiency, Sadi Carnot devised a theoretical engine cycle that is often discussed in, thermodynamics. The details of its working are less important than the conclusions that, he drew from it., Carnot proved that any fully reversible engine, like his, was the most efficient possible., In his argument, this French scientist showed that a more efficient engine cycle would, violate the second law of thermodynamics, which meant it was not possible to construct, such an engine. In an irreversible process, energy is lost from the system to its, environment in unrecoverable ways, such as through friction or sound energy. All real, engines operate irreversibly and are less efficient than a Carnot engine., Earlier, we showed that the maximum efficiency of a heat engine was less than or, equal to one minus the ratio of the temperature of the reservoirs. The closer an engine, cycle comes to having no increase in entropy, the closer its efficiency will be to this limit., , Carnot engine efficiency, , The Carnot cycle attains the theoretical maximum efficiency for any engine functioning, with reservoirs at two particular temperatures. It can be shown that the efficiency of this, cycle equals one minus the ratio of the temperature of the cold reservoir to the hot, reservoir. This important conclusion is shown in Equation 1., Even the ideal Carnot engine is not 100% efficient. To make a Carnot engine operate at, , 390, , Copyright 2007 Kinetic Books Co. Chapter 21
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100% efficiency, its cold reservoir would have to be at absolute zero (which is, theoretically impossible), or its hot reservoir would have to be infinitely hot., , ece = efficiency of Carnot engine, Tc = temperature of cold reservoir (K), Th = temperature of hot reservoir (K), , 21.9 - Otto cycle: internal combustion engine, The internal combustion engine shown above is used, in automobiles and is probably familiar to you. In this, section, we describe the operation and engine cycle, of this engine, and present an equation for its, efficiency. The cycle is called the Otto cycle, after the, German inventor of the “four-stroke” internal, combustion engine, Nikolaus August Otto. A “stroke”, is a movement in or out of a piston. In the complete, Otto cycle, a piston moves in and out twice, making, four strokes., In an internal combustion engine, there is no heat, Internal combustion engine., reservoir in the usual sense. Instead of heat flowing, into the engine, the working gas itself is the source of, energy. It is a mixture of air and gasoline vapor that is ignited inside a cylinder (hence the term “internal combustion”), and expands to drive a, piston that does work. The thermal energy that is used to do work is created by burning the gasoline mixture. You see the cylinder and piston in, the illustrations below., The four steps of the basic thermodynamic cycle together require two strokes of the piston. The steps occur rapidly. Two additional strokes, drive the combustion products out of the cylinder and replace the air and gasoline mixture for the next cycle. For simplicity, we combine these, two strokes into one “exhaust” step., Here are the steps in the Otto cycle., , Otto cycle (internal combustion), ·Air-fuel compressed adiabatically, , First: The piston compresses a mixture of air and gasoline vapor inside the cylinder adiabatically., , Otto cycle (internal combustion), ·Ignition: temperature and pressure, increase, , Second: A spark plug ignites the mixture. This is a constant-volume process: The temperature and pressure of the mixture both increase, but, the volume stays the same., , Copyright 2007 Kinetic Books Co. Chapter 21, , 391
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·Mixture expands, doing work, , Third: The gas expands adiabatically, moving the piston to do work., , ·“Exhaust” expelled, new air/gasoline, injected, , Fourth: Finally, there is an “exhaust” step involving both an up and a down stroke of the, piston where the combusted mixture in the cylinder is replaced through some valves by, a fresh mixture of air and gasoline at lower pressure and temperature. The piston, moves in and out, but it returns to its original position so this is considered a constantvolume process, as shown on the pressure-volume graph. The cycle then begins again., The theoretical efficiency of the engine can be computed as shown in Equation 1 on the, right. Vmax is the volume of the cylinder when the piston is “out”. That is the maximum, volume, and Vmin is the minimum volume. The ratio of the molar specific heats for the, air-fuel mixture CP/CV is indicated by Ȗ (the Greek letter gamma). The ratio Vmax/Vminis, called the compression ratio of the engine. The higher the compression ratio, the more, efficient the engine., In Example 1 on the right, you see an example problem that computes the theoretical, efficiency of a typical car engine. The actual efficiency of real car engines ranges from, 20-26%, less than the equation predicts, due to the heat expelled from the engine and, other factors., You may be familiar with octane ratings for gasoline. Typical octane ratings, as seen on, the pumps at gas stations, are in the range of 85 to 95. The octane rating of a gasoline, indicates the tendency of the gasoline-air mixture to self ignite when it is compressed., The higher the octane rating, the more compression the mixture can withstand before, combusting spontaneously. High compression gasoline engines require high octane fuel, to prevent “knocking” that occurs when the fuel mixture in the cylinder ignites before the, compression cycle is complete, creating a shockwave that collides with the piston., A diesel engine operates in a similar fashion to a gasoline engine, but does not require, a spark plug. Instead, air alone is compressed in the cylinder, causing it to reach very a, high temperature. Then, fuel is injected into the cylinder, where it is ignited due to the, high temperature. To create the high temperature needed to ignite the fuel, diesel, engines have higher compression ratios than typical gasoline engines. This means they, are more efficient than gasoline engines., , 392, , Otto cycle efficiency, , eice = efficiency of internal combustion, engine, Vmax = max volume of cylinder, Vmin = min volume of cylinder, Vmax/ Vmin is "compression ratio", Ȗ = CP/CV, ratio of molar specific heats, , Copyright 2007 Kinetic Books Co. Chapter 21
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The car engine has a, compression ratio of 8.0. The, molar specific heat ratio Ȗ is 1.4., What is the theoretical engine, efficiency?, , eice = 0.56 = 56%, 21.10 - Heat pumps, , Heat pump: A device that transfers heat from a, colder environment to a warmer one., It may sound a bit odd to speak of transferring heat from a colder environment (say the, outdoors in the winter) to a warmer one (the interior of a building), but that is what heat, pumps do: They pump heat opposite to the direction it would normally flow., Because heat pumps are more energy-efficient than typical furnaces, they are, becoming increasingly popular. An additional and very attractive benefit of heat pumps, is that in the summertime, they can be run in reverse to pump heat out of the cool, house into the warm outdoors, so the same appliance does double duty as a furnace, and as an air conditioner., A heat pump can be thought of as a heat engine run in reverse. With a heat engine,, heat flows from the hot reservoir to the engine and then to the cold reservoir, and work, is done by the engine. With the pump, heat flows the opposite direction, from the cold, reservoir to the hot, and work must be done on the engine to accomplish this, since, heat does not flow spontaneously from cold to hot., As mentioned above, a heat pump is in some ways analogous to an air conditioner (or a, refrigerator). In both cases, work is performed on the device, and it is used to transfer, heat in the direction opposite to the direction in which it spontaneously flows. However,, there is a difference between the two appliances. A heat pump moves heat from the, cold outdoors to a warmer indoors, with the purpose of further increasing the, temperature of the indoors. An air conditioner moves heat from indoors to outdoors,, with the purpose of further cooling the interior., , Heat pump, Work moves heat from cold to hot, reservoir, , Coefficient of performance, , COP = coefficient of performance, Qh = energy conveyed to hot reservoir, , Efficiency is analyzed somewhat differently with heat pumps than it is with heat engines., Recall that the maximum efficiency of a heat engine is determined by the ratio of the, W = work done on gas, cold and hot reservoir temperatures (e = 1 í Tc/Th). When the temperature difference, is large, the ratio is small and the engine can be more efficient. In contrast, as you can, see from Equation 2, heat pumps work best when there is little temperature difference. The greater the difference, the harder the pump has to, work to transfer energy from the cold to the hot reservoir. For this reason, heat pumps are rated by the coefficient of performance (COP). You, see the definition of the coefficient of performance in Equation 1 on the right: It is the heat transferred to the hot reservoir divided by the work, done on the gas. Both the heat transfer and the work done on the engine are treated as positive quantities., One way to think of the efficiency of a heat engine is as the ratio of “what you want” (work done) to “what you pay” (heat added). This applies to, the COP as well. The COP for a heat pump equals the ratio of “what you want” (heat flow) to “what you pay” (the work done on the gas). The, larger the COP, the more heat is transferred to the hot reservoir per unit of work done., The maximum COP for a heat pump is determined, as for a heat engine, by the temperatures of the hot and cold reservoirs. An inequality that, , Copyright 2007 Kinetic Books Co. Chapter 21, , 393
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limits the maximum COP is shown in Equation 2 on the right. The right side of the, inequality is the ratio of the hot reservoir temperature to the difference of the hot and, cold temperatures. A Carnot engine operating “in reverse” as a heat pump achieves the, maximum COP. That is, the equality holds., In Example 1, the maximum COP is calculated for a heat pump operating to warm a, house, with an indoor temperature of 21°C (about 70°F) and an outdoor temperature, of 13°C (about 55°F). The theoretical maximum COP is 37; the actual operating COP, would be less. Typical COPs for heat pumps range from about five to 10. A heat pump, with a COP of five would transfer five joules of thermal energy for every joule of work, supplied., A refrigerator acts something like a heat pump, but its purpose is to cool the cold, reservoir, the refrigerator’s interior. The coefficient of performance of refrigerators is, sometimes measured as Qc/W. Again, this is the ratio of “what you want” (cooling the, cold reservoir) to “what you pay” (work done). For refrigerators, COP values from four, to six are typical., , Maximum COP, , Th = hot reservoir temperature (K), Tc = cold reservoir temperature (K), , What is the maximum COP of a, heat pump operating between, these indoor and outdoor, temperatures?, , Th = 21 + 273.15 K = 294 K, Tc = 13 + 273.15 K = 286 K, , COPmax = 37, 21.11 - Sample problem: heat pump in action, For these temperatures, what is the, maximum COP for a heat pump? If, the COP of a heat pump is 5.48, and, 1250 J of work is done on it, how, much heat energy could it maximally, transfer inside from the outdoors?, , If you heat a house using an electric furnace, the electric energy supplied to the furnace is converted into thermal energy and transferred to the, air inside the house, increasing its temperature. If the furnace is supplied with 1250 joules of electrical energy, it can increase the energy of the, air by at most 1250 J. It will be interesting to contrast this value with the amount of heat that a heat pump can cause to flow into the house, , 394, , Copyright 2007 Kinetic Books Co. Chapter 21
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using the same amount of energy., Variables, temperature of cold reservoir, , Tc = 15.0°C, , temperature of hot reservoir, , Th = 19.0°C, , work done on heat pump, , W = 1250 J, , maximum COP of pump, , COPmax, , actual COP of pump, , COP = 5.48, , heat flowing to interior, , Qh, , What is the strategy?, 1., , Calculate the maximum coefficient of performance of the heat pump, , 2., , Use the definition of coefficient of performance with the actual COP to calculate the maximum amount of heat transferred to the interior., , Physics principles and equations, This inequality specifies the maximum possible COP of a heat pump with given temperature reservoirs:, , (Be sure to use Kelvin temperatures.), The definition of COP, , COP = Qh/W, Step-by-step derivation, We first calculate the maximum possible coefficient of performance for the heat pump., , Step, , Reason, , 1., , maximum COP, , 2., , substitute values, , 3., , COPmax = 73.0, , evaluate, , The maximum possible COP is over ten times higher than the actual (and more realistic) COP of 5.48 as stated in the problem. We now, calculate the heat transferred into the house, using this actual value for the coefficient of performance., , Step, , Reason, , 4., , COP = Qh/W, , definition of COP, , 5., , Qh = (COP)W, , solve for Qh, , 6., , Qh = (5.48)(1250 J) substitute values, , 7., , Qh = 6850 J, , evaluate, , The heat pump transfers significantly more heat than the furnace would. As you might expect, this does not violate the principle of conservation, of energy. The device “pumps” thermal energy from outdoors to indoors, further cooling the outdoors in order to warm the indoors., , 21.12 - Interactive summary problem: efficiency of an automobile engine, In the simulation to the right, you get to put an engine to work in a common situation. You are the driver of a car. You specify how much, gasoline you want the engine to consume. When you press GO, the car will accelerate at full engine power until the gasoline is gone, at which, point the simulation will stop and your calculations will be evaluated., Your goal is to specify the amount of gasoline that will allow you to reach a speed of 90.0 km/h. Here, we will consider the engine as supplying, an external force doing work on the car. The engine must do 9.35×10 5 J of work to accelerate the car to 90.0 km/h. This figure is a good, approximation of the amount of work a real engine must do to both increase the car’s kinetic energy to the desired amount and overcome, forces such as air resistance, friction, and so forth., To decide how much gasoline you need, you must first calculate the efficiency of the engine. You can do this by adding any amount of gasoline, and pressing GO to see the resulting heat added and heat expelled while the engine runs. Use these values to calculate the engine’s, , Copyright 2007 Kinetic Books Co. Chapter 21, , 395
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efficiency. Next, calculate how much heat needs to be added to the engine for it to perform the desired amount of work. The gasoline is the, source of the added heat. To find out how much gasoline to add, you need to know that one liter of gasoline generates 1.30×108 J of heat, when combusted in the engine., The correct amount of gasoline will be under a tenth of a liter. Calculate the amount, of gasoline needed to reach 90.0 km/h to three significant digits (to a ten-thousandth, of a liter), enter this value, and press GO to see the results. Press RESET to try, again., If you want, you could calculate the kinetic energy of the car at the final speed and, determine what percent of the work goes to overcoming resistive forces versus, increasing the KE of the car. (The mass of the car is 1280 kg.), If you have trouble getting the right answer, see the section on engine efficiency., , 21.13 - Gotchas, Using Celsius temperatures. All the equations in this chapter that contain a temperature factor require the temperature to be in Kelvin., My friend has an engine that is 101% efficient. This is not possible: It contradicts the second law of thermodynamics. It also contradicts the first, law. Making that claim probably breaks some FTC regulations too., If I clean my room, it will become more ordered and I will be breaking the second law of thermodynamics. Not to dissuade you from cleanliness,, but you are not breaking the law. Your room is not an isolated system; other parts of the system (like the molecules within your body) have, become more disordered., , 21.14 - Summary, The efficiency of a heat engine is the ratio of the net work it does divided by heat, added during an engine cycle. There are several equivalent statements of the, second law of thermodynamics. One states that no heat engine can transform 100%, of the thermal energy supplied to it during a cycle into work., , Efficiency, , e = W/Qh, , Entropy is a property of a system. Entropy increases as heat is transferred into the, system. The change in entropy can be calculated as the heat transferred during a, reversible process (one in which the system can be returned to its initial state, without additional energy) divided by the temperature. Another way to state the, second law is that in any isolated system í such as the universe í entropy never, decreases., The maximum efficiency possible for any heat engine depends on the ratio of the, temperatures of the cold and hot reservoirs. The greater the difference in, temperatures, the lower the ratio and the more efficient the engine can be. A Carnot, engine is a theoretical engine that achieves this maximum efficiency., The internal combustion engine in most automobiles utilizes the Otto cycle. The, efficiency of the internal combustion engine depends on the compression ratio, the, ratio of the maximum to minimum gas volume in the engine., Heat pumps can be used instead of furnaces to warm buildings. Unlike a heat, engine, a heat pump uses work to transfer heat from the cool reservoir (outdoors) to, the hot reservoir (the building interior). Just as we rated heat engines by their, efficiency, for a heat pump the analogous quantity is its coefficient of performance., A heat pump’s coefficient of performance increases when the difference in, temperature between the hot and cold reservoirs decreases., , 396, , Entropy, , In an isolated system, ǻS 0, Coefficient of performance, , Copyright 2007 Kinetic Books Co. Chapter 21
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Chapter 21 Problems, , Conceptual Problems, C.1, , An amount of gas in a well-insulated container is compressed reversibly and adiabatically to half its initial volume. Does the, entropy of the gas increase, decrease, or stay the same?, i. Increases, ii. Decreases, iii. Stays the same, , C.2, , A cup of hot coffee is placed inside a large, well-insulated container. Over a five minute period, the entropy of the cup, decreases by 100 J/K. What can you say about the entropy of the air surrounding the cup?, i., ii., iii., iv., v., , C.3, , It does not change, It increases less than 100 J/K, It increases by exactly 100 J/K, It increases by more than 100 J/K, It decreases, , A refrigerator operates much like a heat pump, doing work to transfer heat from its interior to its exterior. If you leave a, refrigerator running with its door open in the middle of a well-insulated room, does the room temperature go up, go down, or, remain unchanged? Explain., The room temperature, , i. goes up, ii. goes down, iii. is unchanged, , Section Problems, Section 1 - Efficiency, 1.1, , An engine running at 38.3% efficiency receives 287 J of heat from the hot reservoir during each cycle. How much heat is, transferred to the cold reservoir during each cycle?, J, , 1.2, , An engine does 189 J of net work during a cycle, using 329 J of heat in from the hot reservoir in the process. What is the, engine's percent efficiency?, %, , 1.3, , A two-stroke engine runs at 2300 rpm, which means it performs 2300 engine cycles each minute. If the engine does, 9.2×105 J of net work each minute, and has an efficiency of 0.43, how much heat is transferred into the engine during a, cycle?, J, , 1.4, , During one engine cycle, a lawnmower engine does 55.2 J of net work. In the process, it expels 153 J of energy as heat., What is the percent efficiency of the engine?, %, , Section 4 - Entropy, 4.1, , In a reversible process called isothermal expansion, 457 J of heat are transferred into a quantity of gas in an engine from the, hot reservoir, while 352 J of heat are transferred out to the cold reservoir. The temperature of the gas is 473 K at the, beginning and end of the process. What is the change in entropy of the gas?, J/K, , Section 5 - Second law of thermodynamics: entropy, 5.1, , Two large blocks are momentarily placed in contact. Block A has temperature 567 K and transfers 849 J of heat energy to, block B, which has temperature 288 K. The temperatures of the blocks are unchanged. What is the change in entropy of the, system consisting of the two blocks?, J/K, , Copyright 2007 Kinetic Books Co. Chapter 21 Problems, , 397
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5.2, , An ice cube of mass 0.018 kg, at the temperature of the freezing point of water, is surrounded by water at a temperature just, above the freezing point. What is the change in entropy of the ice cube as it melts? (The latent heat of fusion of water at the, freezing point is 3.34×105 J/kg.), J/K, , 5.3, , An ideal gas expands isothermally at 64.0°C in a reversible process, and its entropy increases 2.00 J/K. What amount of heat, is transferred to the gas in the process?, J, , Section 7 - Maximum engine efficiency and reservoir temperatures, 7.1, , An engine operates with a cold reservoir at 148°C. If the design goal is to achieve an operating efficiency of 42.4%, what is, the minimum temperature required for the hot reservoir, in degrees Celsius?, °C, , 7.2, , An engine, built by an advanced alien civilization, operates at the maximum possible efficiency for the temperatures of its hot, and cold reservoirs. The engine's efficiency is 37.2% and the difference between the temperatures of the two reservoirs is, 93.0 K. (a) What is the temperature of the hot reservoir? (b) What is the temperature of the cold reservoir?, (a) hot reservoir, , K, , (b) cold reservoir, , K, , Section 8 - Carnot cycle and efficiency, 8.1, , A Carnot engine operates with a cold reservoir temperature of 194 K and has an efficiency of 74.3%. (a) What is the, temperature of the hot reservoir? (b) If the engine does 8.99×104 J of net work during each cycle, how much heat does it, absorb from the hot reservoir during a cycle?, , 8.2, , (a), , K, , (b), , J, , A Carnot engine absorbs 4.65×103 J from the hot reservoir and expels 3.12×103 J to the cold reservoir during each engine, cycle. If the temperature of the cold reservoir is 489 K, what is the temperature of the hot reservoir?, K, , 8.3, , A Carnot engine operates with reservoirs at 135°C and 378°C, and 2.38e+4 J of heat are transferred into the engine from the, hot reservoir during each engine cycle. How much net work does the engine do per cycle?, J, , 8.4, , 8.5, , 8.6, , A Carnot engine operating with reservoirs at 459 K and 726 K has a power output of 7.35×104 W. (a) How much heat energy, does it absorb from the hot reservoir each hour? (b) How much heat energy does it expel to the cold reservoir each hour?, (a), , J, , (b), , J, , An engine removes 734 J from the hot reservoir, which is at 487 K, and expels 523 J of heat to the cold reservoir, which is at, 228 K. (a) What is the engine's percent efficiency? (b) What would the percent efficiency of a Carnot engine operating with, these reservoirs be?, (a), , %, , (b), , %, , An engine with reservoir temperatures of 387 K and 725 K operates at half the efficiency of a Carnot engine. If the hot, reservoir supplies the engine 4.72e+3 J of heat energy during each cycle, how much net work does the engine do in a cycle?, J, , Section 9 - Otto cycle: internal combustion engine, 9.1, , An internal combustion engine has a compression ratio of 4.50 and the ratio Ȗ of the molar specific heats of the air-fuel, mixture is 1.43. What is the maximum efficiency of the engine, as a percent, if it operates in an ideal Otto cycle?, %, , (a), 9.2, , An engineer is designing an internal combustion engine. The air-fuel mixture used in the engine has a Ȗ ratio of 1.37. Each, cylinder of the engine has a maximum volume of 4.12×10í4 m3. If the desired efficiency of the engine, operating in an ideal, Otto cycle, is 59.3%, what should the compressed volume of the air-fuel mixture in the cylinder be?, m3, , 9.3, , An internal combustion engine has a maximum ideal efficiency of 33.6%, based on its compression ratio and the, characteristics of the air-fuel mixture. (a) If the compression ratio is 6.45, what is the value of Ȗ for the gas? (b) To increase, , 398, , Copyright 2007 Kinetic Books Co. Chapter 21 Problems
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the efficiency to 46%, what should the value of Ȗ be?, (a), (b), 9.4, , In an internal combustion engine, the air-fuel mixture at the end of the ignition phase has volume 7.34×10í4 m3 and pressure, 3.12×106 Pa. The air-fuel mixture expands adiabatically, pushing the piston out, to a volume of 2.82×10í3 m3. Assume the airfuel mixture behaves like an ideal gas. If Ȗ = 1.42 for the air-fuel mixture, what is the final pressure after the expansion?, Pa, , Section 10 - Heat pumps, 10.1 A Carnot engine operated "in reverse" as a heat pump achieves the maximum coefficient of performance. A Carnot heat, pump operates to warm a house, where the outside temperature is 12°C and the inside temperature is 31°C. (a) What is its, COP? (b) How much work must be done on the heat pump for it to deliver 1400 J of heat energy to the interior of the house?, (a), (b), , J, , 10.2 A heat pump on a house operates with a COP that is 35% of the theoretical maximum, between exterior and interior, temperatures of í12°C and 34°C. How much work must be done on the heat engine to supply 2200 J of heat to the interior of, the house?, J, 10.3 A refrigerator operates much like a heat pump, but the desired goal of a refrigerator is to transfer heat from the cold reservoir, as efficiently as possible. For this reason, the coefficient of performance of a refrigerator is measured as Qc/W. A typical, commercial refrigerator has a coefficient of performance of about 5.5. Assume such a refrigerator is used by a market to make, ice cubes from 2.3 kg of liquid water, starting at 0°C. (a) How much work is done on the refrigerator to freeze this water into, ice? (The latent heat of fusion for the water is 3.34×105 J/kg.) (b) Suppose water costs 0.56 cents per kg, and energy costs 10, cents per kWh. If the market charges 99 cents for 2.3 kg of ice, what is its maximum profit on the ice? Express your answer to, the nearest cent., (a), , J, , (b), , cents, , Section 12 - Interactive summary problem: efficiency of an automobile engine, 12.1 Use the information given in the interactive problem in this section to calculate the correct amount of gasoline. Test your, answer using the simulation., L, , Copyright 2007 Kinetic Books Co. Chapter 21 Problems, , 399
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22.0 - Introduction, In this chapter, we begin the study of electricity and magnetism by discussing, electric charge and the electrostatic force. Although you cannot see the individual, charged particles, such as electrons and protons, that cause this force, you certainly, see its effects. Phenomena ranging from the annoying static cling in freshly, laundered clothing to the operation of laser printers are based on the electrostatic, force. In the sections that follow, we will cover the fundamentals of electric charge:, what it means to say that an object is charged and the nature of the forces created, by charged objects., We start with two simulations, shown to the right. The first allows you to experiment, with positively and negatively charged particles and see the forces they exert on, one another. The positively charged particles in this simulation have the same, charge as protons, and the negatively charged particles have the same charge as, electrons., After you launch this simulation, drag particles from the control panel onto the, screen above it. Once there, they will exert forces on each other. The amount and, direction of each force will be shown on the screen. You can drag particles closer, together or farther apart to see how the force they exert on one another relates to, the distance between them (the heavier grid lines are exactly one meter apart). If, you press GO, the particles, which all have equal mass, will be free to accelerate in, response to the forces they exert on each other. Electrostatics is the study of, electric charges at rest, so the simulation is also providing you with an extremely, informal introduction to the topic of electrodynamics, the study of charges in motion., As you use the simulation, take note of the direction of the forces between, say, two, negative or positive particles or between a positive and a negative particle. You can, also place two particles with the same charge next to each other, and see how the, force on a third particle changes. How the electric force changes with both the, distance between the charged particles í frequently just called “charges” í and the, amount of charge is a fundamental focus of this chapter., In the second simulation, you can play “proton golf”. The ball is positively charged,, and you add protons to the putter to make it positively charged as well. The protons, in the putter exert a force, called the electrostatic force, on the ball even when the, two are not touching. You can control both the location of the putter, by dragging it,, and how many protons it contains, by clicking the up- and down-arrows in the console. The moment you load protons into the putter they exert, a force on the ball, but the ball is locked in place until you press PUTT. The grass of the green supplies a frictional force that will cause the ball, to stop rolling., Your mission, as always in golf, is to sink the ball in the hole í in four or less shots, if you can! The important thing (in addition to having fun) is, to observe how the electrostatic force relates to the amount of charge and to distance. Be warned, though: Obstacles do exist! A clump of, protons acts as a hill that causes the ball to roll away from it, while a clump of electrons is a sand trap that will attract the ball. Fore!, , 22.1 - Electric charge, , Electric charge: A property, of the particles that make up, matter. It causes attraction, and repulsion., Electric charge is a property of matter that can cause, attraction and repulsion. In this section, we focus on, electrons and protons, and the role they play in, causing an object to have an electric charge., , This woman's hair is electrically charged. As you will see, the strands of her hair, , repel each other because each one of them carries a negative charge., An electron is defined as having a negative charge, and a proton is defined as having a positive charge. Charge is a scalar, not a vector. A negative charge is not less than zero, just the opposite, of positive. In this book we will represent negative charges as black and positive charges as red., , The amount of charge of an electron or proton is written as e and is called the elementary charge. An electron has a negative charge of íe and, a proton has a positive charge of +e. This amount of charge is the smallest amount that has been isolated. (Subatomic particles called quarks, have charges of +2e/3 or íe/3 but they have not been isolated.), , 400, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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The SI unit for charge is the coulomb. An electron or a proton has a charge of, magnitude e = 1.602 18×10í19 coulombs. This means approximately, 6,250,000,000,000,000,000 electrons or protons are required for a coulomb of charge to, be present. This is a vast number! However, numbers like this are often present in, nature: A bolt of lightning typically contains about 25 C of charge. To provide you with, another idea of the magnitude of a coulomb, approximately 0.8 C of charge flows, through a 100 watt light bulb every second., Some scientists, chemists in particular, use another unit, the esu or electrostatic unit., One esu equals 3.335 64×10í10 C., A small amount of matter contains a large number of electrons and protons. For, instance, a one-kilogram sample of copper contains about 2.75×1026 protons. When an, object has the same number of electrons and protons, it has no net charge and is said, to be electrically neutral., The addition or removal of electrons from an object causes it to become charged. A, negatively charged object has more electrons than protons and a positively charged, object has more protons than electrons. If the kilogram of copper has a charge of +0.1, C, which is a relatively large amount of charge, this means that about 0.000 002 % of its, electrons have been removed., , Electric charge, Property of particles that make up, matter, ·Electrons negative, protons positive, ·They have opposite amounts of charge, , Electron: q = í1.60×10í19 C, Proton: q = +1.60×10í19 C, q is symbol for charge, Units: coulombs (C), , How much charge do these five, electrons have?, qtotal = (5) (charge of 1 electron), electron charge = í1.60×10í19 C, qtotal = (5) (í1.60×10í19), qtotal = negative 8.00×10í19 C, , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 401
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22.2 - Creating charged objects, How does an object become electrically charged?, The answer is that the addition or removal of, electrons creates negatively and positively charged, objects. Except under extreme conditions, protons, stay in place and electrons move., A piece of silk and a glass rod can be used to, demonstrate one manner in which objects can, become charged. We will assume these two objects, start out electrically neutral. In other words, the silk, has equal numbers of protons and electrons, as does, the glass., A charged comb induces electric charges in the paper dots which, , You can transfer electrons from the glass to the silk, cause them to stick together. This phenomenon is called static cling., by rubbing the two materials together. This close, contact results in a net flow of electrons from the glass to the silk and causes the silk to, become negatively charged. It now contains more electrons than protons. In turn, the, glass becomes positively charged, since it now has fewer electrons than protons., You may wonder why rubbing silk and glass together causes them to become charged., The electrons move because the silk molecules have a greater affinity for electrons than, do the glass molecules. Rubbing the two materials together facilitates the transfer of, electrons by providing a greater level of contact between their molecules., The charging process can be reversed. When free to move, electrons will flow from a, negatively charged object to a positively charged one, reducing or ending a charge, imbalance. Lightning provides a dramatic example of such movement, a grand display, of excess electrons moving to a region that is less negatively charged. With lightning,, the electrons may be moving to a positively charged region of a cloud, or to an, electrically neutral region such as the surface of the Earth. Charges take advantage of, any opportunity to reduce an imbalance., , Creating charged objects, Neutral objects become charged by, movement of electrons, Excess electrons: negatively charged, Excess protons: positively charged, , q = ±Ne, q = charge, N = number of excess charges, ·protons positive, electrons negative, e = elementary charge, , 402, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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What is the net charge on the, disk?, N = 7 protons í 2 electrons, N = 5 protons, q = +Ne, q = (5)(1.60×10í19), q = 8.00×10í19 C, 22.3 - Conservation of charge, , Conservation of charge: The, net charge of an isolated, system of objects remains, constant., Electric charge is conserved. The net charge of an, isolated system may be positive, negative or neutral., Charge can move between objects in the system, but, the net charge of the system remains unchanged., , Although lightning transfers a very large amount of charge from, , a cloud to the ground, the total charge remains constant., To illustrate this principle, we again use the example, of a silk cloth and a glass rod to demonstrate how two different objects can become, charged while, at the same time, overall charge is conserved. Let’s assume that the, cloth and the rod are both neutral to begin with. They each become charged when, rubbed together, but their combined charge is unchanged: It remains zero, or neutral. It, is true that electrons have moved between the rod and the cloth, but to the extent that, one object is negative, the other is positive. The cloth and the rod constitute an isolated, system because all the charge moves solely between them and no charge leaves them., If charge flowed to a person holding these objects, that person would become part of, the system as well, and charge would still be conserved., , Early on, scientists such as Benjamin Franklin (yes, that Ben Franklin) suggested this, conservation principle based on experimental data and intuition, but he and his, colleagues were unable to show why it was so. The discovery of electrons showed why, the conjecture was true: It was the movement of electrons that created the charged, objects that Franklin observed. Under ordinary circumstances, these particles are, neither created nor destroyed, and Franklin observed the results of electrons flowing, from one object to another., , Conservation of charge, Charges can move but system’s net, charge is constant, , An object is charged when it has an imbalance of electrons and protons. Charge is said to be quantized: It is always observed as an integer, multiple of e, the magnitude of the charge of an electron or a proton., In extreme circumstances, charged particles can be destroyed. For example, when a positron (an exotic particle that is the mirror image of an, electron, identical in mass but opposite in charge) collides with an electron, the two will annihilate each other and produce gamma rays, a kind, of high-energy radiation. Does this scenario violate the conservation of charge? No, because gamma rays have no net charge. Before the, collision, the system of one electron and one positron has no net charge. After the collision, the system consists of neutral gamma radiation, so, charge is conserved., , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 403
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Is charge conserved in this, system?, Before: 2 excess protons, After: 2 excess protons, Net charge before = net charge after, Charge conserved!, 22.4 - Conductors, insulators, and grounds, , Conductor: An object or, material in which charge can, flow relatively freely., Insulator: An object or, material in which charge, does not flow freely., The lightning rod mounted atop this cupola intercepts, lightning and protects a building during electrical storms., , Ground: Charge flows from, a charged object to a ground, leaving the object, neutral., , You can easily find conductors and insulators (also called nonconductors) in your home, or classroom. If you examine an electrical cord, you will find that it consists of a, conducting core of copper wire surrounded by an insulator such as vinyl plastic., Charge can be moved relatively easily through a conductor such as copper using a, device like a battery. A battery will cause electrons to flow through a copper wire like the, one shown in Concept 1. In contrast, it is difficult to cause electrons to flow in insulators, like rubber or many plastics. This difference explains the design of electrical cords:, Often, they are made of copper wire wrapped with a flexible vinyl insulator so that, electrons flowing through the wire remain within the cord., Insulators do not allow charges to flow when they are subjected to only moderate, amounts of force. When great amounts of force are applied, charge can flow through an, insulator. There are also materials called semiconductors that enable charge to flow in, some circumstances, but not others. Given their role in devices like transistors, they are, an important topic, but lie outside the scope of this section., , Conductors and insulators, Conductor: charge moves freely, Insulator: charge does not readily flow, , A ground is a neutral object that can accept or supply an essentially unlimited number, of charges. The Earth functions as an electric ground. If you touch a conducting,, charged object to the ground, the object will also become electrically neutral í in other, words, grounded. Excess electrons will flow out of a negatively charged object to the, ground, and electrons will flow into a positively charged object from the ground., Charges move to a ground because charges of the same sign move as far away from, each other as possible due to their mutual repulsion. The ground distributes the excess, charge far enough away that it ceases to affect the object., Protecting houses from lightning presents engineers with the need to use conductors, and grounds. A building is not usually a conductor, but lightning can transform a house, into a reluctant conductor, with disastrous, highly flammable results. A lightning rod is a, conductor that protects houses and other structures by providing an easier, alternate, , 404, , Ground, Makes conductors electrically neutral, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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route to the ground. A conducting wire connects the rod to the Earth. The shape of the, rod also increases the likelihood that it will be the preferred target for lightning., , 22.5 - Interactive problem: charged rods, These interactive simulations are versions of a classic game. In the original version, of this game, you are given glasses, some filled with water and others empty. You, are shown or told a final configuration of glasses and water. Your challenge is to, start with the initial configuration, and by pouring water from one glass to another in, a sequence of steps, end with the specified final configuration. For instance, a, simple challenge would be to start with an empty glass, a half-full glass, and a, quarter-full glass, and end up with a three-quarters-full glass. By pouring the half-full, glass into the quarter-full glass, you achieve that goal., In the simulations to the right, the same overall idea applies to electric charge. You, are supplied with a configuration of charges on rods. Some of the rods have no, charge, some have positive charge, and some are negatively charged. All the rods, are the same size and are made of identical conducting material. In this game,, charge flows between rods instead of water flowing between glasses. Charge flows, until equilibrium is reached. For example, if you touch a rod with +4.000, microcoulombs of charge to a rod with no charge, both rods end up with +2.000, microcoulombs of charge., To play the game, click on any rod and drag it to another rod. When you release the, mouse button, charge will transfer between the rods., In the topmost game to the right, you are given a rod with a charge of positive, 10.000 ȝC, a rod with a charge of í3.000 ȝC, and several neutral rods. Your goal is, to produce a rod with a charge of +1.000 ȝC. This can be done in two moves. The, second game requires a greater number of moves and more planning. You can see, the initial configuration to the right. The challenge again is to create a rod with, +1.000 ȝC of charge. However skilled you are at these two games, the main point is, to observe how charge is conserved., Keep in mind that you do not have to be good at the games to practice the physics, you are learning in this chapter. Give it a try! Whether you get the minimum number, of moves or not, the simulations offer a chance to employ the principle of, conservation of charge., If you have any questions about the conservation of charge or about grounds,, review the preceding sections on these topics., , 22.6 - Electrostatic force, , Electrostatic force:, Attraction or repulsion due to, electric charge., Electrostatics is the study of electric forces between, charges at rest. If you have ever visited a science, museum, you may have seen people press their, hands against an electrically charged device, surmounted by a shiny metallic sphere, and then, watched in amazement as their hair stands on end., This device, called a Van de Graaff generator,, amusingly illustrates how electric charge creates a, repulsive force. In the photograph above you see the, spectacular display that can be created by such, forces in a large Van de Graaff generator, as its huge, electrostatic accumulation discharges through the, atmosphere. The Boston Museum of Science states, that it is home to the largest Van de Graaff generator in the world., , This Van de Graaff generator builds up an enormous, electrostatic charge that escapes into the surrounding air., , Clothes dryers provide a more mundane example of electrostatic forces at work. When your socks stick to your pants and then crackle as you, pull them apart, you are witnessing the static cling caused by electrostatic forces. In this case, electrostatic force is causing oppositely charged, pieces of clothing to attract each other. As the clothing is pulled apart, electric charges arc between the clothing items in an attempt to reach a, more balanced state. (Imagine: When you fold your laundry, you can both please your parents and review your physics studies. What a deal! ), The electric charge responsible for that annoying cling in a sock is typically in the range of a few microcoulombs., When objects have opposite charges, like laundry items or a glass rod and a silk cloth, they attract. When objects like the two balloons you see, , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 405
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to the right have the same charge, either positive or negative, they repel each other., The old cliché í opposites attract and likes repel í proves true in physics. When dealing, with issues of attraction and repulsion, it really is important to know your sign., Two charged objects exert equal but opposite forces on each other. In other words, if, they attract, they pull toward each other with the same force. If they repel, they push, against each other with equal force., The forces act along a straight line between the centers of the two charges. For, instance, if they attract, each force points directly toward the other charge, as illustrated, in Concept 2. If they repel, each force points directly away from the other charge., , Charged balloons repel each, other, Charged objects can attract or repel, , Electrostatic force, Opposites attract, Likes repel, , Will the balloons attract or repel?, Same charge, Balloons repel, 22.7 - Inducing an electric charge, , Inducing an electric charge: Creating a charged, object or region of an object without direct, contact., Objects can become electrically charged when they are put into contact with each other,, for example, by rubbing glass and silk together, or by touching a charged rod to a, neutral one. In this process electrons flow from one object to the other., Objects can also become charged without touching. Like gravity, electrostatic forces act, at a distance, so charges cause other charges to move without direct contact. When a, charged object, like the nonconducting sphere shown in Concept 1, is placed near a, neutral object in which electrons are free to move, such as the joined pair of conducting, metal rods to its right, the charged object causes electrons to move in the neutral, object. Charges in the rods, initially evenly distributed throughout the pair, end up in the, asymmetrical configuration you see in Concept 1. The rod pair as a whole is still, , 406, , Inducing an electrical charge, Creating charged objects without direct, contact, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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electrically neutral., To explain how charged objects can be created without direct contact, we use the, sphere and the pair of neutral rods just discussed. First, the negatively charged sphere, approaches the rods. As the sphere repels electrons in the rods, the closer end of the, rod combination becomes positively charged. As a result of the movement of electrons,, the far end of the rod combination becomes negatively charged. Two regions of charge, have been induced without contact by a charged object., , ·Charge on ball moves electrons in rod, pair to create positive, negative regions, , Next, the rods are separated. The closer one will remain positive and the farther one will, remain negative, even after the charged sphere has moved away. This example shows, how two objects can become charged without coming into direct contact with a third, charged object., , Inducing an electrical charge, ·Separating rods completes induction, , 22.8 - Coulomb’s law: calculating electrostatic forces, , Coulomb's law: The electrostatic force a, charged particle exerts on another is, proportional to the product of the charges and, inversely proportional to the square of the, distance between them., Named for Charles Augustin de Coulomb, the eighteenth century French physicist who, formulated it, this law quantifies the amount of force between charged particles. His law, is shown in Equation 1 to the right., The force is measured in newtons. The constant k in the equation has been, experimentally determined. It equals 8.987 55×109 N·m2/C2., The charges are shown with absolute value signs around them, so that two positive, values are multiplied together to calculate the amount of the force. To determine the, direction, the rule “opposites attract, likes repel” is used. Two opposite charges attract,, so both forces pull the charges together. Two like charges repel, so both forces push, the charges away from each other. Recalling Newton’s third law helps to insure the, correct result: The forces are always equal but opposite to each other., , Coulomb's law, Electrostatic force, ·Proportional to product of charges, ·Inversely proportional to distance, squared, , Coulomb’s law means that larger quantities of charge create more force and that the, force weakens with the square of the distance., Electrostatic forces can be added; they obey the principle of superposition. For, example, if there are three charges, the force exerted by two of the charges on the third, equals the vector sum of the forces exerted by each charge. This is similar to other, forces you have studied; if two people are pushing a car, the net force equals the vector, sum of the individual forces exerted by each person., In Coulomb’s law, r is the distance between two point charges, two infinitesimal sites of, charge. If charges are symmetrically distributed on each of two spheres, a principle, called the shell theorem can be used to show that all the charge on each sphere acts as, though it were located at the sphere’s center. In this case, the distance r is the distance, between the centers of the spheres., If you have studied gravity, you may notice that Coulomb’s law is similar to the equation, for calculating gravitational force. Both are inverse square laws: The forces are, inversely proportional to the distance squared. With Coulomb’s law, the force is, proportional to the product of the charges; with gravity, the force is proportional to the, product of the masses. Both are field forces, acting at a distance. Similarities like these, cause physicists to search for one unified explanation of gravitational and electrostatic, forces. Do remember, however, there is a crucial difference between the two forces:, Masses always attract, while electric charges can both attract and repel., , Coulomb’s law, , F = force, k = Coulomb’s constant, q = charge, r = distance between charges, , Constant k = 8.99×109 N·m2/C2, Sometimes Coulomb’s law is expressed in another fashion, using the permittivity, constant İ0. This traditional way of expressing the law can be particularly helpful in your, later studies. The equation expressed with the permittivity constant is also shown to the right, as Equation 2. The permittivity constant is related, to Coulomb’s constant by the equation İ0 = 1/4ʌk, and it equals 8.854 19×10í12 C2/N·m2., Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 407
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Coulomb’s law, permittivity, constant, , İ0 = permittivity constant, Constant İ0 = 8.85×10í12 C2/N·m2, , What is the magnitude and, direction of the force the right, charge exerts on the left charge?, , F = 18.0 N (to the left), 22.9 - Sample problem: electric vs. gravitational force, How many excess electrons must be, added to each neutral lead sphere to, balance the force of gravity between, them?, , The two balls above are floating in deep space, with the only significant gravitational forces acting upon them being the ones they exert upon, one another. If they were electrically neutral, they would drift slowly together due to these forces and, after about an hour, come to rest against, each other., You are asked to determine how many electrons should be added to each sphere so that the electrostatic force exactly counteracts the, gravitational force. You add the same number of electrons to each sphere, and disregard the change in mass of the spheres due to the added, electrons. It is negligible., , 408, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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Variables, sphere 1, , sphere 2, , mass of sphere, , m1 = 1.00 kg, , m2 = 1.00 kg, , charge of sphere, , q1, , q2, , gravitational constant, , G = 6.67×10í11 N·m2/kg2, , distance between spheres, , r = 1.00 m, , gravitational force on sphere, , Fg, , electric force on sphere, , Fe, , Coulomb’s constant, , k = 8.99×109 N·m2/C2, , number of excess electrons, , N, , elementary charge, , e = 1.60×10í19 C, , What is the strategy?, 1., , Calculate the gravitational attraction between the lead spheres., , 2., , Set the gravitational force equal to the repulsive electrostatic force between them and solve for the charge. The amount of charge on, each sphere is the same., , 3., , Convert the charge from coulombs to the equivalent number of excess electrons., , Physics principles and equations, We will use Newton’s law of gravitation, , together with Coulomb’s law, , The charge due to N excess electrons is, , q = íNe, Step-by-step solution, In the first steps, we calculate the charge q needed on the spheres to balance their gravitational attraction., , Step, , Reason, , 1., , Newton’s law of gravitation, , 2., , evaluate, , 3., , set forces equal and use Coulomb’s law, , 4., , solve for q, , 5., , note electron charge negative and evaluate, , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 409
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In the following steps we find the number of excess electrons., , Step, , 6., , q = –Ne, , Reason, , equation for charge, , 7., , solve for N, , 8., , evaluate, , Step 5 shows that a miniscule amount of charge í about a ten-thousandth of the charge you transfer to a balloon when you rub it on your shirt, í is enough to balance the gravitational attraction between two one-kilogram masses separated by one meter. If you were concerned about, whether adding the excess electrons would alter the mass of each sphere enough to require recalculating their gravitational attraction, you can, compute that they add an insignificant mass, about 5×10í22 kg, to each sphere., As an additional exercise, you can use Avogadro’s number, and the atomic weight and atomic number of lead, to find the total number of, electrons in an uncharged kilogram of lead. This calculation is not shown, but the total number is 2.39×1026 electrons. This means that the, excess electrons constitute about 10í16 percent of the electrons in the sphere., , 22.10 - Interactive checkpoint: electric vs. gravitational force, Compute the ratio of the electric to, the gravitational force between the, proton and electron in a hydrogen, atom. Use the average distance, between the two, which is called the, Bohr radius and equals 5.29×10í11 m., , Answer:, , FE / FG =, , 22.11 - Superposition of electrostatic forces, Electrostatic forces obey the principle of superposition. The forces caused by multiple, charges can be added as vectors. For instance, consider the charges shown in Concept, 1. To calculate the net force exerted by the other charges on the charge labeled q1, the, forces exerted by q2 and q3 on q1 are individually calculated and then those two forces, are added as vectors. In the next section, we solve a sample problem involving charges, that requires the use of this principle., You must be careful about the directions of electrostatic forces, especially when, combining forces that may point in opposite directions. The location and signs of the, charges determine the direction of the forces., For example, consider another set of charges, q4, q5, and q6, with all the charges on a, line, and q6 the rightmost charge. Let’s say charges q4 and q5 have opposite signs., Since both are on the same side of q6, the forces they exert upon it will act in opposite, directions. There will be cancellation and the net force will be less than the sum of the, magnitudes of the individual forces. There is no “rocket science” here. Just be careful to, consider the direction of forces before combining them., , 410, , Electrostatic forces: vectors, Electrostatic forces are vector quantities, Net force = vector sum, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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Forces F1, F2, …, Fn, , 22.12 - Physics at work: laser printers, When you print a document created on your, computer, you may use a laser printer. Laser printers, use the electrostatic force as a crucial part of the, process of transferring an image from the computer to, a sheet of paper. The process is known as, xerography or electrophotography., A key component of the laser printer is a rotating, metal cylinder or drum. This drum is coated with a, light-sensitive material called an organic, photoconductor, a carbon-based compound whose, electric properties change when it is exposed to light., In the dark it is an insulator, and electric charges cannot move through it. When it is, exposed to light, it becomes a conductor and charges can flow freely. The organic, photoconductor layer is on the outside surface of the drum. Inside it is a hollow metal, cylinder connected to a ground, which allows any charge trapped in a portion of the, photoconductor to drain away if that portion becomes conductive., The printing process begins with a charging step, shown on the right, where the drum is, given a uniform negative charge by bringing it into contact with a charged roller. Since, the drum is shielded from light in the interior of the printer, the photoconductive layer, acts as an insulator, trapping negative charges on its surface., In the imaging step, shown next on the right, a laser controlled by signals from the, computer directs light at the surface of the drum in a pattern corresponding to the image, to be printed. Areas of the drum exposed to the laser light become conductive, allowing, charge to escape from the drum surface to the ground. In this way the laser draws an, electrostatic image of the document on the surface of the drum. Areas of the drum, struck by the laser will be electrically neutral while the unexposed areas will retain their, negative charge., , Charging, Photosensitive drum negatively charged, , It is interesting to note that the laser itself is stationary. Its beam is projected onto the, drum by a series of movable lenses and mirrors controlled by the printer’s, microprocessor, using data from your computer., The next step is the development step. As the drum continues to rotate, it is brought, close to a container or roller furnished with a toner, consisting of fine particles of ink that, have a negative charge. The printer uses small pulses of electricity to eject the toner, onto the drum surface. The negatively charged toner particles are repelled from the, negatively charged unlit regions of the drum but cling to the neutral areas that were, struck by the laser. The drum surface now holds toner particles in the pattern of the, image to be printed., Then the image is transferred to paper. A sheet of paper is given a positive charge and, pressed against the drum. The negatively charged toner on the drum is attracted to the, positively charged paper. The toner is permanently fused to the paper by a heated roller, that melts the ink particles into the paper fiber. A final step, not shown on the right,, prepares the drum for the next image by flashing it with light, causing the complete, discharge of all charged areas on the drum., , Imaging, Pattern of laser light strikes drum, Exposed areas lose charge, , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 411
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Development, Toner sticks to neutral exposed areas, , Transfer, Positively charged paper captures toner, , 22.13 - Interactive summary problem: proton golf, Above and to the right, you see the 24th century, version of golf. Protons in the putter cause the proton, golf ball to move away. Coulomb’s law is well known,, well loved and well used., To sink the putt in the first game, your putter must, supply an initial force of 1.96×10í25 newtons. This will, cause the ball to be rolling slowly as it reaches the, hole, overcoming the force of friction due to the grass., If you apply too much force, the ball will fly over the, hole. The ball is free to move when you press PUTT., In this game, you cannot move the putter, and you get, only one stroke, but you can play again by pressing, RESET., , Proton Golf Association members at play in the year 2316., , The ball is initially 0.200 meters away from the putter and has a charge of +e. What, should the charge be in the putter? You set the amount of charge by specifying the, number of protons., After you calculate your answer, click on Interactive 1 to launch the simulation. You, use the up and down arrows to set the number of protons in the putter. Select your, value, press PUTT and the golf ball will roll toward the hole. If you need to review, how to calculate the repulsive force between two positive charges, see the section, on Coulomb’s law., The second game is like the game of golf you played at the beginning of this, chapter. You can change both the charge and the position of the putter. You will, almost certainly need several strokes to sink the ball. The challenge of the game is, to do so in as few strokes as you can. Again, the grass supplies a force of friction, that must be overcome. You may feel that you are now more familiar with charges, and the way they behave. Grab your prodigious proton putter and give it another try!, , 412, , Copyright 2000-2007 Kinetic Books Co. Chapter 22
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22.14 - Gotchas, A neutral object that gains electrons is negatively charged. Yes, a neutral object that gains electrons becomes negative; a neutral object that, loses electrons becomes positive., An object has a net charge of negative 10 coulombs. How many electrons does it contain? Do not bother trying to calculate a value. You could, calculate how many excess electrons the object contains í how many more electrons than protons í with the given information. But unless, someone tells you how many electrons the object contained when it was neutral, you cannot answer the question. The point here is: Charge, refers to the number of excess electrons. A neutral object has electrons, too, but they are balanced by an equal number of protons., You use the number of excess protons and electrons in Coulomb’s law. No, the charges q1 and q2 in Coulomb’s law are measured in (what, else?) coulombs. If you are given the number of excess electrons or protons in a problem, you must determine the electric charge in coulombs., I calculated a negative force from Coulomb’s law. Then you erred. The amount (magnitude) of the force is calculated by multiplying the, absolute values of the charges, so it will always be positive. The direction of the force will vary by sign: attraction when the signs are opposite,, repulsion when they are the same., , 22.15 - Summary, Electric charge is a property of matter. It occurs in positive and negative forms. One, unit of charge is e. A proton has a charge of +e and an electron has a charge of, íe. Charge, represented by q, is measured in coulombs (C). The elementary, charge e equals 1.602 18×10í19 C., , Coulomb’s law, , An ordinary object is charged when it has an imbalance of protons and electrons., Charge is always conserved. Though charges may be transferred from object to, object, charge cannot be created or destroyed, and the net charge of an isolated, system will remain the same., Electrons flow more freely in some objects than in others. Conductors allow electrons to move relatively easily, while insulators do not. A, ground can drain away any excess charge from a conducting object. The most common ground is literally the ground: Earth., Charged particles exert an electrostatic force on each other. Unlike gravity, which is always attractive, the electrostatic force can be either, attractive or repulsive. Opposite charges attract each other, while like charges repel., Coulomb’s Law describes the amount of the electrostatic force between two point charges. It is proportional to the product of the charges’, magnitudes and inversely proportional to the square of the distance between them., , Copyright 2000-2007 Kinetic Books Co. Chapter 22, , 413
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Chapter 22 Problems, , Conceptual Problems, C.1, , Two scientists are locked in a bitter dispute about the charge on a particularly famous particle of dust. Maria claims it has a, charge of 2.40×10í19 C. Richard disagrees; he thinks its charge is 3.20×10í19 C. Which scientist should you believe? Explain., Maria, , Richard, , C.2, , You are given an apple and an orange. The apple has a net charge of +3×10í17 C. The orange has a net charge of, í3×10í17 C. With only this information, can you determine which one has more total electrons? Explain., , C.3, , In a process known as beta decay, a neutron in an unstable atomic nucleus becomes a proton, in the process ejecting an, electron and an antineutrino. (a) Use conservation of charge to determine the charge of an antineutrino. (b) Sixty billion, neutrinos (mostly from the Sun) pass through every square centimeter on Earth every second. They are hardly noticeable due, to their negligible mass and weak interaction with matter. When a neutrino and an antineutrino collide, however, they, annihilate each other and produce two (electrically neutral) gamma rays traveling in opposite directions. What is the charge of, a neutrino?, , Yes, , C.4, , No, , (a), , C, , (b), , C, , The Sun generates most of its light through a series of nuclear reactions called the proton-proton chain. The overall process, of these nuclear reactions can be summarized as 4 1H ĺ 2He + 2Ȟ + 2Ȗ + 2? where Ȟ is a small neutral particle called a, neutrino, and Ȗ is a photon (a particle of light), which is also neutral. The pre-subscripts on the element symbols H (hydrogen), and He (helium) indicate the number of protons in the nucleus of each element. What is the charge of the unknown entity, "?" ?, +2e, , C.5, , +e, , 0, , íe, , í2e, , Under typical conditions in a safe home, classify the following materials as conductors or insulators: (a) rubber, (b) iron, (c), copper, and (d) wood., (a) Rubber is a(n), , i. conductor ., ii. insulator, (b) Iron is a(n), i. conductor ., ii. insulator, (c) Copper is a(n), i. conductor ., ii. insulator, (d) Wood is a(n), i. conductor ., ii. insulator, , C.6, , Vanessa has been walking around all day, and electrons that rubbed off from the carpet have caused her to acquire a net, negative charge. Explain why her hair might stand on end., , C.7, , A positively charged eraser is placed near the "0 cm" end of a 10 cm metal ruler. As a result of the induced charge effect,, which end of the ruler becomes positively charged: the "0 cm" end, or the "10 cm" end?, the "0 cm" end, the "10 cm" end, , C.8, , You are locked in a rubber room and given a pair of rubber gloves along with a positively charged bar of gold (marked with a, "+") and two electrically neutral bars of gold. You will be released if you can produce a negatively charged bar of gold, and, you get to keep the gold. Explain how you might accomplish this., , C.9, , Two unequally charged pellets are held apart at a fixed distance. The charge on one of the pellets is halved, and the charge, on the other is doubled. How this will affect the electric force one pellet exerts on the other? Explain., i., ii., iii., iv., v., , 414, , The force will quadruple., The force will double., The force does not change., The force will be halved., The force will be quartered., , Copyright 2007 Kinetic Books Co. Chapter 22 Problems
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C.10 A can of root beer and a can of cola are given different amounts of charge. Consider the cans to be point charges. By what, factor must their separation be changed so that the electric force on the root beer can is the same (a) if the charge on each of, the cans is doubled, and (b) if only the charge on the root beer is doubled?, (a) Multiply original separation by, , (b) Multiply original radius by, , i., ii., iii., iv., v., , i. .500, ii. 1.000, iii. 1.414, iv. 2.000, v. 4.000, ., .500, 1.000, 1.414, 2.000, 4.000, , ., , C.11 Consider an isolated system of three stationary nonzero charges. No other forces are present. (a) Is it possible that any of the, charges is experiencing zero net force? (b) If your answer to part "a" is yes, give an example. If no, explain why not., (a), , Yes, , No, , (b), , Section Problems, Section 0 - Introduction, 0.1, , Using the simulation in the first interactive problem in this section, answer the following questions. What is the direction of the, force between (a) two particles whose charges are the same sign and (b) two particles of opposite sign? (c) What happens to, the magnitude of the force when the distance between two particles is increased?, (a), (b), (c), , i., ii., i., ii., i., ii., iii., , Towards each other, Away from each other, Towards each other, Away from each other, Increases, Stays the same, Decreases, , Section 1 - Electric charge, 1.1, , The nucleus of a helium atom contains two protons, two neutrons, and no electrons. Neutrons have no net charge. What is, the charge of the nucleus?, C, , 1.2, , An electron has a mass of 9.11×10í31 kg. What is the charge of 1.00 grams of pure electrons?, , 1.3, , An initially neutral ball bearing is hooked up to a machine that transfers electrons onto it. Afterwards, the ball bearing is, , C, analyzed and its charge is í2.81×10í9 C. Find the mass increase due to the electron transfer. An electron has a mass of, 9.11×10í31 kg., kg, , Section 2 - Creating charged objects, 2.1, , Chlorine is used in the making of computers and blood bags, and is found in household bleach as well as in the skin of the, Ecuadorian tree frog. (a) A chlorine molecule (Cl2) has 34 protons and 34 electrons. What is its charge? (b) A chlorine ion, (Clí) has 17 protons and 18 electrons. What is its charge?, , 2.2, , (a), , C, , (b), , C, , The game of checkers has two players, black and red. Each player has 12 pieces. Suppose each red piece has a charge of +, 2.10eí15 coulomb and each black piece has a charge of í5.00×10í16 C. Find the net charge of the following systems: (a) a, black piece and a red piece, (b) two black pieces and a red piece, (c) all the pieces., (a), , C, , (b), , C, , (c), , C, , Copyright 2007 Kinetic Books Co. Chapter 22 Problems, , 415
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2.3, , An iron arrowhead has an initial charge of 3.35eí6 C. How many electrons are required to give it a charge of í2.82 μC?, electrons, , 2.4, , A capacitor is a device having two electrodes that can be used to store electric charge. When a battery is attached to the, capacitor, a positive charge moves to one electrode and an equal negative charge collects on the other. The capacitance of, this system is defined as the ratio of the magnitude of the charge on either electrode to the voltage of the attached battery., Capacitance is measured in farads (coulombs per volt). A 1.5 volt D-cell battery is attached to a 2.2 picofarad capacitor. (a), Find the charge that appears on the positive electrode. (b) Find the charge that appears on the negative electrode. (c) Find, the net charge of the entire capacitor., (a), , 2.5, , C, , (b), , C, , (c), , C, , A cloud that is about to unleash a lightning bolt has a charge of +24 coulombs. In stormy weather, powerful updrafts of air, carry electrons from one place in the storm system to another. If the cloud started out electrically neutral, how many electrons, were removed from it to give it this positive charge?, electrons, , Section 3 - Conservation of charge, 3.1, , In an experiment, a particle called a pion (ʌ) is observed to decay into two other particles, a muon and a neutrino. The muon, then decays into an electron and two more neutrinos. Neutrinos are electrically neutral. (a) What is the charge of a muon? (b), Pions come in three types: ʌ+ has a charge of +1.60×10í19, ʌ has a charge of í1.60×10í19, and ʌ0 is electrically neutral., What kind of pion could decay as described in this experiment?, , 3.2, , (a), , í2e, , (b), , ʌ+, , íe, , ʌ, , 0, , e, , 2e, , ʌ0, , In a complex circuit, three colored wires (red, green, and blue) are joined together inside a mysterious black box. An, ammeter, a device that measures the rate of charge flow (current), shows you that charge enters the box at 3 coulombs per, second through the red wire and leaves the box at 2 coulombs per second through the green wire. There is no buildup of, charge inside the box. After 10 seconds has elapsed, how much charge has flowed out of the box through the blue wire?, coulombs, , Section 4 - Conductors, insulators, and grounds, 4.1, , A nine-volt battery has two terminals that are 1.0 cm apart. The space between them is filled by air. How much charge flows, between the two terminals in 1.0 minutes?, C, , Section 5 - Interactive problem: charged rods, 5.1, , Using the simulation in the first interactive problem in this section, produce a rod with a charge of +1.000 μC in two turns., Explain the steps you took to achieve this., , 5.2, , Using the simulation in the second interactive problem in this section, produce a rod with a charge of +1.000 μC in five turns., Explain the steps you took to achieve this., , Section 6 - Electrostatic force, 6.1, , Three indistinguishable balloons are given charges of í 1.1 μC, í2.5 μC, and +2.0 μC respectively. You are given two of, them at random, and you observe that they repel each other. Find the total charge of the balloons you have been handed., C, , 6.2, , Danny has a pile of four metallic marbles. Each marble has a charge of either í0.2 μC or +0.4 μC. He observes that red, attracts blue, blue attracts green, and green attracts black. Red and black are brought in contact with each other so that they, have the same charge. Afterwards, Danny observes that red now repels blue. For each color of marble, use the radio buttons, below to tell the initial charge of the marble., Red, , +0.4 μC, , í0.2 μC, , Blue, , +0.4 μC, , í0.2 μC, , Green, Black, , 416, , +0.4 μC, +0.4 μC, , í0.2 μC, í0.2 μC, , Copyright 2007 Kinetic Books Co. Chapter 22 Problems
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Section 7 - Inducing an electric charge, 7.1, , A pack of 100-dollar bills is charged so that each bill has a charge of í0.0100 μC. The pack is suspended by an insulating, thread inside of a neutral safe made out of a conducting metal. No charge can flow between the bills and the safe, or the safe, and its surroundings. (a) Explain how the induced charge effect will change the distribution of charges on the safe. (b) A, burglar knows that there is no net electrical charge on the safe. If the charge on the outer surface has magnitude í4.00, what is the charge on the inner surface of the safe? (c) How much money is in the safe?, (a), (b), , μ C, , (c), , dollars, , μC,, , Section 8 - Coulomb’s law: calculating electrostatic forces, 8.1, , Two grapes are given equal charges and held apart at a distance of 1.3 m. They experience a repulsive force of 2.2 N. Find, the magnitude of the charge on each grape., C, , 8.2, , Two 2.0 kg plastic garbage cans are sitting 2.6 meters apart on a sticky classroom floor (coefficient of static friction, , μs = 0.40). They are not moving. If the first one has a charge of 10 μC, find (a) the most negative possible charge and (b) the, most positive possible charge for the other garbage can. Assume that the charges can be represented as point charges, located at the cans' centers., (a), , C, , (b), , C, , Section 9 - Sample problem: electric vs. gravitational force, 9.1, , The single electron and the single proton in a hydrogen atom in its lowest energy state are separated by a tiny distance called, the Bohr radius, 5.29×10í11 m. Both particles carry a charge of magnitude e = 1.60×10í19 C. (a) Is the force between them, attractive or repulsive? (b) What is the magnitude of the force?, (a), (b), , 9.2, , i. Attractive, ii. Repulsive, N, , Two steel juggling balls each carry a charge of 2.75 μC. There is a repulsive force between them of 1.45 N. What is the, distance between the centers of the two balls?, m, , 9.3, , Two balls of radius 2.00 mm have a separation between their centers of 5.33 cm. The same electric charge is placed on both, balls so that there is a repulsive force between them of 2.75 N. Assume that the charge is uniformly distributed over each ball., What is the magnitude of the charge?, C, , Section 11 - Superposition of electrostatic forces, 11.1 Goldilocks is visiting the Three Bears' house and she notices three bowls of porridge arranged on the table. Papa Bear's, porridge has a positive charge of +2.0 μC and is too hot. Mama Bear's porridge has a charge of í2.0 μC and is too cold. Baby, Bear's porridge is just right and has a charge of +1.0 μC. If the bowls are in an equilateral triangle (with sides 0.50 m) what is, the magnitude of the net force on Baby Bear's bowl?, N, 11.2 A doorframe is twice as tall as it is wide. There, is a positive charge on the top left corner and, an equal but negative charge in the top right, corner. What is the direction of the electric, force due to these charges on a negatively, charged dust mite in the bottom left corner of, the doorframe?, °, , 11.3 Particles P1 and P2 are located in three dimensional space at the points (2.00, í3.50, 1.75) mm and (í3.50, 2.25, í2.00) mm., These particles carry charges of +3.00 μC and +4.50 μC respectively. What is the magnitude of the net force they exert on, , Copyright 2007 Kinetic Books Co. Chapter 22 Problems, , 417
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another particle P3, with charge í3.50 μC, located at the origin?, N, 11.4 A positively charged ant is crawling through a plastic drinking straw with non-zero charges on both ends. One end is located, at x = 0 and the other is located at x = 10 cm. The ant experiences zero net electric force at x = 3.0 cm. (a) Can you say, whether the ends are positively or negatively charged? (b) Which end has a charge with greater magnitude? (c) Find the ratio, of the charge at the 0 cm end to the charge at the 10 cm end. Express your answer as a decimal number., (a), , Yes, , (b), , The 0 cm end, , No, The 10 cm end, , (c), 11.5 Two positively charged objects and two negatively charged objects are kept in the corners (NE, NW, SW, and SE) of a, square dorm room. The charge on each object has the same magnitude, Q, and no more than one charge can be kept in a, corner. Describe the configuration of charges if the net force on a positive test charge in the center of the room (a) is zero and, the charge in the NE corner is +Q, (b) points north, and (c) points west., (a), NW corner, SW corner, SE corner, (b), NE corner, NW corner, SW corner, SE corner, (c), NE corner, NW corner, SW corner, SE corner, , i., ii., i., ii., i., ii., , +Q, -Q, +Q, -Q, +Q, -Q, , i., ii., i., ii., i., ii., i., ii., , +Q, -Q, +Q, -Q, +Q, -Q, +Q, -Q, , i., ii., i., ii., i., ii., i., ii., , +Q, -Q, +Q, -Q, +Q, -Q, +Q, -Q, , 11.6 A bicycle wheel with radius 35 cm has 73 charges (1.0 μC each) spaced at regular intervals along its edge. A flea with charge, í0.010 μC lands in the center of the wheel. (a) What is the net electric force on the flea? (b) One of the charges is removed., Find the magnitude of the new net electric force on the flea., (a), , N, , (b), , N, , 11.7 A mad scientist is designing a trap for intruders that will lift them up into the air and hold them helpless. The device consists, of an equilateral triangle, 10.0 meters to a side, embedded in his floor. When he flips a switch, the corners of the triangle will, be charged equally by a generator and any negatively charged object above the center of the triangle will be lifted upwards by, the electric force. A computer-controlled system of giant fans keeps the intruder from straying horizontally from the center of, the triangle. While he is testing the system, his cat walks into the trap. She has a net charge of í1.00 nanocoulombs due to, electrons that rubbed off from the carpet. The cat, which has a mass of 5.00 kg, begins to hover 3.00 meters up in the air., Find the charge on each corner of the triangle., C, , Section 13 - Interactive summary problem: proton golf, 13.1 Using the information given in the first interactive problem in this section, how many protons should be loaded onto the putter, to sink the ball in the hole (without banking it off the wall)? Test your answer using the simulation., protons, , 418, , Copyright 2007 Kinetic Books Co. Chapter 22 Problems
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23.0 - Introduction, An electric charge can exert force on another charge at a distance. No direct, contact between the charges is required. The fact that the electrostatic,, gravitational, and magnetic forces act at a distance puzzled early scientists who, studied them. They speculated about the mechanism that would allow one body to, push or pull on another without touching it. To explain how this could occur, the, British scientist Michael Faraday (1791-1867) pioneered the concept of fields. As, time has passed, this concept has assumed an increasingly important role in, physics., Today, physicists say that an electric field surrounds an electric charge, and that the, electric field of one charge exerts a force on another charge. Investigating the, fundamentals of fields is the central topic of this chapter., To begin your study of fields, try the simulation to the right. Here, the charged, particle you see is creating an electric field, represented by the symbol E. The field, is invisible but you can observe it using a field meter., In essence, the field supplies a “road map” to calculate the force that will be exerted, on any second charge that enters the region surrounding the charged particle you see. The stronger the field at a point in space, the greater, the force that will be exerted on a given second charge when it is placed there. The field points in the same direction as the force that would be, exerted on a positive test charge., Locate your mouse pointer anywhere on the screen. When you click, you will see an electric field vector that points in the direction of the field,, and a readout that tells you the strength of the electric field at that point., The initial charge of the visible particle is 1.00 nC, but you can use the controller in the simulation to change this to other values, both positive, and negative., Initially, keep this particle’s charge constant in the simulation, and try to answer the following questions: How does the field strength vary with, the distance from the charge? Does it seem to increase linearly as you move closer, or do you see great increases in the field at points near, the charge? You should see similarities between the electric field and the force that would be exerted on a second, positive charge., Next, try changing the charge of the visible particle in the simulation. Does the amount of its charge affect its field strength at a given point? In, what direction does the field of the charged particle point when it is positive? Does the field point in the same direction if you give the source, particle a negative charge?, Using the simulation, you can experiment with some of the fundamentals of electric fields. In the rest of this chapter, you will continue your, exploration of this topic., , 23.1 - Electric fields, , Electric field: An electric, field describes the nature of, the electric force that a, charge will encounter at a, given location. Fields provide, the model for forces acting at, a distance., The muscles and heartbeats of fish generate telltale electric fields,, perceptible to organs in the wide head of this hammerhead shark., , Electrically charged objects exert forces on other, charged objects at a distance: The forces occur without direct contact. Charged objects exert forces on each other analogous to the, gravitational forces exerted by the Earth and the Moon on one another. Electric, gravitational and magnetic forces all act at a distance. To, describe their behavior, scientists have developed the concept of fields, which have become a fundamental tool for explaining the nature of the, universe., Charges establish electric fields around themselves. A small, positive test charge is often used to establish the nature of an electric field. A test, charge is assumed to be weak enough that it does not alter the field being analyzed. In diagrams, we represent a test charge as a white sphere, with a red plus sign., You see in Concept 1 a test charge placed in a field, which in this case is generated by a negatively charged particle. The direction of the field, is the direction of the force on the test charge. In the diagram, the field is represented as a vector and labeled with E, the symbol for an electric, field. This field exists at every point whether or not a test charge is present., As we explain the nature of electric fields, we first review the fundamentals of the electrostatic force. Coulomb’s law states that the force, , Copyright 2007 Kinetic Books Co. Chapter 23, , 419
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between two charges is proportional to their magnitudes and inversely proportional to, the square of the distance between them. Opposite charges attract one another, and, like charges repel., The strength of the field equals the amount of force divided by the magnitude of the test, charge. The equation for calculating the field in this fashion is shown in Equation 1., Since an electric field is used to describe the nature of the force that a test charge, experiences at a point in space, the field has a specific direction and magnitude at each, point. It is a vector quantity., The strength of the field is independent of the test charge. Dividing out the charge, cancels its effect in determining strength. For example, if the test charge were doubled, in strength, then Coulomb’s law states that the force would be twice as large, but after, dividing by the doubled charge, the field strength is the same. Electric fields are, measured in newtons per coulomb; there is no special name for this combination of, units., The stronger the field at a given point, the greater the force it will exert on any charge at, that point. You can perhaps envision this by considering the nature of the gravitational, field surrounding the Earth. The gravitational field is stronger near the surface of the, Earth than it is at locations farther away. An object with a particular mass will, experience more gravitational force closer to the Earth, where the field is stronger, than, farther away, where the field is weaker. The gravitational field is stronger on the surface, of Jupiter than it is on the Earth’s surface. The stronger field means you weigh more on, Jupiter because there is more gravitational force pulling on you., , Electric fields, Charged object exerts force at a, distance, Field surrounds charged object, Field equals force per unit charge, , In this section, we implicitly focused on fields produced by a single electric charge. This, is an important case, and provides a concrete example of how fields arise. However, as, you advance in your studies, you will often study fields and their effects by themselves,, with less focus on their sources., In the industrialized world, you are constantly surrounded by electric fields. You may, experience electric fields ranging in strength from nearly zero up to 10 N/C due to, electrical appliances and wiring as you walk around your house. Outside your house, you may experience significantly greater electric fields. The field at ground level directly, beneath a power transmission line is about 2000 N/C. This is enough to cause a, fluorescent tube to light up if you hold it vertically in the field. Within atoms, electric, fields are extremely large. The electric field due to a proton at the distance typical of an, electron in a hydrogen atom is 5×1011 N/C., , Electric field, E = F/qtest, E = strength of electric field, F = force on test charge, qtest = positive test charge, Direction of field: same as force, Units: newtons per coulomb (N/C), , What is the electric field at the, location of the test charge?, E = F/qtest, E = (1.2 N) / (1.5×10í6 C), E = 8.0×105 N/C, (same direction as force), , 420, , Copyright 2007 Kinetic Books Co. Chapter 23
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23.2 - Electric fields and Coulomb’s law, You can use Coulomb’s law to calculate the strength of the field around a point charge., First, calculate the amount of force exerted by that charge on a (positive) test charge, using Coulomb’s law. Then divide by the amount of the test charge, because the, strength of an electric field equals the amount of force divided by the charge used to, measure the force., We derive the equation for the field caused by a point charge as shown in Equation 1., We state the definition of the electric field, and then expand the expression for the force,, using Coulomb’s law. The quantity qtest appears in both the numerator and the, denominator of the resulting fraction, so it cancels out. As the equation states, the, strength of the field is proportional to the absolute value of q, the charge causing it,, divided by the square of the distance from this charge., An example problem is shown on the right, as well: You are asked to calculate the field, strength at a point 0.110 m away from a charge of positive 10.5 C., , Electric field, point charge, , Remember: Even after you calculate the strength of the field, your task is not finished,, because an electric field has both magnitude and direction. You must specify the field’s, direction, as we do in the example problem., , E = electric field, F = force, q = charge, qtest = test charge, k = Coulomb’s constant, r = distance between charges, , What are the magnitude and, direction of the electric field at, point A?, , E = 7.80×1012 N/C (away from charge), , Copyright 2007 Kinetic Books Co. Chapter 23, , 421
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23.3 - Interactive checkpoint: electric field, A point charge rests in an otherwise, charge-free region. Locations 1 and 2, are 0.0300 m and 0.0515 m from the, charge, respectively. If the electric, field at location 1 is 3.40×10í5 N/C,, what is the electric field at location 2?, , Answer:, , E2 =, , N/C, , 23.4 - Electric field diagrams, Electric field diagrams are used to illustrate the nature of an electric field. An electric, field diagram consists of electric field lines, as you see illustrated in Concept 1., The left-side diagram in Concept 1 shows the electric force acting on several test, charges at various points around a central positive charge. The vectors representing the, force on each test charge point away from the positive charge, indicating the direction of, the force. The length of each vector is proportional to the amount of force on each, charge., The field diagram, shown to the right of the force diagram, is a common and efficient, way to present similar information. The field lines have arrows pointing in the direction, of the field vector at each point, which is the same as the direction of the force that the, field would exert on a positive test charge at that point. However, the field lines do not, use length to represent magnitude. Rather, field strength is indicated by how close the, lines are to one another. Field strength is roughly proportional to the density of the lines., The illustration for Concept 2 shows how you can determine relative field strength from, a field diagram. We have drawn two boxes at different distances from a negative, charge. The field lines are closer together in the near box than they are in the far box,, which means that the field is stronger in the region defined by the box closer to the, negative charge., , Electric field diagrams, Lines show force on positive test charge, Away from “+” charge, toward “í”, charge, , Field line “density”, Proportional to field’s magnitude, , 422, , Copyright 2007 Kinetic Books Co. Chapter 23
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23.5 - Interactive problem: fields and forces, On the right are two simulations that allow you to explore electric field diagrams,, and the relationship between electric fields and the electrostatic force., In the first simulation, you explore the relationship between force and a field that is, created by two unequal charges: a charge of +90.0 nC and a charge of í45.0 nC. In, this simulation your task is to drag a positive test charge (+1.00 nC) into the field, and observe the direction and magnitude of the force that the field exerts on it., After experimenting with this simulation, consider the answers to three questions., First, is the force on the test charge greater where the field lines are closer together,, or where they are farther apart? Second, what is the relationship between the, direction of the force and the field lines? Hint: The word “tangent” must appear in, your answer. Third, what…is the air-speed velocity of an unladen swallow? (Just, kidding…But Monty Python fans know the consequences of a wrong answer.), The uniform field in the second simulation is not caused by a single charge, or even, by a small number of charges. The strength and direction of such a field are the, same at all points. The field in this simulation points to the right, and its strength, everywhere is 100 N/C., In the simulation, you are given three small charges, having values íq, +q, and, +2q. The negative charge is not, strictly speaking, a test charge, since test charges, have to be positive. However, all three are so small that they do not interact with, each other when they are placed in the uniform field., Drag the three charges into the field, and observe the direction and magnitude of, the force it causes on each one. How does the direction of the force exerted by the, field differ for a positive and a negative charge? Does it exert a greater force on the, +2q charge or on the +q charge? If so, by what factor is the force greater?, If you have trouble with the questions posed by either of these simulations, review, the introduction to electric fields and electric field diagrams in previous sections., , 23.6 - Electric fields caused by multiple charges, The individual fields from multiple charges can be combined to determine the overall, field they create. The net electric field at any point is the vector sum of the individual, fields due to each of the source charges. This additive property of electric fields is an, example of the principle of superposition., To calculate the overall field at a point, you start by calculating the field vector, generated by each charge at that point. You then add the vectors together. The result is, the net electric field at that point due to all the charges. In the illustration of Concept 1, you see a diagram of the combined field generated by two equal, positive charges. We, also show the combined electric field vectors at several points in the field. In the upper, right corner of the diagram, you can see that such a vector is tangent to a curving field, line., This method of combining fields is illustrated in detail in Concept 2 for one of the, combined field vectors. The individual field vectors caused by the two identical positive, charges at a location in space are depicted. We also draw the net field as the vector, sum of the two individual fields. At the location shown, the x components of the, contributing fields cancel out since they point in opposite directions and are of equal, magnitudes. The y components of these two fields point in the same direction. The, result is a combined field that points in the y direction, away from the two positive, charges., , At each location in a field, Enet is sum of fields of each charge, , Electric field of multiple charges, Calculate field vector due to each, charge, then sum vectors to determine, net field, , Copyright 2007 Kinetic Books Co. Chapter 23, , 423
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Electric field of multiple charges, , Enet = net electric field, Ei = electric field of a charge, 23.7 - Drawing field diagrams for multiple charges, Drawing the field diagram of a field generated by multiple charges requires the, application of three rules. To explain them, we use the configuration of two charges, shown in Concept 1. The negative charge is twice as strong as the positive charge., Rule 1: The number of field lines in a diagram that emanate from a positive charge, or, terminate at a negative charge, is proportional to the magnitude of the charge. In the, diagram to the right, 8 lines emanate from q, and 16 terminate at í2q., Rule 2: Other field lines may start from or extend to infinity. The “other” eight field lines, that terminate at the negative charge, the ones that do not emanate from the positive, charge, start at infinity. In other words, they enter from “outside” the diagram. If the, positive charge were +4q, there would be more field lines emanating from it than, terminating at the negative charge. The “extra” field lines would extend to infinity., Rule 3: Field lines never cross. A positive test charge located at such an intersection, would be “confused” as to what force it experienced, since the field would point in two, directions. This rule reflects an important point: At any location, a field has a single, direction and magnitude. This point is emphasized in the Concept 2 illustration., , Field lines, Incident lines proportional to charge, Start at “+” charges, end at “í” charges, ·Excess lines start or end at infinity, , These rules enable you to draw a correct field diagram. They provide a starting point for, drawing a field diagram like the one shown in Concept 1. Considering the nature of the, force experienced by a positive charge at various locations is required to flesh out the, diagram., A last point to keep in mind: Electric field diagrams are commonly drawn in two, dimensions, but electric fields exist in three dimensions. You usually see field diagrams, on flat surfaces, like a computer monitor, a piece of paper or a blackboard. However,, the field exists above and below the surface of the diagram as well., , Field lines, Do not cross, , 424, , Copyright 2007 Kinetic Books Co. Chapter 23
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23.8 - Describing the force exerted by an external electric field, In this section, we shift our attention from the field, caused by a charge to the force exerted on a charge, by an external electric field. Such a field could be, created by just one other charged particle, or it could, be created by a much larger number of charged, particles., Fields are used in many everyday devices. For, example, a field is used in some high-end inkjet, printers to control the direction of charged ink droplets, as they fly toward a sheet of paper. A charged, camera flash can store energy in a field created by, Polluting smokestacks. Electrostatic precipitators could electrically charge this, more than a trillion excess electrons. Given this huge, particulate matter and use electric fields to remove it from the effluent., number, determining the overall field by summing the, individual fields created by each electron would be a bit tedious. Instead, placing a test, charge in the field and observing the force that the field exerts on it can determine the, nature of the field., When a charged particle is in an electric field, the equation in Equation 1 can be used to, describe the amount of force that the field exerts on the particle. This equation comes, from the defining equation for the electric field, solved for the force., A point charge creates a field that diminishes with the distance from the charge. This, means the force it exerts on a test charge will vary by location. However, engineers are, clever enough to create nearly uniform fields, fields whose strength and direction is the, same at all locations within the field. For instance, there is a nearly uniform field in the, center of the region between two large, oppositely charged flat plates separated by a, small distance. In Example 1, you are asked to determine the force exerted on an, electron by such a field., Newton’s second law enables you to determine the acceleration of the electron in this, electric field and we do so in Example 2. The force exerted by the field is divided by the, mass of an electron to determine the acceleration. The result is a large acceleration,, having a magnitude of approximately 350,000 m/s2., This acceleration is caused by a relatively weak field (it is well below the maximum, allowed for human exposure by government safety standards). At the acceleration, stated, the speed of the electron would quickly approach that of light, and the, acceleration would have to diminish due to the relativistic effects predicted by Albert, Einstein., , The force due to an electric field, F = qE, F = force, q = charge, E = electric field, , What is the force on the, electron?, F = qE, F = (í1.6×10í19 C)(2.0×10í6 N/C), F = í3.2×10í25 N (opposite to field), , Copyright 2007 Kinetic Books Co. Chapter 23, , 425
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What is the acceleration of the, electron? Its mass is 9.1×10í31, kg., a = F/m, m = me = 9.1×10í31 kg, a = (í3.2×10í25 N)/(9.1×10í31 kg), a = í3.5×105 m/s2 (opposite to field), 23.9 - Interactive checkpoint: work on a charge, A uniform electric field of strength, 3.00×105 N/C is oriented in the íx, direction. How much work is done by, the field to move an electron 2.60 m, in the +x direction?, , Answer:, , W=, , J, , 23.10 - Physics at work: spacecraft powered by electric fields, In 1998, NASA began testing in space a new type of, rocket drive: the ion propulsion system. A conceptual, diagram (not drawn to scale) of this drive is shown in, Concept 1. Traditional rocket engines use chemical, reactions to expel rocket fuel exhaust from the engine, at high speed. If you have seen video footage of the, rockets that carried astronauts to the Moon, you have, seen a traditional rocket engine., In contrast, the ion drive uses an electric field to, accelerate charged gas particles, or ions. It, accelerates these particles to extremely high, NASA's Deep Space 1 ion drive undergoing testing, velocities, velocities much greater than those, at the Jet Propulsion Laboratory., achieved by chemical rocket exhausts. Why are, higher velocities of interest? Recall the law of conservation of momentum. In this case, the “forward” momentum of the rest of the rocket must, match the “backward” momentum of the rocket exhaust. Momentum is the product of mass and velocity. Since the ion drive accelerates, particles to greater velocities than those of chemical drives, less fuel mass is necessary than with chemical systems to produce the same, amount of momentum., Less massive space probes are important because they require less energy to launch into space. In practice, a space probe is still launched by, a powerful chemical rocket engine, but once it is relatively free of the Earth’s gravity, the probe relies on a less massive ion propulsion system, to maneuver. (Ion systems supply less than one newton of force, a small fraction of the force required to launch a rocket from the Earth’s, surface.) The system’s lesser mass also means it accelerates more for a given force., The first step in the operation of this engine system is to bombard xenon gas with electrons. These fast electrons “knock off” an electron from, each xenon atom, turning it into a positive ion. The ions are then inserted into an electric field created by two charged plates. The plates are, less than a meter in diameter and are placed close together. The electric field between the plates accelerates the ions to a velocity of about, , 426, , Copyright 2007 Kinetic Books Co. Chapter 23
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30,000 m/s (70,000 mi/h). The ions escape through small holes in the rear plate. (To, prevent the ions from being attracted back into the engine, a beam of electrons is fired, into the ion exhaust to neutralize it.), NASA tested this engine system on the Deep Space 1 mission. Deep Space 1 was shut, down on December 18, 2001, after successfully validating the performance of the ion, propulsion system. The drive had provided more than 16,000 hours of thrust, using only, 72 kg of xenon gas. This was less than 10% of the overall mass of the spacecraft. An, equivalent chemical rocket drive would have needed 720 kg of fuel to generate the, same amount of propulsion. This is equivalent to the difference between launching a, person and launching a small car., The ion propulsion system is now in use in satellites that orbit the Earth, where it is, used to “tune” their orbits. Photovoltaic panels supply the energy required to ionize the, gas and charge the plates., , An electric-field rocket engine, Electron bombardment ionizes xenon, gas, Field accelerates ions to 30,000 m/s, Low mass, high velocity system, , 23.11 - Conductors in electrostatic equilibrium, , Electrostatic equilibrium: In, an isolated conductor, excess, charges quickly achieve a, state where there is no net, motion of charge., An isolated, charged conducting object has several, interesting and perhaps unexpected properties. (By, isolated, we mean the object is attached neither to a, source of what is called a potential difference, such, as a battery, nor to a ground.) The excess charge on, such an object will rapidly reach a state called, electrostatic equilibrium, which means there is no, further net motion of the charge. In other words, any, Charge resides on the surface of the inner metallic sphere in this toy., excess charges in the object can be treated as, An electric field extends radially outward, made visible by ionized gas., stationary. Many experiments have confirmed the, existence of electrostatic equilibrium and the properties of conductors described below., What are some properties of a charged conductor in electrostatic equilibrium? First, it, creates an electric field that, just outside the conductor’s surface, is perpendicular to the, surface. You see this in the diagram for Concept 1, which shows a solid conducting, sphere., Why is the field perpendicular? This question can be answered by asking another, question: What if the field were not perpendicular? If there were a component of the, field parallel to the surface, this would mean a charged particle there would have a force, exerted on it by the field, parallel to the surface. This force would cause the particle to, move along the surface, and this motion would contradict the assumption of, electrostatic equilibrium., Second, all excess charge resides on the surface of a conducting object. One way to, explain this property is to note that such a configuration allows like charges to maximize, their distance from one another. If a charge remained in the middle of the conductor, it, would not be maximizing its distance from its fellows., , Electrostatic equilibrium, Excess charge moves to surface, Electric field perpendicular to surface, No field inside material of conductor, , Third, there is no electric field within the bulk of a conductor. Again, consider the, conducting sphere in Concept 1. If there were a field within the sphere, it would cause, movement of excess charges there, meaning the sphere would not be in electrostatic equilibrium., In Concept 2, we show what occurs when a hollow conductor is placed in an external field. In this case, two charged plates cause the field. If, the sphere were not present, the electric field would be uniform, and a field diagram would represent the field with equally spaced, horizontal, lines., When the sphere is placed in the field caused by the plates, an asymmetrical distribution of charge is induced on the outside of the sphere. You, can see the charge distribution and how the sphere alters the external field in the diagram., There is another important point illustrated in the diagram: There is no field within the hollow conductor. This effect is quite useful because it, means a conducting shell will insulate its contents from electric fields. When you see electric or electronic circuits placed within a protective, metal box, one reason is to shield them from nearby electric fields that might distort the circuitry’s operation. The shielding effect occurs only, , Copyright 2007 Kinetic Books Co. Chapter 23, , 427
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with conductors. It cannot occur with insulators because charge is not free to move, within such substances., , Hollow conductor, Shields contents from external fields, , 23.12 - Electric dipoles, , Electric dipole: Two point, charges with equal strength, but opposite signs separated, by a fixed distance., A configuration of two charges that are equal in, strength but opposite in sign is important enough to, merit its own name: the electric dipole. The two, charges are at a fixed distance from each other., , The world's oceans contain about 7.8x1046 electric dipoles: water molecules., , The diagram in Concept 1 shows the basics of an, electric dipole. The equal and opposite charges have magnitude q and are separated, by the distance d. An electric dipole can be described by its electric dipole moment, p = qd, where d is the displacement vector from the negative charge to the positive, charge. The dipole moment vector points in the same direction as the displacement, vector., When an electric dipole is placed in an external electric field, the positive end, experiences a force in the direction of the external field, and the negative end, experiences a force in the opposite direction. As the diagram in Concept 2 shows, if the, dipole moment is not aligned with the field, the dipole will experience a torque that, rotates it into the direction of alignment with the external electric field., One reason why electric dipoles are important is that the individual molecules in what is, arguably the most vital substance for human existence í water í behave as electric, dipoles. The oxygen side of a water molecule tends to be negatively charged, while the, hydrogen side tends to have an equal but opposite positive charge. The result is a, dipole. The fact that water molecules are dipoles accounts for many of water’s crucial, properties. The illustration below shows a conceptual model of a water molecule, together with its electric dipole moment., , A convenient device, the microwave oven, takes advantage of the dipole nature of, water molecules. The oven uses an oscillating electric field to cause water molecules to, rotate back and forth, increasing their energy and the thermal energy of the food that, contains them., , Electric dipole, Two point charges, ·Separated by fixed distance, ·Equal strength, opposite sign, Dipole moment vector p, ·points from “í” to “+” charge, , Electric dipole in an external, field, Moment tends to align with external field, , 428, , Copyright 2007 Kinetic Books Co. Chapter 23
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Dipole moment and electric field, p = qd, p = dipole moment, q = charge, d = displacement, “í” to “+” charge, 23.13 - Gotchas, I increased the amount of the charge of a test charge, and the force exerted on it changed. This means the field I was assessing must have, changed. No. Fields are independent of the test charge. The force increased, in proportion to the test charge, but the original field strength did, not change. The increased force was a result of the increased magnitude of the test charge., Electric field lines are the same as electric field vectors. No. The direction of the field is tangent to the field lines at any point. In a field diagram, for a point charge, the lines are straight so they do point in the direction of the field. But in many fields the field lines are curved, and the field, vector at any location on a field line is tangent to but distinct from it. Also, field lines and vectors use different conventions for representing field, strength. With vectors, the length is proportional to the strength of the field. In a field diagram, the field lines are closer together in a region with, greater field strength. Although a field diagram is used to represent an electric field, field lines and field vectors are not the same thing., , 23.14 - Summary, An electric field is a vector quantity defined throughout a region of three-dimensional, space. It describes the forces that will be experienced by an electric charge if it is, placed at various locations in the field. If a positive charge is introduced, the force, on it equals the field strength times its own charge magnitude, in the direction of the, field. When a negative charge is introduced, the force on it is opposite to the, direction of the field. The field strength is measured in newtons per coulomb (N/C)., A test charge is a small positive charge used to measure a field., , Electric field, , E = F/qtest, Field of a point charge, , Point charges generate electric fields. At any distance from the charge, the strength, of the field is proportional to the magnitude of the charge and inversely proportional, to the square of the distance from the charge., An electric field diagram is a convenient means of representing the direction and, strength of an electric field in a region. The direction of a field line represents the, local direction of the electric field. The strength of the field at a particular location is, indicated by the proximity of the field lines to each other around that location., , Force due to an electric field, , F = qE, , In an isolated, charged conductor the excess charges distribute themselves on the conductor’s surface in a state of electrostatic equilibrium., There is no net motion of these charges. The electric field extends perpendicularly outward from all points on the conductor’s surface, and, there is no field inside the conductor., An electric dipole consists of a positive and a negative charge of equal magnitude, separated by a fixed distance. A vector called the dipole, moment p points from the negative to the positive charge. Its magnitude is equal to the magnitude of either dipole charge times the distance, between the charges. When the dipole is placed in an external electric field, the dipole moment experiences a torque tending to align it in a, direction parallel to the field., , Copyright 2007 Kinetic Books Co. Chapter 23, , 429
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Chapter 23 Problems, , Conceptual Problems, C.1, , Three locations, 1, 2, and 3, are labeled in the, electric field diagram. At which of these, locations is the strength of the field the, greatest?, i., ii., iii., iv., , C.2, , Suppose a charged particle is placed in the, field whose diagram is shown here. (a) If a, positively charged particle is placed at point 1,, in what direction will it accelerate? (b) If a, negatively charged particle is placed at point, 3, in what direction will it accelerate? (c) At, which of the three locations labeled in the, diagram should a charged particle be placed, so that it experiences the greatest, acceleration?, (a), (b), (c), , C.3, , i., ii., i., ii., i., ii., iii., , In the direction of the field, Opposite the direction of the field, In the direction of the field, Opposite the direction of the field, 1, 2, 3, , The electric field created by a uniformly, charged infinite plane is uniform in strength, and perpendicular to the plane. Charged, planes A and B are identical, except that A, has a positive charge and B has the same, magnitude of negative charge. Plane A is, 4.0 cm to the left of plane B, and the two, planes are parallel. (a) What direction is the, combined field to the left of A? (b) Halfway, between A and B? (c) To the right of B?, (a), , (b), , (c), , 430, , 1, 2, 3, All three same strength, , i., ii., iii., i., ii., iii., i., ii., iii., , To the right, To the left, The combined field is zero, To the right, To the left, The combined field is zero, To the right, To the left, The combined field is zero, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems
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C.4, , In the diagram, two charges of +2q and í5q, are placed on a line. (a) There is a point on, the line where the strength of the electric field, due to the two charges is zero. Describe, where the point is, relative to the positions of, the two charges. (b) Is there any point not on, the line, where the strength of the electric field, is zero?, (a), , i., ii., iii., iv., v., , To the left of +2q, At +2q, Between +2q and -5q, At -5q, To the right of -5q, , (b), C.5, , A neutron is a particle with the same mass as a proton, but no electric charge. An alpha particle consists of two protons plus, two neutrons. A proton and an alpha particle decide to have a "drag race" starting from zero velocity in a certain uniform, electric field E. Who wins? Why?, Proton, , Alpha particle, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following question. If the you keep the particle's, charge constant, does the field strength increase linearly as you move closer to the charge, or does it greatly increase at, points near the charge?, Increases linearly, Greatly increases, , 0.2, , Use the simulation in the interactive problem in this section to answer the following question. (a) If you increase the, magnitude of the source charge, does the field strength at a given point increase, stay the same, or decrease? (b) In what, direction does the field point when the source charge is positive? (c) At a given point, does the field point in the same direction, for a negative source charge as it does for a positive source charge?, (a), , (b), (c), , i., ii., iii., i., ii., , Increase, Stay the same, Decrease, Towards the source charge, Away from the source charge, , Yes, , No, , Section 1 - Electric fields, 1.1, , An electric field has a value of 350 N/C at a particular point in space, directed along the x axis. What is the force this field, exerts at this point on a particle of charge (a) 5.00 μC? (b) í8.25 mC? (c) í525 C? (d) 5.34×10í15 C? Make sure to check the, sign of your answer., (a), , 1.2, , N, , (b), , N, , (c), , N, , (d), , N, , A test charge of 2.50 μC is placed at various points in an electric field where it experiences forces in the positive direction of, 25.0 μN at point A, 250 μN at point B, and 2.50 mN at point C. (a) What is the electric field strength at those points? (b) What, forces would the field exert on a particle with charge í2.50 μC at the same points? (c) What forces would the field exert on an, electrically neutral particle at the same points?, (a) A, , N/C B, , N/C C, , N/C, , (b) A, , NB, , NC, , N, , (c) A, , NB, , NC, , N, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems, , 431
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1.3, , The electric field in the copper wire of a household circuit is oriented along the wire, so that it pushes electrons through the, circuit. (In alternating current, the direction of this electric field reverses 120 times a second.) A typical value for the field, strength is 0.00850 N/C. The charge of a mobile electron in the wire is í1.60×10í19 C. What is the magnitude of the force that, the field exerts on the electron?, N, , 1.4, , A hard rubber comb can be given a negative electric charge by rubbing it against a variety of materials, including human hair., Suppose an atmospheric ion having a charge of +1.60×10í19 C is resting at a point near the comb where the strength of the, field is 973 N/C. (a) What is the magnitude of the force that the field exerts on the ion? (b) Will the ion move toward the comb, or away from the comb?, (a), (b), , 1.5, , N, i. Toward the comb, ii. Away from the comb, , The surface of the drum of a laser printer is negatively charged, except in certain "print black" regions, such as letterforms,, that have been discharged by a laser beam. Negatively charged ink particles carrying a charge of 5.0 mC cling to the "print, black" areas but are repelled by an electric field elsewhere on the drum whose strength at the surface equals 120,000 N/C., What is the magnitude of the repelling force felt by these particles?, N, , Section 2 - Electric fields and Coulomb’s law, 2.1, , Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,, and declares he has a field strength of 23.5 N/C. The second friend looks at her meter and realizes she is three times as far, away. What field strength does the second friend's meter read?, N/C, , 2.2, , An ion harvester is scouring deep space for isolated xenon (Xe+) ions. A xenon ion is singly ionized, and has a charge of +e., If the field meter reads 2.93eí7 N/C, how far away is it from the ion?, , 2.3, , In a hydrogen atom in its lowest energy state, the single electron and the single proton are separated by a tiny distance called, , m, the Bohr radius, 5.29×10í11 m. The proton carries a charge e = 1.60×10í19 C. (a) What is the strength of the electric field, generated by the proton at the Bohr-radius distance? (b) In what direction does the field point?, (a), (b), 2.4, , N/C, i. Towards the proton, ii. Away from the proton, , A solid conducting sphere with a radius of 35.4 cm contains a total charge of 5.46 mC, evenly distributed over its surface. (a), What is the direction of the electric field at its surface? (b) What is the strength of the electric field at its surface? Hint: use the, shell theorem, which states that when calculating an electrostatic force or field outside a charged sphere, the sphere can be, treated as though all of its charge resides at its center., (a), , (b), , i. Tangent to the surface, ii. Radially away from the center, iii. Radially toward the center, N/C, , Section 5 - Interactive problem: fields and forces, 5.1, , Use the simulation in the first interactive problem in this section to answer the following questions. (a) Is the force on the test, charge greater where the field lines are closer together, or where they are farther apart? (b) What is the relationship between, the direction of the force and the field lines?, (a), , (b), 5.2, , Closer together, Farther apart, The force is the same all over, The direction of the force is perpendicular to the field lines, The direction of the force is tangent to the field lines, , Use the simulation in the second interactive problem in this section to answer the following questions. (a) At a particular, location, how does the direction of the force exerted by the field on a positive charge differ from the force direction on a, negative charge? (b) How many times larger is the force on the +2q charge than on the +q charge?, (a), , 432, , i., ii., iii., i., ii., , i. Directions are the same, ii. Directions are opposite, iii. Directions are perpendicular to each other, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems
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(b), , i., ii., iii., iv., , They are equal, Twice as large, Four times as large, Half as large, , Section 6 - Electric fields caused by multiple charges, 6.1, , Two charges, q1 = í3.00 mC and q2 = 4.50 mC, are located at í2.00 cm and 3.00 cm on the x axis, respectively. What are, the magnitude and direction of the resulting field at (a) x = í3.00 cm? (b) x = 0 cm? (c) x = 5.00 cm?, , 6.2, , (a), , N/C,, , (b), , N/C,, , (c), , N/C,, , i., ii., i., ii., i., ii., , to the right, to the left, to the right, to the left, to the right, to the left, , A particle with charge +7.88 μC is placed at the fixed position x = 3.00 m in an electric field of uniform strength 300 N/C,, directed in the positive x direction. Find the position on the x axis where the electric field strength of the resulting configuration, is zero., m, , 6.3, , A particle with a charge of + 4.95eí3 C is placed on the y axis at y = 3.50 cm. Calculate the charge on a second particle,, placed at y = í2.00 m, that will result in a field whose strength at the origin is zero., C, , Section 7 - Drawing field diagrams for multiple charges, 7.1, , Two point charges of q and í3q located near each other create an electric field. If a correctly drawn diagram of the field, shows six field lines touching q, describe the electric field lines touching í3q., Two lines going out of í3q, Two lines going into í3q, 18 lines going out of í3q, 18 lines going into í3q, , 7.2, , The illustration shows a diagram for the, electric field created by two charged particles,, q1 and q2. (a) Which charge is positive? (b), Which charge is negative? (c) Which charge, has a greater magnitude? (d) How many times, greater is it?, (a), , q1, , q2, , Neither, , Both, , (b), , q1, , q2, , Neither, , Both, , (c), , q1, , q2, , (d), 7.3, , times greater, , The illustration shows a diagram for the, electric field created by three charged particles, q1, q2 and q3. (a) Which charge(s) is/are, positive? (b) Which, charge(s) is/are negative? (c) List the charges, in order of increasing magnitude (disregarding, sign). (d) Charge q1 is how many times, stronger than charge q3?, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems, , 433
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(a), , q1, q2, q3, , (b), , q1, q2, q3, , (c), , |q1| > |q2| > |q3|, |q1| > |q3| > |q2|, |q2| > |q1| > |q3|, |q2| > |q3| > |q1|, |q3| > |q1| > |q2|, |q3| > |q2| > |q1|, , (d), 7.4, , 7.5, , times stronger, , The three electric field diagrams labeled "1",, "2", and "3" in the accompanying illustration, each represent the field around two charged, particles of equal magnitude. (a) Which, diagram has a positive charge and a negative, charge? (b) Which diagram has two positive, charges? (c) Which diagram has two negative, charges?, (a), , 1, , 2, , (b), , 1, , 2, , 3, , (c), , 1, , 2, , 3, , 3, , Two point charges, +4q and íq, are located near each other. Sketch a diagram of the electric field generated by these two, charges, including some indication of how the field lines behave between the particles as well as at greater distances from, them., , Section 8 - Describing the force exerted by an external electric field, 8.1, , A particle of mass 1.20 kg with a charge of +2.30 mC is shot into a region with an electric field of strength 18.0 N/C, having a, direction opposite the particle's velocity. If the particle has an initial speed of 3.10 m/s, how long does it take to come to a, complete (though momentary) standstill? Assume there are no other forces acting on the particle., s, , 8.2, , Protons carry an electric charge of e = 1.60×10í19 C, electrons carry a charge of íe, and neutrons are electrically neutral., Protons and neutrons have a mass of 1.67×10í27 kg, and electrons have a mass of 9.11×10í31 kg. A uniform electric field can, exert a force on a charged particle that effectively neutralizes the force of gravity on the particle, causing it to be "weightless.", What is the strength of the electric field that will levitate (a) a proton? (b) an electron? (c) an alpha particle (two protons and, two neutrons)?, (a), , 8.3, , N/C, , (b), , N/C, , (c), , N/C, , An electron (charge í1.60×10í19 C, mass 9.11×10í31 kg) is in a uniform electric field of strength 543 N/C in the positive, direction. (a) What is the force on the electron? (b) What is the acceleration of the electron? (c) What is the electron's speed, after 1.05×10í8 s? (d) What fraction of the speed of light (3.00×108 m/s) is your answer to part c? Be careful of the sign in, each of your answers., (a), , N, , (b), , m/s2, , (c), , m/s, , (d), 8.4, , 434, , A proton (charge 1.60×10í19 C, mass 1.67×10í27 kg) is in a uniform electric field of strength 2.51×104 N/C. (a) What is the, magnitude of the force on the proton? (b) What is the magnitude of acceleration of the proton? (c) What is the proton's speed, after 1.25 ns?, (a), , N, , (b), , m/s2, , (c), , m/s, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems
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8.5, , An alpha particle (charge 3.20×10í19 C, mass 6.68×10í27 kg) is in a uniform electric field of strength 7.75×10í3 N/C. What is, the particle's speed after 12.5 ms?, m/s, , 8.6, , A proton with mass 1.67×10í27 kg and charge 1.60×10í19 C is traveling at 6.00×105 m/s in the positive direction when it, enters a uniform electric field with a strength of 1250 N/C in the negative direction. The opposing electric force brings the, proton to rest. Calculate the displacement of the proton while it is coming to rest., m, , 8.7, , A proton with mass 1.67×10í27 kg and charge 1.60×10í19 C accelerates from rest in a uniform electric field of strength, 500 N/C. (a) What is the magnitude of the acceleration of the proton? (b) How long does it take the proton to reach a speed of, 35,000 m/s? (c) What distance has the proton traveled when it reaches this speed? (d) What is the kinetic energy of the, proton at 35,000 m/s?, m/s2, , (a), (b), , s, , (c), , m, , (d), , J, , Section 12 - Electric dipoles, 12.1 Two opposite point charges of magnitude 2.30eí6 C are located near the origin. The positive charge is at (1.50, 1.50) mm, and the negative charge is at (í1.50, í1.50) mm. What is the magnitude of the dipole moment vector for this pair of charges?, C·m, 12.2 The magnitude of the dipole moment of a water molecule in liquid water is 6.2×10í30 C·m. A simplified model of the water, molecule H2O can be constructed by imagining that each of the two hydrogen atoms in the molecule contributes its single, electron to the one oxygen atom, and thereby becomes bound to it. In this model, the molecule is a simple dipole, with the, oxygen atom at one end and the two hydrogen atoms at the other, each end having a charge of magnitude 2e = 3.2×10í19 C., In this model, what is the dipole separation of the charges in the water molecule?, m, 12.3 An electric dipole with moment p = (3.43, í5.63) μC·m consists of two opposite charges of magnitude 1.00x10í6 C. How far, apart are the charges?, m, , Copyright 2007 Kinetic Books Co. Chapter 23 Problems, , 435
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24.0 - Introduction, In this chapter, we discuss electric potential energy and electric potential., Understanding these topics will be crucial to your understanding of how electric, circuits and components work. For instance, when you turn on a flashlight, you are, allowing an electric potential difference to drive a current that causes the light bulb, to glow., In the sections that follow, you will learn how electric charges and fields can create, electric potential energy, electric potential, and electric potential differences. You, may be unfamiliar with the term electric potential difference, but you likely know its, units, volts, and you have probably heard the informal term “voltage” used for it., The simulations to the right allow you to experiment with electric potential energy., Just as a configuration of masses, such as a barbell held above the Earth’s surface,, possesses gravitational potential energy, so too does a configuration of electrically, charged particles possess electric potential energy., The first simulation contains a stationary positive charge (red) and a positive test, charge (white) that you drag around with your mouse. As you move the test charge, from place to place, you can change the electric potential energy of the system of, two charges. The potential energy, which depends on the distance r between the, charges and their strengths, is displayed in the control panel. A graph of the PE is, drawn on the right side of the simulation as you move the test charge around. A grid, in the background of the simulation allows you to estimate the distance in, millimeters between the two charges., As you move the test charge, consider the following questions: What is the sign of, the PE? When is the potential energy the greatest? The least? The graph shows, PE as a function of the distance r between the charges., As with gravitational potential energy, a configuration having zero potential energy, must be defined. In the first two interactives, the potential energy is defined to be, zero when an infinite distance separates the two charges., Interactive 2 is the same as Interactive 1 except that the stationary charge is, negative instead of positive. Experiment with this configuration by moving the, positive test charge around. What is the sign of the PE now? As the distance, increases, does the PE increase or decrease? Can you move the charge in a way, such that the PE does not change?, In the Interactive 3 simulation, you see a test charge that you can move between, two oppositely charged plates. The electric field is uniform between the plates. In, this simulation, the x axis of the graph tells the distance of the test charge from the, negative plate. How does the PE change as you move the test charge away from, the negative plate and toward the positive plate?, , 436, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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24.1 - Electric potential energy, , Electric potential energy:, Potential energy determined, by the configuration of, electric charges., Electric potential energy, PEe, is the measure of the, energy stored due to the configuration of a system of, charges. It reflects the positions (not the motion) of, the charges in the system., , This Honda FCX automobile can cruise up to 350 km by using the electric, , potential energy stored in its capacitors as a configuration of electric charge., The release of electric potential energy can be highly, visible. In a thunderstorm, clouds often accumulate a tremendous number of charged, particles in a configuration that stores huge amounts of electric potential energy. Part of, this energy can be released in the form of a lightning strike, a particularly dramatic, example of the effects of discharging PEe., , Recalling the fundamentals of gravitational potential energy may help you to understand, electric potential energy because the two are analogous. For example, the distance of, an apple above the Earth’s surface directly affects its gravitational potential energy, or, more precisely, the gravitational potential energy of the Earth/apple system. Lifting the, apple higher requires work í a force applied through a distance í and increases the, gravitational potential energy. Lowering the apple reduces the potential energy of the, system., To apply these concepts to electric potential energy, consider the initial and final, configurations of opposite charges shown in the illustration of Concept 1. Initially, the, charges are touching. Then, the positive charge is moved to the right by an external, force while the negative charge on the left is held stationary. The work done on the, system by the force increases its potential energy., , Electric potential energy, Property of a system of charges, Depends on charge separation, strength, , In this chapter, since we are discussing potential energy in the context of electrostatics,, we focus on stationary charges. The charges are stationary in their initial and final, configurations, which means their KE does not change. Any work done on the charges, contributes only to changing their PEe. We state that as the first equation in Equation 1:, Work done on the charges equals the change of PEe. You can think of a direct, gravitational analogy: The work you do to raise an apple from one stationary position to, another equals its increase in PE., Two charges with opposite signs attract each other, just as two masses do. Separate, two such charges, or two masses, and you increase their PEe. However, like charges, repel each other. It takes work to move them closer together, and this means their PEe, increases as they approach each other. The simulations you used in the introduction to, this chapter are designed to emphasize these points, and this may be a good time to try, them again. They show the similarities and differences in the PEe of similar, configurations of like or opposite charges., So far, we have been discussing work done on a system of charges by an external, force. On the other hand, the fields of the particles themselves can do work on each, other, similar to what happens when an apple falls to the Earth under the pull of gravity., This is referred to as work being done by the system. You see an example of this in the, illustration of Equation 1: After positive work done on the system separates the charges, and increases its potential energy, positive work done by the system pulls them back, together and decreases its potential energy again. This is directly analogous to the, relationship of work and gravitational potential energy: The gravitational potential, energy of a system decreases when two objects approach, as when an apple falls to, the Earth. We state this relationship between ǻPEe and work as the second equation in, Equation 1: The change in PEe equals the opposite of the work done by the system., , Changes in energy, ǻPEe = W (work done on system), PEe = electric potential energy, W = work, ǻPEe = íW (work done by system), Charges are stationary at end points, , In either case í work done on the system or by the system í the work equals a force applied through a displacement. The electrostatic force, increases with the strength of the charges, and it decreases with distance. Be careful when you are asked to find the work done in changing, the separation between two charges. The amount of force constantly varies with distance, a fact you must account for when you calculate the, work. Calculus proves a useful tool for this, since it provides a technique for calculating work as the force varies over each small increment of, displacement., The electrostatic force is conservative, and the work done by an electric field on charges is path independent. This means that the work done, by the field as a charged particle moves from one stationary position to another does not depend on the particle’s path between the positions., For example, in the illustration of Concept 1, a horizontal force was exerted to push one particle directly away from the other to increase the, system’s potential energy. If the two particles had been moved from their indicated starting positions to their end positions by other paths, (perhaps involving extravagant zigzags and loops), the net work done and the final potential energy would have been the same. The, , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 437
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conservative nature of the electrostatic force means that the potential energy depends, solely on the configuration, not on how it was arrived at., We conclude this section with a review of the relationship of the signs of work and, potential energy, since they can be confusing. In terms of work done on the system,, separating two charges with opposite signs takes positive work (force in the direction, of displacement) and increases their potential energy. Pushing together two like, charges also takes positive work, and also increases their potential energy. Checking a, computationally based answer against some of your own physics intuition may help, make this clear. For instance, if you have to pull apart opposite charges, it should seem, that you are doing work on the system, increasing its energy, and in fact you are., The proper treatment of signs is summarized in the table below., , It takes 5.0 J of work by the righthand wand to separate the, charges. What is the change in, potential energy?, ǻKE = 0, ǻPEe = W (work done on system), W = 5.0 J, ǻPEe = 5.0 J, , 24.2 - Sample problem: electric potential energy, A particle with a charge of 0.15, coulombs is in a uniform electric field, of strength 52 N/C. An external force, pushes the charge 0.12 meters, directly against the field., What is the change in electric, potential energy?, , In this problem, work is done on the system consisting of the charged particle q and the uniform electric field E. The particle is stationary both, before and after it is moved., Variables, change in electric potential energy, , ǻPEe, , work done to move particle, , W, , change in particle’s kinetic energy, , ǻKE = 0 J, , force moving particle, , F, , charge of particle, , q = 0.15 C, , strength of electric field, , E = 52 N/C, , distance particle moves, , ǻx = 0.12 m, , What is the strategy?, 1., , Calculate the work done in moving the particle from its initial to its final position., , 2., , Since there is no change in kinetic energy, the change in potential energy equals the work done on the system. Using the given values,, calculate this change., , Physics principles and equations, The work done to move the particle against the field is, , W = Fǻx, The force required to move the particle against the field is, , F = qE, Since the charged particle is stationary before and after it is moved by the external force, ǻKE, , 438, , = 0. This allows us to apply the equation, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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relating the change in potential energy to work done on the system,, , ǻPEe = W, Step-by-step solution, We calculate the work done to move the charge q through the displacement ǻx., , Step, , Reason, , 1., , W = Fǻx, , definition of work, , 2., , F = qE, , definition of field, , 3., , W = qEǻx substitute equation 2 into equation 1, , Since ǻKE, , = 0, we may state that ǻPEe equals the work done on the system, and evaluate the result., , Step, , Reason, , 4., , ǻKE = 0, , particle stationary before and after, , 5., , ǻPEe = W, , work done on system, , 6., , ǻPEe = qEǻx, , substitute equation 3 into equation 5, , 7., , evaluate, , The solution shows that the system has a greater potential energy when q is in its final position. The external force did positive work on the, system, increasing its PE., , 24.3 - Electric potential energy and work, For a given configuration of a system to have a certain “absolute” amount of potential, energy, a reference configuration with zero potential energy must be established., Changes from that zero point will then be the measure of the potential energy of the, system., With gravitational potential energy, an infinite separation between two masses is often, defined as the configuration that has zero potential energy. This convention is, frequently used for orbiting bodies. It is also common to use the convention that an, object at the Earth’s surface represents a configuration with zero gravitational potential, energy., When analyzing electric potential energy, physicists usually state that a system of two, charges has zero electric potential energy when an infinite distance separates them., The relationship between ǻPEe and work that was introduced previously can be, combined with this reference value to define a relationship between absolute PEe and, work, shown in Equation 1 on the right., The relationship between the work W done by a system of two charges as one of, them moves in from an infinite separation, and the PEe of the system, is PEe = íW., Why does this equation hold true? We already know that ǻPEe = íW, and when the, initial PEe is zero, then the change in potential energy is just the potential energy of the, system in its final configuration., , Electric potential energy, To determine electric potential energy, ·Set point of zero potential energy, ·Infinite separation often used, , In Equation 2, we show you how to calculate the work done by the system as the, separation between charges changes. This equation can be derived using calculus., Note that the signs of the charges do matter in the formula. If they are opposite,, applying the equation confirms that the system (the field) does positive work as the, particles approach one another. Conversely, the work done by the field is negative if, two like charges are moved closer to one another, as illustrated in the diagram., In Equation 3, we combine the work and potential energy equations using an infinite, separation to define zero potential energy. Since zero is the result of dividing by infinity,, the 1/ri term “disappears” and only the fraction í1/rf remains in the final factor of the, work formula. Its negative sign cancels the one in the potential energy equation, PEe = íW, to yield the equation you see as Equation 3., Note an implication of this equation: When the signs of the two charges are opposite,, the potential energy of the pair is negative. When they are the same, the potential, energy is positive., , Electric potential energy and, work, PEe = íW, , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 439
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PEe = electrical potential energy, W = work done by system as charge, moves in from infinity, , Work by field, , W = work done by field, k = Coulomb’s constant, q = electric charge, ri = initial separation, rf = final separation, , Electric potential energy, When PEe = 0 at infinite separation, , PEe = electric potential energy, k = Coulomb’s constant, q = electric charge, r = separation of charges, , 440, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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What is the potential energy of, this system?, , PEe = 2.16×1010 J, 24.4 - Electric potential, , Electric potential: The electric potential energy, of a test charge at a given point in a field,, divided by the test charge., The electric potential of a location in an electric field is the measure of how much, electric potential energy will be generated by placing a charge at that point. It is, analogous to the concept of electric field. Just as the concept of electric field provides a, way to determine how much electric force any charge will encounter at a given location,, the concept of electric potential is used to determine how much potential energy placing, the charge in a field will generate., The term “electric potential” may seem confusing, since we are already using the, related term “electric potential energy.” The formula in Equation 1 provides a direct way, to calculate electric potential. The electric potential at a point in a field is calculated by, placing a test charge in a field, determining the potential energy of the system created, by introducing the charge at that location, and then dividing by the test charge., , Electric potential, Reflects ability of fields to create PE, Calculated as electric PE / test charge, , The electric potential at any point is unique, which means it has only one value. This, reflects the fact that the electrostatic force is conservative, and any configuration of, charges has just one value for its potential energy. Electric potential has no direction,, only a magnitude, making it a scalar. In Concept 1, you see a graph of the electric, potential around a positive point charge, and in Concept 2, you see the graph of the, electric potential around a negative point charge. Note how the electric potential, increases near the positive point charge, and becomes increasingly negative near the, negative point charge. This reflects how the PEe of a positive test charge will increase, near a positive charge, and take on larger negative values near a negative charge., The graphs on the right provide metaphors for electric potential. The positive charge, creates a potential peak, a mountain to be scaled, while the negative charge creates a, potential well, a hole to be climbed out of., , Electric potential, Potential can be positive or negative, , You should remember that, just like electric fields, real potential peaks and wells exist, around charges in three-dimensional space. They would require four dimensions to properly graph! The three-dimensional graphs in Concepts, 1 and 2 only show the electric potential at locations lying in a plane around a point charge., The equation for calculating the electric potential at locations around a point charge is shown in Equation 2. The equation can be derived from, the equation for the PEe of a pair of charges. The PEe created by placing a test charge in the field is calculated, and then the PEe is divided, by the test charge, which means the test charge factor cancels out of the equation. Like the equation for potential energy from which it comes,, this equation assumes there is zero potential energy when there is an infinite separation between the charges. This is the same as stating that, the electric potential infinitely far away from a point charge is zero, as suggested by the graphs of the potential peak and the potential well., The unit of electric potential is the volt. A volt is defined as one joule per coulomb, or energy per unit charge. It is named after Alessandro Volta, (1745-1827), the Italian inventor of the system that underlies the design of most batteries., , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 441
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Electric potential of location in, field, V = PEe/qtest, V = electric potential, PEe = electric potential energy, qtest = test charge, Units: volts (V), , Electric potential around a point, charge, , V = electric potential, k = Coulomb’s constant, q = point charge, r = distance from charge, , What is the electric potential at, the location of the test charge?, V = PEe/qtest, V = (3.2×10í16 J)/(1.6×10í19 C), V = 2.0×103 V = 2000 V, 442, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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24.5 - Spreadsheet: electric potential graphs, At the right, you see a three-dimensional graph of the electric potential for the points in, a plane surrounding three charges. The graph was generated by a spreadsheet that, you can access from this textbook if you are using a computer that can open Microsoft®, Excel files. The electric potential at any point in the horizontal plane is indicated by the, graph’s distance above or below zero. Two positive charges of different magnitudes, cause the “peaks” you see at the right, while the “well” is caused by a negative charge., The electric potential takes on extremely large positive or negative values at locations, very near the positive or negative charges. We do not show the most extreme values, because it would require a drastic change to the vertical scale of the graph., The graph may remind you of very mountainous terrain. It serves as a map of electric, potential, showing the locations of peaks, plains and wells. The electric PE of a test, charge will increase as it “climbs up” the side of any peak. It requires positive work on, the test charge to cause it to approach the underlying charge generating the peak. On, the other hand, the PE of a test charge that is free to move will naturally “fall into” a, well, as the test charge is attracted by the underlying negative charge., Click here to launch the spreadsheet. You can change the amounts and locations of, the charges, or add additional charges, and see the results. If the file does not open, on, Windows click with the right mouse button and choose the save option. On the, Macintosh, hold down the “control” key and click on the link, then choose the option to, download the file., Click here to see a separate document that explains the programming of the, spreadsheet in case you want to modify it., , 24.6 - Electric potential difference, Electric potential difference is the difference in electric, potential between two points. For example, if point B, has an electric potential of positive five volts, and, point A has an electric potential of positive three volts,, then the potential difference VB í VA is two volts., You see this in Equation 1. Like electric potentials,, electric potential differences are measured in volts., In the three illustrations to the right, you see two, plates. They are both charged: The plate on the left, (plate A) is negatively charged, and the plate on the, right (plate B) is positively charged. In the example, problem, we chose to make the electric potential of, plate A negative 0.5 V and of plate B positive 1.0 V., The potential difference between the plates is 1.5 V., , The potential difference between the terminals of the car battery is 12 V., Between two contacts of a household outlet, ǻV averages 120 V., , The electric potential values just cited for the plates depend on some choice of a, reference point for zero potential. Often, in practical applications, the ground (the, electrically neutral Earth) is defined as having zero electric potential, and all values are, measured relative to this value. In an interactive problem in the introduction to this, chapter, the midpoint between the plates was selected as the location of zero potential., Also at the right, you see a common source of electric potential difference: a battery. Its, two terminals have different electric potentials. Batteries are often described by the, potential difference between their terminals. For instance, the potential difference, between the terminals of a D cell is 1.5 V. This is often referred to as the battery’s, voltage. In the configuration at the right, the potential difference between the battery’s, terminals caused electrons to flow to plate A and away from plate B, thereby charging, the plates., , Electric potential difference, Difference in electric potential between, two points, , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 443
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Potential difference, ǻV = VB í VA, ǻV = electric potential difference, VB = potential at point B, VA = potential at point A, Units: volts (V), , What is the potential difference, between the plates?, ǻV = VB íVA, ǻV = 1.0 V í (í0.5 V), ǻV = 1.5 V, 24.7 - Potential difference, electric potential energy and work, At the right, you see two charged conducting plates. Plate A on the left is negatively, charged, and plate B on the right is positively charged. The electric potential of plate A, is í6.0 V, and that of plate B is +3.0 V. In this section, we will use this configuration to, summarize the relationship of electric potential difference, electric potential energy and, work., The potential difference between the plates is +9.0 V (+3.0 V minus í6.0 V). How does, this relate to potential energy and work? Consider a positive test charge with a charge, of one microcoulomb that we cause to move between the plates. The charge is initially, stationary at plate A. We move it to the positive plate B, and hold it still against this, plate. The work we do changes only the charge’s PEe: It is stationary at its initial and, final positions, so it has zero KE at both plates. The movement of the charge between, the two plates is indicated in Concept 1., There is an electric field between the plates. It points from B to A, opposite to the, direction we move the positive charge. Plate B repels the charge and plate A attracts it., This means we have to apply a force to move the charge from A to B. Since we are, applying a force through a displacement, we are doing work on the system. The force, we apply is in the same direction as the displacement of the test charge, so we are, performing positive work. The force through a displacement is shown in Concept 2., , Potential difference, Change in energy per unit charge, , V = PEe/q , and, ǻV = ǻPEe/q, , The (positive) work we do increases the electric potential energy of the positive test charge. It has more potential energy at B than at A., If you like, you can use a gravitational analogy. Consider the work required to roll a ball from an area of low gravitational potential, a valley, up, an inclined ramp to an area of high gravitational potential, a mountaintop. Moving the ball uphill requires work, and the positive work done on, the ball increases its potential energy. This resembles the positive test charge’s trip from plate A to plate B. Conversely, if the ball rolls back, , 444, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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down the hill (or the test charge is allowed to coast back to plate A), its potential energy, decreases., Now we discuss the same points, using equations. Electric potential can be calculated, as the electric potential energy created by placing a test charge in a field, divided by the, test charge. The electric potential difference is the change (difference) in electric, potential energy divided by the test charge. These two equations are shown in Concept, 1., What causes this change in potential energy? The work done on the system. Taking the, previous equation and replacing ǻPE with work W,we find that the electric potential, difference is W/q, or work per unit charge. This is shown in Concept 2., In all this discussion, a positive test charge has been used. This is the standard way to, determine the electric potential. However, negative charges (like electrons) can move,, too. When free to move, a negative charge will naturally move to an area of higher, potential (the positive plate in this example). This will decrease the system’s potential, energy. With positive work, it can be forced to return to the negative plate, and this will, increase the potential energy of the system. (As a form of shorthand, you may, sometimes see this referred to as the PE of the charge, but it is important to remember, that the charge only has PE because of the field caused by other charges.), , PE, work and potential difference, W = ǻPEe , so, ǻV = W/q, , The electron volt, a common unit of energy in many areas of physics study, is based on, the relationship of energy, charge and potential difference. One electron volt (eV), equals the change in the potential energy of an elementary charge (1.60×10í19 C) when, it moves through a potential difference of one volt. We can use a rearrangement of the, second equation in Concept 1, ǻPEe = qǻV, to calculate that an electron volt equals, 1.602 18×10í19 J. Note that the electron volt equals charge multiplied by electric, potential, which equals electric potential energy. In spite of its name this unit refers to, electric potential energy, not electric potential as volts do., , What is the potential difference, between the plates?, W = Fǻx, ǻV = W/q, ǻV = Fǻx/q, ǻV = (3.0 N)(5.0 ȝm)/(7.5 ȝC), ǻV = 2.0 V, 24.8 - Interactive checkpoint: lead acid cell, A typical car battery consists of six, lead-acid cells. Each cell has a, potential difference of 2.05 V across, it. The negative electrode is at a, potential of í1.685 V, while the, positive electrode is at a potential of, +0.365 V. What is the change in, electric potential energy for one, electron that is moved from the, positive to the negative electrode of, one of the cells?, , Answer:, , ǻPEe =, , J, , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 445
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24.9 - A comparison of electric and gravitational fields, In this section, we compare the uniform electric field, produced by a large negatively charged plate with the, gravitational field created by the Earth. The goal is to, use the gravitational field í one you experience all the, time í to help you better understand the electric field., In the gravitational field, the test objects in the, illustrations to the right are people: Individuals of, varying masses can walk up or down the stairs of a, building to change their height above the Earth’s, surface. In the electric field, the test objects are, positive test charges that can be farther from or closer, Points in space around Jupiter have more potential in its gravitational field, to the charged plate. Their distance from the plate is, than would points the same distance from the Earth in its gravitational field., indicated by d. In this discussion, the surface of the, plate, and the Earth, are assumed to be points of zero potential energy., Let’s now use the analogy. First, consider two people on the stairs. One person has a, mass of 50 kg, the other 100 kg. Say they are both on the 12th floor of a building. Each, person’s potential energy is a function of his mass and height above the ground, and, the acceleration due to Earth’s gravitational field. Specific examples of this are, illustrated in the Concept 2 diagram to the right: A 100 kg person on the 12th floor has, twice as much gravitational potential energy as a 50 kg person on the same floor., An analogous situation exists for a positive test charge above the center of the large, negatively charged plate. The plate attracts the charge because it creates a field with, the same downward orientation as the gravitational field. The electric potential energy of, the charge is a function of its magnitude, its distance from the plate, and the strength of, the field. In the Concept 2 diagram, two charges of different magnitudes are the same, distance from the plate. The stronger charge will have more electric potential energy,, just as the more massive person at a given height has more gravitational potential, energy., Now we will observe how position relates to potential energy. This is shown in the, Concept 3 diagram. The same person is shown on the 12th and 6th floors. (To simplify, the discussion, we use the European convention that the ground level is the “zeroth”, floor, not the first, so that the 12th floor is twice as far above the ground as the 6th.) The, person’s gravitational potential energy is twice as much on the 12th floor as on the 6th, floor. We also show two test charges of the same magnitude at different distances from, the plate. The test charge at twice the distance from the plate has twice the potential, energy., Finally let’s consider the role of the field. Potential energy is a function of the strength of, the field-producing object. For example, if we moved the building to Jupiter, the same, people on the same floors would have greater gravitational potential energy than they, do on Earth. Jupiter’s stronger gravitational field would account for the greater, gravitational PE. If the plate had more charge and therefore a stronger field, test, charges at the same positions would have greater electric potential energy. With a, mass like Earth or with an electrically charged plate, you can think of fields as, surrounding these entities, waiting to exert forces on nearby objects., In both fields, the work done on an object equals its change in potential energy. For, instance, if an elevator picks up a person at the 6th and drops him off at the 12th floor,, the work done by the elevator equals the change in his potential energy. Similarly, if you, exert a force to move a test charge farther from the plate, the work you do equals the, change in the system’s potential energy. (The person and charge are both stationary at, the beginning and end of their journeys, so the work only changes their PEe. Their KE, at the beginning and end is zero.), , Electric and gravitational fields, Compare electric and gravitational, effects, ·Use charged plate, tall building, ·Consider distance, ·Use test object’s charge or mass, , Potential energy, Depends on charge or mass, , Now consider the concept of gravitational potential. As illustrated in the Concept 4, diagram, gravitational potential depends only on height above the Earth’s surface. The, gravitational potential is twice as great on the 12th floor as it is on the 6th floor., We can also compare positions 12 floors above the ground on Jupiter and on the Earth,, two planets of different masses. The gravitational potential 12 floors above the ground, is greater on Jupiter than it is on the Earth. To put it another way, any object you place, on the 12th floor on Jupiter will have more potential energy than the same object would, on the 12th floor on the Earth. This is because Jupiter’s gravitational field is stronger, than the Earth’s. The important distinction between potential energy and potential is that, the potential is independent of the test object. Jupiter’s mass generates more, gravitational potential at a given height above its surface than the Earth’s mass does., , Potential energy, Also depends on location of test object, , The concept of gravitational potential is analogous to electric potential. You see this also in the Concept 4 diagram. A location one centimeter, above the plate has twice the electric potential of a location only half a centimeter above the plate. The distance above the plate determines the, , 446, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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electric potential. It is independent of the strength of the test charge. Any charge,, regardless of its magnitude, placed in an area of greater electric potential will create a, system with more electric potential energy than when it is placed in an area of lower, electric potential., On the other hand, a location one centimeter from a highly charged plate with a, stronger electric field has more potential than a location the same distance from a less, charged plate. Again, this idea directly parallels the idea of gravitational potential. The, statement for charged plates is true no matter what test charge you might use to assess, the electric potential., For the last idea, turn your attention to the illustration for the Concept 5 diagram. If you, want to calculate the potential difference between the 6th and 7th floors, subtract the, value of the potential at the 6th floor from its value at the 7th floor. Since we can treat the, Earth’s gravitational field as essentially uniform near its surface, the gravitational, potential difference between the 6th and 7th floors is the same as it is between the 12 th, and 13th floors., , Potential, Depends solely on location, , You should think of electric potential difference in a similar way. If you know the electric, potentials at two points, you can subtract them to calculate the potential difference. With, electric potential difference, the actual potentials at the points do not matter, only the, change in potential between two points. The concept of electric potential difference is, used frequently in everyday devices. For instance, if you are told you have a nine-volt, battery, you are being told the difference between the terminals, not the electric, potential of either terminal. This is analogous to how you might think if you were told to, run up the stairs of a skyscraper. It does not matter whether you run from the 5th floor to, the 15th, or from the 18th to the 28th. In either case, you have to run up 10 flights of, stairs., , Potential difference, Change in potential between two points, , 24.10 - Equipotential surfaces, , Equipotential surface: A, surface with the same electric, potential everywhere., As its name indicates, an equipotential surface is one, along which the electric potential of a field is, everywhere the same. The boundary of a sphere, centered about a point charge q is an equipotential, surface. All the points on this surface are the same, distance r from q. This means the electric potential,, which in this case can be calculated with the formula, kq/r, is the same at all points on the surface., , The mountainous contour map above displays curves of constant altitude., Its “equi-altitude” curves are like equipotential surfaces in an electric field., , Such an equipotential surface is shown in the first diagram to the right. In the, illustration, we represent the equipotential surface around q with a circle instead of a, sphere for the sake of visual clarity., No work is needed to move a charge from one resting place to another along an, equipotential surface, because the potential energy neither increases nor decreases as, the charge moves. If work were needed to change the charge’s position, its electric, potential energy would change, which means it would be in a location with a different, electric potential., The second diagram shows a particular example of a relationship that holds true in, general: Electric field lines that intersect an equipotential surface are always, perpendicular to the surface where they intersect it., , Equipotential surface, , Surface with constant electric potential, This is true because any motion of a charged particle from one place to another along, such a surface must be in a direction perpendicular to the force exerted by the field., No work to move charge along surface, Work equals the component of the force parallel to the motion, multiplied by the, displacement. No parallel component means no work occurs. In turn, no work means no change in potential energy, and no change in potential, energy means no change in potential., You can explore equipotential surfaces in the simulations in the introduction to this chapter. If you move a test charge along an equipotential, surface, you will see that its electric potential energy stays the same., , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 447
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Equipotential surface, Electric field perpendicular to surface, , 24.11 - Electric potential and a uniform electric field, The concepts of electric potential and electric field are linked. In Concept 1, you see, their relationship illustrated in the context of a uniform electric field. Plate A is negatively, charged and plate B is positively charged. Assume that the plates are large enough, compared to their spacing that the electric field between them is of uniform strength, and direction, pointing from plate B toward plate A., The electric potential between the plates increases in the opposite direction, from, plate A to plate B. Since it takes positive work to push a test charge against the field, from left to right, the potential energy of the charge increases as it moves from plate A, to plate B. Since the strength of the field is everywhere the same, the electric potential, increases at a constant rate as the charge moves from left to right. The potential’s, change per unit displacement parallel to the field is constant., Another way of expressing the opposite orientations of the field and the direction of, increasing potential is to say that the field is always directed from regions of higher, potential to regions of lower potential., We state the proportionality between the change of potential and the displacement as, the first equation in Equation 1, and derive it below. The variable ǻs measures a, displacement parallel to the field lines. The negative sign in the equation reflects the, fact that the potential difference is negative for a displacement ǻs in the same direction, as the field. Movement perpendicular to the field lines results in no change in electric, potential. Again, we stress that the field must be uniform., , Displacement parallel to uniform, field, Potential V changes at constant rate, Field directed from higher to lower V, , In many applications later in this book, we will only be interested in the magnitude of the, potential difference between two points in a uniform field that are separated by a, distance d parallel to the field, and we will write ǻV = Ed., The second equation in Equation 1 is just the first equation, solved for the electric field., It proves to be a very useful formulation. It shows how the electric field strength can be, determined when the potential difference between two points in the field is known. This, form of the equation is applied in the example problem below., Derivation. We will show that for a uniform electric field E, and a displacement ǻs, measured in the direction of the field, the potential difference between two points, separated by ǻs is given by the equation ǻV = íEǻs. This is shown in the illustration, of Equation 1., , Potential difference in uniform, field, , Variables, uniform electric field, , E, , two points in field, , P 1, P 2, , test charge, , qtest, , force, , F, , work, , W, , distance parallel to field, , ǻs, , potential difference between points, , ǻV, , ǻV = potential difference, E = electric field strength, ǻs = displacement parallel to field, , We use the subscripts field and ext in the steps below to distinguish the forces exerted and the work done by the field and by an external force, that moves a test charge against the field., , 448, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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Strategy, 1., , Calculate the work done by an external force to move a test charge against the, constant force of the field., , 2., , State the work done on the field in a different way, using the relationship of, potential difference, work done on the field, and charge., , 3., , Set the two expressions for work equal to each other and simplify to get the, desired equation., , Physics principles and equations, The work done to move a test charge from point P1 to point P2 , opposite to the, direction of ǻs, is F·(íǻs). Since F is constant and parallel to the displacement, between the points, this dot product can be written as the scalar product Fǻs cos, or just, , 0°,, , W = Fǻs, The force exerted on the test charge by the electric field is, , What are the magnitude and, direction of this uniform electric, field?, Field is directed to the right, ǻs directed from left to right, , F = qE, A relationship stated and derived earlier for the relationship between potential difference, and work done on a system is, , ǻV = 10 V í 15 V = í 5.0 V, , ǻV = W/q, Step-by-step derivation, We find two expressions for the work W and set them equal to each other. We derive, this using a positive charge being pushed against the field. The work done by the, external force is positive in this case; work done by the field is negative., , Step, , E = í (í 5.0 V)/(0.20 m) = 25 N/C, E = 25 N/C to the right, , Reason, , 1., , W = Fǻs, , definition of work, , 2., , Ffield = qtestE, , electric field force on test charge, , 3., , Fext = – qtestE, , external force on test charge, , 4., , Wext = – qtestEǻs substitute equation 3 into equation 1, , 5., , Wext = qtestǻV, , potential difference and work, , 6., , ǻV = – Eǻs, , equate steps 4 and 5 and simplify, , Although we derived Equation 1 for the case of a positive charge being moved against a uniform field, the equation is equally true for negative, charges, and for charges moving in either direction., , 24.12 - Interactive checkpoint: an accelerated proton, In a uniform electric field, the potential, changes from 14.0 V to 32.6 V in a, displacement of 0.130 m (the, direction is opposite to the field, as, shown). What is the magnitude of the, acceleration of a proton (charge, 1.60×10í19 C and mass 1.67×10í27, kg) that is placed in this field?, , Answer:, , a=, , m/s2, , Copyright 2000-2007 Kinetic Books Co. Chapter 24, , 449
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24.13 - Gotchas, Electric potential energy and electric potential are the same. No, they are not even measured in the same units. Potential energy is a property, of a specific configuration of charges, while electric potential is a property of a point in space. Electric potential is measured by assessing the, potential energy with a test charge, and then dividing by the magnitude of that charge., I quadrupled the strength of a test charge, but I think the electric potential at the location is unchanged. You are correct. Like an electric field,, electric potential is independent of the test charge., There is electric potential surrounding a single charge. Yes, there is. There is a field. Add a second charge, and you will have created a system, with electric potential energy., Where does an electron go when it is free to move: to a location of higher or lower potential? Toward a location of higher potential. Electrons, tend to move toward positive charges and away from negative charges, which means they move toward regions of higher electric potential,, minimizing the electric potential energy. A positive test charge (or proton) would do the opposite, moving toward a region of lower potential,, which also minimizes the electric potential energy., I calculated a negative potential difference between two points. Is that possible? Quite possible. If point A has an electric potential of 9.0 volts,, and point B an electric potential of 6.0 volts, the difference VB í VA is í3.0 volts. It is perhaps an unstated convention that when calculating a, potential difference, VA is subtracted from VB., , 24.14 - Summary, Electric potential energy is conceptually similar to gravitational potential energy. It is, determined by the configuration of a system of charges. As with gravitational, potential energy, external positive work that is done on a system results in a, positive change in the sum of the system’s PE and KE. On the other hand, positive, work done by the system on something outside the system results in a negative, change in the sum of the system’s PE and KE., Since in this chapter we often assume that charges move from one stationary, configuration to another, there is no change in KE due to work, and positive work, done on a system increases its PE. Negative work done on a system, as when two, opposite charges are moved closer together, decreases its PE., Electric potential is a scalar quantity defined throughout an electric field in a region, of three-dimensional space. It describes the electric potential energy that will be, possessed by an electric test charge í or more properly by the system of charges í, when the test charge is placed at various locations in the field. When a test charge, is introduced, its electric potential energy equals the electric potential at its location,, times its own charge value. The potential is measured in joules per coulomb, or, volts (V), where 1 V = 1 J/C., , Potential energy of 2-charge system, , Electric potential, , V = PEe/ qtest, , Potential difference in a uniform field, , ǻV = –Eǻs, , Electric potential difference is more commonly measured and used than the electric, potential itself. It is the difference in electric potential between two points. Batteries, are categorized by the potential difference between their terminals, which is 1.5 V, for commonly sold small batteries., The electric potential difference between two points also equals the work required to move a test charge from one point to the other, divided by, the value of the charge., An equipotential surface in a field is a surface that has the same electric potential at all points. Because there is no potential difference, between any two points in the surface, no work is required to move a charge along the surface. The electric field is perpendicular to such a, surface at every point on the surface., , 450, , Copyright 2000-2007 Kinetic Books Co. Chapter 24
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Chapter 24 Problems, , Chapter Assumptions, Unless stated otherwise, the reference configuration for zero electric potential and zero electric potential energy is one in which there, is infinite separation between charges., , Conceptual Problems, C.1, , A positively charged particle and a negatively charged particle are located near one another in space. Their masses, the, magnitudes of their charges, and the distance between them are such that their gravitational potential energy equals their, electric potential energy. (The reference configuration for zero gravitational PE, like that for zero electric PE, is an infinite, separation of the particles.) (a) If the distance between the particles is halved, what happens to the electric potential energy of, the configuration? (b) Is the gravitational potential energy of the new configuration still equal to the electric potential energy?, (c) Are they still equal if one of the charges reverses in sign but keeps the same magnitude?, (a), , i. It halves, ii. It doubles, iii. It stays the same, , (b), , Yes, , No, , (c), , Yes, , No, , C.2, , Explain the difference between electric potential energy and electric potential. How are the two related?, , C.3, , Electric field strength can be measured in N/C, but in view of the equation E = |ǻV/ǻs|, it is often written with the units V/m., Using only metric units and their definitions, show that 1 N/C = 1 V/m., , C.4, , (a) Can equipotential surfaces with different electric potentials ever intersect each other? (b) Why or why not? (c) Do any two, equipotential surfaces have to be a constant distance apart? Explain., (a), , Yes, , No, , (b), (c), , Yes, , No, , Section Problems, Section 0 - Introduction, 0.1, , Using the simulation in the first interactive problem in this section, answer the following questions. (a) What is the sign of the, PE? (b) Is the magnitude of the potential energy greater when the test charge is closer to, or farther away from the stationary, charge?, (a), , (b), 0.2, , Closer to, , Farther away, , Using the simulation in the second interactive problem in this section, answer the following questions. (a) What is the sign of, the PE? (b) As the distance between the charges increases, does the PE increase or decrease? (c) In what type of path, should you move the charge in order to keep its PE constant?, (a), , (b), , (c), , 0.3, , i. Always positive, ii. Always negative, iii. Both positive and negative, , i., ii., iii., i., ii., iii., i., ii., iii., iv., , Always positive, Always negative, Both positive and negative, Increases, Decreases, Stays the same, A straight line, A circle, A parabola, A square, , Using the last interactive in this section answer the following questions. (a) How does the potential energy change as you, move the test charge away from the negative plate and toward the positive plate? (b) In what type of path should you move, the test charge in order to keep its PE constant?, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems, , 451
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(a), , (b), , i., ii., iii., i., ii., iii., iv., , It increases, It decreases, It stays the same, A circle, A horizontal line, A vertical line, A square, , Section 1 - Electric potential energy, 1.1, , A system consisting of two positive particles is altered by an outside force so that its potential energy increases by 5.4 μJ., There is no change in kinetic energy between the initial and final states of the system. (a) What is the work done by the, outside force? (b) What is the work done by the electric force? (c) Are the particles closer together or farther apart after the, system's rearrangement?, (a), (b), (c), , 1.2, , i. Closer together, ii. Farther apart, iii. No change, , A system consisting of one positively charged particle and one negatively charged particle is altered by an outside force so, that its potential energy increases by 6.25 mJ. There is no change in kinetic energy between the initial and final states of the, system. (a) What is the work done by the outside force? (b) What is the work done by the electric force? (c) Are the particles, closer together or farther apart after the system's rearrangement?, (a), (b), (c), , 1.3, , μ J, μ J, , mJ, mJ, i. Closer together, ii. Farther apart, iii. No change, , A system consisting of two positive particles is altered by an outside force so that its potential energy decreases by 52.3 J., There is no change in kinetic energy between the initial and final states of the system. (a) What is the work done by the, outside force? (b) What is the work done by the electric force? (c) Are the particles closer together or farther apart after the, system's rearrangement?, (a), (b), (c), , J, J, i. Closer together, ii. Farther apart, iii. No change, , Section 2 - Sample problem: electric potential energy, 2.1, , A system consists of a uniform electric field of 73.5 N/C, directed in the positive x direction, together with a particle having a, positive charge of 3.46×10í5 C. The particle is moved from the location x = 3.50 m to the location x = 1.50 m in the field, and it, is stationary before and after it is moved. (a) What is the work done on the system by the external force? (b) What is the work, done by the field on the particle? (c) What is the change in the potential energy of the system?, (a), , 2.2, , J, , (b), , J, , (c), , J, , A system consists of a uniform electric field of 42.6 N/C, directed in the positive x direction, together with a particle having a, positive charge of 7.87×10í6 C. The particle is moved from the location x = 1.50 m to the location x = 3.50 m in the field, and it, is stationary before and after it is moved. (a) What is the work done on the system by the external force? (b) What is the work, done by the field on the particle? (c) What is the change in the potential energy of the system?, , 2.3, , 452, , (a), , J, , (b), , J, , (c), , J, , A system consists of a uniform electric field of 2.50 N/C, directed in the positive x direction, together with a particle having a, negative charge of 5.50 mC. The particle is moved from the location x = 1.50 m to the location x = 3.50 m in the field, and it is, stationary before and after it is moved. (a) What is the work done on the system by the external force? (b) What is the work, done by the field on the particle? (c) What is the change in the potential energy of the system?, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems
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(a), , 2.4, , J, , (b), , J, , (c), , J, , This problem explores the relationship between electric potential energy and work done on a system (by a non-conservative, force) when there is a change in the KE of the system after the work is done. A system consists of a uniform electric field of, 73.5 N/C, directed in the positive x direction, together with a particle having a positive charge of 3.46×10í5 C and a mass of, 2.42×10í9 kg. An external force moves the initially stationary particle from the location x = 3.50 m to the location x = 1.50 m in, the field, and when it stops "pushing" the particle is moving at 346 m/s to the left. (a) What is the change in the kinetic energy, of the system? (b) What is the change in the potential energy of the system? (c) What is the work done on the system by the, external force? (d) Is the change in the potential energy of the system still equal to the work done on the system?, (a), , J, , (c), , J, , (d), 2.5, , J, , (b), Yes, , No, , This problem explores the relationship between potential energy and work done by a system (by a conservative force) when, there is a change in the KE of the particle after the work is done. The system consists of a uniform electric field of 42.6 N/C,, directed in the positive x direction, together with a particle with a positive charge of 7.87×10í6 C. As part of an experiment, the, field is allowed to accelerate the particle starting from rest at x = 1.50 m. The experiment ends when the particle passes, x = 3.50 m, at which time it is moving to the right. (a) What is the change in the kinetic energy of the system? (b) What is the, change in the potential energy of the system? (c) What is the work done by the field on the particle? (d) Is the change in the, potential energy of the system still equal to the negative of the work done by the conservative force?, (a), , J, , (b), , J, , (c), , J, , (d), , Yes, , No, , Section 3 - Electric potential energy and work, 3.1, , Two positive charges of 3.6 μC and 2.5 μC are separated by a distance of 0.031 m. What is the electric potential energy of, this configuration?, , 3.2, , A 6.3 μC point charge and a í1.7 μC point charge are separated by a distance of 0.00032 m. What is the electric potential, energy of this system?, , J, , J, 3.3, , The nucleus of a helium atom consists of two protons, each with a charge of +e, and two electrically neutral neutrons. The, distance between the repelling protons is, on the average, about 1×10í15 m. They are constrained there by a force called,, appropriately enough, the strong force. For each of the following questions, it is sufficient to report your answer with just one, significant digit, multiplied by a power of 10. (a) In a classical, non-quantum model of the helium nucleus, what is the average, strength of the repulsive force between the two protons? (b) What is the average electric potential energy of the nucleus?, , 3.4, , (a), , N, , (b), , J, , The Bohr model of the hydrogen atom describes it as a proton (charge +e) orbited by a single electron (íe). When the atom, is in its lowest energy state, the orbital radius of the electron is 5.29×10í11 m, a distance called the Bohr radius. What is the, PEe of a simple system consisting of a proton and an electron separated by that distance?, J, , 3.5, , The gravitational potential energy of all the water stored behind a very large dam is 1.2×1016 J, using the base of the dam as, the reference point for zero PE. Imagine you could store the same amount of potential energy in a system consisting of two, particles, each with a charge of 0.0020 C. How far apart should you place the charges?, m, , 3.6, , Two small iron spheres each have a charge of 0.250 C. (a) How much work does it take to move the spheres from a, stationary configuration 2.00 m apart to another stationary configuration 1.00 m apart? (b) How much work does it take to, move the spheres from a stationary configuration 12.0 m apart to another stationary configuration 11.0 m apart? (c) How, much work does it take to move the spheres from a stationary configuration 22.0 m apart to another stationary configuration, 21.0 m apart?, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems, , 453
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(a), , J, , (b), , J, , (c), , J, , Section 4 - Electric potential, 4.1, , (a) Calculate the potential 1.00 m from an isolated proton. (b) Calculate the potential 1.00 m from an isolated electron. (c), Which potential is greater? (d) Which potential has a greater magnitude?, (a), , V, V, , (b), (c), , (d), , 4.2, , i., ii., iii., i., ii., iii., , Near the proton, Near the electron, Potentials are equal, Near the proton, Near the electron, Magnitudes are equal, , In a science fiction story, a microscopic black hole is given an enormous positive charge by firing an un-neutralized ion drive, exhaust (consisting of positively charged xenon ions) into it for six months. The idea behind the charging process is to be able, to confine and manipulate this dangerous object with powerful electric fields. Suppose the charge on the black hole is 5740 C., At what distance from it is the electric potential equal to 1.09e+3 V?, m, , 4.3, , A positively charged, small region in a thundercloud creates an electric potential of 67.6 mV at ground level, 321 m below., What is the charge of the region?, , 4.4, , A nonconducting sphere of radius R = 5.00 cm contains a total charge of 4.25×10í8 C, uniformly distributed throughout its, volume. What is the potential due to this charge distribution at a point at distance r (measured in meters) from the center of, the sphere, where r > R? Express your answer in terms of r. (Hint: According to a version of the shell theorem, the force, exerted by the sphere on a test charge at any external point is the same as if all the charge of the sphere were concentrated, at its center.), , C, , 382/(0.05+r), 382/r, 382/(r-0.05), , Section 6 - Electric potential difference, 6.1, , The negative (black) terminal of a 12 V automobile battery is grounded to the frame of a car. If we arbitrarily take the, automobile frame to be at an electric potential of 20 V, what is the electric potential of the positive (red) terminal of the, battery?, V, , Section 7 - Potential difference, electric potential energy and work, 7.1, , An electron with charge í1.60×10í19 C moves a distance of 3.17 mm in the negative x direction in a uniform electric field of, 125 N/C that points in the positive x direction. (a) How much work does the field do on the electron? (b) What is the change in, the potential energy of the electron? (c) Through what potential difference does the electron move?, (a), , 7.2, , J, , (b), , J, , (c), , V, , The electric potential difference across the cell membrane of a human cell is 0.080 V, with the higher potential being outside, the cell. In one cellular process a sodium ion Na+, with charge +e, is "pumped" through a channel in the membrane to be, ejected from the cell. (a) How much work must be done by the cell to eject the ion? (b) A candy bar contains 270 Calories, or, 1.13×106 J of food energy. How many sodium ions could the cells of your body eject using the energy obtained in one candy, bar?, , 7.3, , (a), , J, , (b), , ions, , A planet far, far away was impacted by a one-kilogram ball of pure "protonium" containing 6.00×1026 protons, and nothing, else. This excess charge eventually distributed itself evenly through the planet, which had previously been electrically neutral., (a) If the radius of the planet, like Earth's, is 6.37×106 m, what is the magnitude of the electric field at the planet's surface?, , 454, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems
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(Hint: According to a version of the shell theorem, the force exerted by a spherically symmetric charge distribution on a test, charge at any external point is the same as if all the charge were concentrated at its center.) (b) What is the potential differnce, between a location 500 m above the surface of the planet and a location at ground level?, , 7.4, , (a), , N/C, , (b), , V, , A certain 4.8-volt cellular telephone battery has a capacity of 600 mAh. (One mAh equals 3.6 coulombs.) When the battery's, charge is too low to run the phone any longer (assume it is zero), the battery must be recharged. (a) How much work is done, by the recharger? (b) If electricity costs you 9.0 cents per kWh, how much does one recharge cost you?, J, , (a), (b) $, 7.5, , A charged particle of mass 8.39×10í4 kg starts from rest and accelerates through a potential difference of +25,000 V to reach, a speed of 1.26e+3 m/s. What is the charge on this particle?, C, , Section 10 - Equipotential surfaces, 10.1 A point charge of 8.40 mC is surrounded by an equipotential surface with a radius of 0.298 m. What is the electric potential, on the surface?, V, 10.2 What is the radius of the 2.5×10í9 V equipotential surface that surrounds an isolated proton?, m, 10.3 A positive point charge is surrounded by two equipotential surfaces. (a) What is the shape of the surfaces? (b) On which of, the surfaces is the electric potential greater? (c) If the larger of the surfaces has 10 times the area of the smaller, what is the, ratio of the potential on the larger to the potential on the smaller?, (a), , (b), , i., ii., iii., i., ii., , Cube, Pyramid, Sphere, The surface closer to the point charge, The surface farther from the point charge, , (c), , Section 11 - Electric potential and a uniform electric field, 11.1 A uniform electric field of 2.0×105 N/C is oriented in the negative x direction. How much work must be done to move an, electron 3.0 m in the positive x direction, if it begins and ends at rest?, J, 11.2 In a typical television or in an older computer picture tube (CRT), the electrons that strike the screen to form an image are, accelerated between two charged plates that have a potential difference of 2.5×104 V. The plates are separated by a distance, of 1.3 cm. Assuming that there is a uniform electric field between the plates, calculate the strength of the field., N/C, 11.3 Points 1 and 2 are located in a uniform electric, field of strength 325 N/C. The displacement, vector from point 1 to point 2 points in the, same direction as the field, and has a length of, 15.4 cm. The electric potential at point 1 is, 9.15 V. What is the potential at point 2?, V, , 11.4 Starting at a point in a uniform electric field of strength 9.50 N/C, how far and in what direction (relative to the field) must one, travel to experience an increase in potential of 13.6 V?, m, , i. In the direction of the field, ii. Opposite the direction of the field, iii. Perpendicular to the field, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems, , 455
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11.5 In the diagram you see the spark gap between, the contacts of an automobile spark plug., When the contacts are charged to a potential, difference great enough to cause a spark to, jump across the gap, the electric field between, them approximates a uniform field of strength, 3.1×106 N/C. What is the magnitude of the, potential difference that causes the spark?, V, , 11.6 An important factor in many biochemical cell processes is the potential difference across a cell membrane. The inside of a, cell is at a lower electric potential than the outside, with a potential difference of 0.080 V between them. The membrane, consists of two layers of chainlike lipid molecules packed together like the blades of grass in a lawn, but it is only 7.5 nm thick., What is the strength of the electric field inside the membrane?, N/C, 11.7 A uniform field has strength 66.8 N/C and points in the positive x direction in three-dimensional space. The electric potential, at the point (2.50 m, 0 m, 0 m) is í17.3 V. (a) What is the shape of the equipotential surfaces for this field? (b) Write an, equation to describe the set of points for which the electric potential is zero volts. (c) Write an equation to describe the set of, points for which the electric potential is +24.8 V., (a), , Spheres, Planes parallel to the xy plane, Planes parallel to the yz plane, Planes parallel to the xz plane, , (b), , x=0m, z=0m, x = 2.24 m, x = 2.50 m, x = 2.78 m, , (c), , x = 2.48 m, y = 2.48 m, z = 2.48 m, x = 1.87 m, x = 2.50 m, , 456, , Copyright 2007 Kinetic Books Co. Chapter 24 Problems
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25.0 - Introduction, We began our study of electricity and magnetism with electrostatics, which focuses on the forces and fields created by stationary charges., Now we will concentrate on charges that are in motion, a branch of physics called electrodynamics. We will discuss the flow of charge (electric, current), resistance to current, and the relationship among potential difference, resistance, and current., In this chapter, we focus primarily on the fundamentals of these topics rather than their applications. We will not concern ourselves too much, about the source of the potential difference, nor worry too about where the current may be flowing or why one would want it to flow. Once we, have discussed the essentials, we can apply these concepts to electric circuits, and consider typical sources of potential difference, such as, batteries, and common sources of resistance such as resistors and light bulbs., , 25.1 - Electric current, , Electric current: Amount of net charge passing, through a surface per second., Current is the rate at which net charge passes through a hypothetical surface, the, number of coulombs of charge per second. As a practical matter, currents are often, measured as they pass through a wire, so we use this configuration to explain currents, in the illustrations to the right. In this section, we focus solely on the current, not its, cause., To measure the amount of current, we can place an imaginary surface across the wire, shown in Concept 1 and count the net number of electrons moving through it each, second. The electrons we show are the charge carriers, the charges that make up the, current. In this example, they are moving from right to left., Counting the electrons as they pass by is a useful start, but electric current is charge, per second, so we need to multiply by the charge of an electron. An electron has a, charge of í1.6×10í19 C, so if there are five electrons flowing by every second, the, electric current is 8.0×10í19 coulombs per second. For reasons discussed below, the, flow of current here is considered positive., , Electric current, Rate of net flow of charge, Typically composed of moving electrons, , The equation in Equation 1 states that current is the net charge passing through a, surface divided by time. The ampere (A) is the unit for electric current. It is named after, the French scientist André-Marie Ampère. One ampere equals one coulomb per, second. A coulomb of charge is equivalent to 6.2×1018 electrons, so one ampere equals, that number of electrons passing through every second. The letter I represents current., Using water as an analogy may help you understand electric currents. The rate of water, flow is measured in several settings. You may have seen water flow ratings for shower, heads. Newer shower heads allow a water flow of nine liters per minute, about half the, rate of flow of older models. Boaters also keep a close eye on water flow. For a, particular river, a rafter might measure how much water is flowing, in cubic feet per, second, to judge whether it is safe to run the rapids., , Direction of conventional current, Current arrow in direction of positive, charge flow, , Similarly, the rate of flow of electrons can be measured. Many electrons move in the, currents used in household devices. About 2.5×1018 electrons pass through the light, bulb in a typical household flashlight every second, which is 0.4 amperes., Current is a scalar quantity. It states the rate of flow of charge, not the direction of the, flow of charge. Again, think of water. Liters per second tells you how much water is, flowing, but not in which direction. Often, it is important to know the direction of an, electric current: which way the charge is flowing. Confusingly, the arrow used to indicate, a current’s direction points opposite the way you might expect. It does not point in the, direction that electrons flow. Instead, it points the way positive charges would be flowing, if they were the charges moving in an electric current. This is shown in Concept 2. The, arrow indicates the direction of what is called conventional current, the direction in, which positive charge carriers creating the current would move. The flow of electrons, discussed above would be described as a positive current flowing to the right, not a, negative current flowing to the left., , Electric current, , Why is typical electric current shown as though positive charge carriers flow, even if, they do not? Scientists began studying electric currents before they knew about, electrons and protons. One of the early explorers of electricity, Benjamin Franklin,, established the convention that the current points in the direction of positive charge, flow. More than a hundred years later, when the scientist Edwin Hall determined that, , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 457
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current is actually the movement of negative charge carriers (electrons), the positive, charge flow convention had already been established, and it remains in place to this, day., In basic configurations like the segment of conducting wire shown to the right, electric, current is stated as a positive quantity and its direction is indicated with an arrow. In, other circumstances, such as alternating current circuits where the direction of current, flow changes constantly, signs can be used to indicate the direction of current., , I = current, ǻq = charge passing through surface, ǻt = elapsed time, Units: amperes (A), , In 2.0 seconds, 3.0 coulombs of, charge flow past this charge, counter. What is the current?, , I = 1.5 A, 25.2 - Drift speed, , Drift speed: Average speed of charge carriers, that make up a current., Whether or not there is an electric current in a wire, the electrons in the wire move, randomly at high speed, on the order of a million meters per second, due to their, thermal energy. This random motion does not create a current because it causes no net, motion of charge over time., When there is an electric current, there is a net motion of charge. The average speed at, which charge moves in a current is called its drift speed. To illustrate this concept, we, use the simplified diagram of the wiring in a car shown in Concept 1. In the diagram, a, battery is shown; the potential difference between its terminals causes electrons to, move counterclockwise along the wire. The average speed at which they move along, the wire is the drift speed., , Drift speed, Average speed of charges in current, In typical wiring, about 0.1 mm/s, , The magnitude of the drift speed may surprise you; it is on the order of 10í4 m/s. That is, less than a meter per hour, slower than a snail slimes across the ground. This may, seem surprising because when you flip the car’s headlight switch, the lights illuminate, almost instantaneously. It may seem as if electrons are flowing from near the battery to, the headlights at an incredible speed., Why do the lights go on so quickly? When the switch is flipped, the electric field created, by the car’s battery travels through the wires at nearly the speed of light, 3.00×108 m/s., It causes electrons everywhere in the wire, including those in the filaments of the, headlights, to start to move almost immediately. It is like when you turn on a water, faucet: water flows out very quickly. You receive water near the faucet; you do not have, to wait for water to flow all the way from the reservoir to your tap., Why do the electrons move along the wire so slowly? If the electrons were in a vacuum,, the electric field would accelerate them until they moved at near the speed of light,, Electrons, since there would be no force opposing their motion. But in a wire, the electrons collide, ... but move, with the atoms that make up the wire. The combination of the collisions and the electric, field cause them to move in the erratic zigzag pattern you see in Concept 2. (We, overstate the curved element of their paths to emphasize the effect of the field on their motion.), , 458, , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , zigzag ..., along wire
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The motion of these electrons is like that of a ball in play in a pinball machine. The ball, moves fairly randomly as it rebounds from bumpers and other obstacles. However, the, surface of a pinball table slants downward, so despite the erratic motion caused by the, bumpers, a component of the gravitational force pulls the ball down toward you over, time. (In short, the maker of the pinball game uses gravity to ensure that the ball “drifts”, downward and “drains”.), There is no simple relationship between drift speed and the amount of current in various, conductors. A great number of charge carriers moving slowly can result in a larger, electric current than a smaller number of charge carriers flowing quickly., There is a final point to be made about the erratic motion of electrons in the current., Energy transfers from the electrons to the atoms during the collisions. The collisions, increase the random (thermal) motion of the atoms. This explains why the temperature, of the filaments in the car headlights increases when a current flows through them:, Energy is being transferred from the electrons in the current to the atoms that make up, the filaments. The filaments become hot and emit light. The other parts of the wiring, increase in temperature as well, but the system is designed to maximize the, temperature increase in the filaments and minimize it elsewhere., , If the switch is 1.0 meter away, from the headlights, and charge, carriers move at the speed, shown above, how long will it, take for the headlights to light, once the switch is thrown?, Nearly instantaneous, ·Field acts along entire wire very quickly, , 25.3 - Ohm’s law and resistance, , Resistance: The ratio of the, potential difference across a, conductor to the current, through it., Resistor: An electrical, component often used to, control the amount of current, flow., , Resistors wired into a computer circuit board. Each resistor, has a color code that indicates its resistance in ohms., , Resistance is defined as the potential difference across two points on a conductor, divided by the current flowing through the conductor. At the right, we use a common, electrical component called a resistor to illustrate this concept. There is a potential, difference across the resistor and current flowing through it. Divide the potential, difference by the current and you have calculated the resistance of this resistor., The resistance of a resistor such as those shown above or to the right is constant. The, resistor is made of ohmic materials, which are empirically known to obey Ohm’s law., Increase the potential difference and the current increases proportionally. The, resistance does not change. The linear relationship between potential difference and, current is shown as the first equation in Equation 1., Since resistance equals potential difference divided by current, its unit, the ohm, is volts, per ampere. Resistance is represented by , the Greek letter omega., , Resistor, Component used to regulate current, , Resistors are not the only components that resist the flow of current. The filament in a, light bulb or the coils of an electric hot plate both function as resistors. The term, “resistor” broadly refers to any element that is a source of resistance., Again, water is a good analogy. The potential difference driving current in a wire, resembles the pressure exerted on water in a pipe. Increase the pressure, and the, water flow increases. Different pipes have different amounts of resistance. For instance,, one with a rough interior wall would have greater resistance to water flow than one with, a smooth wall., Materials that do not obey Ohm’s law are called non-ohmic. Many components used in, modern circuitry are made of non-ohmic materials. For instance, a diode has little, resistance to current flow in one direction, and great resistance to current flow in the, other., , Resistance, , Georg Ohm published his major work, including what we now know as Ohm’s law, in, Potential difference divided by current, 1827. His theories were greeted with skepticism and his career was slow to progress., Why it took so long for his work to be appreciated is hard to say. Perhaps it is because, the law is empirical as opposed to a fundamental law of nature. It is fair to note that many major leaps forward in physics were met with, , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 459
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skepticism and opposition., , Ohm’s law: potential difference, and current, ǻV = IR, , ǻV = potential difference, I = current, R = resistance, Units of resistance: ohms (),, volts/ampere, , What is the resistance of this, resistor?, , R = 100 , 25.4 - Interactive problem: Ohm’s law, The interactive problem on the right shows a tanning bed, an electrical appliance, through which current can flow., Your challenge is to configure the tanning bed so that 2.4 amperes of current flow, through it. If you accomplish this goal, Joe Pasty will get a perfect tan. Anything less, and he will remain pasty. Anything more and it will be red and toasty for Joe., In the simulation, the potential difference applied to the lamps of the tanning bed is, 240 volts. (Assume that this potential difference is unchanging.), You can control the tanning bed’s resistance using the up and down buttons in the, control panel. The resistance can vary from 10 to 200 ohms., Use Ohm’s law to calculate the tanning bed resistance that will result in 2.4, amperes of current flowing through the device. Set the resistance to the nearest 10, ohms, then press GO to see if you were correct. A gauge located on the tanning, bed will display the amount of current flowing through the device., , 460, , Copyright 2000-2007 Kinetic Books Co. Chapter 25
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25.5 - Resistivity, Resistivity is used to quantify how much a material resists the flow of current. It is, represented by the Greek letter ȡ (rho). A material considered a good conductor, such, as copper, has low resistivity. Materials used as insulators, such as glass or rubber,, have high resistivity., Resistivity is the inverse of conductivity (which quantifies how well a material conducts, current). Current does not readily flow through materials with very high resistivity. As the, table in Concept 2 shows, there is an enormous range in the resistivity of materials., The table includes nichrome, an alloy of nickel and chromium. Due to its high resistivity,, Nichrome is commonly used in hot plates, toasters and other electrical appliances that, generate intense heat. Its large amount of resistivity causes electrons in the current to, lose energy to the nichrome atoms, increasing their thermal energy., Extension cords, designed to allow current to flow safely from one point to another, take, advantage of the relative resistivities of materials. The cords typically are composed of, copper wire, a material with low resistivity, surrounded by rubber or vinyl, materials with, high resistivity. Electrical current flows readily through the copper, but not through the, material that surrounds it., , Resistivity, Inverse of conductivity, ·Conductors have low resistivity, ·Insulators have high resistivity, , Resistivity is a property of a material. Its units are the ohm·meter (·m). In contrast,, resistance is the property of a component like a resistor or the filament in a light bulb., Resistivity is one factor that determines a resistor’s resistance; the other factor is the, geometry of the resistor. The longer a resistor, the greater its resistance; the wider it is, (the greater its cross sectional area), the less its resistance. Again, water can be used, to provide an analogy: a long, thin pipe resists the flow of water more than a short, fat, one., The equation used to calculate resistance as a function of a resistor’s geometry and, composition is shown in Equation 1., The problem in Example 1 asks you to compute the amount of resistance found in a, segment of copper wire. The resistance of resistors that are sold in electronics shops, runs from the tens of ohms to the thousands of ohms and beyond. When electric circuits, are analyzed, the resistance of wire is often ignored because it is relatively insignificant., , Table of resistivities, ȡ, , The low resistivity of wire, along with Ohm’s law, helps to answer a question physicists, like to pose: Why can a bird safely stand on an unshielded high voltage line? Why does, the bird not get injured as current flows up one of its feet, through its body, and down, the other foot?, The answer is that the potential difference is very small across the segment of wire, involved because the wire’s resistance is quite low. Its resistance is low because the, segment is wide and short and is made of a material with low resistivity. Ohm’s law, states that the potential difference is the product of the current and the resistance. With, low resistance, there is very little potential difference. With very little potential difference,, effectively no current flows through the bird., Do NOT test this out yourself. High voltage lines are enormously dangerous. If you, touch such a line while standing on the ground, you are providing a path for current to, flow from the wire to the Earth. The potential difference between these points is, enormous, and the resulting current could kill you., , Resistance, Depends on resistivity, configuration of, material, , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 461
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R = resistance of component, ȡ = resistivity of material, L = length of component, A = cross-sectional area, Units of resistivity: ·m (ohm-meters), , The lamp cord is 0.75 meters, long and contains copper wire., What is the wire’s resistance?, , R = 0.0018 , 25.6 - Resistivity and temperature, The resistance of many conductors increases with temperature. As the temperature of, the metal coils of the hot plate shown on the right increases, so do the resistivity of the, metal and resistance of the coils. Other materials, such as semiconductors, have, decreasing resistivity with temperature., Equation 1 shows two equations that reflect the relationship of resistivity and resistance, to temperature. Both include the temperature coefficient of resistivity, represented by Į, (the Greek letter alpha). A table of these coefficients is shown in Concept 2 for some, materials. These coefficients are empirically determined and apply over a specific range, of temperatures. To apply the resistivity equation, the material’s resistivity must be, known at one temperature, T1. Its temperature coefficient of resistivity must also be, known for the temperature T1. These values are used to calculate the resistivity at, another temperature., The second equation is used to determine the change in resistance of a resistor. If the, resistance of a resistor is known at one temperature, the equation can be used to, calculate its resistance at another temperature. This equation can be derived from the, first., , In many materials:, Resistivity and resistance vary linearly, with temperature, , In Example 1, we use this equation to calculate the change in resistance of a nichrome, coil in a hot plate that is heated from 25°C to 375°C., , Temperature coefficient of, resistivity, Į, 462, , Copyright 2000-2007 Kinetic Books Co. Chapter 25
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Positive Į indicates resistivity increases, with temperature, Negative Į indicates resistivity, decreases with temperature, , ȡ2 = ȡ1[1 + Į(T2 íT1)], R2 = R1[1 + Į(T2 íT1)], ȡ = resistivity, Į = temperature coeff. of resistivity, T = temperature, R = resistance, , The hotplate contains a, nichrome wire with a resistance, of 15 at 25°C. What is its, resistance at 375°C? For this, nichrome alloy, Į = 4.0×10í4 °Cí1, at 25°C., R2 = R1[1 + Į(T2 íT1)], R2 = R1[1 + (4.0×10í4 °Cí1)(T2 íT1)], R2 = 15 [1 + (4.0×10í4)(350)], R2 = 17 , 25.7 - Electric power, Power is defined as work per unit time. Power also can be stated as the amount of energy transfer per unit time, and this is often a useful, formulation when considering electric circuits., Electrical devices are often rated on how much energy they use per second. A 100-watt light bulb requires an average of 100 joules every, second. As it operates, the light bulb’s filament warms up and emits heat. This flow of heat from a resistor is called joule heating., Power can also be calculated as the product of current flowing through a resistor and the potential difference across it. This is the equation in, the first line in Equation 1. Ohm’s law can be used to then restate this equation in terms of resistance and current, or in terms of potential, difference and resistance. We show these equations in the second line in Equation 1., In the example problem, we simplify things by using an average or constant value for current and potential difference í likely this burner is, powered with an alternating current where these values vary. Alternating current (AC) is a topic later in the textbook., , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 463
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Below we derive the first equation, P = IǻV. Before deriving it, we review why a, resistor consumes power. The essential component of many household devices í, toasters, light bulbs, electric burners í is a resistor., As electrons move through any resistor, they collide with the atoms that make up the, resistor. The electrons lose energy in these collisions and the atoms gain it, which, increases the temperature of the resistor. More current through a given resistor means, more collisions, and a warmer resistor. This analysis confirms one aspect of the power, equation: Power increases with the amount of current., To derive the power equation, we will consider the work done on the electron by the, electric field. We will state the work in terms of the potential difference: It equals the, charge times the potential difference. We will use that equation to derive the power, equation., , Electric power, , Variables, , Function of current, potential difference, , work, , W, , total charge moving across resistor, , q, , potential difference across resistor, , ǻV, , time interval, , ǻt, , power consumed by resistor, , P, , current passing through resistor, , I, , Strategy, 1., , State the equation that relates work, charge and potential difference., , 2., , Divide by the time interval ǻtto calculate the work per unit time, which is power., , Equations, Here, work is related to charge and potential difference by, , W = ǻPE = qǻV, We will also use the definitions of power and current., , P = IǻV, P = I 2R = ǻV 2/R, P = power, I = current, ǻV = potential difference, R = resistance, , Step-by-step derivation, We state the relationship between work and potential difference, then divide by the time, interval ǻt. This converts work to power and total charge to current, resulting in the first, formulation of the power equation., , Step, , 1., , W = qǻV, , work and potential difference, divide by ǻt, , 2., 3., , Reason, , P = IǻV, , definitions of power and current, , Ohm’s law is used to derive the other equations for power in the second line of Equation, 1. For instance, ǻV = IR, so we can substitute for ǻV and conclude that P = I 2R. The, equation P = IǻV holds true for all electrical devices, while the other equations apply, solely to power (energy) dissipated as heat by resistors., , How much power is being, supplied to this hot plate?, P = IǻV, P = (8.00 A) (120 V), P = 960 W, , 464, , Copyright 2000-2007 Kinetic Books Co. Chapter 25
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25.8 - Sample problem: solar panels, This public school uses solar power., Using information supplied below,, determine how many watt-hours of, electricity the solar panels will provide, during their lifetime., What will the cost per kW·h of the, solar panel system be over its, lifetime? Suppose the cost of, traditional energy will average $0.11, per kilowatt-hour over the next 25, years. Will the school save money?, , A school in Los Angeles is deciding whether or not to install solar panels on its roof. The photovoltaic system under consideration provides 680, watts of power under direct sunlight, costs $3500, and is expected to last 25 years, according to the experts at Stewart Solar Systems. In Los, Angeles the average insolation, or equivalent hours of direct sunlight reaching the Earth’s surface, is 5.62 hours per day., (a) How many watt-hours of electricity will the solar panels provide during their lifetime? (A watt-hour is a unit of energy just like the Joule; it is, calculated by multiplying the power and the amount of time that power is produced.), (b) Suppose the cost of electricity coming from the local utility company is predicted to average $0.11 per kilowatt-hour (kW·h) over the next 25, years. What is the cost per kW·h of the solar panel system over its lifetime? Should the school buy the solar panels?, Variables, generation power of solar panel system, , P = 680 W, , lifetime of system, , t = 25 y, , average equivalent hours of sunlight per day, , s = 5.62 h/d, , equivalent hours of sunlight over system lifetime, , S, , energy produced over system lifetime, , E, , price of utility electricity per kW·h, , p = $0.11, , one-time cost of solar panel system, , C = $3500, , cost of solar panel system per kW·h, , c, , What is the strategy?, 1., , Find the total number of hours of direct sunlight the solar panels will receive over their lifetime., , 2., , Calculate the number of watt-hours the solar panels will produce over their lifetime. Convert to kW·h., , 3., , Calculate the price per kW·h from the solar panels and compare it with the utility company’s price., , Physics principles and equations, Power stated as energy used over a period of time, , P = ǻE/ǻt, Step-by-step solution, First, find the total amount of energy the solar panels will produce over their lifetime., , Step, , Reason, , 1., , S = st, , hours of sunlight over lifetime, , 2., , S = (5.62 h/d)(25 y)(365 d/y), , enter values, , S = 51,283 h, 3., 4., , E = PS, , energy produced over lifetime, enter values, , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 465
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For part (b), calculate the cost per kW·h., , Step, , Reason, , 5., , 6., , convert to kW·h, , c = C/E, , cost per kW·h, enter values, , 7., , This is less than the predicted cost of electricity from the utility company, so the school will save money in electricity costs if they install the, solar panels. Even better, by using renewable energy they will contribute to a less polluted future., On the web, you can find out more about solar energy and find out the average insolation anywhere in the world., , 25.9 - Interactive checkpoint: running a dishwasher, Suppose you measure the current, flow to a dishwasher and find that the, average current is 9.50 A over the, 45.0-minute dishwasher cycle. The, dishwasher is powered by a potential, difference of 120 V. If electricity costs, $0.0900 per kilowatt-hour, what is the, cost of running one dishwasher load?, , Answer:, , p=$, , per dishwasher cycle, , 25.10 - Sample problem: using a hair dryer, The label on an electric blow dryer, says it uses 1100 W and is designed, for 120 V outlets. How much current, flows through the hair dryer? What is, its resistance?, , Here, you will treat the hair dryer as though it were a direct current (DC) appliance., Variables, power rating, , P = 1100 W, , potential difference, , ǻV = 120 V, , current, , I, , resistance, , R, , 466, , Copyright 2000-2007 Kinetic Books Co. Chapter 25
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What is the strategy?, 1., , Use the equation for power as a function of current and voltage to find the current through the hair dryer., , 2., , Use Ohm’s law to find the hair dryer’s resistance., , Physics principles and equations, Power in an electric circuit, , P = IǻV, Ohm’s law, , ǻV = IR, Step-by-step solution, We use the power equation to find the current through the hair dryer., , Step, , 1., , Reason, , P = IǻV, , power equation, solve for I, , 2., 3., , I = (1100 W)/(120 V) = 9.2 A evaluate, , Now we can use the current to find the resistance of the hair dryer., , Step, , 4., , ǻV = IR, , Reason, , Ohm’s law, , 5., , solve for R, , 6., , evaluate, , 25.11 - Sample problem: power transmission, A power plant provides 450 MW of, power through transmission lines at a, potential difference of 500 kV. What, current flows through the lines?, For a city 300 km away from the, power plant the total resistance in the, transmission lines is 60 ohms. What, fraction of the generated power is, dissipated as “line loss”?, If the generated power is transmitted, at a potential difference of 175 kV, instead, what fraction is dissipated?, , Variables, generated power, , P = 450×106 W, , transmission line potential difference, , ǻV, , resistance, , R = 60 , , dissipated power, , PD, , What is the strategy?, 1., , Find the current in the wire using the equation for power in terms of the potential difference and the current., , 2., , Calculate the power dissipated by the power lines using the equation for power dissipated by a resistor., , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 467
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3., , Divide the dissipated power by the generated power to find the fraction of the power dissipated by the lines., , 4., , Repeat this calculation for the last question asked above., , Physics principles and equations, Power as a function of potential difference and current, , P = IǻV, Power dissipated by a resistor, , P = I 2R, Power in an electric circuit, , P = (ǻV)2/R, Step-by-step solution, First we calculate the current in the transmission lines when the potential difference is 500 kV. We do this using the power for the power plant, and the potential difference across the wires, which are both stated above., , Step, , Reason, , 1., , power equation, , 2., , solve for I, , 3., , evaluate, , First, we find the power dissipated as heat by the wire. Then we divide that by the amount of power generated by the plant to determine what, fraction is wasted., , Step, , Reason, , 4., , power dissipated by a resistor, , 5., , evaluate, , 6., , divide by generated power, , For the second case we repeat the same calculations with a lesser potential difference., , Step, , Reason, , 7., , evaluate equation in step 2, , 8., , power dissipated by resistor, , 9., , evaluate, , The differing rates of power consumption explain why power companies transmit power at high potential differences. The potential difference is, lowered at transformers near the consumer. (Transformers are discussed further in the chapter on electromagnetic induction.) The potential, difference is lowered to 120 V, the standard potential difference of a power outlet in a house, just before it enters the house or building using, the power., , 468, , Copyright 2000-2007 Kinetic Books Co. Chapter 25
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25.12 - Gotchas, Do conventional current arrows reflect the flow of negative or positive charge? Positive charge. Although in a typical current found in household, appliances and such, it is negatively charged electrons that flow, the convention is that the arrow indicates the direction of positive charge flow., If electrons flow to the left, the conventional current arrow points to the right., Current decreases when passing through a resistor and then increases again upon exiting. No. The current before, in and after the resistor is, the same., A wire with a current has a net electrostatic charge. No. Although there is a net charge flowing by any given point, the overall wire is neutral., A person walks by. The person contains electrons. Therefore, there is a current. No, the person is electrically neutral. Current is the flow of net, charge. In this case, there is no movement of net charge and therefore no current. If the person were electrically charged, there would be a, current as the person passed by., , 25.13 - Summary, Electric current is the rate at which charge flows through a conductor. The symbol, for current is I and it is measured in amperes (A), where 1 A = 1 C/s., Current usually consists of moving electrons, which have negative charge., However, current is almost always represented in descriptions and diagrams as, conventional current, which is a flow of positive charges that would constitute the, same current, so the direction of conventional current is opposite to the actual, movement of electrons. Even though drawings often show the direction of current, with an arrow, current is a scalar., Individual electrons in a wire travel much more slowly than the electric field that, propels them. Collisions with the atoms that make up the wire prevent the electrons’, continued acceleration, and cause them to follow a meandering zigzag path. The, drift speed is the average net speed of electrons along the wire. In household, wiring, it is about the speed of a snail., Ohm’s law states the relationship between potential difference, current and, resistance. A resistor is an electrical component that can be used to regulate, current. A particular resistor is characterized by its resistance. Resistance is, measured in ohms (). 1 = 1 V/A., , Definition of current, , Ohm’s law, , ǻV = IR, Electric power equations, , P = IǻV, P = I2R = ǻV2/R, , Resistivity is a measure of how much the material resists current. For a simple wire made of a single material, the resistance is the resistivity of, the material times the wire’s length, divided by its cross-sectional area., The temperature of a material can affect its resistivity., Energy is dissipated, often as heat, when an electric current passes through a resistance. Electric power measures the amount of energy, consumed per unit time. It is equal to the current through a circuit element times the potential difference across it. Sometimes, as when a, battery is being charged, the energy consumed is stored rather than dissipated. For resistors that obey Ohm’s Law, the two equations shown in, the last line on the right may also be used to calculate power., , Copyright 2000-2007 Kinetic Books Co. Chapter 25, , 469
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Chapter 25 Problems, , Chapter Assumptions, For any problems in this chapter that involve a household electrical outlet, assume that the voltage supplied is fixed at 120 V. While, this is not true for actual outlets, whose voltage varies and periodically exceeds 120 V, the simplification will not affect the accuracy, of your answer., , Conceptual Problems, C.1, , C.2, , C.3, , The three parts of the illustration show positive, and negative charges moving in the indicated, directions, at the rates shown. (a) Which, scenario depicts the greatest current? (b), Which scenario depicts the smallest current?, (a), , 1, , 2, , 3, , (b), , 1, , 2, , 3, , 3 amperes of current flow through a wire. (a) What net charge flows past a point in the wire each second? (b) What is the net, charge on the wire?, (a), , C, , (b), , C, , A current of 2 amperes flows to the left in a stationary wire. If the wire also starts moving to the left, what happens to the, current as measured by a stationary observer? Why?, i. The current increases, ii. The current decreases, iii. The current stays the same, , C.4, , Two like charges of one coulomb each, separated by one meter, repel each other with a force of 8.99×109 N, equivalent to, the weight of 10 battleships. A flashlight bulb may have a current of 0.40 A flowing through it, which is equivalent to one, coulomb of charge flowing through every 2.5 seconds. But a flashlight circuit is clearly not subject to tremendous electrostatic, forces -- or even small ones. Why not?, , C.5, , A section in the textbook that explains drift speed displays a diagram that shows the zigzag path followed by a black electron, as it pursues an average path to the right through a section of conducting wire, but collides with many yellow atoms in the, wire on its way. Suppose that the black circle in the diagram were a steel ball instead of an electron, and that the yellow, circles were bumpers in a pinball game instead of atoms. That is, suppose this were a diagram of a pinball machine in play. In, a pinball game, what field is acting analogously to the uniform electric field that propels electrons through a wire?, , C.6, , The section in the textbook that explains drift speed displays a diagram that shows the zigzag path followed by an electron as, it pursues an average path to the right through a section of conducting wire, but collides with many atoms in the wire on its, way. The diagram correctly shows the zigs and zags as being curved rather than straight line segments. (a) Why is this? (b), What is the shape of the curved line segments?, (a), (b), , C.7, , A resistor has 2.0 A of current flowing through it when the potential difference across it is 6.5 V. When the potential difference, is increased to 19.5 V, the current increases to 5.5 A. What kind of material is this?, Ohmic, , C.8, , Two wires are made from materials, such as copper and aluminum, that have different resistivities. The wires have the same, length. Is it possible for them to have the same resistance? Explain., Yes, , C.9, , Non-ohmic, , No, , Would you expect there to be any materials that have both high resistivity and high conductivity? Explain., Yes, , No, , C.10 Equations are given in the textbook that allow you to calculate how the resistivity of materials and resistance of objects vary, with the temperature. The discussion of the equations does not specify whether the temperatures in the equations should be, measured in kelvin or in degrees Celsius. Does it matter? If it matters, deduce which should be used. If it doesn't matter, tell, , 470, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems
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why not., i. It matters, ii. It does not matter, C.11 You buy a light bulb in the U.S. labeled 75 watts. Since you bought it in the U.S., the power rating assumes that the potential, difference across the light bulb will be 120 volts. Instead, you provide it with a potential difference of 220 volts (the potential, difference provided by wall outlets in most European countries). Will the light bulb dissipate 75 watts of power? Explain., Yes, , No, , C.12 Solar panels are priced based on the number of kilowatts they generate. You pay the utility an amount based on the number, of kilowatt-hours you use. What is the connection between kW and kW·h?, C.13 You receive your bill from the electric company, and it states that you have used 3470 kW·h (kilowatt·hours) of electricity this, month, and that due to the energy crisis in your town your bill is an astronomical $832.80. What is it, exactly, that you are, paying for?, i., ii., iii., iv., , Electric power, Electric energy, Electric potential, Some hotshot energy trader's vacation in Tahiti, , Section Problems, Section 1 - Electric current, 1.1, , A current of 1.00 mA passes through a copper wire for 60.0 s. (a) How many coulombs of charge pass any point in the wire, during this time period? (b) How many electrons pass any point in the wire during this time period? (c) How many protons, pass any point in the wire during this time period?, (a), , 1.2, , 1.3, , C, , (b), , electrons, , (c), , protons, , In a Crooke's Tube, a stream of electrons boils, off a negative electrode, or cathode, and flies, through a vacuum toward a positive electrode,, or anode. Often, the anode is given a, distinctive shape, like a cross, so that the, electrons that miss it and fly past will create a, dramatic shadow on the far end of the glass, tube. The stream of electrons is flowing past a, certain point between the cathode and the, anode at a rate of 3.50 A. If 95.0% of the, electrons reach the anode, (a) what amount of, charge reachs the anode during a period of 15.0 s? Be careful of the sign of your answer. (b) How many electrons reach the, anode during a period of 15.0 s?, (a), , C, , (b), , electrons, , A mouse sits next to a cell phone charger cord for 4.10 seconds, debating whether it is something worth chewing on. If the, cord carries a current of 3.00 amperes, what is the magnitude of the charge that goes by the mouse while it thinks?, C, , 1.4, , According to the Bohr model, a hydrogen atom in its lowest energy state has a nucleus consisting of a single proton, which is, orbited by a single electron. The speed of the electron is 2.19×106 m/s and the radius of its orbit (the "Bohr radiuis") is, 5.29×10í11 m. What is the magnitude of the "current" flowing around the nucleus in the Bohr model?, A, , 1.5, , A wire spans two telephone poles. A tightrope walker picks up a sphere with a charge of 5.3 microcoulombs at one end and, carries it to the other end in 7.2 seconds. What is the average current from one end of the wire to the other?, A, , 1.6, , A wire etched onto a computer chip carries a current of 4.1eí8 A. In 1.8 milliseconds, how many electrons flow past a point, on the wire?, electrons, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems, , 471
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1.7, , A vacuum cleaner runs on a current of 12 amperes. How long will it take for 5.0 microcoulombs of charge to flow through it?, s, , 1.8, , Physicists often demonstrate the power of electricity by simulating lightning with a device called a Tesla coil. Charge is, deposited onto a metal sphere until it amasses so much electric potential that it discharges through the air in the form of a, spark. With larger Tesla coils, this spark resembles a miniature lightning bolt. Find the average current into a Telsa coil that, amasses 1.38×10í6 C of charge in 5.72 s., A, , 1.9, , The conducting sphere on top of a Van de Graaff generator is charged by a silk conveyer belt that picks up electrons at the, bottom of the generator's column, from grounded brushes consisting of certain substances (for example, fur), and releases, them at the top of the column to another set of brushes consisting of say, hard rubber. Suppose the belt of a certain Van de, Graaff generator is 10.0 cm wide, and it travels upward at 45.0 m/s, carrying a "current" of 175 mA. (a) What is the surface, charge density of the electrons on the belt? (b) Is the conventional "current" in the Van de Graaff generator flowing up or, down?, C/m2, , (a), (b), , Up, , Down, , Section 2 - Drift speed, 2.1, , In one of Bill Cosby's comedy routines, he talks about being so tired that "I turned off the light switch, and I was in my bed, before the light went out!" Suppose the drift speed of electrons in the wiring of the Cosby household is 3.0×10í4 m/s, and that, there are 3.2 m of wiring between the light switch in his bedroom and the lamp. (a) In a direct current circuit, where the current, flows one way, how long would it take a conduction electron to go from the switch to the lamp? (b) Even though Cosby, couldn't really have gotten into bed before the light went out, could he at least have gotten into bed before one electron had, traveled from the switch to the lamp?, (a), , s, , (b), 2.2, , In a typical household current, the electrons in the wire may have a drift speed of 1.3×10í4 m/s. Actual household current is, not a direct current, but instead is an alternating (oscillating) current. (If you have ever received an electric shock from an, outlet, you have felt this rapid alternation as a painful vibration. A safe way to see the alternation is to wave your fingers, rapidly in front of a fluorescent light, and observe the stroboscopic stutter-silhouette that they form.) You can model the motion, of each conduction electron as simple harmonic motion, with a frequency of 60 Hz. The drift speed is the maximum speed,, what the speed would build up to if the electric field were applied continually, as in the case of direct current. What is the, approximate amplitude of an electron's oscillation?, m, , Section 3 - Ohm’s law and resistance, 3.1, , A resistor has a potential difference of 5.65 volts applied across it, and a current of 345 mA flows through it. What is its, resistance?, , 3.2, , A device found at some amusement parks asks you to "test your personal force" by grasping two conducting bars, one in, each hand, and tolerating a muscle-clenching flow of direct current through your body. Assuming that your body is an ohmic, , ȍ, , material, the resistance of your body between your two hands (if you are not sweating!) is about 1.00×104 ȍ. What voltage, does the machine have to apply to the conducting bars to drive a current of 30.0 mA through your body, causing strenuous, muscle contraction?, volts, 3.3, , The resistance across your (wet) tongue, from one side to the other near the tip, is 375 ȍ (assume that your tongue is an, ohmic material). If you connect two thick copper wires of relatively negligible resistance to the terminals of a 1.5 volt flashlight, battery, and place them on opposite sides of your tongue, you will be able to "taste" the electricity as it flows across,, producing a sour metallic flavor. What current is flowing through the saliva that coats your tongue?, A, , 3.4, , A certain lightbulb filament when hot has a resistance of 205 ȍ. The potential difference across the filament at a certain, instant is standard household voltage, 120 V. (a) What current is flowing through the bulb at that time? (b) If the filament of the, bulb is a tungsten wire having a radius of 475 μm, what is the magnitude of the current density in the filament?, , 3.5, , 472, , (a), , A, , (b), , A/m2, , You find a used battery and discover that it is no longer strong enough to power your CD player. Out of curiosity, you decide, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems
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to find out what potential difference it actually supplies. You connect it to a 1.25 kȍ resistor and an ammeter (a device that, measures current) and discover that the current through the circuit is 1.03 mA . What is the potential difference?, V, , Section 4 - Interactive problem: Ohm’s law, 4.1, , Use the information given in the interactive problem in this section to calculate the resistance required to achieve the desired, current and give Joe Pasty a nice tan. Test your answer using the simulation., , ȍ, , Section 5 - Resistivity, 5.1, , An automotive junkyard crushes old vehicles into cubes 0.40 m on a side. Assume that these cubes are made of iron. Find, the resistance you would expect to encounter if you connected one wire to the center of a face and one wire to the center of, the opposite face, so that the two faces acted like the ends of a resistor., , 5.2, , A length of nichrome wire (composed of nickel and chromium) is 75.0 cm long and has a rectangular cross section measuring, , ȍ, 0.700 mm by 0.500 mm. The resistivity of nichrome is 1.08×10í6 ȍm. If a potential difference of 225 V is applied between the, two ends of the wire, how much current flows through it?, A, 5.3, , You have a fine wire with a cross sectional area of 9.4×10í10 m2 and a length of 1.1 m. You wish to determine what it is made, out of, so you connect it to a standard AA battery (1.5 V) and measure a current of 0.048 A. What material is the wire most, likely made out of? (Refer to the table in the text.), i., ii., iii., iv., , 5.4, , 5.5, , Silver, Copper, Aluminum, Iron, , 12-gauge copper wire has a radius of 1.03×10í3 m. (a) What is the resistance of 11.5 m of this wire at 20°C? (b) How much, current will pass through the wire if a potential difference of 9.00 V is applied across it?, (a), , ȍ, , (b), , A, , A section of wire 10.0 m long has a resistance of 4.00 ȍ. (a) What is the resistance of 25.0 m of this wire? (b) What is the, resistance of a 10.0 m section of wire of the same material with half the diameter? (c) What is the resistance of a wire whose, length is 20.0 m and diameter is twice as large as the wire in part a?, , ȍ, ȍ, ȍ, , (a), (b), (c), 5.6, , The graphite in a pencil has a resistivity of 7.84×10í6 ȍm. The graphite's diameter is 0.700 mm and its length is 40.0 mm. (a), Find the resistance of this graphite "wire." (b) Find the diameter of a cylindrical copper wire of the same length (having, resistivity 1.70×10í8 ȍm) that would produce the same resistance., (a), , ȍ, , (b), , m, , Section 6 - Resistivity and temperature, 6.1, , A tungsten wire, initially at room temperature (20.0°C) with a resistance between its ends of 41 ȍ, is placed in an oven. What, will the temperature of the wire be when the resistance is 110 ȍ? The temperature coefficient of resistivity of tungsten is, 0.0050°Cí1 at 20.0°C., °C, , 6.2, , A thermometer can be made from a brass wire (temperature coefficient of resistivity 0.00150°Cí1 at 20.0°C) which has a, potential difference of 1.50 V between its ends and is connected to an ammeter. A measurement determines that the current, flowing through the wire is 347 mA when the wire is at room temperature (20.0°C). (a) What is the resistance of the wire? (b), What change in current will accompany a decrease in temperature of one degree Celsius (or one kelvin)? Be careful of the, sign of your answer. (c) How much current flows through the wire when the temperature is 35.0°C?, (a), , ȍ, , (b), , A, , (c), , A, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems, , 473
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6.3, , You have a resistor made of an unknown material. You note that at a temperature of 20.0°C its resistance is 529 ȍ. Upon, lowering its temperature to 0°C, the resistance drops to 476 ȍ. (a) Find the temperature coefficient of resistivity of the, material at 20.0°C. (b) Use the table in the text to make a guess as to the material the resistor is made of., ° Cí1, , (a), (b), , 6.4, , i., ii., iii., iv., v., , Silver, Carbon, Tungsten, Germanium, Silicon, , A piece of iron has a resistance of 103 ȍ at 40.0°C. Iron has a temperature coefficient of resistivity of 0.00640°Cí1 at 20.0°C., What is the resistance of the iron at 20.0°C?, , ȍ, 6.5, , At 20.0°C, a mainly silicon resistor has a resistance of 585 ȍ, and a tungsten resistor has a resistance of 752 ȍ. At this, temperature, the silicon resistor has a temperature coefficient of resistivity of í0.070°Cí1 and the tungsten resistor has a, coefficient of 0.0050°Cí1. At what temperature will their resistances be equal?, °C, , 6.6, , A circuit with silver wiring operates at 20.0°C. If you were to replace the silver with copper wiring of the exact same, dimensions, at what temperature would you have to run the circuit in order to have the resistance of the copper circuit be the, same as that of the silver circuit? (Use the resistivities measured at 20.0°C found in the table in the text.), °C, , 6.7, , A conductor of an unknown substance has a resistance of 43.5 ȍ at 40.0° C, and a resistance of 55.4 ȍ at 50.0° C. What is, the resistance of the conductor at 47.0° C? Hint: When applying the equation in the text, the temperature coefficient of, resistivity must be measured at the temperature T1., , ȍ, , Section 7 - Electric power, 7.1, , Instead of using batteries, a certain type of flashlight has a built in hand pump which you can squeeze to generate the power, necessary to light the bulb. Suppose that squeezing the pump at a rate of twice per second generates 16 watts. If the light, bulb has a resistance of 0.60 ohms, and you squeeze twice per second, what will be the potential difference across the light, bulb?, , 7.2, , Many 120 V household circuits are rated at 15 A. This means that, if the current in the circuit exceeds this amount, there is a, danger of overheated wires causing a fire. Such circuits are protected by circuit breakers that trip or fuses that blow and, interrupt the current if it exceeds 15 A. (a) Is it safe to plug a 1400 W space heater into such a circuit? (b) Is it safe to plug two, such space heaters into the same circuit, so that the power requirement is 2800 W?, , V, , (a), , Yes, , No, , (b), , Yes, , No, , 7.3, , How many 60 W light bulbs can you use in a 10 A, 120 V household circuit before tripping the circuit breaker and attempting, to finish this homework problem in total darkness? The number of bulbs, times the individual power, equals the total power for, the circuit., , 7.4, , A certain appliance running at 240 V consumes 325 W of power. How much energy does it use in one hour?, , light bulbs, , J, 7.5, , A simple heater for a cup of water consists of an insulated resistive coil that plugs directly into a 120 V electrical outlet. The, heater is placed into a coffee mug that contains 0.550 kg of water and heats it to boiling (100°C) from room temperature (20°, C) in 180 s. If the specific heat of the water is 4200 J/kg·K over this temperature range, what is the resistance of the coil?, Ignore any changes in the resistance of the coil due to the increase in temperature., , ȍ, 7.6, , In a graphing calculator, there are four AAA batteries, which together provide a potential difference of 1.5 V and a total of, 6.0 W·h of energy. If the batteries last 73 hours before they need to be replaced, what was the average current through the, calculator?, A, , 7.7, , 474, , A 12 V car battery powers a light bulb. The bulb consumes 60 watts of power. (a) Assuming that the bulb is the only, significant source of resistance in the circuit, find its resistance. (b) How much power would be consumed by a light bulb, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems
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whose resistance was 1.0 ȍ less?, (a), , ȍ, , (b), , W, , Section 8 - Sample problem: solar panels, 8.1, , The power output of the Sun is 3.91×1026 W. At an energy cost of $0.11/kW·h, what is the value of the energy produced by, the Sun in one minute?, $, , 8.2, , The intensity of sunlight at the distance of the Earth's orbit is 1380 W/m2. An Earth-orbiting satellite has a solar panel that, measures 1.35 m by 4.86 m, which converts solar energy to electrical energy with an efficiency of 26%. In one hour, how, much electrical energy does the panel produce? Assume that the satellite's attitude control jets keep the panel oriented, perpendicular to the incoming sunlight., J, , Section 11 - Sample problem: power transmission, 11.1 Power transmission lines often use a form of electric current called alternating current, but in many regions, such as the, Province of Quebec, high-voltage direct-current lines are used instead. Direct current is the kind of electric current you are, studying in this chapter. A certain direct-current power transmission line has a resistance of 0.255 ȍ/km. 812 kV of potential, drives the current from the generating station to a city located 125 km from the plant. What is the power loss due to resistance, in the line?, W, , Copyright 2007 Kinetic Books Co. Chapter 25 Problems, , 475
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26.0 - Introduction, Capacitors store charge, and in doing so, they store, energy. They are used to store energy in devices, ranging from camera flashes to defibrillators, the, systems used to “shock” the human heart back into its, proper rhythm., Capacitors are employed for other purposes as well., Because it takes a fixed amount of time to charge or, discharge a given capacitor, capacitors are used in, circuits where timing is essential, such as the tuner of, a radio., At the right, you see the design of a basic capacitor. It, The "capacitor tree" (left) at the Fermilab National Accelerator Laboratory, consists of two conducting plates separated by air., stored huge amounts of energy for high-energy accelerator experiments., There is a wire connected to each plate, and these, wires are attached to a source of potential difference that has caused the plates to, become charged. The amounts of charge on the plates are equal in magnitude but, opposite in sign., You can launch the simulation and try an experiment that demonstrates one of the, capacitor’s fundamental properties. In the simulation, you can change the potential, difference across the plates with the ǻV controller next to the capacitor. When you, change the potential difference, note what happens to the amount of charge stored, by the capacitor. The approximate charge is displayed visually on the capacitor, plates; for a more precise value you can rely on the charge readout gauge displayed, below. Is the charge proportional to the potential difference?, , 26.1 - Capacitors, , Capacitor: A device with, two conducting plates that, can hold equal but opposite, amounts of charge., Capacitance: The ratio of, the charge on one of the, capacitor’s plates to the, potential difference between the plates., , Various capacitors, , The simplest capacitors consist of two parallel metal plates separated by a narrow gap., We use this configuration, a parallel-plate capacitor, for the definitions above. A battery, (or other source of potential difference) causes the plates to become electrically, charged: The plates contain equal but opposite amounts of charge. An insulator, which, can be as simple as an air gap, separates the plates., The capacitance of a capacitor tells how much charge it can store for a given potential, difference between the plates; specifically, it equals the value of the positive charge on, one plate in coulombs, divided by the potential difference in volts, a relationship shown, in Equation 1. The amount of charge on one plate is represented by the letter q. The, potential difference is also measured as a positive value, so charge, potential difference, and capacitance are all positive values., The geometry of a capacitor and the nature of its insulator are the two factors that, determine its capacitance. Capacitors with larger plates, or with plates separated by a, narrower distance, have greater capacitance., , Capacitors, Composed of two conductors, Separated by insulator, Equal, opposite charges on conductors, , The other factor that can affect capacitance is the nature of the insulator between the, , 476, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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plates. This insulator is called a dielectric. Air is one dielectric, but other dielectrics can, be used to increase the capacitance., The unit of measure of capacitance is the farad (F), named after the physicist Michael, Faraday. A one-farad capacitor has a great deal of capacitance. It can store a lot of, charge. Many commonly encountered capacitors have capacitances ranging from, microfarads (10í6 F) down to picofarads (10í12 F). However, capacitors with greater, capacitances certainly exist: car audio enthusiasts often boast systems employing onefarad capacitors., The movement of electrons creates the charges on the two plates of a capacitor. They, move away from one plate, leaving a net positive charge behind, and go toward the, other plate, which takes on a negative charge. Note that the electrons do not move, across the gap between the plates. Rather, they move along the wires whose ends you, see in the illustrations., A source of potential difference such as a battery exerts a force on the electrons,, causing them to move. This force through a distance constitutes work. The amount of, work done during the charging process equals the electric potential energy stored by, the plates. The greater the charge on its plates, the greater the amount of energy a, particular capacitor is storing., , Capacitance, Relationship between charge, potential, difference, , C = capacitance, q = positive charge on one plate, ǻV = potential difference across plates, Units: farads (F), , What is the capacitance of this, capacitor?, C = q/ǻV, C = 0.231 C / 330 V, C = 0.000700 F = 700 ȝF, , Copyright 2000-2007 Kinetic Books Co. Chapter 26, , 477
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26.2 - Interactive checkpoint: the fuel-cell car, How much charge does an 8.00-farad, ultracapacitor accumulate when it is, charged with a potential difference of, 200 V?, , Honda’s FCX fuel-cell automobile uses a custom built, 8.00-farad ultracapacitor that provides power in faster bursts than a fuel cell can deliver, on its own, allowing the car to accelerate quickly., The ultracapacitor is charged by both the car’s fuel cell and energy recovered each time the car brakes. Together, these energy sources can, be used to provide a potential difference of 200 V across the capacitor electrodes (plates)., , Answer:, , q=, , C, , 26.3 - Physics at work: capacitors and computer keyboards, Your computer keyboard may contain capacitors. In, some types of keyboards, each key sits above its own, capacitor. Although more expensive to produce than, those based on other designs, capacitor based, keyboards are more reliable. The bottom part of the, capacitor is fixed, and the top is a movable plate. A, springy insulating material separates the plates. As, you press down on a key, the two plates of its, capacitor are pushed closer together. Since this, changes the geometry of the capacitor, it also, changes the capacitance: Specifically, it increases it., A microprocessor in the keyboard interprets the, change in capacitance as a signal that the key has been pressed., The keyboard microprocessor relays this information to the computer’s central, processing unit, which responds to it according to which application is running. For, example, a word processor might display the keystroke as a character, while a game, might interpret it as a command for a player to shoot a basketball., , Computer keyboards, Keystroke reduces distance between, plates, Microprocessor detects changed, capacitance, , 478, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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26.4 - Energy in capacitors, Capacitors store energy. A battery or other device performs work as its electric field, pulls electrons from the positive plate of a capacitor and pushes them to the negative, plate. The force of this field drives electrons away from the attractive force of the, positive plate and toward the repulsive force of other electrons already on the negative, plate. This work increases the electric potential energy of the capacitor., To the right you see two equations that express the energy stored in a capacitor. The, first equation is stated in terms of the charge on the capacitor; the second is stated in, terms of the potential difference across the capacitor. We can derive these equations by, calculating the work required to move each charge from one plate to the other., The example problem asks you to find the amount of energy stored in a capacitor with a, potential difference of 25,000 V across its plates. This is representative of the voltages, found in television sets. The stored energy in a TV capacitor can give you a dangerous, shock, even after the set is unplugged. This is why the back panels of televisions are, labeled with warnings against attempting to disassemble them yourself. We recommend, that you follow this advice!, , Energy in a capacitor, Battery’s field causes electrons to move, Charges create electric field, Electric field stores energy, , PEe = electric potential energy, q = charge on capacitor plate, C = capacitance, ǻV = potential difference, , What is the electric potential, energy stored in this capacitor?, PEe = C(ǻV )2 /2, PEe = (0.0015 ȝF)(25,000 V)2 /2, PEe = (1.5×10í9 F)(2.5×104 V)2 /2, PEe = 0.47 J, Copyright 2000-2007 Kinetic Books Co. Chapter 26, , 479
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26.5 - Interactive checkpoint: energy in a fuel-cell car, An FCX fuel-cell car brakes and stops, at a traffic signal. During this time, its, 8.00-farad ultracapacitor gets fully, charged by a combined potential, difference of 200 V generated by the, brakes and the fuel cell. When the, light turns green, the 1680-kilogram, car initially uses only energy from the, capacitor to accelerate., How fast will the car be moving when, the capacitor is fully discharged?, , In this problem, assume that road friction, air resistance, and electrical resistance use up 20.0% of the energy available, and that the rest of the, electric potential energy stored in the capacitor’s electric field is transformed into the kinetic energy of the car., State your answer in miles per hour (1.00 m/s = 2.24 mph)., , Answer:, , v=, , mph, , 26.6 - Physics in medicine: defibrillator, Although electricity can be harmful if misused, doctors, have used it for medicinal purposes since the days of, the ancient Greeks. For example, early physicians, used discharges from the electric torpedo fish to, relieve aches and pains., Electricity continues to be used in medicine. If you, have ever watched the television show ER, chances, are you have become well acquainted with a device, called a defibrillator. The heart relies on electrical, impulses to cause its muscles to contract. When the, muscle cells of the heart begin to contract out of, synchronization, or to fibrillate, a defibrillator can be, used to jolt them back into a synchronous rhythm., Contrary to popular belief, defibrillators are not used to “restart” a heart that has, stopped beating. You may get the impression from TV shows that doctors use, defibrillators as frequently as stethoscopes, but that is just one of the exaggerations of, television. However, these instruments are becoming standard equipment in, ambulances and on airplanes., , Cardiac defibrillator., , A conceptual diagram of a defibrillator is shown on the right. The source of potential, difference on the left charges the capacitor. When the switch on the right is closed, the, capacitor rapidly discharges, sending an electrical current through the patient’s heart., The example problem shows how to calculate the average current when the discharge, time, capacitance and potential difference are known. The current is large, but it lasts, for just a few milliseconds, providing a powerful but brief shock., , Defibrillator, Capacitor used in heart defibrillator, ·Battery charges capacitor to 2500 V, ·Discharges in milliseconds, sending, large current through heart, , 480, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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The defibrillator capacitor, discharges in 11 ms. How much, current does it send through the, heart?, C = q/ǻV, q = CǻV = (68 × 10í6 F)(2.5 × 103 V), q = 0.17 C, I = ǻq/ǻt, I = (0.17 C) / (0.011 s), I = 15 A, 26.7 - An insulator (dielectric) in an electric field, The simplest capacitor is made up of two plates separated by a vacuum. In this section, we will discuss what happens when an insulating material, a dielectric, is placed, between the plates., Like a vacuum, this insulating material is intended to prevent the flow of charge, between the plates, but it has another effect, as well: It reduces the overall strength of, the field between the plates, and this proves to be a desirable effect., To understand how a dielectric reduces the net field, let’s first consider what happens, when an atom is placed in the electric field generated by two charged plates. It might, seem that nothing would happen because the atom is electrically neutral., However, an atom is made up of a positive nucleus surrounded by negative electrons. If, the field is very strong, it can ionize atoms, separating electrons from the atoms. At this, point, the dielectric material becomes a conductor: Current can flow between the plates,, causing the capacitor to break down. Although an interesting phenomenon to witness,, this outcome is not relevant to the rest of this section., Less extreme fields can polarize an atom, in essence stretching it so that its electrons, tend to be on one side and its positive nucleus is on the other. This turns the atom into, a dipole, a body with positively and negatively charged regions., , Dielectric, When plates are charged:, Dipole alignment creates field in, dielectric, ·Dielectric field opposes field of plates, , Some kinds of molecules, such as water molecules, are always dipoles. This type of, molecule (called a polar molecule) has regions of positive and negative charge based, on its structure. When no external electric field is present, the dipoles of substances, such as water are randomly aligned, and they create no net electric field., When a substance containing dipoles is placed in the electric field between a charged, capacitor’s plates, the field will cause some of the dipoles to align so that their positive, poles point toward the negative plate and their negative poles point toward the positive, plate., Many, but by no means all, of the dipoles that compose the substance will align in this, manner with the external field. Their random thermal motion provides a constant, counterbalance to the organizing tendency of the field. The effect of the alignment that, does occur is to weaken the overall field between the plates. There are two ways to, understand the weakening of the capacitor field in the presence of a dielectric., , Dielectric effect, Dielectric weakens overall capacitor, field, , First, consider the net effect of the individual fields of all the aligned dipoles. Each, aligned dipole has its own electric field, oriented in the direction opposite to the field, created by the plates. This is shown in Concept 1. The aligned dipoles’ fields reduce the overall field between the plates because they point, oppositely to the capacitor’s field., You can also look at the diagram in Concept 2 to understand why the overall field is diminished. In the bulk of the dielectric the net charge is, zero. This is because every aligned positive pole there is next to an aligned negative pole on a neighboring dipole, which it balances out., However, on the far left there is an unbalanced layer of negative charge, and on the far right there is an unbalanced layer of positive charge., , Copyright 2000-2007 Kinetic Books Co. Chapter 26, , 481
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This means there is a layer of negative charges adjacent to the positive plate of the capacitor and a layer of positive charges adjacent to the, negative plate., For this reason the entire dielectric material can be said to be polarized (to have positive and negative regions). This polarization decreases the, net charge in close proximity to the surface of each capacitor plate. The net charge of each plate is now the charge on the plate surface minus, the charge on the adjacent face of the dielectric. Less effective charge on each plate surface means the overall field between the plates is, weaker., , 26.8 - Dielectrics, , Dielectric: An insulating substance placed, between the plates of a capacitor to increase its, capacitance., Dielectric constant (ț): Measures reduction in, electric field caused by a dielectric. It equals the, ratio of the field in a vacuum to the field in the, dielectric., Dielectrics are insulating materials used in capacitors to increase their capacitance., Effective dielectrics make possible the manufacture of small, high-farad capacitors., , Dielectric, Insulator that increases capacitance, , The materials discussed in this section are linear dielectrics. The dielectric field of a, linear dielectric is linearly proportional to the strength of the external electric field., The dielectric constant, represented by ț (Greek letter kappa), is a property of a, material. It equals the ratio of two electric fields: the electric field inside a capacitor with, a vacuum separating the plates and the field strength for the same capacitor charge, with the dielectric present. This is stated in Equation 1. The greater the dielectric field, within a dielectric, the more it diminishes the field caused by the plates, and the greater, the value of the dielectric constant., Dielectrics increase the capacitance of capacitors. As is stated in Equation 2, the, capacitance with the dielectric present equals the capacitance without it times the, dielectric constant., The table in Equation 3 lists some dielectric constants. As you can see, the dielectric, constants for vacuum and air are quite close. Strontium titanate, a substance used in, commercial capacitors, has a far greater dielectric constant, especially at low, temperatures, than a vacuum or air., , Dielectrics and electric fields, Dielectric diminishes electric field, , Dielectrics also can be classified by their dielectric strength. The dielectric strength, characterizes the field strength at which the dielectric becomes a conductor and charge, will flow through it. This is not a desirable effect, so capacitors are often labeled with, their maximum safe field strength (in V/m). Like many stress properties, the dielectric, strength of a material can be difficult to determine precisely, so the stated dielectric, strength should be an approximate, conservative value. The dielectric strengths of, selected materials are also shown in the table., Lightning is a very visible example of the breakdown of a dielectric í air í between, oppositely charged objects, such as two clouds or a cloud and the nearby surface of the, Earth. When the electric field between them becomes strong enough, the separating, atmosphere becomes a conductor and current flows through it as a lightning bolt., , Definition of dielectric constant, ț = Evacuum/Edielectric, ț = dielectric constant, E = field in vacuum or dielectric, , 482, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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Capacitance with dielectric, present, Cț = țC, Cț = capacitance with dielectric, ț = dielectric constant, C = capacitance with vacuum, , Table of dielectric constants, , The capacitors have identical, dimensions and charges. What is, the dielectric constant of the, dielectric on the right?, ț = Evacuum/Edielectric, ț = (4.0 × 103 V/m) / (2.8 × 103 V/m), ț = 1.4 (dimensionless), , Copyright 2000-2007 Kinetic Books Co. Chapter 26, , 483
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26.9 - Sample problem: dielectric and potential difference, A battery causes the plates of a, capacitor to charge so that the, potential difference between them is, 9.0 V. The capacitor is then isolated, so that its charge remains constant, and the battery no longer acts on it., After a dielectric (Ȁ = 16) is inserted, between the plates, what is the, potential difference between them?, , Variables, initial potential difference, , ǻV = 9.0 V, , initial capacitance, , C, , charge on capacitor, , q, , dielectric constant of material, , ț = 16, , capacitance with dielectric present, , Cț, , potential difference with dielectric present, , ǻVț, , Strategy, 1., , Use the definition of capacitance, and the equation for capacitance with a dielectric present, to determine the potential difference when, the dielectric is present., , Physics principles and equations, The definition of capacitance is, , C = q/ǻV, The relation between capacitances with and without a dielectric present is, , Cț = țC, Charge in any isolated system is conserved., Step-by-step solution, We use the definition of capacitance twice; once when the dielectric is not present, and again when it is. The amount of charge remains, constant since the capacitor is isolated., , Step, , Reason, , 1., , q = CǻV, , definition of capacitance, , 2., , Cț = țC, , capacitance with dielectric, , 3., , definition of capacitance, , Now we combine the above expressions to find a relation between the potential differences with and without a dielectric present. We evaluate, the resulting equation to answer the problem., , Step, , Reason, , 4., , substitute equations 1 and 2 into equation 3, , 5., , simplify, , 6., , evaluate, , The potential difference between the plates after inserting the dielectric turns out to be the original potential difference divided by the dielectric, constant. The higher the dielectric constant, the more the potential difference will be decreased after the dielectric is introduced., , 484, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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This result is interesting when applied to the potential energy PEe, energy becomes, , = ½C(ǻV)2 stored by a capacitor. After you insert a dielectric, the potential, , In other words, the potential energy decreases when you insert a dielectric between the plates of a capacitor. Why? The capacitor is doing, work on the dielectric as you insert it, rotating many of its small dipoles into alignment., , 26.10 - Interactive checkpoint: dielectric in a capacitor, A parallel plate capacitor has a, capacitance of 3.50 ȝF when a, vacuum separates its plates. You, insert an insulator with a dielectric, constant of ț = 5.60 and apply a, potential difference of 145 V across, the plates., How much charge builds up on a, plate?, , Answer:, , q=, , C, , 26.11 - Physics at work: commercial capacitors, Capacitors in electronic equipment, like a stereo, system, that you might have at home are often made, from two layers of metallic foil backed by thin sheets, of a dielectric material. When these layers are rolled, up, the result is a multilayer cylindrical capacitor with, alternating positive and negative cylindrical, conductors, separated by the dielectric material., Several capacitors of this type are shown in Concept, 1., Higher voltage capacitors can consist of sets of thin, parallel metallic plates immersed in insulating silicone, A variable capacitor used for tuning a radio., oil. Other capacitors, capable of storing large, amounts of charge, are made with insulators, consisting of ceramic materials that can have very high dielectric constants., A variable capacitor, like the one in the picture above, features interleaved metal plates., You change the capacitance by rotating the knob, increasing or decreasing how much, of the area of each plate is close to the neighboring plates on either side. Devices like, this can be used to change the capacitance of a circuit in a radio tuner, altering what is, called the resonant frequency of the circuit and enabling it to “tune in” a radio station, broadcasting at a particular frequency., Electrolytic capacitors can operate with very high potential differences across their two, conducting surfaces. An electrolyte is a nonmetallic conductor, or a substance that, when dissolved in a suitable solvent becomes a conductor. In an electrolyte, current is, composed of ions, not electrons., , Commercial capacitors, , In an electrolytic capacitor, one of the conductors is a solid or porous metal electrode,, Rolled foil capacitors, the dielectric is an insulating metal oxide coating on the surface of the electrode, and, the other conductor is an electrolytic liquid or solid. Since the distance between the, “plates” is the thickness of the insulating metal oxide, which is quite small, tiny electrolytic capacitors can have extremely high capacitance. The, capacitors used in certain fuel-cell automobiles are a form of electrolytic capacitor., , Copyright 2000-2007 Kinetic Books Co. Chapter 26, , 485
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26.12 - Gotchas, The total amount of charge on the plates of a capacitor is designated by q. No, the amount of charge on one plate of the capacitor is q. Each, plate has the same amount of charge: positive on one plate and negative on the other. The total charge is zero., Capacitors have a net charge. No, the two plates have equal and opposite amounts of charge, so the capacitor is electrically neutral. When you, say “the capacitor has a charge” or the like, it means each plate is charged., , 26.13 - Summary, A capacitor is a device with two conductors that are placed in close proximity, and, which store equal but opposite amounts of charge., , Definition of capacitance, Capacitors are characterized by their capacitance, represented by the symbol C,, which equals the amount of charge on one conductor divided by the potential, difference between the conductors. The unit of capacitance is the farad (F)., 1 F = 1 C/V., By storing charge, capacitors also store electric potential energy. A battery or other, device causes charge to accumulate on the conductors of the capacitor. The, opposite charges on the conductors create an electric field between them. Electric, potential energy is stored in the electric field., , Potential energy in a capacitor, , Between the conductors of a capacitor there may be a dielectric, an insulator that, can increase the capacitance by decreasing the field strength between the, conductors for a given charge. The factor characterizing the increase in capacitance, (or the decrease in field strength) is called the dielectric constant, ț., , Dielectric constant, , ț = Evacuum/ Edielectric, Cț = țC, , 486, , Copyright 2000-2007 Kinetic Books Co. Chapter 26
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Chapter 26 Problems, , Conceptual Problems, C.1, , If you increase the potential difference, between the plates of a capacitor, what, happens to (a) the amount of charge on each, plate and (b) the net charge of the capacitor?, (a), , (b), , C.2, , i., ii., iii., i., ii., iii., , It increases, It decreases, It stays the same, It increases, It decreases, It stays the same, , A dielectric is inserted between the plates of an isolated, charged capacitor, replacing the vacuum that was there before. Its, dielectric constant is greater than one. (a) Has the capacitance increased, decreased, or stayed the same? (b) Is there more,, less, or the same amount of charge on the capacitor?, (a), , (b), , i., ii., iii., i., ii., iii., , Increased, Decreased, Stayed the same, More charge, Less charge, Same amount of charge, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to determine whether the charge on a capacitor is proportional to, the potential difference., i. Charge is proportional to potential difference, ii. Charge is not proportional to potential difference, , Section 1 - Capacitors, 1.1, Suppose you disassemble a cell phone and remove a capacitor inside labeled " 6.5eí11 F." If you apply a potential difference, of 5.0 volts across the capacitor, how much charge would you expect to be on the positive plate?, C, 1.2, , When a 9.0 V battery is connected to a capacitor, it puts 6.0eí4 C of charge on the capacitor. When a different battery with, an unknown voltage is connected to the same capacitor, it puts 1.5×10í4 C of charge on the capacitor. What is the unknown, battery's voltage?, V, , 1.3, , A capacitor accumulates 6.4eí5 C of charge when connected to a 9.0 volt battery. What is its capacitance?, F, , 1.4, , A 6.1eí6 F capacitor accumulates 51 μC of charge when connected to a battery. What is the battery's voltage?, V, , 1.5, , When a battery is connected to a 2.0 microfarad capacitor, it places an equal but opposite charge on each plate, of, magnitude 4.0×10í5 coulombs. How much would the same battery charge a 3.6eí6 farad capacitor?, C, , Section 4 - Energy in capacitors, 4.1, , In an electronic circuit, you need a capacitor to store 4.50eí9 J of energy. You have 1.50 volts available to charge it with., What capacitance should you choose?, F, , 4.2, , Suppose a capacitor in a video camera has a capacitance of 26.0 μF, and is currently holding 5.20eí5 C of charge. How, , Copyright 2007 Kinetic Books Co. Chapter 26 Problems, , 487
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much potential energy is stored in the capacitor?, J, 4.3, , How much potential energy is stored in a 2.10eí6 F capacitor charged by a potential difference of 651 V?, J, , 4.4, , A warning label on a 56.1 μF capacitor says that the capacitor can safely hold no more than 5.00 J of energy. What is the, maximum potential difference you can safely apply to the capacitor?, V, , 4.5, , A capacitor that initially holds a charge of 3.31×10í3 C on its plates fully discharges, in the process doing 5.50 J of work., What is its capacitance?, F, , 4.6, , A 79.1 μF capacitor is initially uncharged. After you expend 263 J of energy to charge it up, how much charge does it hold on, each plate?, , 4.7, , A 6.40 nF capacitor is initially charged by a potential difference of 1.40×103 V. Later, the potential difference is increased so, , C, that 5.10 μC of charge are added. What is the final electric potential energy stored in the capacitor?, J, , Section 6 - Physics in medicine: defibrillator, 6.1, , A certain defibrillator sends 0.12 C of charge through the heart. The capacitor responsible for holding this charge before it is, released has a capacitance of 95 μF. How much energy is released when it discharges?, J, , Section 8 - Dielectrics, 8.1, , The dielectric paste used in a tiny 360 nF capacitor has a dielectric constant of 50. What would the capacitor's capacitance, be without the dielectric?, nF, , 8.2, , An isolated parallel-plate capacitor with no dielectric has an electric field of 8.4×103 V/m. When you inject dielectric paste, from an unlabeled jar, the field goes down to 1.4×103 V/m (the charges on the plates of the capacitor stay the same). What is, the dielectric constant of the unlabeled paste?, , 8.3, , A tiny capacitor found in a clock radio uses a dielectric with a dielectric constant of 262. The electric field strength between, the "plates" of this capacitor is 6.11×105 V/m. If the dielectric were replaced with a vacuum, what would the strength of this, electric field become, assuming the charges on the plates remain the same?, V/m, , 8.4, , When you insert a sheet of wood (ț = 2.00) between the plates of a capacitor, the capacitance becomes 22.0 nF. What, capacitance will it have if you insert a block of glass (ț = 6.00) instead?, nF, , Section 11 - Physics at work: commercial capacitors, 11.1 A variable capacitor has a range of capacitances from 2.0 nF to 20 nF. What is the difference between the most and least, charge that can be stored on the variable capacitor when it is connected to a 9.0 V source of potential difference?, C, , 488, , Copyright 2007 Kinetic Books Co. Chapter 26 Problems
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27.0 - Introduction, Electric circuits include components such as batteries, resistors and capacitors., These basic elements can be combined in a myriad of ways. How these, components function, by themselves and in combination, defines the fundamental, operation of electric circuits. In this chapter, we examine direct current electric, circuits, circuits in which the current always flows in the same direction., In this chapter, you will learn how to analyze direct current circuits. On the right is a, simulation in which you can build your own electric circuits. Initially, the circuit, consists of a battery, a light bulb and wires that connect these components. The, simulation also contains additional wire segments and light bulbs., You can place light bulbs in various places in the circuit. Pay special attention to the, brightness of the light bulbs: The brighter the light, the more power the circuit is, supplying to it. You can also use two devices in the control panel to study the circuit., One, a voltmeter, measures the potential difference across components, while the, other, an ammeter, measures the current flowing through the wire at any location in, the circuit., In this simulation, the battery and wires effectively have zero resistance. The simulation includes a total of five light bulbs, each with a, resistance of 50 ohms. You add and remove light bulbs and wires by dragging them; the components will snap into place. Only one light bulb, can be placed on each wire segment., The purpose of this simulation is for you to experiment with the electric components in a circuit. One way to start is by assessing the circuit in, its initial state using the voltmeter and ammeter. How does the potential difference across the battery compare with the potential difference, across the light bulb? What about the current? Is it the same everywhere or does it differ from place to place?, Now add another light bulb above the first one in the circuit: Snap in two wire segments in a vertical orientation, and then put a segment, containing a light bulb between them. How does the potential difference across the battery now compare to the potential difference across each, light bulb? Is the current still the same everywhere? This time you should find that the current can differ by location., You can also use the voltmeter to confirm Ohm’s law. Since you are told the bulb’s resistance (50 ohms) and can measure the potential, difference across it using the voltmeter, you can use the law to calculate the current flowing through the wire segment containing the bulb. You, then can verify your calculation using the ammeter., You are probably thinking that this introduction has asked you to answer a lot of questions! If you cannot answer all the questions above, that is, fine. This chapter is dedicated to preparing you to address them., , 27.1 - Electric circuits, In this section, we provide an overview of the components of a basic circuit. All these, components merit more discussion, but here we want to provide some context on how, they function together in a circuit., A flashlight provides an example of a circuit. The flashlight we show contains two, batteries, a switch, a light bulb, and some metal wires that connect these components., In the flashlight, the batteries are the source of the energy that causes the net motion of, charge in the circuit. The ends of a battery are at different electric potentials. There is a, potential difference across the two terminals that are at opposite ends of the battery. A, typical potential difference for a battery like those shown is 1.5 volts. The terminal with, the greater electric potential is marked with a plus (+) sign, and the terminal with the, lower electric potential is marked with a negative (í) sign. Putting together two batteries, as shown creates a potential difference across the two batteries of approximately 3.0, volts. The batteries are used to create an electric field that causes a net flow of, electrons: a current., , Electric circuit, Set of electric components connected, by wires, , Current will only flow when there is a complete path: a loop, as opposed to a dead end., (Circuit comes from the Latin word circumire, to go around). When the switch is in the “off” position, it creates a gap in the circuit, and current, cannot flow. When the switch is pushed to “on”, the gap is closed and a current can flow., We have highlighted the circuit inside a flashlight in Concept 2. Let’s trace the direction of conventional current around the circuit. A positive, charge starts at the positive terminal of the battery on the right, and moves through a coiled wire called the filament in the light bulb. It exits the, filament and moves through the wire that contains the switch. The charge then flows through the batteries and starts its round trip over again., There is resistance inside the batteries, in the wires and in the light bulb. The resistance inside the batteries and in the wires is minor compared, to that of the filament in the light bulb, and it is often reasonable to ignore these minor resistances. The light bulb is the major source of, resistance, and it supplies what is called the load resistance of the circuit. The resistance of this component and the potential difference across, the batteries determine the amount of current in the circuit., The flashlight creates light when current flows through the filament. Modern day filaments often are made of very thin tungsten wires. As a, , Copyright 2000-2007 Kinetic Books Co. Chapter 27, , 489
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current passes through the filament, electrons lose energy in collisions with the atoms, of the filament, and the filament’s temperature increases. It becomes hot (up to, 3000°C), and energy leaves the filament in the forms of heat and light., Although tungsten is an effective filament, when it is hot the tungsten vaporizes and the, filament becomes thinner. This increases its resistance, which means it shines less, brightly. Eventually, when it “burns out”, it becomes so thin and brittle that it ruptures., The flashlight contains all the essential elements of a circuit. It has a source of energy,, the batteries. They create a potential difference that causes electrons to move. As the, electrons move through the light bulb, they encounter resistance, which causes this, resistor to dissipate energy., The current flows only in one direction in this circuit, which makes it a direct current, circuit. Direct current circuits are the topic of this chapter. We study these circuits, primarily when the current has reached a steady state í a constant flow í not in the, brief moments when the current changes, such as immediately after the flashlight is, switched on or off., Scientists and engineers use symbols to represent components in circuit diagrams, as, shown in Concept 3. The thin black lines represent wires. The battery and resistance, symbols are labeled in the circuit. The switch is “on”, so it is in its closed position in this, diagram. The red arrow indicates the direction of flow of the conventional current., , Circuits usually contain:, An energy source, A load resistance, Wires connecting it all in a loop, A switch, , Drawing electric circuits, Use circuit diagrams with symbols, , 27.2 - Electromotive force, , Electromotive force (emf, Ǜ):, Maximum potential, difference from an energy, source such as a battery., A potential difference applied across a conductor, causes a current to flow. Common devices such as a, flashlight need a continuing source of potential, difference í an emf í so that current will keep flowing, and the flashlight’s bulb will stay illuminated., A battery often supplies the emf. The symbol for emf, is Ǜ. Like any potential difference, an emf is, Batteries are a common source of emf., measured in volts. There are many sources of emf:, electric generators, solar photovoltaic cells and so on. Even living creatures can be an, emf source. Humans rely on emfs generated in the body to cause electric currents in, nerves., The term “electromotive force” is misleading because an emf is not a force. It is a, potential difference and its unit is the volt. It is a well-established term in physics,, however, and we will use it too. In any case, since we typically write it in its abbreviated, form as emf, you should not too often be confused by seeing the word “force”., A battery is the typical emf source for direct current circuits. Chemical reactions within, the battery cause one terminal of the battery to be positively charged, and the other to, be negatively charged. These terminals are marked with plus (+) and minus (í) signs., Batteries are classified by their emf. A typical battery used in a flashlight has an emf of, 1.5 volts, while a car battery has an emf of 12 volts. If a battery has an emf of 1.5 volts,, this means that the electric potential of its positive terminal is 1.5 volts higher than that, of its negative terminal., , emf (Ǜ), A potential difference, , Sometimes batteries are referred to as “charge pumps.” They increase the potential energy of the charge flowing through the circuit. The unit of, , 490, , Copyright 2000-2007 Kinetic Books Co. Chapter 27
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emf, the volt, equals joules per coulomb. You can think of a nine-volt battery as doing nine joules of work on each coulomb of charge that flows, through it., Batteries also are defined by how much energy they can supply over their lifetimes. AAA and D batteries are both 1.5-volt batteries, but the, larger D battery supplies more energy over its lifetime. This total available energy is measured in watt-hours. A small battery for watches will, have about 0.1 watt-hours of total energy, while a car battery has a total available energy of about 500 watt-hours., When current flows through a battery, it encounters resistance. This is called a battery’s internal resistance. This means that when placed in a, circuit, a battery’s emf and the potential difference across its terminals are not the same. The emf is greater than the potential difference, (unless the battery is being charged, in which case the emf can be greater). The internal battery resistance is usually minor, however. Many, times in this chapter we will treat it as zero, and consider the emf and the potential difference to be the same., , 27.3 - Energy and electric potential in a circuit, We will use the simple circuit shown to the right as our starting point for discussing how, to analyze circuits. Specifically, it is important to correctly determine the changes in, electric potential across various components in the circuit., The circuit in Concept 1 contains a resistance-free battery and a resistor: a flashlight, bulb. As the diagram reflects, the resistor has a resistance of 5.0 ohms, and the battery, has an emf of 1.5 volts. We want to determine the potential difference across the, resistor., To analyze the changes in electric potential occurring around the circuit, consider some, charge conducting a hypothetical journey around the circuit loop, as shown in Concept, 2. Imagine 0.5 coulombs of positive charge traveling one complete loop in the direction, of conventional current (clockwise in this case). What happens to the potential energy of, this charge as it passes through each component í the battery and the resistor í in, completing a round trip around the loop?, , Battery-resistor circuit, , For starters, the potential energy must be the same at the beginning and end of the, closed path, because the electrostatic force is a conservative force. Just as your, gravitational PE is the same after you take a round trip walk, even if you go uphill and, down on the way, so too the charge’s electrostatic PE is the same when it returns to its, starting point., To determine the changes in potential energy as the charge makes its journey through, each component, we will use an equation that relates the change in potential energy to, charge and potential difference: ǻPE= qǻV., We will start with the charge at the negative terminal of the battery, and have it flow, through the battery to the positive terminal. The change in potential energy for a, positive charge traveling across the battery in this direction is positive. The battery must, do work on this positive charge to move it toward the positive terminal. The charge’s, change in PE is the product of the charge and the potential difference, (0.5 C)(1.5 V),, which equals +0.75 J. The electric potential energy of the charge increases and the, chemical potential energy of the battery decreases correspondingly., Next, the charge flows through the wire around the circuit. We assume that the wire has, negligible resistance. To complete a circuit, the charge must pass through theresistor., This is the one location in this simple circuit where the charge loses energy before, returning to its starting point. The resistor heats up, giving off this energy as heat. The, charge’s change in potential energy in the resistor must equal í0.75 joules, because the, sum of the changes in potential energy around the complete loop must equal zero., , Circuit energy analysis, PE of charge is unchanged after any, round-trip, ·Energy increases in battery, ·Energy decreases in resistor, , Now that we have assessed the potential energy changes around a circuit loop, we can, also assess the changes in electric potential. Concept 3 shows the changes in electric, potential around a complete circuit loop. The changes in electric potential can be, determined using the equation ǻV = ǻPE/q., Concept 3 again shows a clockwise path around the circuit. Moving in this direction, the, electric potential increases by +1.5 V as the battery is traversed. The potential, difference across the battery traveling in this direction is equal to the change in PE,, +0.75 J, divided by the amount of charge, 0.5 C. Across the resistor, since the change, in PE is í0.75 J in this direction, the change in potential is í1.5 V. The electric potential, is greater on the “upstream” end of the resistor than on the “downstream” end. The, charge moves across the resistor from a region of a higher electric potential to one of, lower electric potential., , Electric potential changes in, circuit, Potential changes around circuit sum to, , zero, What we have shown is fundamental and important. In this two-component circuit, the, change in electric potentialacross one component í the battery í is equal to but, opposite the change in electric potentialacross the other component: the resistor. The changes sum to zero. We have shown this in a specific, case, but the rule holds in general for any closed loop around a circuit., The direction we chose to travel the circuit was arbitrary. Had we chosen to travel the circuit in a counterclockwise direction, the change in, electric potential would be negative as we crossed the battery and positive as we crossed the resistor. We also chose to consider a positive, , Copyright 2000-2007 Kinetic Books Co. Chapter 27, , 491
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charge rather than a negative one. In any case, clockwise or counterclockwise, positive or negative, the principle holds that the changes in, electric potential around a complete circuit loop sum to zero., This section can be used to illustrate a more general principle: As you traverse a circuit, you can add the potential differences across the, components. For instance, if you traversed two resistors, the potential difference across the two resistors would equal the sum of each, resistor’s potential difference., , 27.4 - Measuring current and potential difference, , Ammeters and voltmeters, , Ammeter: A device that measures current in a, wire., Voltmeter: A device that measures the potential, difference between two points., Ammeters and voltmeters are common tools for analyzing circuits. Both are shown, above., Ammeters measure current. An ammeter is inserted in the circuit, as shown in Concept, 2, so that all the current in that part of the circuit flows through it. Because current flows, through them, ammeters are built to have extremely low resistance in order to minimize, their impact on the circuit. An ideal ammeter would have no resistance., , Ammeter, Measures current, , A voltmeter measures the potential difference (voltage) between two points in a circuit., In Concept 3, the potential difference across the two terminals of a battery is being, measured. To use the voltmeter, a lead is placed on each side of the battery., Voltmeters are also designed to have minimal effects on a circuit. In the case of the, voltmeter, this means they are designed to have a high resistance, so little current flows, through them, and the rest continues to flow through the circuit component being, measured., Ammeters and voltmeters often are combined into one instrument called a multimeter. A, multimeter may also contain a device for measuring resistance, called an ohmmeter., , Voltmeter, Measures potential difference, , 27.5 - Series wiring, , Series wiring: Circuit wiring in which the, components are placed one after another. All of, the current flows through each component., The diagram to the right shows an example of a series circuit. The same amount of, current passes through the battery and through each light bulb. The circuit has no, branches; there are no places where the current can split to follow another path. The, light bulbs and the battery are said to be connected in series., Place an ammeter anywhere in this circuit and you will measure the same value for the, current. The same amount of current passes through the battery, the first light bulb, the, second light bulb and the wires that connect these components., If the current in a series circuit is interrupted anywhere, it is interrupted everywhere., Because there are no alternative routes, if the circuit is broken at any point, then there, , 492, , Series wiring, Current has one path through, components, , Copyright 2000-2007 Kinetic Books Co. Chapter 27
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is no detour path for the current to follow., This phenomenon is explored in more detail in Example 1. Two light bulbs are wired in, series, similar to old-fashioned holiday lights. The filament of each light bulb is part of, the circuit. When a light bulb burns out, its filament breaks, and there is no path for, current through that bulb. The example problem asks what happens to the second light, bulb in a series circuit if the first bulb burns out., , Current same at all points, , If light bulb A burns out, will light, bulb B remain lit?, No í a break in a series circuit causes, all current to stop, 27.6 - Resistors in series, , To calculate the equivalent resistance of, resistors in series: Add each resistor’s, resistance., Analyzing circuits with multiple components can be complex. The task can sometimes, be simplified by treating several components of the same type as if they were one. You, can simplify a circuit by calculating what is called the “equivalent resistance” or, “equivalent capacitance” of multiple resistors or capacitors. You can then treat the, components as one, using their equivalent resistance or equivalent capacitance., Our first case for doing this is resistors in series. Once we determine their equivalent, resistance, we can treat them as though they were one component, and determine how, much current flows through the circuit in Example 1. In the case of resistors wired in, series, the equivalent resistance is the sum of the resistances. We show this in, Equation 1., Consider the circuit shown in Example 1 that contains two resistors in series. The, question asks for the amount of current in this circuit. You can determine the current, using Ohm’s law, ǻV = IR, but what should you use for R?, , Resistors in series, Requiv = R1 + R2 +…+ Rn, Equivalent resistance = sum of, individual resistances, , First, determine the series circuit’s equivalent resistance by summing the individual, resistances. The 6 ȍ resistor plus the 4 ȍ resistor equals an equivalent resistance of, 10 ȍ. The circuit in the example has a 20-volt battery. Since the two resistors create an, equivalent resistance of 10 ȍ, you can use Ohm’s law to calculate the current through, them. It equals 20 V divided by 10 ȍ, or 2 A., Derivation. Why can resistances in series be added? We now derive the series rule for, resistors, for n resistors in series., Variables, potential difference across all resistors, , ǻVtotal, , potential difference across ith resistor, , ǻVi, , current through circuit, , I, , current through ith resistor, , Ii, , equivalent resistance of circuit, , Requiv, , resistance of ith resistor, , Ri, , What is the equivalent resistance, of the resistors and what is the, current?, Requiv = R1 + R2, Requiv = 6 ȍ + 4 ȍ = 10 ȍ, , Strategy, 1., , Find the total potential difference across all the resistors combined., , 2., , Use Ohm’s law to rewrite potential difference in terms of current and resistance., An algebraic simplification gives the series rule for resistors., , Copyright 2000-2007 Kinetic Books Co. Chapter 27, , 493
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Physics principles and equations, With components in series, the total potential difference equals the sum of the potential differences across each component., Ohm’s law relates potential difference, current and resistance:, , ǻV = IR, With components in series, the same amount of current flows through each component., Step-by-step derivation, , Step, , Reason, , 1., , ǻVtotal = ǻV1 +…+ ǻVn total ǻV is sum of series ǻV’s, , 2., , IRequiv = I1R1 +…+ InRn Ohm’s law, , 3., , I = I1 = I2 = … = In, , current the same, , 4., , Requiv = R1 +…+ Rn, , divide by equal current, , 27.7 - Interactive problem: series wiring, The circuit to the right contains two light bulbs and a 12-volt battery. One of the light, bulbs, R1, has a resistance of 75 . Your task is to determine the resistance of the, R2 bulb., You have an ammeter at your disposal. You drag this tool onto the circuit to, measure the current., Use the ammeter, Ohm’s law and your knowledge of equivalent resistance in series, circuits to determine the resistance of the R2 bulb. In order to solve the problem you, may want to first consider what the equivalent resistance of the two resistors must, be given the reading of the ammeter and the voltage shown for the battery., Type your answer in the space provided. Press CHECK, and a message will, indicate whether your answer is correct., You can try again by entering a new answer and pressing CHECK again., If you have trouble solving this problem, review Ohm’s law and the section on the, equivalent resistance of resistors in series., , 27.8 - Interactive checkpoint: voltage divider, A voltage divider is often used in, circuits where the potential difference, available is too large for certain, components. A resistor is placed in, series with the sensitive component,, “using up” some of the potential, difference available. In the circuit, shown, a 5.00 V battery is connected, to a 20.0 resistor and a sensitive, 15.0 light bulb. What is the, potential difference across the light, bulb? Use equivalent resistance to, solve the problem., , Answer:, , V=, , 494, , V, , Copyright 2000-2007 Kinetic Books Co. Chapter 27
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27.9 - Parallel wiring, , Parallel wiring: Circuit wiring that branches., The same potential difference exists across each, branch., In parallel wiring, there are junctions where multiple wires come together. A current can, flow into a junction and then divide along different paths. A path between two junctions, is called a branch., Consider electrons just to the left of junction A in Concept 1. The electrons are moving, to the right when they reach the junction. As they reach it, some turn left into the branch, along the middle wire, while others continue moving straight into the branch around the, outer wire loop. The two flows of electrons, the two currents, rejoin at the junction, labeled B., Because charge is conserved around a circuit, the sum of the currents flowing into a, junction equals the sum of the currents flowing out. This is an important principle known, as Kirchhoff’s junction rule., , Parallel wiring, Current has more than one path, ·Current may vary from loop to loop, Potential difference same across, parallel branches, , The potential difference is the same across the end points of parallel branches in a circuit. This is a crucial concept required for understanding, the functioning of parallel circuits., In the circuit on the right, a battery is connected in parallel with the light bulb in the branch AB, and with the light bulbs in the branch CD. The, potential difference is identical across the battery and these two branches. The potential differences do not sum as in a series circuit., In this circuit, the battery is a 1.5-volt flashlight battery. If you placed a voltmeter’s leads on either side of the battery, you would read a value of, 1.5 volts. You would also read the same value if you placed the leads across the middle branch or across the CD branch. The potential, difference across all three branches is identical. This confirms that all three branches are in parallel., , 27.10 - Resistors in parallel, , To calculate the equivalent resistance of, resistors in parallel: Add the reciprocal of each, resistor’s resistance. The reciprocal of this sum, equals the equivalent resistance., The equation on the right shows how to calculate the equivalent resistance of resistors, in parallel. First, take the reciprocal of each resistance. Those values are summed. The, reciprocal of that sum is the equivalent resistance of the parallel resistors., Another equation, also shown to the right, allows you to quickly calculate the equivalent, resistance when just two resistors are wired in parallel. This equation can be derived, from the first with a little algebra., The example problem on the right shows two resistors, one of 4.0 , the other of 6.0 ,, wired in parallel. To calculate the equivalent circuit resistance of these two resistors,, first invert each value and add these reciprocals. Then, invert that sum. The equivalent, resistance is 2.4 ., We can then calculate the current in the circuit. We must specify the current’s location,, because the current is not the same in all parts of the circuit. In the example problem,, we specify that we are calculating it near the battery., , Resistors in parallel, , Reciprocal of equivalent resistance =, sum of reciprocals of resistances, , In this circuit, there is less current in the branches containing the resistors than in the, For two resistors:, branch that contains the battery. You use Ohm’s law to calculate the current in the, branches with the resistors. The potential difference across each branch must be the, same as that of the battery, 20 volts. Using Ohm’s law, you determine that there is a, 5.0 A current in the branch that contains the 4.0 resistor. (The current equals the, potential difference, 20 V, divided by the resistance, 4.0 .) A similar process enables, you to determine that the current in the middle branch is 3.3 A. Since charge is, conserved, you add these two currents to determine that 8.3 A flows in the branch that contains the battery., , Copyright 2000-2007 Kinetic Books Co. Chapter 27, , 495
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Derivation, Variables, current through circuit, , Itotal, , current through ith resistor, , Ii, , potential difference across all resistors, , ǻV, , potential difference across ith resistor, , ǻVi, , equivalent resistance of circuit, , Requiv, , resistance of ith resistor, , Ri, , Strategy, 1., , Find the total current flowing through all the resistors combined., , 2., , Use Ohm’s law to rewrite the current in terms of potential difference and, resistance. An algebraic simplification gives the parallel rule for resistors., , What is the equivalent resistance, of these resistors and what is the, current at the point shown?, , Physics principles and equations, Since charge is conserved, the current flowing out of a junction equals the sum of the, currents flowing into the junction., Ohm’s law, , ǻV = IR, The potential difference is the same across all parallel branches., Step-by-step derivation, , Step, , 1., , Itotal = I1 + I2 + … + In, , Reason, , conservation of charge, , 2., , Ohm’s law, , 3., , potential differences equal, , 4., , divide by equal potential, difference, , 27.11 - Interactive problem: a parallel circuit, The circuit contains a battery and three light bulbs. The resistance of two of the light, bulbs is known. Your task is to determine the resistance of the third light bulb., You have an ammeter. It can be placed anywhere in the circuit to determine the, current at that point. Use the ammeter, along with your knowledge of parallel circuits, and Ohm’s law, to determine the resistance R., Type your answer in the space provided. Press CHECK to see if your answer is, correct. You can try again by entering a new value and pressing CHECK again., Central to solving this problem is the nature of potential difference and equivalent, resistance in parallel circuits. You also need to calculate the equivalent resistance, for resistors arranged in series, and apply Ohm’s law. Review the sections of the, textbook on these topics if you are having trouble., , 496, , Copyright 2000-2007 Kinetic Books Co. Chapter 27
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27.12 - Circuits with series and parallel wiring, Circuits commonly include some components that are wired in series and some that are, wired in parallel. The diagrams to the right show an example of such a circuit., You can use the current to determine what is in series and what is in parallel. If the, current flows along a single path from one component to another, then the two, components are in series. If the current divides, then the components are in parallel., In the circuit on the right, for example, the two resistors on the far right wire of the, circuit, labeled R2 and R3, are in series with one another. There is no junction between, them, no place for the current to split. Since there is only a single path for the current, between the resistors, the resistors are in series, as emphasized in Concept 2., This pair of series resistors is wired in parallel with the resistor in the middle, R1. Both, branches are in parallel with the battery. Consider current that exits the battery. It, encounters a junction at A where it can flow to R1 or to the R2R3 combination. The, current then recombines at B after flowing through those resistors. The same potential, difference exists across R1 as across the R2R3 combination. This means R1 is in, parallel with the R2R3 combination., , Mixed circuits, A circuit can be part series, part parallel, , Components in series, No junction separates series, components, (Single current path between them), , Part of circuit is parallel, Junctions separate parallel components, (Current differs, potential difference the, same), , 27.13 - Sample problem: circuits with series and parallel wiring, Which resistors are wired in series, and which are wired in parallel?, , The diagram shows a circuit with three resistors and asks which resistors are wired in series and which are wired in parallel., What is the strategy?, 1., , Use the definitions of series and parallel wiring., , 2., , Once two or more resistors are identified as series or parallel, simplify the circuit by determining the equivalent resistance. Then see, , Copyright 2000-2007 Kinetic Books Co. Chapter 27, , 497
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whether the resulting equivalent resistance is in series or parallel with other resistors. Continue this process until the circuit cannot be, simplified any further., Physics principles and equations, Resistors are wired in series if there is only a single path for current to flow through them., Branches are wired in parallel if the potential difference across them is the same., Step 1, We start by checking for resistors in series. As the circuit is depicted above, none of these resistors are wired in series. The current running, through any of the three resistors branches before reaching any of the other resistors., Step 2, , Next we check for parallel resistors. The resistors R2 and R3 are wired in parallel., Step 3, , Now, we reexamine the circuit having made a parallel combination out of R2 and R3. The resistor R1 is in series with the equivalent resistance, of the R2R3 combination. This may be tricky to see, but applying the tests for series wiring confirms it. The same current flows through R1 as, flows through the equivalent combination of R2 and R3. Another way of saying this is that the sum of the currents going to R2 and R3 is equal to, the current through R1., (Note: It is possible to create circuits that are irreducible, in which the components are neither in parallel nor in series. If you think you have, simplified a circuit as much as you can, you may be right.), , 27.14 - Gotchas, Components in series must have the same potential difference across them. No, they have the same amount of current flowing through them., The potential difference across components in parallel is the same. Yes., , 27.15 - Summary, An electric circuit is a set of electric components such as batteries, capacitors and, resistors that are connected directly or through wires., , Resistors in series, Current will only flow through a circuit that is closed, which means that it makes a, loop. To cause current to flow continuously through a circuit, a source of emf, Ǜ,, such as a battery, must be present., One basic type of circuit wiring is series wiring. Two circuit components are said to, be in series when current must go through both components: there is only one, possible path. The same amount of current flows through both components, and the, sum of the potential differences across the components is the net potential, difference across the combination., , Requiv = R1 + R2 +…+ Rn, Resistors in parallel, , Resistances in series add to give the equivalent resistance of a combination of, resistors., The other basic type of wiring is parallel wiring. Components are connected so that they have the same potential difference across them, but, may have different currents through them., The sum of the reciprocals of resistances wired in parallel equals the reciprocal of the equivalent resistance of a combination of resistors., , 498, , Copyright 2000-2007 Kinetic Books Co. Chapter 27
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Chapter 27 Problems, , Conceptual Problems, C.1, , A cheap flashlight is dropped into seawater. It is "short-circuited". What does this expression mean?, , C.2, , Why do batteries rated at the same emf sometimes differ so much in size? For example, consider a 1.5 AA penlight battery, compared to a D size battery., , C.3, , Why would a voltmeter with a low amount of resistance be a mediocre tool?, , C.4, , Consider a circuit containing a battery and several resistors. You measure the current flowing out one end of the battery, and, also measure the current entering the other end. Do you expect these measurements to be the same or different? Explain., The same, , Different, , C.5, , A battery is connected to either (1) three 5 ȍ resistors wired in series or (2) one 10 ȍ resistor. Which configuration will have, the most current flowing through it?, , C.6, , You have five identical light bulbs. You build one circuit with a battery and two light bulbs in series and build another circuit, with an identical battery and three light bulbs in series. The circuits are otherwise identical. In which circuit do you expect the, light bulbs to shine brighter? Explain., , Configuration (1), , Configuration (2), , i. Circuit with two light bulbs, ii. Circuit with three light bulbs, C.7, , Four 1.5 V batteries wired "in parallel" power an electronic device. If all but one of the batteries are removed, under what, circumstances (if any) will the device continue to function normally?, , C.8, , Are the light fixtures in a typical building wired in series or in parallel? How do you know?, Series, , C.9, , Parallel, , In some strings of holiday lights, especially cheaper or older ones, if one light "burns out", all the other lights in the string also, go dark. (a) Are these types of lights wired in series or parallel? (b) Why does one light burning out cause the rest to go dark?, (a), , Series, , Parallel, , (b), C.10 Why might an engineer design a circuit with components in parallel?, C.11 (a) Which resistors in the diagram are wired in, series? (b) Which are wired in parallel?, (a), , R1 and R2 are in series, R1 and R3 are in series, R2 and R3 are in series, , (b), , R1 and R2 are in parallel, R1 and R3 are in parallel, R3 is in parallel with the combination of, R1 and R2, , C.12 (a) Which resistors are in series? (b) Which, are in parallel?, (a), , R1 is in series with R2, R1 is in series with R3, R1 is in series with the combination of, R2 and R3, R2 is in series with R3, , (b), , R1 is in parallel with R2, R1 is in parallel with R3, R1 is in parallel with the combination of, R2 and R3, R2 is in parallel with R3, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems, , 499
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C.13 There are three light bulbs in parallel, powered by a single battery. A fourth bulb is added in parallel. (a) What effect does this, have on the three other light bulbs? (b) What effect is there on the battery that is powering the circuit?, (a), (b), C.14 R1 and R2 are identical. At first, the switch is, open. When the switch is closed, what, happens to the power dissipated by R2? Why?, i., ii., iii., iv., , It will increase, It will stay the same, It will decrease to a positive value, It will decrease to zero, , Section Problems, Section 0 - Introduction, 0.1, , Leave the simulation in the interactive problem in this section in its initial state to answer the following questions. (a) How, does the potential difference across the battery compare with the potential difference across the light bulb? (b) Is the current, the same everywhere, or does it differ from place to place?, (a), , They are equal, , (b), , It is the same everywhere, , They are unequal, It differs from place to place, , Section 2 - Electromotive force, 2.1, , A boat's battery has 325 watt-hours of available energy. How many joules of energy will it supply over its lifetime?, J, , 2.2, , A battery has an emf of 7.5 V and 16 A of current flows through it. What power is this battery supplying?, W, , Section 3 - Energy and electric potential in a circuit, 3.1, , You have a battery with unusual packaging: It is rated by how much work it does on charge. It says it does 6.6 J of work on, every 2.2 C of charge. What is the emf of this battery?, V, , 3.2, , How much work does a 1.50 V flashlight battery do on an electron as it passes through?, J, , 3.3, , 3.4, , A battery charger is connected to a "dead" battery and causes 4.4 A of current to flow through the battery for 1.5 hours. It, maintains a potential difference of 1.5 V across the terminals of the battery. (a) What is the power the charger delivers? (b), How much energy does it supply to the battery during the 1.5 hours?, (a), , W, , (b), , J, , There are two light bulbs in a circuit, powered by one battery. One light bulb is a 100 watt light bulb and the other a 200 watt, light bulb. How much power must the battery be supplying for the lights to be running at full power?, W, , 3.5, , A battery in a circuit has an emf of 8.8 V. The only other component in the circuit is a resistor. 4.4 A of current flows in the, circuit. What must the resistance of the resistor be?, , ȍ, 3.6, , 500, , While the repair shop works on your voltmeter, they loan you an unusual replacement: Rather than measuring the potential, difference across a resistor, it measures the amount of charge that flowed, and the change in electric potential energy of that, charge. You use the device on a resistor and it tells you that 2.5 C of charge flowed through it, and that the electric potential, energy of that charge decreased by 0.40 J. What is the potential difference across the resistor? (Report the answer as a, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems
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positive number.), V, 3.7, , A specialized 8.4-volt battery will supply 430 C of charge over its lifetime. Assume that the potential difference it can maintain, between its terminals does not change as it ages. It is wired into a simple circuit with a 350 ohm resistor. What is the resulting, lifetime of the battery?, s, , Section 5 - Series wiring, 5.1, , Many flashlights are powered by a stack of batteries that are slid into the end of the flashlight one after another. Suppose you, have a flashlight that takes four ideal 1.5 V batteries. If the resistance of the bulb is 11 ohms, what is the current through the, bulb when the flashlight is turned on? The fact that the batteries are stacked end to end means they are wired in series., A, , Section 6 - Resistors in series, 6.1, , A portable electric heater is powered by a 45 V battery. It has four identical heating elements inside, wired in series. If a, current of 11 A runs through them when the heater is turned on, what is the resistance of each heating element?, , 6.2, , A 12 ohm, 14 ohm and 16 ohm resistor are connected in series. What is the equivalent resistance?, , ȍ, ȍ, 6.3, , Three hundred 14 ohm resistors are connected in series (think of a chain of cheap, old-fashioned holiday lights). What is the, equivalent resistance?, , 6.4, , There are six identical resistors in series. A multimeter measures the total potential difference across all the resistors as, 331 V. 4.56 A of current flows through the resistors. (a) What is the equivalent resistance of this group of resistors? (b) What, is the resistance of each resistor?, , ȍ, , ȍ, ȍ, , (a), (b), 6.5, , The total potential difference across two resistors in series is 34.0 V. The current through them is 6.50 A. One resistor's, resistance is 2.00 ȍ. What is the other resistor's resistance?, , ȍ, 6.6, , 6.7, , A 55 ȍ and 65 ȍ resistor are wired in series in a circuit that contains a 12 V battery. (a) What is the equivalent resistance of, the two resistors? (b) How much current is flowing in the circuit?, (a), , ȍ, , (b), , A, , The ammeter in the diagram displays 1.6 A, when the switch is open. What will it display, when the switch is closed?, A, , 6.8, , A circuit contains a single resistor of 12.0 ohms and an ideal emf source (one with zero internal resistance). The current is, 5.50 A. A second resistor is added in series, and the current falls to 4.23 A. What is the resistance of the second resistor?, , ȍ, 6.9, , A 15.0 ȍ and 25.0 ȍ resistor are in series with a battery. The potential difference across the 15.0 ohm resistor is 4.50 V., What is the potential difference across the battery?, V, , 6.10 Three silver wires of equal length are connected in series with a 1.50 V battery. Their cross-sectional areas are 1.00, 5.00, and 10.0 cm2. (a) What is the potential difference across the narrowest wire? (b) The medium wire? (c) The widest wire?, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems, , 501
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(a), , V, , (b), , V, , (c), , V, , Section 7 - Interactive problem: series wiring, 7.1, , Use the simulation in the interactive problem in this section to calculate the resistance of the R2 bulb., , ȍ, , Section 10 - Resistors in parallel, 10.1 The four heating elements on an electric stove are wired in parallel so that any combination of them can be on at the same, time. When an element is on, it behaves as a 6.70 ȍ resistor. When all of the heating elements are on, what is the equivalent, resistance of the stove?, , ȍ, 10.2 A 6.00 ȍ and 42.0 ȍ resistor are in parallel. What is their equivalent resistance?, , ȍ, 10.3 Forty 7.00 ȍ resistors are in parallel. What is their equivalent resistance?, , ȍ, 10.4 The potential difference across points C and, D is 9.2 V. The current flowing past C is 4.6 A., The two resistors are identical. (a) What is the, resistance of each one? (b) What is their, equivalent resistance?, , ȍ, ȍ, , (a), (b), , 10.5 This circuit, called a Wheatstone bridge, is, useful for making precise resistance, measurements. The potential difference, measured by the voltmeter in the middle, depends on the relative resistances of the, four resistors. If that potential difference is, zero, the Wheatstone bridge is said to be, balanced. Suppose you balance the, Wheatstone bridge in the figure by adjusting, the variable resistor to 51.0 ohms. What is the, resistance of R1? (Hint: Because the bridge is, balanced, the potential difference across R1 equals that across R2. The same relation holds between Rvariable and R3.), , ȍ, 10.6 You have access to many 12.5 ȍ and 50.0 ȍ resistors. (a) How can you create a 62.5 ȍ equivalent resistor? (b) A 10.0 ȍ, equivalent resistor? (c) A 60.0 ȍ equivalent resistor?, (a), (b), (c), 10.7 What is the equivalent resistance of the circuit, shown?, , ȍ, , 502, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems
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10.8 You have ten 4.0 ȍ resistors. How can you create an equivalent resistance of 5.0 ȍ? (There is more than one correct, solution, and you do not have to use all the resistors.), 10.9 When the 10.0 ȍ resistor starts to dissipate, 100 W of power, it will rupture. What minimum, emf can be applied to this circuit to cause the, resistor to rupture?, V, , 10.10 Many lamps have a switch that lets you, choose three levels of brightness. The bulbs, in this type of lamp have not one but two, filaments inside, which are wired in parallel, and have their own switches. One of the, filaments is designed to give off more light, than the other when connected to the average, 120 V emf provided by a wall outlet. In the, dimmest position, only the least bright, filament is on. In the next position, only the, brighter filament is on. In the brightest, position, both filaments are on. Suppose one of the filaments consumes 55.0 watts and the other consumes 70.0 watts. What, is their equivalent resistance in the brightest position (when both are turned on)?, , ȍ, 10.11 You have 3 identical resistors. How many equivalent resistances can you create, using all three resistors?, 1, , 2, , 3, , 4, , 5, , 6, , 10.12 There are n identical resistors in a section of a circuit. The potential difference across this section of the circuit is ǻV. The, resistors can be placed either all in series, or all in parallel. (a) Which configuration "consumes" more power, series or, parallel? (b) By what factor must you multiply the power consumed by the low-power configuration to get the power, consumed by the higher-power configuration?, (a), , Series, , (b), , 1/n, , Parallel, 1/n2, , n, , n2, , 10.13 In parallel, two resistors have an equivalent resistance of 100 ohms. In series, they have an equivalent resistance of 625, ohms. (a) What is the resistance of the resistor with greater resistance? (b) The lesser resistance?, , ȍ, ȍ, , (a), (b), , Section 11 - Interactive problem: a parallel circuit, 11.1 Use the simulation in the interactive problem in this section to calculate the resistance of the unknown resistor in the parallel, circuit., , ȍ, , Section 12 - Circuits with series and parallel wiring, 12.1 Four identical 5.00 ȍ resistors are joined and, form the four sides of a square. (a) What is the, resistance between points A and B? (b) What, is the resistance between A and C?, (a), (b), , ȍ, ȍ, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems, , 503
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12.2 (a) What will the ammeter read? Report this as, a positive number. (b) How much power is, dissipated by one of the 10.0 ȍ resistors?, , 504, , (a), , A, , (b), , W, , Copyright 2007 Kinetic Books Co. Chapter 27 Problems
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28.0 - Introduction, Humankind has long been familiar with magnets, objects possessing “north” and, “south” poles that can attract or repel certain other objects. The ancient Greeks, understood their properties, and the word “magnet” itself likely originates from the, Greek region of Magnesia, where naturally occurring magnets are found. Early, navigators learned to steer their ships with the aid of magnetic devices that were the, forerunners of today’s compasses. The natural world also takes advantage of, magnetism. A notable example of this is Aquaspirillum magnetotacticum, a, bacterium that synthesizes tiny magnets that help it determine which way to move., We are a little worried about these creatures: Since the Earth’s magnetic field, changes its orientation every several hundred thousand years or so, the, microorganisms may find themselves unintentionally heading away from dinner one, day., Much more complex creatures, namely scientists, employ magnets as well. In the, 16th and 17th centuries, they learned to create their own magnets and used them to, study the Earth’s magnetic field. In the 19th century, scientists began the crucial, work of piecing together a more complete picture of the relationship between, magnetic fields and electric currents., Despite centuries of practical use, magnets and their fields still pose mysteries. For, instance, scientists cannot definitively establish the cause of the Earth’s magnetic, field, and they are only able to speculate about why its direction periodically, changes. In addition, physicists continue their quest for the magnetic monopole: a, magnet with just one pole. As you proceed in your studies in this chapter, remember, that you are in the good company of other fine minds who have found the workings, of magnets to be an area of continuing fascination., You can begin your exploration of magnetic fields by launching the simulations to, the right that show the effect of a magnetic field on the motion of a charged particle., These two simulations are the same except for the initial viewing angle, the angle at, which you view what is occurring in the simulation. In both, you control the initial, velocity of a positively charged particle that moves in a magnetic field, represented, by magnetic field lines. In the illustration for Interactive 1, you see that the magnetic, field points straight down the screen and the initial velocity vector points to the right,, perpendicular to the magnetic field. As the particle moves you will see, represented, as a purple vector, the force exerted on it by the magnetic field., Clicking on Interactive 2 launches the same simulation but with the viewing angle rotated 90°. Here, the magnetic field points directly toward, you, and you are seeing the heads of the field lines. With either simulation, you can change the viewing angle by using the slider provided, and, see either of these points of view. In the simulations you will also see a magnetic field meter that is there principally to help you understand the, changing perspective as you manipulate the viewing angle slider., Launch the upper simulation and conduct some experiments. Does the moving particle travel along a straight line or a curve? To answer this, question, you will need to change the viewing angle, in the process seeing why a three-dimensional view of a charged particle moving through, a magnetic field is so useful., As you study the path, answer two more questions. First, is the particle’s speed changing? Second, is it accelerating? For the second question,, recall that acceleration measures the change in velocity, which is a vector., You can also consider the relationship of the directions of the various vectors you see. What is the relationship between the force and velocity, vectors? Are they parallel or perpendicular? What is the angle between the force vector and the magnetic field? Again, the viewing angle slider, proves a useful tool. You can only change the direction of the initial velocity when the viewing angle is set to the far right so that the magnetic, field is pointing straight down the screen; this makes it easy to see the angle between the velocity vector and the magnetic field vector., As another experiment, set the velocity to zero. Does the magnetic field exert a force on the particle? If the magnetic field exerts a force, the, particle will accelerate. How does this compare to what would occur if the stationary charged particle were in an electric field? (Note: We ignore, other forces, notably gravity and air resistance, in these simulations.), We have asked a lot of questions above. Answer as many as you can now, and prepare to explore magnetism in depth in this chapter., , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 505
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28.1 - Magnet fundamentals, , Magnet: An object that creates a magnetic field and exerts a magnetic force., All known magnets have two poles., Just like electrical charges, magnets create fields and, exert forces. All magnets are dipoles, meaning they, have two poles: a north pole and a south pole. As, with electrical charges, magnetic opposites attract, and likes repel; a north pole attracts a south pole, and, a pair of like poles, such as two south poles, repel, each other., A rectangular magnet like the one shown to the right, is called a bar magnet. It has a magnetic pole at each, end. If you were to bend this magnet into a “U,” you, would create a horseshoe magnet like the one the, boy is holding in the photo above. Even though you, have changed the magnet’s shape, the poles remain, at the two extremities., , This boy is holding a U magnet. The magnet suspends a few paperclips., , The attraction and repulsion of magnetic poles can be demonstrated with a pair of bar, magnets. If you have two such magnets, position them so that their opposite poles are, adjacent. What happens? They attract each other. You can see this illustrated in, Concept 1. Now place them so their like poles are next to each other; the magnets will, repel each other. This is shown in Concept 2., The strongest and most frequently found form of magnetism is called ferromagnetism., Only certain types of material (iron is a notable example) exhibit this form of magnetism., Everyday videotapes reveal a common application of ferromagnetism: A movie is, encoded as a pattern of small magnets on the videotape, and the VCR “reads” this, data. Computer hard drives function in a similar way., You may have noticed that magnets can “stick” to surfaces that do not initially exhibit, magnetism. For instance, the exterior of your refrigerator is typically not magnetic, and, yet if it is made of a ferromagnetic metal, magnets will stick to it. A magnet sticks, because it is able to induce a temporary magnetic field in the refrigerator’s surface. The, permanent and temporary magnets then attract each other., , Magnet fundamentals, Have two poles, north and south, ·Opposite poles attract, , Another way to observe this phenomenon is to attach a metal paper clip to a magnet, and then attach a second clip to the first clip. The two clips will attract each other. When, you remove the original magnet, however, the clips will no longer attract one another, because the fields of the paper clips disappear when they are removed from the, influence of the magnet’s field., No one has created a magnet with just one pole (a monopole). Magnetic poles always, come in pairs. If you take a bar magnet and cut it in half, you will create two magnets,, each with a south and a north pole. Cut each of those in half and you will have four, magnets, each with two poles. In principle, if you could cut the magnet into its, constituent atoms, each atom would have its own magnetic field., , Magnet fundamentals, ·Like poles repel, , 28.2 - Magnetic fields, Two magnets can attract or repel each other without, touching: They exert a force at a distance. Magnetic, fields surround magnets. Like an electric field, the, magnetic field is a vector field. It has a strength and a, direction at every point. The letter B represents the, magnetic field and the unit for magnetic field strength, is the tesla (T)., The photographs above show the alignment of iron, filings gathered around the poles of two pairs of, magnets held close together. The filings in both, photos align with the magnetic fields between the, Iron filings line up with the magnetic field., poles. The photograph on the upper left shows the, filings between a south pole and a north pole, which, attract each other. The filings are connected and aligned in the same direction as the magnets because the magnetic field lines pass directly, from the north pole to the south pole. In the photograph on the upper right, two south poles are repelling each other, which causes the filings to, , 506, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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separate., Like electric fields, magnetic fields can be diagrammed with field lines. We have, superimposed a drawing of magnetic field lines on the illustrations to the right. You can, see how the orientations of bits of iron filings reflect the direction of the field at a number, of places around the magnet., As the diagram shows, the field lines run outside the magnet from the north to the south, pole, and they always form closed loops. As with an electric field, the closer the lines,, the stronger the field. The strength of the field decreases with distance from the, magnet., The table in Equation 2 shows approximate magnetic field strengths near some, magnetic objects. The magnetic fields you encounter every day are much less than one, tesla. The most powerful lab magnets have field strengths of tens of teslas. However,, human technology cannot create magnetic fields anywhere near the field strengths, exhibited by a class of highly magnetized neutron stars called magnetars., , Magnetic field, Region where magnet exerts force, External lines point from north to south, , B is symbol for magnetic field, Units: teslas (T), , Magnetic field strength near, magnetic objects, , 28.3 - Physics at work: lodestones, , Lodestone: A naturally occurring magnetic, rock found in various regions of the world., The Chinese first used lodestone magnets in their navigational compasses in the twelfth, century. Later explorers, including Columbus, carried lodestones to “recharge” their, iron-based compasses. To do so, they would pass the needle of a compass near the, lodestone to strengthen its magnetic properties., A lodestone consists of two materials, one with a high saturation magnitude. Saturation, magnitude is the measure of the maximum strength of the magnetic field that can be, induced in a substance. Iron, for example, has a high saturation magnitude. The other, material has high coercivity, which is the measure of how well a substance retains its, magnetic field. Pure iron loses its magnetic properties relatively easily, while a, substance such as magnetite does not., The combination of these two materials results in a strong and durable magnetic, source. One substance supplies the strength, the other supplies durability., , Lodestones, Natural magnetic rock combines, materials with, ·Strong magnetic field, ·Long-lasting magnetic field, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 507
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28.4 - The Earth and magnetic fields, The Earth acts like a huge magnet with an accompanying magnetic field. Compasses, take advantage of this field: Their needles are mounted so that they can rotate freely, and align with it. Once a compass has established the north and south directions, other, directions can be established relative to them., The Earth’s magnetic south pole is located near its geographic North Pole. This, means that the north pole of the magnet in a compass points north, since the south pole, of the planetary magnet attracts it. The geographic poles lie on the planet’s axis of, rotation., Scientists still do not know exactly how our planet produces its magnetic field, though, the prevalent theory is that Earth’s magnetism arises from processes that occur in its, core. Scientists have arrived at this conclusion because of the fundamental differences, between the cores of the Earth and the Moon. The Earth has a molten metallic core and, a magnetic field, while the Moon has a solid core and no magnetic field. Many scientists, believe the large-scale liquid flow that occurs inside the Earth’s core causes its, magnetic field., , The Earth and magnetic fields, Earth is huge magnet, ·North Pole § south pole of magnet, ·South Pole § north pole of magnet, , While scientists are still puzzled by the “why” of Earth’s magnetic field, they do know, that the locations of its magnetic north and south poles change. The magnetic south pole is currently about 1900 km away from the geographic, North Pole of the Earth. The magnetic north pole resides in the ocean south of Australia, not even in the continent of Antarctica! These, positions are not fixed; shifts of several degrees in their locations have been measured over the last century. The poles do more than wander,, they also reverse their orientation. The last switch occurred a few hundred thousand years ago, when magnetic north became magnetic south,, and vice versa., Scientists deduce this change in orientation by analyzing parts of the ocean floor that originate as magma (molten rock) emerging from cracks, in the Earth’s crust. Over a period of millions of years, vast quantities of solidified magma have been deposited on the ocean floor. As magma, congeals at any point in time, the iron it contains “records” the orientation of the Earth’s magnetic field. New magma flows force earlier deposits, apart, which separates them and enables scientists to establish their sequence. By examining different sections of the rock, scientists can, deduce the direction of the magnetic field at various times throughout Earth’s history., The current strength of Earth’s magnetic field is about 5×10í5 teslas, and that strength is decreasing at about 0.07 percent per year. If it, continues to weaken at this rate, it will be reduced to only 1 percent of its present value in 6,500 years. Since the magnetic field helps to shield, our biosphere from cosmic rays and charged particles from the Sun, this could be a matter of concern., Scientists theorize that the poles “flip” alignment after the field passes through a state with zero magnitude. Because there are no signs of, massive mutations in the fossil record that date from the period of the last “flip,” perhaps the results are not as severe as one might fear. No, one knows exactly how long the zero-field condition exists, whether five years, 50 years, or 1000 years. As the NASA website above has, stated, “Stay tuned…”, , 28.5 - Physics at work: compasses and the Earth, A magnetized compass needle aligns with the Earth’s, magnetic north and south poles. The needle’s north, pole, sometimes distinguished by a bright color as in, the photograph above, points toward the Earth’s, magnetic south pole. However, the Earth’s magnetic, poles are not at the same locations as its geographic, poles. For instance, in 1831, British explorer James, Clark Ross located the magnetic south pole off the, coast of Canada (remember, the magnetic south pole, corresponds to the geographic North Pole). He tried, to duplicate this feat for the magnetic north pole near, Antarctica, but failed due to weather and ice., Nonetheless, many regions of that continent (such as, the Ross Sea) are named after him., In mountains near Seattle, Washington, the magnetic declination is about 20°., The angle between the directions to the magnetic and, geographic poles is called the magnetic declination., Many topographic maps provide the local angle of declination so that hikers and others can compensate for it and orient their compasses to, geographic north. The first sketch on the right shows the declination for Seattle, Washington: It is approximately 20°., , Scientists and navigators also found that compasses could be used to estimate latitude (north/south position on the planet). Although today’s, compasses are designed to move in a horizontal plane, the needles in some early instruments were allowed to rotate freely in all directions, (like a bar magnet suspended by a string tied around its middle). When carried to a magnetic pole of the Earth, the magnetic needles of these, compasses would point straight up or down; conversely, when near the equator, they would point almost horizontally. By measuring the angle, of the needle from the horizontal, navigators could estimate their latitude., , 508, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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Declination, Angle between magnetic, geographic, pole, , Latitude, Can be estimated with a dip needle, , 28.6 - Ferromagnetism, , Ferromagnetism: A strong, magnetic effect exhibited by, the atoms of certain elements,, notably iron. It is the cause of, the magnetic field of, commonly used magnets., Ferromagnetism is the basis of the most familiar type, Color image: magnetic domains in an amorphous ferromagnet., of magnetic devices. When you see a magnet affixed, Inset: magnetic domains in heat-treated carbon steel., to a refrigerator door, you are witnessing the results, of ferromagnetism. “Ferro” comes from the Latin word for iron, since iron is the most, common element to exhibit this property. Ferromagnetism is present in the magnets, that can be purchased in toy or hardware stores., Nickel and cobalt are two other common elements with atoms that exhibit, ferromagnetism, though to a lesser extent. Gadolinium and dysprosium are two exotic, elements that can be used to make strong magnets that perform well at low, temperatures., In order to explain the source of ferromagnetism, we have to pay a quick visit to a, concept developed in quantum theory. The classic model of atoms í electrons revolving, about a central nucleus í does not suffice. Ferromagnetism is caused by a property of, electrons called spin. For the purposes of this discussion, we say that spin is used to, describe the angular momentum of an electron due its rotation, akin to the angular, momentum of the Earth due to its rotation about its axis. (This is only an analogy. A, more detailed, much less mechanistic description requires the context of quantum, theory.) This is one form of angular momentum possessed by the electrons., , Inside a domain, Electron spins align due to exchange, coupling, ·Result is magnetic domains, , They also possess orbital angular momentum due to their motion about the nucleus., Both contribute to the magnetic moment of the electron, but only spin is relevant to the discussion of ferromagnetism. The magnetic moment, vector helps quantify how much torque a magnetic dipole will experience in a magnetic field; the larger the moment, the greater the torque. The, moment points from the south pole to the north pole of the dipole., Spin means that each electron has its own magnetic field. It can be considered as acting like a bar magnet, a dipole with north and south poles., , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 509
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In many materials, the spin of each electron cancels out that of another electron in the, same atom with which it is “paired,” resulting in no net magnetic field. Ferromagnetism, results from the net magnetic field created by unpaired electron spins. For example,, each iron atom contains four electrons with uncancelled spins, giving the iron atom a, net dipole moment., In ferromagnetic materials, the spins of the electrons of one atom interact with the spins, of electrons of neighboring atoms. This interaction, called exchange coupling, is, quantum mechanical in nature. As a result of exchange coupling, the magnetic dipoles, of atoms within a ferromagnetic material tend to align in the same direction., When the magnetic dipole moments of ferromagnetic atoms align, the result is a, magnetic domain, a region in a material where there is a net magnetic field. Magnetic, domains can be seen under powerful microscopes, often with the aid of a technique, called magnetic force microscopy. In the color micrograph above, domains are given, false colors to be more easily seen in the image. Typical domains have a diameter, about one-third that of a human hair. This can be best seen in the small inset, micrograph that shows the magnetic domains in a sample of carbon steel. Even small, amounts of magnetic materials contain vast numbers of domains. A domain may, contain 1012 to 1015 atoms, but since a cubic centimeter of iron contains about 2.5×10 19, atoms, there are still a large number of domains., , Random domains, Unaligned domains mean no net, magnetic field, , There is a good question here: Why are not all iron objects magnets? The reason is that, although domains have magnetic fields, they point in random directions unless they, have been subjected to an external magnetic field. This is shown in Concept 2. The, random nature of their directions means there is no net magnetic field in the substance., However, when the ferromagnetic material is placed in an external magnetic field, a, process occurs that results in the material gaining its own overall magnetic field. This, happens because the external field exerts a torque on the magnetic dipoles., The magnetic field overcomes the tendency of the dipoles to stay aligned with their, neighbors, and causes them to align with the field. Unaligned dipoles bordering, domains that are aligned with the field steadily become “converted” to the new, orientation. This causes the domains aligned with the external field to expand. This is, shown in Concept 3, where the domains shown in Concept 2 have combined to form, three domains., , In external magnetic field, In external magnetic field, B0, ·domains aligned to external field grow, , Since there is a tendency for the dipoles to remain aligned, when the external magnetic, field is removed, the domains (and the dipoles that make them up) remain aligned. This, means the material now has its own magnetic field. In short, a magnet has been, created., An external magnetic field can cause non-ferromagnetic materials to develop their own, magnetic field. These other forms of magnetism are called diamagnetism and, paramagnetism. However, ferromagnets are distinct and quite useful for two reasons., First, a ferromagnet retains its magnetic field after the external field is removed. A, refrigerator magnet, or the information stored on a VCR tape, both rely on the longevity, of ferromagnets. The other forms of magnetism disappear after the external magnetic, field is removed., Second, the magnetic field created by ferromagnetic materials is typically orders of, magnitude stronger than that found in the other forms of magnetism in most materials., This makes ferromagnets useful for a wide range of everyday applications., , Magnetic field removed, Substance maintains magnetic field, , 28.7 - Magnetic fields and charged particles, Magnetic fields exert forces on moving, electrically, charged particles., This phenomenon makes for a good demonstration in, a physics class. A current-carrying wire is placed near, a magnet. The magnet exerts a force on the electrons, moving in the wire, which causes the wire to move, toward or away from the magnet. When the current is, turned off, the magnetic field stops exerting a force on, the wire., The images on the screens of traditional televisions, and computers are the result of electrons being, accelerated by an electric field, subsequently being, TV tube. A gun (far left) accelerates electrons across a potential difference., “steered” by magnetic fields, and then striking the, Electromagnets (brown) steer the moving charges to light up screen pixels., screen to create light of different colors at specific, locations. If you were to place a magnet close to such a system, you would distort the image. However, we do not recommend you do this, as it, could cause expensive or irreversible damage!, , 510, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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Four factors determine the amount of force exerted by a magnetic field on a moving, particle. They are the particle’s charge and speed, the strength of the magnetic field and, the angle of intersection between the particle’s velocity and the magnetic field. The, force is greatest when these two vectors are perpendicular, and zero when they are, parallel., When a charged particle is surrounded by an external electric field, the electric force on, the charge is exerted along the field lines. The electric field exerts a force on the charge, whether it is moving or stationary. In contrast, magnetic fields only exert a force on, moving charged particles, pushing them neither in the direction of the particle’s motion, nor along the lines of the field. The force exerted on a moving charge by a magnetic, field is perpendicular to both the particle’s velocity and the direction of the field., You see this illustrated to the right, and you also experienced it when you used the two, simulations in the introduction to this chapter. In Concept 1, the same phenomenon is, shown from two different vantage points., In both illustrations, a positive charge is moving through a magnetic field. In the view, labeled “side view,” the magnetic field points away from you. This is depicted with ×’s,, which represent the field lines viewed from behind. This view is used to best show the, direction of the force: It is perpendicular to both the field and the velocity vectors., , Magnetic field, charged particle, Magnetic field exerts force on moving, charge, ·Force perpendicular to velocity, field, , In the view labeled “front view,” your viewpoint has been rotated 90° so the field, appears parallel to the screen. The force vector in the front view points toward you and, is represented by a dot. You are looking at “the business end” of the vector’s arrow., A right-hand rule is used to determine the direction of the force. The front view makes it, easier to see, so we use that in Concept 2. To apply the right-hand rule, the fingers of a, flat hand start out pointing in the same direction as the velocity vector of the charge., Then they curl, so that the fingers point in the same direction as the magnetic field. The, fingers wrap from the velocity vector to the magnetic field vector., For a positive charge, the thumb points in the direction of the force exerted on the, moving charge. The thumb points opposite to the direction of the force for a negative, charge. In other words, with a negative charge, you apply the same rule, but reverse, the results., The equation to determine the force is shown to the right. The force equals the charge, times the cross product of the velocity and magnetic field vectors. The cross product is, calculated with the sine, as shown to the right., , Right-hand rule, Determines direction of force, ·Fingers curl from v to B, ·Thumb shows force on positive charge, , To determine the amount of force, multiply the absolute value of the charge (its positive, value), the charge’s speed, the field strength, and the sine of the angle between the, velocity and field vectors. This angle is shown in the illustration for the equation. When, calculating the force magnitude, you use the smaller, positive angle between the, velocity and magnetic field vectors. For instance, in the Equation 1 diagram the angle is, 90°, not í90° or 270°., As mentioned earlier, the unit for magnetic field strength is the tesla. One tesla equals, one newton·second per coulomb·meter, or N·s/C·m. In other words, one coulomb of, charge traveling at one meter per second through a magnetic field having a strength of, one tesla experiences one newton of force. A tesla is a rather large unit (remember that, one coulomb is a lot of charge), so the smaller unit gauss (G) is fairly common. Ten, thousand gauss equals one tesla. As mentioned earlier, the Earth’s magnetic field is, about 5×10í5 T, which equals 0.5 G., , Force on charge moving through, magnetic field, F = qv × B, F = |q|vB sin ș, F = magnetic force on particle, q = charge, v = velocity, B = magnetic field, ș = angle between velocity and field, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 511
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The particle is moving to the, right. What are the magnitude, and direction of the force on it?, ș = 90°, F = |q|vB sin ș, , F = 2.3×10í5 N, directed down (opposite to thumb), 28.8 - Interactive checkpoint: a charge in a magnetic field, An electron is traveling in the yz plane, at an angle of 60.0° from the positive, z axis and 30.0° from the positive y, axis as shown. Its speed is a constant, 365 m/s when a uniform magnetic, field of 1.22×10í3 T is turned on,, pointing in the negative y direction., State the acceleration of the electron, in the instant after the magnetic field, is turned on as a vector in rectangular, notation., , Answer:, , a=(, , ,, , ,, , ) m/s2, , 28.9 - Interactive problem: charged particle moving in a B field, In this simulation you can change two properties of a particle: its charge and its, velocity (both speed and direction). We ask you to alter the velocity of the particle in, order to achieve certain forces., In the initial view of the magnetic field it is directed straight down, and the charge, starts near the middle of the screen. You can use the slider in the control panel to, change the viewing angle. Press GO to start the simulation, and press RESET, whenever you want to change the particle’s initial velocity or charge., You can only change the direction of the initial velocity when the viewing angle is, set to the far right and the magnetic field is pointing straight down. This orientation, makes the angle between the velocity vector and the magnetic field vector easy to, see. You change the direction of the velocity vector by dragging it with the mouse., Your first task is to set the direction of the particle’s initial velocity so that no force is, exerted on the charge by the field. No shortcuts! You could solve this by setting the, velocity to 0 m/s, but we want you to solve the problem by setting the velocity, vector’s direction correctly. When the field exerts no force, both the direction and, speed of the particle will remain constant., Note that after you change the charge or the velocity vector (speed or direction), but before you press GO, the force meter tells you what the, force on the particle will be, not what it is. The force is zero whenever the particle is not moving, but we thought you would appreciate seeing, immediate feedback on your changes, without having to launch the particle every time., , 512, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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What is the relationship between the directions of the field and the velocity vector when the force equals zero? (Note: There are actually two, directions in which you can set the velocity vector to achieve this, one 180° opposed to the other.), For your next challenge, set the parameters in the simulation to maximize the amount of force on the particle. As with your first task, there are, two directions that result in a maximum force on the charge. You will find yourself clicking the up and down arrows on parameters like “Charge”, until you reach their maximum value in the simulation. The maximum amount of force you can achieve in this simulation is 1.68×10í8 newtons., For your final task, change the charge from positive to negative while keeping the initial velocity the same. What effect does this have on the, magnitude and direction of the force?, If you are surprised by the results of any of these configurations, refer back to the section that discussed the effect of a magnetic field on a, moving charged particle., , 28.10 - Determining the strength of a magnetic field, A magnetic field exerts a force on a moving charged particle. When the charge, velocity, and field are known, the direction and magnitude of that force can be calculated. This, process can also be reversed: Just as an electric field can be measured using a test, charge, the strength of a magnetic field can also be determined using a test charge., The process is similar, but with a couple of twists., In an electric field, a stationary positive test charge is used to determine the, electrostatic force at a given point. The amount of force is then divided by the test, charge to determine the electric field strength. The direction of the force on the test, charge indicates the direction of the field., In a magnetic field, the force must be determined using a moving charge. Physicists, fire a charged particle through a magnetic field and measure how the force exerted on it, by the field alters its velocity. They then use Newton’s second law, together with the, observed acceleration and the mass of the particle, to calculate the force the field exerts, on it., The angle between the velocity and the magnetic field vectors must also be known to, determine the field. (A compass can be used to determine the direction of the field.), Once the force and angle are known, they can be used in the equation shown on the, right to find the strength of the magnetic field. This is the same equation shown earlier, to determine the amount of force exerted by a given magnetic field, but here it is solved, for the field strength., , Measuring a magnetic field, Use a moving test charge, Measure acceleration of charge, Newton’s 2nd relates acceleration, force, With force known, field can be found, , B = magnetic field strength, F = force, q = charge, v = speed, ș = angle between velocity and field, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 513
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The force is 4.0×10í5 N and the, charge q is 2.0 ȝC. What is B?, , B = 0.80 T, 28.11 - Interactive problem: B field strength and particle motion, In this simulation, you want the magnetic field to exert a force of 1.16×10í8 newtons, on the positively charged particle. Set the particle’s speed v (the maximum possible, value is 520 m/s) and the magnetic field strength B (the maximum possible value is, 3.00 T) to cause the required amount of force, using the controllers provided in the, control panel. You will find that you need to adjust both the speed and the field, strength upward from their initial settings to achieve this force., If you have any trouble achieving the desired force, review the discussion in the, previous section on determining the strength of a magnetic field. If you have any, questions about how to use the simulation, see the previous interactive problems in, this chapter for complete instructions., , 28.12 - Physics at work: velocity selector, A velocity selector is a device that allows charged particles moving only at a specified, speed to pass through it. It uses a combination of an electric field and a magnetic field, to “trap” particles moving at other speeds. In addition to being a useful tool, a velocity, selector provides a good way to explore the contrasting effects of electric and magnetic, fields., To the right is a conceptual diagram of a velocity selector for charged particles. Perhaps, a scientist wants only electrons moving at 50,000 m/s to pass through. The electric field, of the velocity selector points down, and will exert an upward force on an electron., (Remember that the force is opposite to the direction of the electric field because, electrons are negatively charged.), A magnetic field is used to counter the effect of the electric field. The magnetic field, strength is set so that the field exerts an equal but opposite force on any electron, moving at 50,000 m/s. Since the force exerted on the electron by the electric field is, upward, the magnetic field force must point down. Electrons with the required speed will, travel horizontally and pass through an aperture at the right end of the selector. Slower, or faster ones will experience a net force and be forced up or down. (We ignore the, force of gravity, whose effect would be minor for a particle moving at this speed.), , Velocity selector, Charges of specific speed pass through, Uses electric and magnetic fields, Electric force balances magnetic force, , The sum of the forces on the particle is qE + (qv × B). This equation is called the Lorentz force law. The strength of the electric force equals, |q|E. The strength of the magnetic force equals |q|vB (since the velocity is perpendicular to the field). When the forces sum to zero, these two, expressions are equal in magnitude. On the right, we solve for the speed and see that it equals the ratio of the electric to the magnetic field, strength. As you can see, the charge cancels out., Can you predict the direction in which this device’s magnetic field needs to point using a right-hand rule? The electric force on the electron, points up, so we want the magnetic field force to point down. Since the charge is negative, the thumb will point in the opposite direction of the, force. This means you want the thumb to point up. The fingers must wrap from the velocity vector to the magnetic field vector. You can, , 514, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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conclude that the magnetic field vector must point away from you, which is how it is, depicted in the diagram., In the setup shown here, if a negatively charged electron is moving too fast, the force, exerted by the magnetic field will be greater than that exerted by the electric field, and, as a result it will pull it down. If the electron is moving too slowly, the electric field will, win the contest and pull it up. The example problem to the right shows how to determine, the strength of the magnetic field that permits an electron moving at 53,000 m/s to pass, undeflected through a uniform electric field with a magnitude of 4.0 N/C., It is worth emphasizing that the name of the device is a velocity selector, not a speed, selector. Direction matters. You may want to consider what magnetic force would be, exerted on a charged particle that was moving horizontally, but from right to left as it, passed through the selector at the critical speed. Would the magnetic force still cancel, out the electric force?, , At equilibrium:, FE + FB = 0, so |q|E í |q|vB = 0, and, , FE = force of electric field, FB = force of magnetic field, q= charge, v = speed of charge, E = electric field strength, B = magnetic field strength, , This electron passes through the, velocity selector with its velocity, unchanged. What is B?, v = E/B, B = E/v, B = (4.0×103 N/C)/(5.3×104 m/s), B = 0.075 T, 28.13 - Circular motion of particles in magnetic fields, When the velocity of a charged particle is perpendicular to a uniform magnetic field, it, causes the particle to move in a circular path. In this section we discuss why this, occurs, and we state some properties of that motion., At the right is a diagram showing a magnetic field (pointing directly into the screen) and, a positively charged particle moving in a circular path. Its motion started when the, particle was fired into the field along the surface of the page., Why is the motion circular? First, recall that the magnetic force is always perpendicular, to the velocity vector. This means it neither increases nor decreases the speed of the, particle. It only changes the direction of its motion., The force always points toward the center of the circle. You can use the right-hand rule, at any point of the circle to confirm this. (The fingers wrap from v to B, so the thumb, points toward the center.), The magnitude of the force is constant. The quantities that determine the force, q, v, B,, and ș, do not change as the particle moves. Even as the particle’s velocity changes, direction, the angle ș between the velocity and magnetic field stays constant at 90°., , Circular motion, Velocity perpendicular to magnetic field, ·Force perpendicular to velocity, ·Magnitude of force constant, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 515
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If you recall your studies of uniform circular motion and centripetal forces, this may all, sound familiar. A constant force that points toward the center causes uniform circular, motion. In other words, the force exerted by the magnetic field is a centripetal force., , ·Circular motion results, , An interesting aspect of this kind of motion is that the field does no work on the particle., One way to conclude this is by noting that the particle’s energy does not change. Its, speed is constant (which means its KE is constant) and its PE does not change in this, uniform field. Since there is no change in energy, no work occurs. This is quite distinct, from the situation of a charged particle in an electric field: Electric fields can do work on, charged particles., Now we summarize the equations on the right. In Equation 1, we take the centripetal, force in uniform circular motion to be the magnetic force, and use Newton’s second law, to state that this must equal the mass of the particle times its centripetal acceleration, , v2/r. We then solve for the radius r., We can state other equations that further describe the motion of the particle. First, we, calculate the period T of the particle’s motion: The result is shown in Equation 2. We, derive this equation by starting with the equation for the period of an object in circular, motion (the circumference divided by the speed). Then we substitute the formula for the, speed of the particle in the magnetic field (obtained by solving the radius equation in, Equation 1 for v)., Notice the interesting fact that the period is not a function of the speed, but only of the, mass and charge of the particle, as well as the strength of the magnetic field. A faster, moving particle moves in a circle of greater circumference, but the period does not, change., We have also included in Equation 3 the equations for the frequency f and the angular, frequency Ȧ of the particle’s motion. These can be derived using the relationship of, period and frequency, and of frequency and angular frequency., , Radius, |q|vB = ma = m(v2/r), so, , r = radius, m = mass, v = speed, q = charge, B = magnetic field strength, , Without discussing it further in this section, we do note that if the velocity of the charged, particle is not perpendicular to the field, but has a component parallel to the field, then it, will move in helical motion. If you would like to observe helical motion, go to the, simulation in the introduction to this chapter, move the particle’s velocity vector to a, direction that is not perpendicular to the field, press GO, and use the viewing angle, slider to watch the resulting motion from various vantage points., , Period, , T = period, , Frequency, angular frequency, , 516, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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f = frequency, Ȧ = angular frequency, , m = 2.5×10í7 kg and B = 0.50 T., What is the angular frequency of, the particle’s motion?, , Ȧ = 2.0 rad/s, 28.14 - Sample problem: proton in a magnetic field, This proton is moving in a circular, path in a magnetic field. Its angular, velocity is 2.00×107 rad/s., What is the strength of the magnetic, field?, , Variables, angular velocity, , Ȧ = 2.00×107 rad/s, , speed, , v, , radius of circular path, , r, , magnetic field strength, , B, , mass of proton, , mp = 1.67×10í27 kg, , charge of proton, , q = 1.60×10í19 C, , What is the strategy?, 1., , Find the speed of the circling proton in terms of the radius and angular velocity of its motion., , 2., , State the equation for the radius of the circular motion of a charged particle in a magnetic field. For the particle’s speed, use the, expression from strategy step 1., , 3., , Solve for the magnetic field and evaluate., , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 517
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Physics principles and equations, The relationship between angular velocity and tangential speed is, , v = rȦ, The radius of a particle’s circular motion in a magnetic field is, , Step-by-step solution, , Step, , 1., , v = rȦ, , Reason, , angular velocity and speed, , 2., , radius of circular motion in a magnetic field, , 3., , substitute equation 1 into equation 2, , 4., , simplify and solve for B, , 5., , evaluate, , 28.15 - Physics at work: mass spectrometer, Mass spectrometers are used to determine the masses of atoms or molecules, or their, relative abundance in a sample. They are used in a range of settings from surgery (to, determine the mixture of gases in a patient’s lungs) to space missions (to analyze the, atmospheres or soils of planets and other celestial bodies). Chemists also frequently, use them to analyze materials., A conceptual diagram of a mass spectrometer is shown to the right. To use the device,, the substance being analyzed is first vaporized if it is not already a gas. It is then, ionized: An electron is removed from each particle so that it has a net positive charge, equal to +e. The ionized particles are then accelerated across a potential difference, between two charged plates., The particles all have the same charge, and the experimenter keeps the potential, difference between the plates, and the resulting electric field strength, at a constant, value. This means the force exerted on various particles by the electric field does not, change; their accelerations depend only on their masses., Each moving, charged particle passes through an entry port into a uniform magnetic, field. The field is perpendicular to the velocity of the particle, which means the particle, will move in a circular path. The magnetic field is kept constant so that the radius of the, path is a function of the particle’s speed and mass., , Mass spectrometer, Device that measures mass of particles, Electric field accelerates ions, Radius measured to determine mass, , Each particle traverses a circular path and the point where it strikes the sensing plate of, the device is recorded. The radius of the path is one-half the distance from the entry, port. The particle’s mass can be determined using the equation shown to the right., Determining the mass of a particle (Derivation). In the following derivation we, assume that the ion has a charge of +e, so that it is accelerated from the positively, charged plate to the negatively charged plate, and then curves to the right after it enters, the magnetic field, which is directed out of the screen toward you., , Mass of a particle, , 518, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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Variables, change in kinetic energy, , ǻKE, , m = mass, q = charge, , mass of ion, , m, , r = radius, , accelerated speed of ion, , v, , B = magnetic field strength, , radius of circle in magnetic field, , r, , ǻV = potential difference, , charge of positive ion, , q = +1.6×10í19 C, , magnetic field strength, , B, , change in potential energy, , ǻPE, , potential difference across plates, , ǻV, , Strategy, 1., , Express the change in the kinetic energy of the ion during the acceleration, phase in terms of its mass and quantities that can be controlled or measured by, the experimenter. By “acceleration phase,” we mean the linear acceleration in the, electric field., , 2., , Express the change in the potential energy of the ion during the acceleration, phase in terms of some of the same quantities, including the potential difference, between the plates., , 3., , Use the conservation of energy to relate kinetic energy to potential energy, and, solve for the unknown mass of the charged particle., , This mass spectrometer is, testing a hydrogen molecule, (H2+) having charge +e. What is, its mass?, q = 1.6×10í19 C, , Physics principles and equations, We use the definition of kinetic energy., , KE = ½ mv2, The radius of the circular path of a charged particle moving perpendicular to the, magnetic field is, , r = mv/qB, The potential difference equals the change in PE per unit charge., , ǻV = ǻPE / q, , m = 3.3×10í27 kg, , The principle of the conservation of energy applied to mechanical energy states that, , ǻKE + ǻPE = 0, Since the magnetic field changes the particle’s direction but not its speed, its KE is not changed by the magnetic field., Step-by-step derivation, In the first stage of the derivation we write ǻKE in terms of the mass m and the speed v of an ion as it leaves the electric field and enters the, magnetic field. Neither m nor v can be directly observed. We use another equation to replace v by quantities that can be controlled or observed, in the laboratory., , Step, , 1., , ǻKE = ½mv2, , Reason, , change in KE for particle starting at rest, , 2., , radius of positive ion’s path in magnetic field, , 3., , solve equation 2 for v, , 4., , substitute equation 3 into equation 1, , 5., , simplify, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 519
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Now we write ǻPE in terms of q and the potential difference ǻV across the accelerating plates. Finally, we write an equation for ǻKE and, ǻPE based on the conservation of energy, substitute the expression found above, and solve for the unknown mass m of the charged particle., , Step, , Reason, , 6., , ǻPE = qǻV, , change in potential energy, , 7., , ǻKE = –ǻPE, , conservation of energy, , 8., , substitute equations 5 and 6 into equation 7, , 9., , solve for m, , Although we derived the mass equation for a positively charged particle, we could equally well do so for a negatively charged particle (the, charged plates in the mass spectrometer would have to be reversed to accelerate it in the correct direction)., , 28.16 - Helical particle motion in magnetic fields, In order for a charged particle to move in a circular, path in a uniform magnetic field, it must enter the field, with a perpendicular velocity. But what if its velocity, has a component parallel to the field?, The result is shown to the right: It is called helical, motion. The particle traces out circles that wind, upward (or downward) in a fashion similar to the, motion of a car navigating the “corkscrew” ramps, found in many multistory parking garages. The, particle moves both in circles and up or down at the, same time. You can use the interactive simulation in a, following section to observe helical motion., , Aurora Borealis display in the northern sky., , To explain why helical motion occurs, we need to decompose the velocity into, components perpendicular and parallel to the magnetic field. The two diagrams in, Concept 1 do this. In the left-hand diagram, the magnetic field is viewed obliquely (but, not fully parallel to the surface of the screen) and you can see both components of the, particle’s velocity., The component vperpendicular is perpendicular to the magnetic field. This component, accounts for the circular motion of the particle. The other component of the velocity,, vparallel, is parallel to the magnetic field. Since there is no magnetic force exerted on a, charge moving parallel to a magnetic field, this velocity component does not change. It, accounts for the constant upward or downward motion of the particle., The result of the forces exerted on the particle is that the particle moves in circles in a, plane perpendicular to the magnetic field and at a constant speed in a direction parallel, to the magnetic field. The sum of these motions is helical motion., As the particle moves in a helical path, the vertical spacing between the loops of the, helix, known as the pitch, remains constant since the vertical velocity component does, not change., , Helical motion, Velocity components perpendicular,, parallel to field, ·Perp: force causes circular motion, ·Parallel: force zero, vparallel constant, Helical motion results, , Helical motion can arise in a nonuniform magnetic field, as well. A magnetic field that is, stronger at its outer edges can cause a particle to become trapped in a magnetic bottle., The particle moves in a helical fashion, spiraling up and down inside the “bottle.” Physicists construct bottles like this as three-dimensional, systems that can contain charged particles indefinitely., The Earth creates a magnetic bottle of this type. Its magnetic field is stronger near the poles. Electrons and protons are trapped in the bottle, created by the Earth. They oscillate back and forth over a short distance every few seconds, resulting in what are called the Van Allen radiation, belts., Such belts, as well as solar flares, are responsible for auroras, the glorious bands of light visible in the sky at high latitudes at certain times of, the year. The auroras result from solar flares that shoot ionized particles, primarily electrons and protons, into the Earth’s atmosphere. These, particles get trapped in the Van Allen belts. As the particles collide with oxygen and nitrogen molecules from the atmosphere, they emit green, and pink light respectively. From a great distance, you may perceive a faint aurora as white light., , 520, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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28.17 - Interactive problem: helical particle motion, In this simulation you will have a chance to observe a charged particle following a, helical path through a magnetic field., If you use the initial velocity provided by the simulation, which is perpendicular to, the field, the particle will move in a circle. You can cause its path to be a helix by, supplying a component to its initial velocity that is parallel to the field., To do this, first make sure the magnetic field lines are viewed as pointing straight, down. Now you may drag the tip of the velocity vector arrow to set the initial speed, and direction of the particle. Once you have the initial velocity you want, change the, viewing angle by moving the slider provided for this purpose to a position near the, middle of its range. Press GO to observe the helical motion of the particle., , 28.18 - Magnetic force on a current-carrying wire, A magnetic field exerts a force on a wire carrying a current. Since the moving charges, in the wire í electrons in this case í cannot escape from it, the wire as a whole will, react to the magnetic force on them, just as a large net full of helium balloons will rise, due to the balloons’ individual buoyancies., At the right, we depict a configuration that shows how to determine the magnetic force, exerted by a uniform magnetic field on a straight, current-carrying wire. We show only a, segment of the wire, and not the entire circuit loop that allows current to flow., The right-hand rule can be used to determine the direction of the force. Since the, current shown is conventional (positive), the thumb points in the direction of the force, when the fingers wrap from the direction of current flow to the magnetic field. (If the, current were shown as flowing electrons, the thumb would point in the direction, opposite to the current.) In this case the force points out of the screen, toward you., To calculate the magnitude of the force exerted on a given length of wire, use the, second equation shown in Equation 1. The amount of force increases with the amount, of current, the length of the wire and the strength of the magnetic field. Since more, current means either more electrons flowing, or the electrons moving faster, this, relationship follows from the equation for force on a single charge, F = qvB sin ș (or to, state it using cross-product notation, F = qv × B). A greater length of wire will also, experience more force since it contains more moving charge., , Magnetic force on a wire, Uniform magnetic field exerts force on, current-carrying wire, Proportional to current, length of wire, Direction of force found with right-hand, rule, , Variables, We list below the variables that are used in the derivation, but which do not already, appear in Equation 1. The vector L appearing in the equation is the “directed length” of, the wire segment. That is, L is parallel to the segment, in the direction of the current,, and its magnitude L equals the length of the segment., amount of free charge in wire segment, , Q, , velocity of a charge carrier in segment, , v, , time for carrier to move through segment, , ǻt, , Strategy, , IL = qv., , 1., , Use the definition of the current flowing through the segment to find that, , 2., , Substitute this equation into the cross-product formula for the force exerted on a, moving charge by a magnetic field to get Equation 1., , Physics principles and equations, , F = ILB sin ș, F = force, I = current, , The force exerted on a moving charged particle by a magnetic field is,, , F = qv × B, , F = IL × B, , L = directed length of wire segment, B = magnetic field, ș = angle between wire and field, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 521
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Step-by-step derivation, We use the definition of the current flowing through the wire to obtain an equation, relating IL and qv. This enables us to make a substitution into the equation for the, force exerted by a magnetic field on a moving charge., , Step, , Reason, , 1., , I = q/ǻt, , definition of current, , 2., , IL = qL/ǻt, , multiply both sides by L, , 3., , IL = qv, , L/ǻt equals velocity of charge carrier, , 4., , FB = qv × B equation for force on charged particle, , 5., , FB = IL × B substitute equation 3 into equation 4, , What is the force on the wire?, F = ILB sin ș, F = (5.0 A)(1.5 m)(0.010 T)(sin 40°), F = 4.8×10í2 N straight towards you, , We state the equation proved above as a cross product. It is the same as saying the, amount of force is FB = ILB sin ș., , 28.19 - Physics at work: direct current electric motor, Direct current motors power a range of familiar, everyday devices, from hairdryer fans to handheld, electric drills. In these appliances, a few fundamental, physics principles and clever engineering combine to, create a motor that yields a constant amount of, torque., To understand how electric motors work, recall that a, magnetic field exerts a force on a wire that is, conducting electricity perpendicular to the field. When, the wire is in the shape of a loop, one side of the loop, will experience a force in one direction, and the other, in the opposite direction. This means there will be a, torque on the loop and it will rotate., , DC electric motor. The sparking copper commutator is visible, between the rotors and the red plate at the top of the image., , The magnetic moment of a coil of loops is a vector, perpendicular to each loop, used to, describe how much torque the coil will experience in a magnetic field. You can think of, the coil as acting like a bar magnet. When the coil’s moment is parallel to the field, it, experiences no torque, since the “magnet” is lined up with the field. When it is, perpendicular to the field, it experiences maximum torque., At its simplest, a direct current motor consists of a current-carrying wire coil, wrapped, around a metal armature, inside a uniform magnetic field. The external field creates a, torque on the coil when current flows through it. That torque is used to rotate, something: a fan, a drill bit, the blades of a blender and so forth., However, a motor has two requirements that this configuration alone does not meet., First, it needs to rotate continuously in the same direction. The simple wire coil with a, current will not do this: It will rotate in one direction until the field exerts no net torque on, it, continue on beyond that due to its momentum, and then rotate back., The second requirement for the motor is that it should provide a nearly constant torque., The torque on a simple coil varies as the angle of the coil in the field changes., In Concept 1, you see a schematic diagram of a simplified direct current motor. Two, permanent magnets form the circular outer edge of the motor. A magnetic field is, directed from the north poles of these two magnets to the south poles. Inside these, magnets is a wire coil, which is connected to a direct current, such as the current from a, battery. This assembly is called a rotor., , Electric motor, Rotor: bar surrounded by wire coil, Current flows through coil, Current creates magnetic moment, External field exerts torque on coil, , The problem of rotating the rotor in a constant direction is solved with a commutator. A, commutator consists of two sets of contacts that supply current to the coil. During onehalf of a rotation, the coil is in contact with one set and the current flows in one, direction. During the other half turn, the coil is in contact with the other set and the, current flows in the opposite direction. In Concept 2, you can see that the current keeps, reversing direction, which alters the direction of the coil’s magnetic dipole moment. This, reversal of current causes the rotor to keep experiencing a counterclockwise torque., The problem of supplying constant torque is addressed with multiple rotors as you see, in Concept 3. At any moment in time, current flows through two of the rotors, which are, at different angles to the permanent field. Although the torque on any one rotor varies, , 522, , Commutator, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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as it rotates, the sum of the torques on all three rotors stays nearly constant. This, combination of current-reversing commutators with multiple rotors enables the motor to, provide a steady source of torque., , Changes direction of current, Reverses magnetic moment, , Rotors with multiple arms, More arms = smoother torque, , 28.20 - The magnetic field around a wire, Currents produce magnetic fields. In this section, we analyze the orientation of the, magnetic field created by a current flowing through a wire. In the Concept 1 and 2, diagrams, the electrons that make up the current are moving to the right. Conventional, current flow is indicated (as always) with an arrow pointing in the opposite direction., The Concept 1 diagram uses two compasses to show the direction of the magnetic field, due to the current at two different points. Each needle points in the direction of the, magnetic field at its location., As the diagram shows, the magnetic field wraps around the current. The arrows indicate, its direction. The field is strongest near the wires. As with electric field diagrams, the, density of lines reflects the strength of the field. This means the lines are drawn close, together near the wire., How do you determine the direction of the magnetic field? You use the right-hand rule, for currents. Wrap your hand around the wire so that your thumb points in the direction, of the (conventional) current. Your fingers wrap around, just like the concentric circles in, the diagram, and they point in the direction of the magnetic field. You see this shown in, Concept 2. If the direction of the current were to reverse, your thumb would point in the, opposite direction, and your fingers would wrap in the other direction. When the, direction of the current changes, so does the direction of the magnetic field., , Magnetic fields and wires, Currents generate magnetic fields, ·Magnetic field “circles” around wire, , Right-hand rule for currents, Right-hand rule tells direction of field, ·Point thumb in direction of conventional, current, ·Curled fingers show direction of, magnetic field, , Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 523
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28.21 - Two wires and their magnetic fields, To illustrate the nature of the fields and forces produced by currents, we use the, example of two wires placed parallel to one other and conducting currents flowing in the, same direction, as illustrated in Concept 1. The two wires attract each other., Let’s determine why they attract each other by first considering the orientation of the, magnetic field created by the wire on the bottom. We use the right-hand rule for, currents to determine this orientation. This is shown in Concept 1., Next, we determine the direction of the force exerted by the magnetic field on the, electrons moving in the upper wire. The direction of electron movement is shown with, velocity vectors in Concept 2. The right-hand rule for a charge moving in a magnetic, field is used to determine the direction of the magnetic force on the electron. To apply, the rule, point your fingers in the direction of the velocity vector and wrap them in the, direction of the magnetic field. As the diagram shows, when it intersects the upper wire,, the field is pointing away from you. Wrapping your fingers from the velocity vector to the, field vector will cause your thumb to point up. Since an electron is a negative charge,, your thumb will point in the direction opposite to the magnetic force on it. The result is, that the force points down, toward the lower wire., , Currents and their magnetic, fields, Each current creates a magnetic field, ·Field intersects other wire, , The right-hand rule for force can be used again to determine the force on the moving, charges in the lower wire. If you apply it, you can confirm that they experience an, upward force. This analysis explains why the two wires attract each other. We show the, forces acting on both wires in Concept 3., If the currents flow in opposite directions, the two wires repel each other. This scenario, is illustrated in Concept 4. Here, the current in the bottom wire now flows in the opposite, direction than in the prior discussion, which means its field has the opposite orientation., The force on the electrons in the upper wire now pushes them and the wire that, contains them up, as the diagram reflects. The field of the upper wire pushes the, electrons in the lower wire down. The two wires repel each other., , Currents in same direction, Field exerts force on moving charges, , Currents in same direction, The wires attract, , Currents in opposite directions, The wires repel, , 524, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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28.22 - Strength of the magnetic field around a wire, The strength of the magnetic field around a current-carrying wire can be calculated with, the equation to the right. The field strength is a function of two variables: the amount of, current and the distance from the wire. The field increases with current and decreases, with distance. The equation includes ȝ0, the permeability constant of free space. This, constant equals 4ʌ×10í7, , N/A2., , This equation is derived under the assumption that the wire is long and thin. It provides, accurate values when the distance from the wire is significantly less than the wire, length., The direction of the magnetic field is determined using the right-hand rule for currents., To review: The conventional current runs to the right, so the thumb points to the right, and the fingers wrap in the direction of the magnetic field., To gain a sense of the strength of the magnetic field around a typical wire, you can look, at the example problem to the right. Calculations show that the magnetic field strength, 15.0 cm away from a current of two amperes is 0.0267 G., , Magnetic field around a wire, Strength of magnetic field, ·Increases with amount of current, ·Decreases with distance from wire, , The field at this distance from the wire is much less than the average magnetic field, strength of the Earth, which is about 0.5 G. On the other hand, at a point 1.5 cm from, the wire, the field would be 0.267 G, or about half as strong as the Earth’s field., , B = magnetic field, ȝ0 = permeability of free space, I = current, R = distance from center of wire, Constant ȝ0 = 4ʌ×10í7 N/A2, , What is the magnetic field, strength and direction at point, P?, , B = 2.67×10í6 T = 2.67×10í2 G, Field points into screen at point P, Copyright 2000-2007 Kinetic Books Co. Chapter 28, , 525
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28.23 - Gotchas, Compass needles point north because Earth’s magnetic north pole is located near the geographic North Pole. No, the north end of the needle, on a compass is attracted to the magnetic south pole of the Earth. Compass needles point approximately to the geographic North Pole, because the magnetic south pole of the Earth is near its geographic North Pole., Magnetic force vectors on charged particles point in the same direction as the magnetic field. No, the magnetic force vectors which act on, moving, electrically charged particles are perpendicular to the magnetic field, and to the particles’ velocities, as well., Magnetic fields exert a force on all moving electrically charged particles. Almost true. The charges have to be moving for there to be a force,, but if they are moving parallel to or opposite to the field, there will be no force. There must be at least some component of the velocity, perpendicular to the field for a force to exist., When I use the right-hand rule for any charged particle moving in a magnetic field, my thumb points in the direction of the magnetic force. No,, the right-hand rule gives the direction of the force on a positively charged particle. For negative particles, the thumb points opposite to the, direction of the magnetic force., , 28.24 - Summary, A magnet is an object that creates magnetic fields and can exert a magnetic force, on other magnets or on moving charged particles. Magnets always have two poles,, called north and south poles. As with electrical charges, opposite poles attract each, other and like poles repel., The exterior field lines of the magnetic field generated by a magnet are directed, from its north pole to its south pole. The symbol for a magnetic field is B, a vector., Magnetic fields are measured in teslas (T). 1 T = 1 N·s / C·m. Magnetic fields í, especially weaker ones í are also measured in smaller units called gauss (G)., 1 G = 10í4 T., The Earth has its own magnetic field, which is why compasses work on its surface., The Earth’s magnetic south pole is near the geographic North Pole. Because the, two do not coincide, when using a compass you need to know the declination, the, angle between the magnetic and geographic poles at your location. Compasses can, also help you estimate your latitude if they are allowed to orient in three dimensions., , Force on charge moving in B field, , Motion of a charge in a B field, , Magnetic force on a wire, , F = ILB sin ș, , Materials that are able to become permanent magnets, like iron, are said to exhibit, ferromagnetism. Ferromagnetic materials contain small regions with their own magnetic fields called domains. Domains are in turn made up of, electrons whose spins have a net alignment., A magnetic field exerts a force on a moving charge. The force is perpendicular to both the velocity of the charge and the magnetic field. If you, wrap the fingers of your right hand from the velocity vector to the magnetic field vector, your thumb points in the direction of the force on a, positive charge. Your thumb is pointing in the direction opposite to the force on a negative charge., A charged particle that is moving perpendicularly to a uniform magnetic field will move in a circular path. If it has a velocity component parallel, to the field, that component will cause helical motion: The particle will move in a circular path (in two dimensions) while moving at a constant, velocity in the third dimension., A device called a mass spectrometer takes advantage of the circular motion caused by a magnetic field to separate moving particles by their, mass-to-charge ratios., Since an electric current consists of moving charges, a current-carrying wire can have a force exerted on it by a magnetic field. The strength of, the force is proportional to the magnitude of the current, the length of the wire, the magnetic field strength, and the sine of the angle between, the wire and the field., An electric motor takes advantage of the torque exerted by a magnetic field on a loop of current to spin a rotor, which turns a shaft to perform, useful work., Magnetic fields exert a force on moving charges, but moving charges also create magnetic fields of their own., A current-carrying wire creates a magnetic field that circles around the wire. The strength of the field decreases as you move farther from the, wire, and increases as the current increases. The direction of the field lines is given by the right-hand rule: Your thumb points in the direction of, current and your fingers wrap around the wire in the direction of the field lines., When two current-carrying wires are placed parallel to each other, the magnetic field of each one affects the other. When the currents in the, wires are in the same direction, the wires attract. When the currents are running in opposite directions, the wires repel each other., , 526, , Copyright 2000-2007 Kinetic Books Co. Chapter 28
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Chapter 28 Problems, , Chapter Assumptions, Elementary charge, e, , = 9.11×10–31 kg, , Mass of electron, me, Mass of proton, mp, , = 1.60×10–19 C, , = 1.67×10–27 kg, , Unless stated otherwise, use 5.00×10–5 T for the strength of the Earth's magnetic field at its surface., , Conceptual Problems, C.1, , While you are on a visit to Santa Claus at his workshop in the Far North, he gives you a special compass whose needle can, rotate freely in three dimensions. You try it out by standing directly above the nearby magnetic pole of the Earth. Which, direction does the north end of the compass needle point? Explain., i., ii., iii., iv., v., vi., , C.2, , North, South, East, West, Up, Down, , Two parallel wires conduct equal currents in opposite directions (as is the case with ordinary household electric cords). At a, point in the plane of the wires midway between them, (a) what is the direction of the combined magnetic field generated by, the two currents? (b) Do the individual fields of the wires cancel or reinforce each other there? (c) At a point in the plane of the, wires but far away to one side, do the individual fields of the wires tend to cancel or reinforce each other? Explain your, answer to part c., (a) The field is, , C.3, , i. parallel, ii. perpendicular, iii. at some other angle, , (b), , Cancel, , Reinforce, , (c), , Cancel, , Reinforce, , to the plane of the wires., , A charged particle moves in a straight line through a region of space. (a) Can there be a nonzero magnetic field in this region, of space? (b) If not, explain why not, and if so, explain in what direction(s) it must point., (a), , Yes, , No, , (b), C.4, , C.5, , Two parallel wires conduct equal currents in the same direction. (a) At a point in the plane of the wires midway between, them, do the individual fields of the wires cancel or reinforce each other? (b) At a point far away from the wires, do their, individual fields tend to cancel or reinforce each other?, (a), , Cancel, , Reinforce, , (b), , Cancel, , Reinforce, , Suppose you are "chirally challenged" (like one of the authors of this textbook, and, as it happens, the physicist Heinrich, Hertz): All too often, you cannot accurately recall which is right and which is left. Not only do your driving friends refuse to let, you "navigate," but you occasionally use your left hand by mistake when attempting to find the direction of the force exerted, on a moving proton by a magnetic field. (a) Do you get the right direction anyway? (b) For which of the following particles, does your left thumb point in the direction of the force?, (a), (b), , C.6, , Yes, i., ii., iii., iv., v., , No, , Positron, Neutron, Electron, Neutrino, Proton, , Do perpendicular straight wires that carry a steady current, and are almost touching, exert a net force on each other? Explain, whether the wires influence each other in any way at all., Yes, , No, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 527
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C.7, , Can the force from a steady magnetic field cause the speed of a moving charged particle to change? Explain., Yes, , No, , C.8, , Is your electric blanket safe? It has been theorized that excess nighttime exposure to electromagnetic fields may reduce the, production of the hormone melatonin in certain women. Furthermore, reduced melatonin levels might lead to an increased risk, of breast cancer. Scott Davis at Seattle's University of Washington School of Public Health and Community Medicine was one, of the first to investigate links between electromagnetic fields and cancer (particularly leukemia). He has been quoted as, saying, "Electric blankets were pretty effectively re-engineered a number of years ago. Modern-day electric blankets í, anything you buy in the store now í (has a different) configuration of the heating elements in the blanket so that it doesn't, produce any magnetic fields." How could the wiring in an electric blanket be designed to minimize or eliminate the production, of external magnetic fields?, , C.9, , Can the force from a steady magnetic field cause a stationary charged particle to start moving? Explain., Yes, , No, , C.10 An up-and-coming entrepreneur, long on marketing skills but short on physics knowledge, develops a "new and improved", device for detecting current-carrying wires inside household walls. This is something a do-it-yourselfer really needs to know, before picking up the old saber-saw and starting to cut out that new doorway! The "Mag-Flag Juicefinder," which costs, $79.99, is actually nothing but a magnetized needle free to rotate. The advertising materials explain that the wires in the walls, generate magnetic fields, and that the needle will align with these fields in a direction perpendicular to the hidden wires. Why, will this device not work? Hint: Household wiring cables inside the walls contain three wires bundled close together. One is a, "ground," and the same amount of current flows in opposite directions through the other two., C.11 Each of the diagrams (a), (b), (c), and (d), represents a positively charged particle, moving at velocity v through a magnetic field, B whose field lines are shown. In each case,, tell the direction of the force experienced by, the moving particle in the field., (a), , (b), , (c), , (d), , 528, , i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., , Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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C.12 Each of the diagrams (a), (b), (c), and (d), represents a negatively charged particle, moving at velocity v through a magnetic field, B whose field lines are not shown. In each, case the field exerts a force F on the particle, in the direction shown. In what direction must, the field be pointing?, (a), , (b), , (c), , (d), , i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., i., ii., iii., iv., v., vi., , Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, Right, Left, Up, Down, Toward you, Away from you, , C.13 In each part of this problem, you know a charged particle is moving at a constant velocity, and you are asked to draw, conclusions about the electric and magnetic fields around the particle. Consider each of the cases independently, and ignore, any gravitational effects. In each case, check all of the configurations consistent with the situation, and explain your answer., (a) There is no magnetic field present. (b) There is no electric field present. (c) There is a magnetic field present, with a, component perpendicular to the particle's motion., (a), , There is no electric field, There is an electric field parallel to the velocity, There is an electric field perpendicular to the velocity, None of these are possible, , (b), , There is no magnetic field, There is a magnetic field parallel to the velocity, There is a magnetic field perpendicular to the velocity, None of these are possible, , (c), , There is no electric field, There is an electric field parallel to the velocity, There is an electric field perpendicular to the velocity, None of these are possible, , C.14 The picture shows the paths of some neutral, and charged particles in a bubble chamber,, which contains an undisturbed liquid heated, above its boiling point. The idea is that, bubbles begin forming on nucleation centers, such as surface irregularities, or particles, moving through the liquid. These bubbles, allow you to see the trails of the particles., Particles enter the chamber from the left. A, uniform magnetic field is directed out of the, screen toward you. A certain incoming particle, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 529
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(whose trail is not visible) splits into 3 other particles, including those that follow paths A and B. (a) What is the sign of the, charge of the particle that follows path A? (b) What is the sign of the charge of the particle that follows path B? (c) Explain, your answers., (a), , i., ii., iii., i., ii., iii., , (b), , Positive, Negative, Neutral, Positive, Negative, Neutral, , (c), C.15 Some computer monitors are made with cathode ray tubes (CRTs) like a traditional TV screen. The glowing image on a CRT, monitor is created by a steerable stream of electrons that travel from the back of the tube to strike tiny phosphors on the, inside of the screen surface. When a bar magnet is placed near a CRT computer monitor, it causes severe (and possibly, lasting) distortion of the image displayed on the screen. Describe what causes the distortion., C.16 A focused magnetic field of variable strength is, used to "steer" electrons in a television tube, (CRT). The electrons emerge in a stream from, a "gun" that has accelerated them across a, potential difference. A magnetic field can, deflect the beam of electrons in various, directions. They then travel to the tube's front, inside surface on the TV screen, where they, strike red, green, or blue phosphors and cause, them to glow, creating part of one pixel in a, picture display (actual color tubes use three, guns, one for each color). Use the orientation of the components as shown in the illustration to answer these questions. (a) In, what direction should the magnetic field point to deflect the electron beam upward, toward the top of the tube? (b) A TV tube, plays the electron beam over its picture screen in a raster pattern, moving it from one side of the screen to the other at the, very top of the tube, then all the way across the screen at a lower point, and again and again at still lower points, until the, whole screen is painted with sideways sweeps. A second magnetic field of varying strength, which is not shown in the, illustration, deflects the beam from side to side at each one of its vertical positions. Suppose a beam is heading straight, toward the left in the illustration (it is not deflected vertically). In what direction would the second field have to point to deflect, the beam toward you?, i., ii., iii., iv., v., vi., , i., ii., iii., iv., v., vi., , Right, Left, Up, Down, Toward you, Away from you, , Right, Left, Up, Down, Toward you, Away from you, , C.17 A charged particle is moving freely through a uniform field. How can you tell if the field is an electric field or a magnetic field?, C.18 Cosmic "rays", which are charged particles approaching the Earth from space, strike the planet more frequently at the poles, than at the equator. Why is this?, C.19 A current-carrying wire is placed in a uniform magnetic field, but there is no magnetic force on it. How is this possible?, , Section Problems, Section 0 - Introduction, 0.1, , 530, , Use the simulation in the first interactive problem, and the initial conditions, in this section to answer the following questions., (a) Does the moving particle travel in a straight line or a curve? (b) Is the particle's speed changing? (c) Is the particle, accelerating?, (a), , Straight line, , (b), , Yes, , No, , (c), , Yes, , No, , Curve, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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0.2, , Use the simulation in the first interactive problem, and the initial conditions, in this section to answer the following questions., (a) Are the force and velocity vectors parallel or perpendicular to each other? (b) What is the angle between the force vector, and the magnetic field?, (a), (b), , Parallel, i. 0, ii. 45, iii. 90, , Perpendicular, degrees, , Section 7 - Magnetic fields and charged particles, 7.1, , An alpha particle (q = +3.20×10í19 C) moves at a velocity of 2.45×104 m/s in the positive x direction through a magnetic field, of 0.0775 T that points in the negative z direction. What are (a) the magnitude and (b) the direction of the magnetic force on, the particle?, (a), (b), , 7.2, , N, i., ii., iii., iv., v., vi., , Positive x, Positive y, Positive z, Negative x, Negative y, Negative z, , direction, , A charged particle moving at a speed of 175 m/s due east, through a magnetic field of 5.00 mT pointing straight up toward, the sky, experiences a force of 4.25 μN directed due north. What is the charge on the particle? Be sure to indicate the sign., C, , 7.3, , An electron (q = í1.60×10í19 C) is traveling through the Earth's magnetic field at a location on the Earth's surface where the, field has a strength of exactly 5.00×10í5 T, directed due north. At a particular instant, a magnetic force of 6.44×10í21 N,, directed straight down, acts on the electron. If the electron is traveling in a cardinal compass direction, what are (a) the, magnitude and (b) the direction of the electron's velocity at that instant?, (a), (b), , 7.4, , m/s, i., ii., iii., iv., , North, South, East, West, , A boy shuffles across an outdoor nylon carpet, acquiring a positive static charge, and then fires a BB gun in the direction of, the setting sun. The BB travels horizontally at a speed of 105, , m/s, carrying away a charge of +3.43 μC. If the Earth's local, N on it, what is the magnitude of, , magnetic field is perpendicular to the velocity of the BB, and it exerts a force of 1.80×10–8, the magnetic field at his location?, T, 7.5, , An electrically charged bullet of mass 0.010 kg leaves the barrel of a gun traveling parallel to the ground and perpendicular to, the Earth's magnetic field at 1.10e+3 m/s. In theory, what charge could the bullet carry in order to remain at a constant, height? Assume that the Earth is flat, and that the magnetic field (0.500 gauss) is horizontal and points left if you are looking, in the direction of the bullet's motion., C, , 7.6, , An F-15 fighter aircraft is flying at 850 m/s in a direction that makes a 30.6° angle with the Earth's magnetic field, which has a, strength of 5.03×10í5 T. The plane has acquired a static charge of 343 μC. What is the strength of the force exerted by the, field on the aircraft?, N, , 7.7, , A youth shuffles across an outdoor nylon carpet, acquiring a positive static charge, and then fires a BB gun due east. The BB, travels horizontally at a speed of 98.4 m/s, carrying away a charge of +8.66 μC. If the Earth's local magnetic field has a, strength of 5.04×10í5 T, is purely horizontal, and exerts a force of 4.10×10í8 N on the BB, what is the declination of the field, at his location? Express your answer as a small positive angle between due north and the direction of the magnetic field., °, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 531
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Section 9 - Interactive problem: charged particle moving in a B field, 9.1, , Use the information given in the interactive problem in this section to answer the following questions. Test your answers, using the simulation. (a) How many angles between 0° and 180°, inclusive, will result in a force of 0 N? (b) What angle in that, range will give a force equal to 0 N? (If there is more than one, give the largest.) (c) What angle between 0° and 180°,, inclusive, will maximize the resulting force?, (a), , i., ii., iii., iv., v., , 0, 1, 2, 3, 4, ° (c), , (b), , °, , Section 10 - Determining the strength of a magnetic field, 10.1 A proton is moving at 2.00×105 m/s and begins to circle at a radius of 5.00eí3 m. What is the strength of the perpendicular, magnetic field acting on the proton?, T, 10.2 A proton travels with a speed of 8.34×105 m/s at an angle of 32.3° with the direction of a magnetic field. The field exerts a, force of 3.00×10í17 N on the proton. What is the strength of the field?, T, 10.3 A charged body has a charge of 3.33 μC. It is moving with a speed of 8320 m/s at an angle of 42.0° with a uniform magnetic, field. The field exerts a force of 0.0117 N on the body. What is the strength of the magnetic field?, T, 10.4 A television screen is illuminated using a beam of electrons traveling at 7.50×106 m/s in a 1.75×10í2 T magnetic field. If an, electron is moving in the positive x direction and the magnetic field is oriented 55.0° from the z axis in the first quadrant of the, xz plane, what are the (a) magnitude and (b) direction of the acceleration of the electron?, m/s2, , (a), (b), , i., ii., iii., iv., v., vi., , Positive x, Positive y, Positive z, Negative x, Negative y, Negative z, , 10.5 A uranium atom is ionized seven times, losing seven electrons in the process. Its resulting mass is 3.90×10í25 kg and it is, given a kinetic energy of 3000 eV. If it circles in a magnetic field at a radius of 3.00 m, what is the strength of the magnetic, field?, T, 10.6 An alpha particle (m = 6.68×10í27 kg, q = 3.20×10í19 C) is moving perpendicularly to a uniform magnetic field at a speed of, 7.88×106 m/s. Because of the force exerted on it by the magnetic field, it is accelerating at 5.67×1012 m/s2. What is the, strength of the magnetic field?, T, , Section 11 - Interactive problem: B field strength and particle motion, 11.1 Use the information given in the interactive problem in this section with an initial speed of 400 m/s to find the B field strength, needed to achieve the desired amount of force. Test your answer using the simulation., T, , Section 12 - Physics at work: velocity selector, 12.1 The electric field in a velocity selector is generated by two parallel plates separated by 5.00 cm with a potential difference of, 90.0 V. If the magnetic field is 6.00eí2 T, at what speed will particles pass through undeflected?, m/s, 12.2 A velocity selector is tuned to let charges with a speed of 325 m/s pass through. If the strength of the magnetic field is, 0.250 T, what is the strength of the electric field?, N/C, , 532, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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12.3 The electric field strength of a velocity selector is 345 N/C. The magnetic field strength is 1.34 T. Charged alpha particles, having a mass of 6.68×10í27 kg pass through the selector. What is the final kinetic energy of each one?, J, 12.4 The oppositely charged plates of a velocity selector are separated by 4.5 cm. The uniform magnetic field is supplied by a, current-carrying solenoid that encloses the selector, and it has a strength of 0.35 T. What potential difference needs to be, applied across the charged plates in order to select a particle velocity of 3.2×104 m/s?, V, 12.5 A velocity selector of length 0.25 m has a magnetic field of 5.0×10í3 T and an electric field of 50 N/C. Electrons are injected, into the apparatus and must pass through a slit at the far end which is exactly 1.0 cm across. (a) What is the velocity of a, particle that goes through the velocity selector undeflected? (b) By how much can the velocity be increased or decreased, from the undeflected value and still make it through the velocity selector? Assume the following: (1) The electrons begin their, journey aligned with the center of the slit. (2) The magnetic force always acts in the same direction (that is, it causes an, acceleration that we can approximate as linear). This is justified so long as the slit width is much smaller than the length of the, velocity selector. (3) The ratio of the ideally selected velocity and the difference between the maximum and minimum, velocities is very small. This will allow you to simplify some of your calculations. The final answer will demonstrate that this, assumption is also justified., (a), , m/s, , (b), , m/s, , Section 13 - Circular motion of particles in magnetic fields, 13.1 A proton moves perpendicularly to a uniform magnetic field of strength 0.0585 T at a speed of 15,400 m/s. What is the radius, of the circle it describes?, m, 13.2 A charged particle moves in a concentric circular orbit inside a solenoid made of very thin wires, with a diameter of 10.4 cm,, just missing the inner surface of the solenoid coils. The charge on the particle is 3.00 μC, its speed is 550 m/s, and the, strength of the field inside the solenoid is 0.734 T. What is the mass of the particle?, kg, 13.3 How fast must an alpha particle (m = 6.68×10í27 kg, q = 3.20×10í19 C) travel perpendicularly to the Earth's magnetic field in a, laboratory (B = 5.02×10í5 T) for it to describe a circular path of radius 2.34 cm?, m/s, 13.4 A mad scientist has designed a new line of highly stylish, Italian-cut Kevlar body armor suits. In order to test one of these, suits he wishes to shoot himself, using a .22 caliber rifle that fires rounds weighing 3.56×10í3 kg with a velocity of 762 m/s. To, shoot himself, he will fire the gun into a vertically oriented magnetic field of strength 1.25 T, hoping that the bullet will follow a, semi-circular path, emerge from the field, and fly straight toward him, striking the body armor. In order for the bullet to hit him, he needs to have the bullet circle through a small radius of 2.00 cm. Finally, he knows that in order to be deflected by the, magnetic field, the flying bullet needs to be electrically charged, so the firearm has an insulating rubber grip and is wired to a, static electricity generator. What magnitude of charge must be placed on the bullet for the scientist's mad scheme to, succeed? (The answer will be impractically large, because the scientist is truly mad.), C, 13.5 What is the period of the orbit for a particle of mass 7.00×10í6 kg and charge 2.00×10í6 C circulating in a 6.00eí2 T, magnetic field?, s, 13.6 A particle with mass 1.65×10í12 kg and electric charge í8.22×10í8 C moves perpendicularly to a uniform magnetic field of, strength 5.45×10í3 T. The angular momentum of the particle about the center of its circular orbit is 3.20×10í10 kg·m2/s. (a), What is the radius of the circular orbit? (b) What is the speed of the particle?, (a), , m, , (b), , m/s, , 13.7 An ionized hydrogen atom (that is, a free proton) is moving through the magnetic field of the Milky Way galaxy, at a location, distant from any star. The kinetic energy of the proton is 5.60 keV, and it moves through a circular path having a radius of, 0.0317 astronomical units. An astronomical unit (AU) equals the average radius of the Earth's orbit around the Sun,, 1.50×1011 m. What is the strength of the galactic magnetic field in this region of space?, T, 13.8 A sodium ion with mass 2.32×10í26 kg and charge +1.60×10í19 C is moving with velocity 3.00×105 m/s in the positive x, direction. An identical particle is moving with the same speed in the negative x direction, but displaced by +1.50 m in the y, direction. When the x coordinates of the atoms are equal, a magnetic field perpendicular to their motion is turned on so that, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 533
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the atoms collide head-on. What are (a) the, magnitude and (b) the direction of the, magnetic field?, (a), (b), , T, i., ii., iii., iv., v., vi., , Positive x, Negative x, Positive y, Negative y, Positive z, Negative z, , Section 15 - Physics at work: mass spectrometer, 15.1 Two elements whose atoms have the same mass number (the number of protons plus neutrons) are called isobars. The only, two stable isobars with mass number 58 are 58Fe with a mass of 57.933277 u and 58Ni with a mass of 57.935346 u, (1 u = 1.6605402×10í27 kg). The mass spectrometer in a laboratory can resolve spatial displacements of 1.00×10í4 m after a, particle's 180° flight. (a) If the particles are singly ionized and enter the spectrometer at 3.00×104 m/s, what magnetic field, strength will just allow these two isobars to be resolved? (Hint: for this problem, you must keep all of the significant figures for, the masses of the two types of atom as you work through the problem and round your answer at the end to the correct, number of significant figures. Why? Try rounding at the beginning of your work and see what happens.) (b) If you want to, increase the final distance between the isobars, should you increase or decrease the magnetic field strength?, (a), (b), , T, Increase, , Decrease, , 15.2 A mass spectrometer has an adjustable magnetic field. When the magnetic field is 3.00eí2 T, a singly ionized hydrogen atom, (a proton) will strike the detector at a particular location. Suppose a triply-ionized carbon 12 atom, whose mass is by definition, 12 u (where 1 u = 1.66×10í27 kg) enters the spectrometer at the same speed as the proton. What is the strength of the, magnetic field that will cause it to strike the detector at the same location as the proton?, T, 15.3 A neon atom of mass 3.82×10í26 kg is triply ionized. That is, three electrons are stripped from the atom so that its charge is, +3e, or 4.80×10í19 C. The atom is accelerated across a potential difference of magnitude 212 V, and enters a mass, spectrometer whose magnetic field has a strength of 1.25 T. (a) What is the radius of the atom's trajectory in the magnetic, field? (b) What is the radius of the trajectory of a sulfur ion, also triply ionized, but having mass 5.64×10í26 kg? (c) What is the, distance between the points on the detector plate struck by these two atoms?, (a), , m, , (b), , m, , (c), , m, , 15.4 When the ionized particles entering a mass, spectrometer are simply accelerated across, the potential difference between two charged, plates, there is always the chance of an, uncontrolled variation in their velocities due to, their random thermal motion before they, undergo linear acceleration. An elaboration of, the basic mass spectrometer design better, controls the incoming velocity of the ions by, passing them through a velocity selector, before they enter the magnetic deflection, chamber. Suppose that the same uniform magnetic field, having a strength 0.750 T, is used in both the velocity selector and, the deflection chamber. In this chamber, singly ionized argon ions with a mass of 6.63×10í26 kg are to be deflected through, semi-circular arcs of radius 12.0 cm. (a) If the oppositely charged plates of the velocity selector are separated by 2.00 cm,, what is the required potential difference between them? (b) If you were to add a velocity selector to the diagram of the mass, spectrometer, as shown in the illustration, which plate would be positively charged with respect to the other?, (a), (b), , 534, , V, The right plate, , The left plate, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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15.5 One of the steps in the manufacture of nuclear, weapons is the separation of the fissile isotope, of uranium, 235U (235.05 u), from its cousin, (238.05 u). The purity of samples of, separated uranium is regularly checked in a, nuclear weapons lab by vaporizing and singly, ionizing the uranium, accelerating the ions and, passing them through a velocity selector into a, mass spectrometer deflection chamber., Photographic film is placed in the position of, the recording plate. The two isotopes of, uranium will strike the film plate at different locations, and their relative abundance will be indicated by the size and darkness, 238U, , of the two spots created on the film. The velocity selector is set to pass ions with a speed of 3.43×105 m/s, and the magnetic, field in the deflection chamber is 0.750 T. What is the distance d between the locations where the two ions of uranium strike, the photographic plate? (Note that 1 u = 1.66×10í27 kg.), m, , Section 16 - Helical particle motion in magnetic fields, 16.1 A particle of charge q = 2.73×10í12 C and mass m = 3.32×10í17 kg enters a uniform magnetic field B of magnitude, 9.41×10í5 T with a velocity v of magnitude 8.34 m/s, and moves in a helix while in the field. The angle between v and B is, 84.0°. (a) What is the magnitude vperp of the component of v perpendicular to B? (b) What is the radius of the helix? (c) What, is the magnitude vpar of the component of v parallel to B? (d) What is the pitch of the helix, the distance between adjacent, loops (measured parallel to B)?, (a), , m/s, , (b), , m, , (c), , m/s, , (d), , m, , Section 17 - Interactive problem: helical particle motion, 17.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) If the velocity vector is, initially perpendicular to the magnetic field lines, what is the shape of the resulting path of the particle? (b) If the velocity, vector is initially parallel to the magnetic field lines, what is the shape of the resulting path of the particle?, (a), , (b), , i., ii., iii., iv., i., ii., iii., iv., , A helix, A zigzag, A circle, A straight line, A helix, A zigzag, A circle, A straight line, , Section 18 - Magnetic force on a current-carrying wire, 18.1 A 15.0 ampere current flows through a 3.75 m section of wire, perpendicular to a 0.250 T magnetic field. What is the, magnitude of the force exerted by the field on the wire section?, N, 18.2 A straight current-carrying wire carries a current of 0.250 A. It makes an angle of 30° with a uniform magnetic field of strength, 0.0255 T. What is the magnitude of the force per unit length that the field exerts on the wire?, N/m, 18.3 A segment of insulated wire 0.032 m long carries a current perpendicular to a uniform magnetic field. The strength of the field, is 0.0055 T. The wire is attached to a spring that lies in a direction perpendicular to both the wire and the field. When a current, is passed through the wire, it stretches the spring by an additional amount ǻx = 4.6×10í4 m. The spring constant is 0.87 N/m., What current is passing through the wire?, A, 18.4 A long horizontal wire with linear mass density Ȝ = 47.5×10í3 kg/m carries a current of 0.534 A. The wire is situated so that it, is perpendicular to a uniform, horizontal magnetic field that is just strong enough to levitate the wire against its weight. What is, the strength of the field?, T, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 535
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18.5 A wire strung across a rotating table carries a current of 0.250 A, and a uniform magnetic field of 0.500 T is applied in the, plane of the table. Consider a straight section of the wire 75.0 cm long. As the table is rotated, the magnetic force exerted on, this section will vary. What angle does the wire segment make with the field at the instant when the force is 0.0540 N? Report, the angle that is less than 90 degrees., °, , Section 20 - The magnetic field around a wire, 20.1 The four diagrams in the illustration show a, current-carrying electric wire from various, viewing angles. In which diagram or diagrams, is the orientation of the resulting magnetic field, correctly drawn? Check all that apply., (a), (b), (c), (d), , Section 22 - Strength of the magnetic field around a wire, 22.1 A long power line has a steady current of 125 A flowing through it. What is the strength of the resulting magnetic field at a, distance of 1.25 m from the wire?, T, 22.2 A typical power transmission line carries a current of 1000 A about 30.0 meters above the ground. The magnetic field, generated at ground level by the current is a matter of concern to those who live nearby. (a) What is the strength of the field, at ground level below one such wire? (b) The strength of the Earth's magnetic field is about 5×10í5 T at sea level in middle, latitudes. How close would you have to be to the power line to experience from it a magnetic field of equal strength?, (a), , T, , (b), , m, , 22.3 When a lightning bolt strikes a lightning rod, its current is typically conducted to the ground by a ground rod. When the, current in the rod exceeds 20.0 kA, it can no longer be confined by the rod and will arc laterally through the air to the ground., What is the strength of the magnetic field 3.50 m from a ground rod that is conducting current at the limit of its capabilities?, Treat the ground rod as a long, straight wire., T, 22.4 A continuous stream of electrons flows along a long straight path through a piece of laboratory equipment: 4.25e+18 of them, pass through the equipment each second. What is the strength of the magnetic field created by this stream at a distance of, 0.250 m?, T, 22.5 You are holding a compass 3.50 m from a long straight electrical supply wire that is carrying a direct current perpendicularly, to the local direction of the Earth's magnetic field. Instead of orienting to any consistent "north" direction, the compass needle, swings aimlessly no matter how you tilt or shake the instrument. The strength of the Earth's magnetic field at your location is, 5.12×10í5 T. What is the magnitude of the current carried by the wire?, A, 22.6 A long, straight current-carrying wire generates a magnetic field of 1.00 μT at a distance of 1.00 m from the wire. (a) At what, distance is the field strength equal to 0.100 μT? (b) At what distance is the field strength equal to 1.00 mT? (c) What current, through the wire creates a 1.00 mT magnetic field at a distance of 1.00 m?, (a), , m, , (b), , m, , (c), , A, , 22.7 A straight horizontal wire carries a current of 3.55 μA. A proton is moving above the wire in a direction parallel to it but, opposite to the direction of the conventional current in the wire, at a constant velocity of 8.00×106 m/s. Assume that the force, of gravity acts on the proton and that the proton is moving parallel to the Earth's magnetic field, so that the only magnetic, force on it comes from the field generated by the current in the wire. (a) What is the magnitude of the upward force that the, magnetic field must exert on the proton to keep it at a constant velocity? (b) How far is the proton above the wire?, , 536, , (a), , N, , (b), , m, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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22.8 A straight horizontal wire carries a current of 6.50 μA. An electron is moving above the wire a direction parallel to it and in the, same direction as the conventional current in the wire, at a constant velocity of 5.41×105 m/s. Assume that the force of gravity, acts on the electron and that the electron is moving parallel to the Earth's magnetic field, so that the only magnetic force on it, comes from the field generated by the current in the wire. (a) What is the magnitude of the upward force that the magnetic, field must exert on the electron to keep it at a constant velocity? (b) How far is the electron above the wire?, (a), , N, , (b), , m, , 22.9 An electron is initially moving at a speed of 1.79e+3 m/s directly toward a long wire that has 2.00 A of current flowing through, it. The electron is 2.50 m away. What are the direction and amount of the force exerted by the magnetic field on the electron?, N, in the, , i. opposite, ii. same, , direction as the current, , 22.10 Four parallel wires are arranged as shown in, the diagram, at the vertices of a square. In, two of them, a 325 mA current flows toward, you, and in the other two the same current, flows away from you. What is the strength of, the magnetic field at the point P, equidistant, from all four wires?, T, , 22.11 Two parallel wires are 1.50 m apart. The same amount of current will flow through each. You want a magnetic field of, 4.00×10í5 T at the midpoint between them. (a) Must the currents flow in the same direction, or in opposite directions, through, these wires? (b) How much current must flow in each wire?, (a), , Same direction, , (b), , Opposite directions, , A, , 22.12 Three parallel wires are arranged in a partial, square configuration as shown in the diagram., In two of the wires, a 5.00×10í4 A current, flows toward you, and in the third wire a, 7.50×10í4 A current flows away from you., What is the strength of the magnetic field at, the point P, at the "missing corner" of the, square?, T, , 22.13 A long wire is placed in a uniform magnetic field whose strength is 0.500 mT. 22.0 A of current flow through the wire, which is, perpendicular to the magnetic field. At point P, the net magnetic field equals zero. How far is point P from the wire?, m, 22.14 Two parallel long straight wires are oriented in a north-south direction at the same height, 10.0 m apart. The "western" wire, carries 6.00 A of current to the north. The "eastern" wire carries 3.00 A of current, also to the north. There is an imaginary, north-south line somewhere between the wires í not halfway! í where their magnetic fields exactly cancel out. How far is this, imaginary line from the western wire?, m, 22.15 Two long, straight wires lie on the x and y axes. They conduct currents of 3.00 A and 7.00 A in the positive x and y directions, respectively. What is the direction and strength of the magnetic field generated by the two currents at the point, (í5.00 m, 7.00 m)?, Direction:, Strength:, , +x, , íx, , +y, , íy, , +z, , íz, , T, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems, , 537
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22.16 Two long parallel wires are separated by a, distance of 2.00 m. The two points P1 and P2, divide that distance into three equal parts., Each wire carries a current in the same, direction. The current I2 is 3.00 A, as shown in, the illustration, and the current I1 is "tuned" so, that the strength of the magnetic field at the, point P1 is zero. (a) What is the strength of the, resulting magnetic field at P2? (b) Suppose, the current I1 were re-tuned to produce zero, magnetic field at P2. What would be the, resulting magnetic field at P1?, , 538, , (a), , T, , (b), , T, , Copyright 2007 Kinetic Books Co. Chapter 28 Problems
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29.0 - Introduction, It was a chance occurrence that led the British scientist Michael Faraday (17911867) to discover electromagnetic induction., It was 1831. Scientists already knew that an electric current could be used to create, a magnetic field. Faraday and others were trying to achieve the opposite: They, wanted to use a magnetic field to create an electric current., Faraday was conducting an experiment in which he wrapped two lengths of, insulated wire around a soft iron ring. One of the lengths was part of a circuit that, included a battery. The second was part of a different circuit containing an ammeter, that could measure any current passing through it. The wires were insulated so that, no current could flow through the iron ring between the circuits., Faraday knew he could create a magnetic field in the iron ring by running electricity, through the first coil. His goal was to use this magnetic field to create a current in, the second coil. The illustration to the right shows a modern-day recreation of his, experiment. The upper coil is the part of the circuit that in Faraday’s experiment, contained the battery; we have replaced the battery with a slider control labeled, “Current” so that you can control the amount of current in this circuit. The coil of the second circuit is the lower one in the illustration; we have, included a light bulb to make it easier to see when a current flows in the second circuit., Faraday had always connected the battery before he connected the ammeter, and he detected no current in the second circuit. However, on, the morning of August 29, Faraday connected the ammeter first and then connected the battery. To his delight, he detected a momentary, current in the second circuit; connecting the battery after the ammeter meant that it was measuring what happened as the current changed in, the upper circuit. Today, scientists would say he created a basic transformer, and they understand that it was the change in the magnetic field, caused by the change in current in the first circuit that caused the current in the second., Faraday rapidly pushed his work ahead. In the next few months, he discovered that by moving a wire in a magnetic field he could also generate, a current in a circuit. Today, this principle is employed in the machinery that generates most of the electricity we use., Faraday’s simple lab equipment yielded powerful insights that engineers continue to utilize today. Electric generators, microphones, VCRs, and, induction stoves all rely on Faraday’s discovery that a current-producing emf can be induced by changing the strength of a magnetic field, or by, moving a wire in a magnetic field., In the simulation to the right, you can recreate Michael Faraday’s groundbreaking experiment. The simulation contains an experimental setup, similar to the one Faraday used in 1831. You use a slider to control the current in the upper circuit on the left. By setting the slider’s position,, you determine the amount and direction of the current. You will see the magnetic field lines created by the current of this circuit; the more, intense their color, the stronger the magnetic field., When you launch the simulation, you will see that an oscilloscope, rather than Faraday’s ammeter, is attached to the bottom circuit. It displays, the potential difference across the light bulb that is part of the bottom circuit., Experiment by changing the current in the top circuit and observing what happens in the bottom circuit. By moving the slider back and forth,, you can continuously change the current. Is there a current in the bottom circuit if the current in the top circuit is steady and unchanging? What, if the current in the top circuit is changing? Does the rate at which the current in the top circuit changes have any effect on the potential, difference you measure in the second circuit?, , 29.1 - Motional electromagnetic induction, , Motional electromagnetic induction: Moving a, wire through a magnetic field induces an emf., Michael Faraday demonstrated two ways to generate a current in a circuit: by changing, the strength of a magnetic field passing through a wire coil or by moving a wire through, a magnetic field. This section examines the current generated by moving a wire through, a field. The phenomenon is called motional electromagnetic induction, or just motional, induction., We start our explanation of motional induction by making sure the diagram on the right, is clear to you. The vertical segment of conducting wire is pushed from left to right,, sliding across the horizontal wires connected to the light bulb. The wires connect to, form a complete circuit. The sliding vertical wire moves through an external magnetic, field, represented by ×’s, that points directly into the computer screen. The sliding wire, moves perpendicularly to this magnetic field., , Motional electromagnetic, induction, Wire moves through magnetic field, Motion induces an emf, , The result is called an induced emf (Ǜ). You studied another source of emf, a battery,, earlier. The process shown here creates an emf, “induced” by sliding the vertical segment of wire through the field. Since this segment is part, , Copyright 2007 Kinetic Books Co. Chapter 29, , 539
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of a circuit, the emf causes an induced current to flow. Remember that the units of emf are volts, the same as for potential difference. We will, often measure the amount of induced emf by measuring the potential difference it causes across a component like a light bulb., Why does current flow? It flows because the motion of the wire causes the electrons in it to be moving in a magnetic field, and the magnetic, field exerts a force on the moving electrons that is directed along the wire., In the illustration on the right, you see how the motion of the sliding wire causes the bulb to glow. You may wonder: Where does the energy to, illuminate the bulb come from? The answer is that it comes from whatever is doing the work of pushing the wire through the field., , 29.2 - Interactive problem: motional induction, In this simulation, you can drag a wire left and right, perpendicular to a magnetic, field that is pointing directly into the screen. This is the same configuration that was, used to explain motional induction in the previous section. If your efforts induce an, emf, current will flow and the light bulb will light., Conduct some experiments: Drag the wire slowly, and then drag it very fast. Does, the speed of the wire through the magnetic field affect the amount of potential, difference across the light bulb? How do the two relate? How does changing the, direction you move the wire change the current? How does it change the potential, difference?, You can also use the magnetic field strength control to make the magnetic field in, the simulation stronger (or weaker, or oppositely directed). Does changing the field, strength change the results of your efforts?, To help you to answer these questions, the simulation has an oscilloscope that, measures the potential difference across the light bulb. The oscilloscope in this, simulation is functionally similar to real-world ones. You can change the output, scale by clicking on its dial. Initially, it is set so that one box of the display grid equals 0.5 volts, but you can change that so one box equals 0.1, volts, 10 volts, and other values shown on the dial as well. (We chose not to show the oscilloscope’s connection to the circuit in this simulation, for the sake of visual simplicity.) An output gauge shows the amount of current; we show its value as positive or negative to indicate direction., There is also a slider control that lets you change the viewing angle of the simulation. This may allow you to better see the orientation of the, magnetic field, and the wire moving through it., , 29.3 - Induction: a coil and a magnet, At the right, you see an apparatus often used to, demonstrate induction. A bar magnet passes through, a coil of wire loops and the bulb lights up., A current flows when the magnet passes by the wire,, or when the wire moves past the magnet. It does not, matter which one is described as moving: The change, in the magnetic field inside the coil as one moves past, the other induces an emf that causes a current., The emf induced in this demonstration will vary,, based on several factors:, 1., , Motional induction demonstration: The magnet is dropped through the coil, The strength of the magnetic field. The stronger, of wire, and the resulting induced emf is displayed on the oscilloscope., the field, the greater the change in field, strength as the loops move by, and the greater the induced emf., , 2., , The speed of the wire relative to the magnetic field. The faster one passes by the, other, the greater the emf., , 3., , The area of the loops. The greater the area enclosed by each loop, the greater, the emf., , 4., , The number of loops of wire. Increasing the number of loops increases the total, area through which the field passes. This, too, increases the induced emf., , The list of factors above includes field strength and surface area, two of the factors that, determine a quantity called magnetic field flux. We will discuss magnetic field flux, shortly; it is analogous to electric field flux. Later, we will discuss how it is the rate of, change of this flux that determines the amount of induced emf., Although this is a classic way of illustrating motional induction, it is difficult to calculate, the actual emf induced in this configuration. The strength and orientation of the, magnetic field that intersects the coil both change as the bar magnet approaches the, wire loops. It is easier to calculate the induced emf for other configurations, like a, straight segment of wire moving in a uniform magnetic field., , 540, , Induced emf depends on:, Strength of magnetic field, Speed of magnetic field past wire, ·Or wire past field, Area of loop, Number of loops, , Copyright 2007 Kinetic Books Co. Chapter 29
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29.4 - Physics and music: electric guitars, Electric guitars use electromagnetic induction in a, component called a pickup to produce music. In the, pickup, the vibrations of the strings are converted into, an electrical signal so they can be amplified and then, played over speakers., The pickup consists of a permanent magnet, surrounded by a coil of wire. You see a conceptual, diagram of a pickup in the illustration to the right., Guitars typically have two or three pickups under, every string, each pickup designed to be maximally, sensitive to a particular frequency., The pickups (raised black modules) on this electric guitar convert, , The permanent magnet in the pickup serves to, the vibrations of the strings into a varying electric current., magnetize the nearby guitar string. A musician plucks, the string, making it vibrate. With this vibration the magnetized string moves back and, forth near the coil, inducing an emf in the wire. This emf causes a current in the wire., The emf and the current oscillate at the same frequency as the string., The signal is transmitted through a circuit to an amplification system, which increases, its strength, or energy. The system then sends the signal to a loudspeaker and the, audience hears the music., , Physics and music: electric, guitars, String magnetized by permanent, magnet in pickup, Guitar player makes string vibrate, Motion of magnetized string induces, current in coil, Current flows to amplifier/speaker, , 29.5 - Motional induction: calculating the potential difference, The diagram to the right shows a vertical wire segment moving through a uniform, magnetic field. This field has a constant strength and points into the screen. The wire, moves at a constant velocity to the right, which means its motion is perpendicular to the, magnetic field. In this section, we show how to calculate the potential difference induced, between the ends of the wire segment by its lateral motion., The potential difference is caused by the force of the magnetic field on mobile electrons, in the moving wire. The electrons move together with the wire as it is pushed through, the field. Because charged particles experience a force when they move through a, magnetic field, the mobile electrons get pushed to one end of the wire, leaving a, positive charge on the other end. (Which end the electrons move to can be determined, by a right-hand rule.) As more and more electrons accumulate on one end, leaving, more and more positive charge on the other, there is an increasing potential difference, across the wire., , Calculating potential difference, Potential difference induced across wire, , The wire’s motion causes the formation of two charged regions, one positive and the, moving in a magnetic field, other negative. Now, let’s consider the electron that is shown in the center of the wire in, Electric, magnetic forces act on charge, Concept 1. The motion in the magnetic field causes it to experience a downward, ·Forces balance, magnetic force FB, and the charged regions cause it to experience an upward electric, force FE. When these two forces balance, the electron experiences no net force and, stays in place. The system is in equilibrium and the equation on the right can be used to determine the potential difference across the wire., , In a simulation a few sections ago you saw how, when a moving wire segment like this is part of a circuit, it becomes an emf source that can be, used to illuminate a light bulb. The amount of the potential difference in this scenario equals the amount of the motional emf in the former one., To derive an expression for the equilibrium potential difference, we start with an equation stating that the magnetic force on the charged, particle equals the electric force. The other variables we use are shown in Equation 1 or defined in the strategy steps below., Strategy, 1., , State that the forces are in equilibrium: FB, , 2., , Substitute expressions for FB and FE, the magnetic and electric forces respectively, in terms of the quantities they depend on., , = FE., , Copyright 2007 Kinetic Books Co. Chapter 29, , 541
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3., , Use the relationship between field strength, potential difference and distance in a, uniform electric field to rewrite E, and then solve for the potential difference, between the ends of the wire segment., , Physics principles and equations, The strength of the force exerted by a magnetic field on a charge moving perpendicular, to it is, , We will also use the definition of an electric field., , The equation that relates potential difference to electric field strength and distance in a, uniform electric field is, , Potential difference across wire, segment, For motion perpendicular to field:, ǻV = LvB, , In this case, the distance d is the wire length L., Step-by-step derivation, , ǻV = potential difference across wire, , We employ the strategy and equations mentioned above and use algebra to solve for, the potential difference., , L = length of wire, , Step, , Reason, , 1., , FB = FE, , forces in equilibrium, , 2., , qvB = Eq, , substitute expressions for forces, , 3., , vB = E, , divide by q, , 4., , solve given equation for E, , 5., , substitute equation 4 into equation 3, , 6., , ǻV = LvB, , v = speed, B = magnetic field strength, , solve for ǻV, , As the equation indicates, longer wires, higher velocities, and stronger magnetic fields lead to greater induced potential differences. This, derivation considered the case of motion perpendicular to the magnetic field. If the motion were not perpendicular, we would use trigonometry, to determine the component of the motion that was perpendicular to the field. In general, we would conclude that ǻV = LvB sin ș, just as the, magnetic force equals qvB sin ș., , 29.6 - Magnetic flux, Magnetic flux is analogous to electric flux. Specifically, magnetic flux is the product of a, surface area with the component of a magnetic field passing perpendicularly through, the surface, just as electric flux is the measure of how much electric field passes, perpendicularly through a surface. The unit for magnetic flux is the weber (Wb). One, weber is one tesla·m2, the units for magnetic field strength times those for area., With both electric and magnetic flux, the cosine of the angle between the field and the, area vector is used to measure the component of the field passing perpendicularly, through the surface. The area vector is normal to the surface and equal in magnitude to, its area., In Equation 1, you see the equation for magnetic flux. It states that magnetic flux equals, the dot product of the magnetic field and area vectors, which is calculated as the, product of the magnetic field strength, the surface area, and the cosine of the angle ș., This angle is shown in the diagram., You will use the concept of magnetic flux, and changes in magnetic flux, to further your, understanding of how changing magnetic fields induce emfs., , 542, , Magnetic flux, Amount of field passing through surface, Depends on:, ·Field strength at surface, ·Amount of surface area, ·Angle between field, area vector, , Copyright 2007 Kinetic Books Co. Chapter 29
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ĭB = B · A, ĭB = BA cos ș, ĭB = magnetic flux, B = magnetic field, A = area vector, ș = angle between field, area vector, Units: webers (Wb = T·m2), , What is the magnetic flux, through this surface?, ĭB = BAcos ș, ĭB = (2.0 T)(3.0 m2)(cos 60°), ĭB = 3.0 Wb, 29.7 - Faraday’s law, Faraday discovered two ways to induce an emf. The first was by moving a wire through, a magnetic field; the second was by changing the strength of the magnetic field passing, through a stationary wire coil. In the latter case, the magnetic flux changed because flux, is proportional to the strength of the magnetic field., In general, since flux is the product of the magnetic field strength, the surface area, and, the cosine of the angle between the field vector and the area vector, changing any of, these three factors changes the flux and induces an emf., To consider changes in magnetic field strength, we use the apparatus shown to the, right. It is similar to Faraday’s equipment. Two wire coils are wrapped around a piece of, iron. Both coils are insulated so that no current can flow directly between them. The coil, on the left is connected to a variable current source, a device that can cause a current, that changes over time to flow through the coil. As the current in this coil changes, so, will the magnetic field that it creates., The iron core facilitates the transmission of the magnetic field from within the left-hand, coil to within the coil on the right. We use two illustrations of the same configuration to, show what occurs. In Concept 1, you see the overall configuration: the variable current, source attached to a coil on the left, the changing magnetic field passing through the, iron core, and the current that is induced in the coil on the right. In Equation 1, we show, the view looking down the coil on the right. The increasing magnetic field points away, from you down the coil in this view, resulting in a counterclockwise induced current., , Changing magnetic flux and, induced emf, Change in magnetic field strength yields, change in magnetic flux, Flux change induces emf, emf drives induced current, , Let’s discuss in more detail what is happening in this system. The iron core ensures that the magnetic field passes essentially unchanged from, within one coil to the other. Since the field is perpendicular to the loops of the coil on the right, the magnetic flux passing through this coil, , Copyright 2007 Kinetic Books Co. Chapter 29, , 543
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equals the product of the magnetic field strength, the surface area of a loop, and the, number of loops in the coil., The current flowing through the coil on the left creates a magnetic field. As that current, changes over time, so does the magnetic field it generates, which means the magnetic, flux passing through the coil on the right changes. That change in magnetic flux induces, an emf in the coil on the right, which in turn causes the current in the circuit on the right., This process is used to illustrate a general principle, called Faraday’s law: A change in, magnetic flux induces an emf. In Equation 1, you see this expressed in mathematical, form. Faraday’s law states that the induced emf equals the negative of the rate of, change of magnetic flux., Faraday’s law is often applied to a coil having N loops, as we do in this section, and we, state this version in Equation 2. In this case, the flux refers to the flux passing through, each loop, and the total flux equals the flux passing through each loop times the, number of loops., The negative sign appearing in both equations indicates that the induced emf acts to, “oppose” the change in magnetic flux that causes it. What this means is explained in, more depth in a later section., , Faraday’s law, total flux, , Ǜ = induced emf, ĭB = total magnetic flux through, circuit, t = time, , Faraday’s law, coil of N loops, , ĭB = magnetic flux through one loop, N = number of loops, , The field passes through six, loops. What is the induced emf?, , A = ʌr 2 = ʌ(0.53 m)2 = 0.88 m2, , 544, , Copyright 2007 Kinetic Books Co. Chapter 29
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Ǜ = í(6 loops)(0.15 V) = í0.90 V, 29.8 - Interactive problem: Faraday’s law, In this simulation, a magnetic field is passing through a solenoid. The solenoid is, part of a circuit that contains a resistor. The potential difference across the resistor, equals the magnitude of any emf induced in the solenoid., You control the rate of change of the magnetic field passing through the solenoid., Your task is to use Faraday’s law to calculate the time rate of change of magnetic, flux through the solenoid that will induce an emf having a magnitude of 12.5 V and, cause current to flow in the circuit. In turn, this will create a potential difference of, 12.5 V across the resistor., The solenoid has 20 loops, and each loop has an area of 0.0100 m2. An, oscilloscope measures the potential difference across the resistor., The magnetic field starts at 0.500 T and will decline linearly to 0 T during a time, interval you specify. You have a controller that sets the time interval during which, the change in field strength will occur to values from 5.00 to 20.00 milliseconds. The, field will then continue to vary back and forth between 0 T and 0.500 T during, alternating time intervals of the same length. The rate of change will alternate, between positive and negative values, but it will have a constant magnitude based on the duration of the time interval you select., Specify the time during which you want the field to decline to zero. (You change it in increments of 0.10 milliseconds.) Press GO to see if the, changing field induces an emf of magnitude 12.5 volts. If not, press RESET, redo your calculations and try again., If you have trouble answering this problem, review the section on Faraday’s law. The oscilloscope displays a graph of the potential difference, across the resistor, with the potential difference plotted on the vertical axis and elapsed time on the horizontal axis. By clicking on different, values on the oscilloscope’s control knob, you can specify what you want the vertical measure, in volts, of one grid square to be., , 29.9 - Lenz’s law, , Lenz’s law: An induced current flows so that, the magnetic field it creates opposes the change, in magnetic flux that causes the current., Lenz’s law, named after the Russian scientist Heinrich Lenz (1804-1865), is used to, determine the orientation of the current induced by a change in magnetic flux., The law states that the magnetic field of the induced current opposes the change in, magnetic flux that causes the current. To apply the law, you first note the change in, magnetic flux and then determine the orientation of the magnetic field that will oppose, that change in flux. The current will flow in the direction that causes this magnetic field., , Lenz’s law, , To illustrate this law being applied, we use the wire loop shown to the right. The, Determines, external magnetic field is pointing into the page and it is increasing in strength. This will, cause a current. But in which direction does the current flow, clockwise or counterclockwise?, , direction of induced current, , To answer this question, consider the direction of the magnetic field created by the induced current. Lenz’s law says this field will oppose the, change that caused it. The external field is increasing, so the magnetic field of the induced current points in the direction that opposes this, change. This means it will point toward you, as the diagram in Concept 2 shows. (In this diagram we have dimmed the external field to make, the opposing field easier to see.), The right-hand rule for currents dictates that the current must be flowing counterclockwise through the wire loop. If you apply the rule, your, fingers inside the loop must point toward you, opposing the increase of the external magnetic flux that is pointing away from you. This means, the thumb points up on the right side of the loop, as illustrated in Concept 2, and down on the left side, as you can imagine, indicating the, direction of conventional current. “Up on the right” and “down on the left,” means the current flows counterclockwise., (Perhaps it is useful to state a right-hand rule for induction loops: “The thumb points in the direction of the induced magnetic field, and the, fingers wrap around the loop to indicate the direction of the current.”), If the external field were decreasing, then the induced magnetic field would oppose the decrease. It would point in the same direction as the, external field, and this means the current would flow clockwise., The example problem asks you to determine the direction of the current in a loop when a decreasing external magnetic field passes through it,, , Copyright 2007 Kinetic Books Co. Chapter 29, , 545
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directed toward you. Remember: The current must flow so that it creates a magnetic, field inside the loop that opposes this decrease in flux., , Lenz's law, Magnetic field of induced current, opposes change in flux, , In what direction will the current, flow?, Counterclockwise, 29.10 - Eddy currents, , Eddy currents: Electric currents created when a, solid conductor moves through a magnetic field., We have used a relatively simple configuration to analyze the current created by, moving a wire in a magnetic field. The moving wire was a straight segment, and an, overall rectangular circuit contained three other straight segments., Here, we use eddy currents to illustrate a significant point: Even when a circuit is not, well defined, currents can still be caused to flow in a conductor at rest in a time-varying, magnetic field or passing through a nonuniform magnetic field. For instance, the, rectangular block to the right is made of a conducting material. When it moves into or, out of the magnetic field shown, currents flow in it, but not in straight lines. These types, of current flow are called eddy currents, and they flow in a complex pattern that, resembles a whirlpool. The diagram in Concept 1 visually approximates the eddy, currents by depicting them as circular., , Eddy currents, Caused by changing flux in solid, conductors, Eddy currents damp motion, , The existence of eddy currents can be vividly demonstrated with two differing pieces of, conducting material. One piece is a solid block attached to a string, swinging in and out, of a magnetic field. This motion generates eddy currents in the block. The magnetic, fields of the eddy currents oppose the change in flux that causes them. In essence, the, block becomes an electromagnet whose motion is opposed by the external magnetic, field. The opposing force quickly damps (reduces) the motion., A second block, with slots cut through it, swings through the same field. Because the, slots prevent the formation of significant large-scale eddy currents, a much weaker, opposing magnetic field is created. There is less force to damp the motion, and the, block swings relatively freely., Like any phenomenon that removes mechanical energy from a system, causing it to, dissipate as heat, eddy currents can be either an undesirable source of inefficiency, or, in the right circumstances, a valuable tool. One useful application of eddy currents is to, the braking of trains., If the upper half of a spinning train wheel is subjected to a strong magnetic field, each, portion of the wheel will rotate into and out of this field. The changing magnetic flux in, that portion of the wheel will induce eddy currents that apply a countertorque to the, , 546, , Slotted block, Slots minimize eddy currents, Oscillation minimally dampened, , Copyright 2007 Kinetic Books Co. Chapter 29
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whole wheel, acting to slow it down. Since the magnitude of the countertorque is proportional to the angular velocity of the wheel, the braking, effect diminishes as the train slows, allowing it to come to a smooth stop. Eddy-current brakes are promoted as being quieter and less prone to, slippage than train brakes that rely on mechanical friction., , 29.11 - Interactive problem: a generator, In this simulation your task is to establish how the angle between a magnetic field, and the velocity vector of a horizontal segment of a rotating loop relates to the emf, induced in the loop., When you open the simulation by clicking on the diagram to the right, you will see a, loop of wire suspended in a magnetic field. Start the loop spinning by pressing GO., An oscilloscope displays the emf induced in the loop. For simplicity’s sake, we omit, showing the oscilloscope’s connection to the generator circuit. We also show the, current induced in the visible portion of the circuit., A black vector represents the tangential velocity of a horizontal loop segment. You, control how fast the loop rotates by setting its angular velocity. You can also change, the angle at which you view the simulation. The side view of the magnetic field is, particularly helpful for this exercise., Watch the loop spinning in the field. Observe the orientation of the velocity vector to, the magnetic field lines. (The angle ș displayed in the control panel is the angle, between the field and the velocity vector as viewed from behind the crank.) With, what orientation of the velocity vector is the emf a maximum? A minimum? Can you explain why?, Also, observe the oscilloscope trace of the emf as the generator spins faster or slower. This always appears to be a sinusoidal wave. Does its, frequency change with angular velocity? Does its amplitude? To answer the second question, consider the relationship between speed and the, induced emf when a wire moves in a straight line through a magnetic field., , 29.12 - An AC generator, its emf, and AC current, In this section, we consider the implications of adding an AC (alternating current), generator to a circuit. We start with the circuit you see on the right, one that contains, only a generator and a resistor. The AC generator is represented by a blue circle with a, wave inside., At its simplest, an AC generator consists of a wire loop that rotates in a magnetic field. It, generates an emf that varies sinusoidally with time., The emf created by the generator can be described mathematically as the product of its, maximum emf and a sine function, as shown in Equation 1. The factor Ȧ in the, argument of the function is the angular frequency of the current. In the United States, and East Asia, alternating current has a frequency of 60 cycles per second (Hz),, meaning the emf reaches its maximum positive value 60 times per second. The, angular frequency is 2ʌ times this value, about 377 rad/s., In the United States, the maximum emf of alternating current is approximately 170 volts., Utilities strive to provide a potential difference that, calculated in a specific way,, averages to 120 volts., , AC generator, Ǜ = Ǜ max sin Ȧt, IR = Imax sin Ȧt, Ǜ = emf, IR = current through resistor circuit, Ȧ = angular frequency, t = time, , Copyright 2007 Kinetic Books Co. Chapter 29, , 547
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29.13 - Interactive problem: alternating current, In this simulation, there is a circuit that contains a signal generator and a resistor., The emf of the signal generator causes a current to flow. Your task is to use an, oscilloscope to determine the maximum current Imax, the maximum potential, difference across the resistor ǻVmax and the signal frequency f., You will see sinusoidal graphs in the oscilloscope similar to those you see in, illustrations in other sections of this chapter., One probe from the oscilloscope is used to measure current. The current is the, same everywhere in the circuit. The two ends of the potential difference (“voltage”), probe in this exercise are placed so it measures ǻV across the resistor., The oscilloscope displays information in the form of two graphs. In the upper graph,, the current is displayed on the vertical axis and time is on the horizontal axis. In the, lower graph, ǻV is displayed on the vertical axis, and time once again is on the, horizontal axis., Sliders allow you to change the scales of the graphs. The base unit is a single gray, square in the oscilloscope window. For example, if you set the current slider to, “8.00×10í2 A,” then the height of a gray box represents 8.00×10í2 amperes. You can set the scale for the time in an analogous fashion. All this, resembles the working of typical laboratory oscilloscopes., You set the scale by trial and error. If you cannot see any signal, move the slider so the units are smaller until you can see the graph. If the, waves you see are so large that you cannot see their peaks, increase the scale until their peaks “shrink” into view., When you launch the simulation, the signal generator is on and all the probes are attached. The peak of the wave in the current display is the, maximum current, Imax. The peak of the wave in the voltage display is ǻVmax., You also have to analyze the wave to determine the frequency of the signal. To do so, identify two adjacent matching points in the wave (such, as two peaks) in order to identify a cycle. The time it takes to complete this cycle is the period of the wave, and the reciprocal of the period is, frequency. For instance, if the time between two peaks is 1.0×10í3 seconds, then the frequency is 1/(1.0×10 í3), or 1,000 hertz (Hz)., A couple of tips may help here. Since the horizontal axis measures time, you can determine the period by counting how many gray boxes there, are between the corresponding points on the wave (be careful to note what the time scale is). It can be easier to use locations where the wave, crosses the horizontal axis, since more precise values can be noted there. But if you use those points, remember that matching points are, where the graph is in both cases going down, or in both cases going up. Two adjacent points where the wave crosses the horizontal axis, but in, opposite directions, represent only half a cycle., When you have analyzed a signal and determined Imax, ǻVmax and the signal frequency, enter these values in the boxes in the simulation., Press CHECK to see if you are correct., To measure the second signal, press the Signal Two button on the signal generator. The signal generator will create a signal with a different, potential difference and frequency. You can also analyze this signal, enter your conclusions and check them by pressing CHECK., , 29.14 - Mutual induction, , Mutual induction: The induction of an emf in, one circuit by a changing current in another, circuit., Currents create magnetic fields. A changing current in a circuit creates a changing, magnetic field. This changing magnetic field will induce an emf and a current in a, nearby second circuit, in a process called mutual induction., At the right, you see an illustration of the steps involved in mutual induction:, 1., , A switch is closed in the left-hand circuit. This circuit contains a battery., , 2., , A current in the left-hand circuit starts to flow after the switch is closed, increasing, toward its steady state value., , Mutual induction, Changing current in one circuit creates, current in other, ·Changing current causes changing, magnetic field, ·Changing magnetic field induces, current, , 3., , The increasing current on the left generates a changing magnetic field., , 4., , The changing magnetic field causes a changing magnetic flux through the circuit, on the right, which induces an emf in this circuit., , 5., , The induced emf in the right-hand circuit causes a current there, lighting the light, bulb for a brief time., , 6., , Once the current on the left reaches its steady state, the light bulb on the right goes out, because there is no longer a changing, magnetic field to induce a current there., , An important point to remember is that currents are induced by a changing magnetic field. The closing of the battery circuit starts a current that, in turn creates a magnetic field where before there was none. It is this change in magnetic field that induces the emf and the current in the, second circuit. When the current caused by the battery reaches a steady state, which occurs quite quickly, then its magnetic field will also be, , 548, , Copyright 2007 Kinetic Books Co. Chapter 29
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constant and it will cease to induce an emf or a current in the circuit on the right., If the current on the left-hand side were an alternating instead of a direct current, it, would induce an undiminishing alternating current on the right-hand side. Why? The, magnetic field due to the changing left-side current would constantly change in strength, and orientation. The constantly changing magnetic field would continuously induce an, emf and current in the circuit on the right, and the light bulb would stay lit., , Mutual induction, When current is constant, no induction, occurs, , 29.15 - Transformers, , Transformer: A device used, to increase or decrease an, alternating potential, difference., The way in which electricity is transmitted and, distributed requires conversions of potential, differences across an enormous range of values., Long distance power transmission lines typically, Power transmission lines conduct AC current at high voltages., operate at about 350,000 V, while city power lines, Transformers convert this current from one voltage to another., near a home function at 15,000 V. Before the, electricity enters a house, the potential difference between a current-carrying and a, “neutral” wire is further reduced to 120 V. All of these currents are alternating, and, transformers are used to convert potential differences in alternating currents. (One, reason why alternating current is used in power systems is the ease with which, transformers can change its potential difference.), The illustration in Concept 1 shows a transformer: two coils of wire wrapped around an, iron core. The wires are insulated so that no current flows directly between them. The, iron core allows a magnetic field to be efficiently transmitted from one coil to the other:, Less than 5% of the energy transformed in a typical transformer is dissipated as heat. In, this section, we focus on the characteristics of an ideal transformer, one in which no, energy is lost., The wire wrapped around the iron core on the left-hand side is called the primary, winding. In our scenario, it is directly attached to an AC generator. The current passing, through this coil generates a magnetic field. Since the current continually changes, so, does the magnetic field it creates., The field passes through the second coil on the right. This coil is called the secondary, winding. The change in magnetic flux through this coil induces an emf that causes a, current in the secondary circuit. The secondary circuit powers a resistive component,, such as a light bulb, that is called the load. The transformer changes the emf from that, created by the AC generator to that required by the load., , Transformers, Increase or decrease potential, difference, AC creates changing magnetic field, Induces potential difference on right, , There are a different number of loops, or turns, in the primary and secondary windings., This is crucial to the operation of the transformer. The ratio of the number of primary, loops to the number of secondary loops is called the turns ratio., The transformer “transforms” the potential difference created by the generator: The, potential difference across the secondary winding differs from that across the primary, winding. In the transformer to the right, the potential difference decreases. This is a, step-down transformer. Step-up transformers increase potential difference. The, transformer symbol that is used in circuit diagrams is shown in Concept 2., The change in the potential differences depends on the turns ratio. The proportion in, Equation 1 states that the ratio of the potential differences equals the turns ratio. To put, it another way: If the number of loops on a given side is reduced, the potential, difference on that side decreases., , Transformer, Symbol used in circuit diagrams, , We derive this “turns equation” below using Faraday’s law. The loops in each coil are assumed to span an equal surface area, and the same, amount of magnetic field passes through each one. This means the ratio of the changes in flux in the primary and secondary windings is, proportional solely to the number of loops in each coil., , Copyright 2007 Kinetic Books Co. Chapter 29, , 549
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Variables, potential difference across primary coil, , ǻV1, , potential difference across secondary coil, , ǻV2, , number of loops in primary coil, , N1, , number of loops in secondary coil, , N2, , time interval, , ǻt, , change in magnetic flux through one loop during ǻt, , ǻĭB, , Strategy, 1., , State Faraday’s law twice, for the emf induced across each coil by the changing, magnetic flux. The emf equals the potential difference across the components., , 2., , Divide the two equations., , Potential differences ratio equals, turns ratio, , Physics principles and equations, Faraday’s law stated in terms of potential difference, and for a coil of N loops, is, , ǻV1 = primary potential difference, ǻV2 = secondary potential difference, , With transformers, it is traditional to refer to potential differences rather than emfs in the, primary and secondary windings., , N1 = primary number of loops, N 2 = secondary number of loops, , Step-by-step derivation, We state Faraday’s law for each coil, then divide the two equations and simplify the, result., , Step, , Reason, , 1., , Faraday’s law, , 2., , Faraday’s law, , 3., , divide, , 4., , simplify, , Currents ratio is inverse of turns, ratio, , I1 = primary current, I2 = secondary current, , Transformers can increase a potential difference, but by now you have learned that in, physics, you do not get something without also giving something up. The principle of, conservation of energy must apply to transformers. How does this precept apply? It, requires that the power, or energy produced or consumed per unit time, is the same on, both sides of an ideal transformer. Power equals the current times the potential, difference. If you boost the potential difference, you decrease the current., The ratio of the currents is the inverse of the turns ratio. This is shown in Equation 2 ., The example problem asks you to analyze a type of transformer commonly sold in, travel stores. This transformer steps down European voltage (240 V AC) to USA, standard voltage (120 V AC), and you may need one if you intend to use any of your, personal appliances on a trip abroad., , 550, , What should the turns ratio of, the transformer be?, , Copyright 2007 Kinetic Books Co. Chapter 29
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29.16 - Sample problem: primary winding in a transformer, A video game console operates at a, potential difference of 9.00 V. The, potential difference supplied by the, power receptacle is 120 V AC. A, transformer is used to convert one, voltage to the other. If the secondary, coil of the transformer (the 9.00 V, side) has 1000 loops, how many, loops does the primary coil have?, , Low-voltage devices like the game console above typically run on direct current, so the power converter would not only transform the potential, difference, but also convert AC to DC. Here, we only consider the AC transformer part of the power converter., Variables, primary potential difference, , V1 = 120 V, , secondary potential difference, , V2 = 9.00 V, , primary number of loops, , N1, , secondary number of loops, , N2 = 1000, , What is the strategy?, 1., , State the turns ratio equation., , 2., , Solve for the number of primary loops and evaluate., , Physics principles and equations, The turns equation for transformers is, , Step-by-step solution, We solve the turns equation for N1, then evaluate., , Step, , Reason, , 1., , turns equation, , 2., , solve for number of primary loops, , 3., , evaluate, , 29.17 - Interactive problem: configuring a transformer, Here, we simulate a transformer., An alternating potential difference across the primary, winding on the top creates a magnetic field that, passes through the iron core to the secondary, winding on the bottom. This magnetic field induces an, emf in the bottom coil. The value of the potential, difference across the resistor in the bottom,, secondary circuit (equal to the induced emf) is, displayed on an oscilloscope. You can turn the dial on, the oscilloscope to change the scale of the display., The height of a grid box equals the setting on the dial., The large cylindrical transformers on the left of this photo convert electricity from, , Your task is to determine the maximum positive, 350,000 V to 15,000 V at an electrical substation., potential difference in the top, primary circuit using, information you glean from the simulation. We use the word “maximum” because, , Copyright 2007 Kinetic Books Co. Chapter 29, , 551
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the potential difference of the alternating current varies sinusoidally with time., There are ten loops in the top coil. You can vary the number of loops in the bottom, coil from 20 to 50, in increments of 5. Pick a number of loops and then use the, oscilloscope to determine the maximum potential difference across the resistor in, the secondary circuit. You can use that value and the ratio of loops to determine the, maximum potential difference in the primary circuit., Enter this value in the box provided in the control panel and press CHECK to see if, you are right. If not, redo your calculations, enter a new answer, and press CHECK, again., If you need help, review the section on transformers., , 29.18 - Gotchas, Any magnetic field induces an emf in a loop of wire. No, a changing magnetic field induces an emf in such a loop. A uniform, constant, magnetic field alone induces no net emf in a loop of wire. However, a potential difference can be induced across a segment of wire by moving it, through a magnetic field., Transformers change potential differences. Power is a function of potential difference, so they change power, as well. No, the current changes, in inverse proportion to the potential difference, so the power stays constant., , 29.19 - Summary, A changing magnetic field can generate an emf in a loop of wire. This phenomenon, is called electromagnetic induction., , Wire segment moving in B field, The motion of a coil in a nonuniform magnetic field can also induce an emf. The, same effect is achieved whether you move the coil through the field, or move the, field past the coil., Magnetic flux, ĭB is the amount of magnetic field passing perpendicularly through a, surface. The amount of flux can be computed as the dot product of the field and, area vectors, which is the product of the field strength, the area, and the cosine of, the angle between them. The unit of magnetic flux is the weber (Wb). 1 Wb = 1, T·m2., , ǻV = LvB, Magnetic flux, , ĭB = BA cos ș, Faraday’s law, , Faraday’s law states that the induced emf in a circuit is proportional to the rate of, change of magnetic flux through the circuit., When a change in the magnetic flux due to an external magnetic field induces a, current in a circuit, the induced current creates its own magnetic field. Lenz’s law, states that the induced magnetic field always opposes the change in flux that, caused it. This helps you to deduce the direction of the induced current., , Transformer, , A loop of wire rotating in a magnetic field is the basis of an electric generator. A, torque causes the loop to rotate in the field, constantly moving its wires through the, field and inducing an emf and an electric current in it. The induced current reverses, direction every half turn of the loop, and is called an alternating current (AC). Both the induced emf and the current vary sinusoidally with time., Currents induced in large solid pieces of conducting material are not like those induced in a wire. When a solid conductor moves through a, nonuniform magnetic field, eddy currents flow through it in complex “whirlpools.” Eddy currents oppose the motion of the conductor through the, nonuniform field., Mutual induction occurs when a changing current in a circuit generates a changing magnetic field, which in turn induces a current in a second, nearby circuit., A transformer is a device that increases or decreases potential difference from one alternating-current circuit to another. Both circuits contain, coils wrapped around the same iron core, but with a different number of loops in each coil., , 552, , Copyright 2007 Kinetic Books Co. Chapter 29
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Chapter 29 Problems, , Chapter Assumptions, Unless stated otherwise, use 5.00×10–5, Elementary charge, e, , T for the strength of the Earth's magnetic field at its surface., , = 1.60×10–19 C, , Conceptual Problems, C.1, , The textbook displays a photograph of an experiment where a bar magnet is dropped through a coil of wire. The changing, magnetic flux through the coil induces an emf that in turn drives a current through the wire. The emf is displayed on an, oscilloscope. After the magnet goes through the coil does it hit the pillow with the same kinetic energy as it would if there were, no coil? Explain., Yes, , No, , C.2, , A large spool holding a coil of hundreds of loops of telephone cable is loaded on a truck at the factory in New York, and, transported to a utility in Georgia. During the trip, a meter which is attached to the coil registers a tiny intermittent current in, the coil. If there is no battery attached to the coil, what is the cause of the current?, , C.3, , A circular loop of wire is located in a constant, uniform magnetic field. How can an emf be induced in the loop?, , C.4, , A wire loop lies in the xy plane. A magnetic field points in the positive z direction. If the radius of the loop begins to shrink, in, what direction will the current in the loop flow? Give your answer from the perspective of looking down on the xy plane, and, explain your answer., , C.5, , Suppose you are looking down on a wire loop, perpendicular to the plane it lies in. A magnetic field is pointed toward you and, is increasing in strength. Will the induced current flow clockwise or counterclockwise? Explain your answer., , Clockwise, , Clockwise, , Counterclockwise, , Counterclockwise, , C.6, , "Superman - The Escape" is a thrill ride at Six Flags Magic Mountain in Valencia, CA. A 3000 kg car is accelerated to a, height of over 100 meters. On the return trip the car is brought to a stop using electromagnets which induce eddy currents in, the large wheels of the car. Since energy must be conserved, where does the nearly three million joules of gravitational, potential energy go?, , C.7, , Carefully observe the animation of the swinging block in the textbook. (a) What is the direction of the eddy currents when the, block swings into the magnetic field? (b) What is the direction of the eddy currents when the block swings out of the magnetic, field? (c) Explain your answer to part a using Lenz's law., (a), , Clockwise, , Counterclockwise, , (b), , Clockwise, , Counterclockwise, , (c), C.8, , The poles of a powerful magnet are oriented to oppose one another, with a gap in between them, so that the field lines from, the north to the south pole are horizontal. You drop a copper penny so that it falls through the gap. (a) Describe what happens, to the acceleration of the penny as it passes between the poles. (b) The effect described in part a would be different for a coin, like a dime, which is close to the size and weight of a penny, but is made of a metal with a different resistivity. Describe a, practical application of this difference in the coins, in a type of machine you have probabaly used., (a), (b), , C.9, , We have seen that it is possible for a changing current in one loop to cause a current to flow in an adjacent loop. (a) If the, loops are lying flat on a surface, side by side, and the current in the first loop suddenly increases, does the induced current in, the second loop flow in the same direction (by "direction" we mean clockwise or counterclockwise) as the original current, or, in the opposite direction? Explain your answer. (b) Describe a configuration for the two loops which gives the opposite result., (a), , The same direction, , The opposite direction, , (b), , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Will a current flow in the, bottom circuit if the current in the top circuit is steady and unchanging? (b) Is there current in the bottom circuit if the current in, , Copyright 2000-2007 Kinetic Books Co. Chapter 29 Problems, , 553
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the top circuit is changing? (c) Does the rate at which the current in the top circuit changes have any effect on the potential, difference you measure in the second circuit?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , Section 2 - Interactive problem: motional induction, 2.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Does the induced current, change direction when the wire moves in a different direction? (b) What happens to the induced emf when you move the wire, faster? (c) For a given wire speed, what happens to the induced emf when the strength of the magnetic field increases? (d) A, controller in the simulation allows you to change the magnetic field strength, but only stepwise rather than continuously., Describe what happens to the induced emf when you impose a sudden jump in field strength., (a), , Yes, , No, , (b) The induced emf, , (c) The induced emf, , i., ii., iii., i., ii., iii., , increases, when the wire moves faster., decreases, stays the same, increases, when the field strength increases., decreases, stays the same, , Section 5 - Motional induction: calculating the potential difference, 5.1, , A wire is moving perpendicularly through a magnetic field of strength 0.100 T with speed 60.0 m/s. If the potential difference, between the ends of the wire is 3.00 V, what is the length of the wire?, m, , 5.2, , A sailboat with an aluminum mast 22.5 m high is sailing eastward along the equator at a speed of 3.66 meters per second., The Earth's magnetic field at the ship's location is 5.50×10í5 T, horizontal and directed due north. What is the induced, potential difference between the top and bottom of the mast?, V, , 5.3, , A Boeing 777-200 airliner with a wingspan of 64.0 m is flying perpendicular to the Earth's magnetic field at its cruising speed, of 250 m/s. The strength of the field is 5.05×10í5 T. What potential difference is induced between the plane's wingtips?, V, , 5.4, , During your biking tour of the Pacific Northwest, you decide to outdo those latte-guzzling ecotopians at their own game., Instead of the four toxic and disposable 1.5-volt batteries suggested for your walkabout CD player, you will place a 2.0 m, conducting wand across your handlebars. Pedaling furiously in a direction perpendicular to the local magnetic field, you wish, to go fast enough to induce a potential difference of 6.0 V (equivalent to four batteries in series). (a) The strength of the, Earth's magnetic field at your location is 5.0×10í5 T. How fast will you need to go? (b) Is this speed street-legal? (c) The, magnetic declination at your location is 19°. At what angle from geographic north should you travel to maximize the potential, difference of your environment-friendly arrangement?, (a), (b), (c), , m/s, Yes, i., ii., iii., iv., , 0, 19, 71, 90, , No, degrees, , Section 6 - Magnetic flux, 6.1, , A 0.150 T magnetic field B crosses the xy, plane perpendicular to the x axis and at an, angle of 30.0° to the y axis. A square surface, with sides of length 7.00 m lies mainly in the, xy plane, but a 1.50 m section of it at one end, is bent 90° in the negative z direction. What is, the total magnetic flux through the surface?, Choose directions for the area vectors so that, the flux through each surface is positive., Wb, , 554, , Copyright 2000-2007 Kinetic Books Co. Chapter 29 Problems
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6.2, , A wire circle of radius 0.050 m is embedded in a solenoid of length 0.20 m with 1000 turns that carries a current of 0.50 A. If a, vector that is normal to the plane of the circle makes a 40° angle with the axis of the solenoid, what is the magnetic flux, through the loop?, Wb, , Section 7 - Faraday’s law, 7.1, , A magnetic field increases steadily in strength, starting at 0 T. The induced emf in a loop that is resting in the field is 2.72 V., After 4.75 s, what is the magnitude of the magnetic flux through the loop?, T · m2, , 7.2, , A coil with multiple loops is in a magnetic field such that the magnetic flux through the coil increases at a rate of 0.0240, Tm2/s. The induced emf in the coil is í3.36 V. How many loops are in the coil?, loops, , 7.3, , A wire has a resistance of 60.0 ohms, and is bent into a square loop with sides of length 15.0 cm. The loop is perpendicular, to a magnetic field which is changing at a rate of 3.00eí3 T/s. What is the current induced in the wire?, A, , 7.4, , You hold a thin gold wedding ring with an inside radius of 1 cm, such that the plane of the ring is perpendicular to the Earth's, magnetic field. At your location it has strength 5.5×10í5 T. What is the emf induced in the ring if you turn it in one-tenth of a, second to be parallel to the field (that is, you make the area vector of the ring perpendicular to the field)?, V, , Section 8 - Interactive problem: Faraday’s law, 8.1, , Use the information given in the interactive problem in this section to calculate the time interval required to induce the desired, emf in the coil. Give your answer in milliseconds. Test your answer using the simulation., ms, , Section 11 - Interactive problem: a generator, 11.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) With what orientation of the, velocity vector is the emf a maximum? (b) With what orientation of the velocity vector is the emf zero? (c) Does the frequency, of the emf change with angular velocity? (d) Does the amplitude of the emf change with angular velocity?, (a) The emf is a maximum when the velocity vector is, (b) The emf is zero when the velocity vector is, (c), , Yes, , No, , (d), , Yes, , No, , i. Perpendicular to the magnetic field., ii. Parallel, i. Perpendicular to the magnetic field., ii. Parallel, , Section 12 - An AC generator, its emf, and AC current, 12.1 An AC generator generates a maximum emf of 35 V and runs at a frequency of 12 Hz. At a time t, the emf is at a maximum., How many seconds after t does it take for the emf to fall to zero?, s, 12.2 An AC generator is connected in series with a resistor in a simple loop circuit. At a particular time t, the current through the, resistor has its maximum value of 12.3 A, and then begins to fall. Just 0.00573 s later, the current through the resistance has, fallen to 10.7 A. What is the angular frequency of the generator?, rad/s, , Section 13 - Interactive problem: alternating current, 13.1 Use the oscilloscope in the simulation in the interactive problem in this section to determine (a) Imax, (b) ǻVmax and (c) the, frequency f of the first signal., (a), , A, , (b), , V, , (c), , Hz, , Copyright 2000-2007 Kinetic Books Co. Chapter 29 Problems, , 555
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13.2 Use the oscilloscope in the simulation in the interactive problem in this section to determine (a) Imax, (b) ǻVmax and (c) the, frequency f of the second signal., A, , (a), (b), , V, , (c), , Hz, , Section 15 - Transformers, 15.1 A transformer with an input current of 0.500 A has 6000 primary turns. How many secondary turns must the transformer have, if an output current of 2.00 A is desired?, turns, 15.2 A machine that makes transformers can only wind the coils with a single specific number density. The number of turns is, therefore controlled by the length of the coils. A transformer is designed to transform a 120 V input potential difference to a, 9.0 V output. If the input coil is 2.8 cm in length, what is length of the output coil?, m, 15.3 The long distance power transmission lines reaching a certain power substation operate at 375,000 V. This potential, difference has to be "stepped down" for local distribution through city power lines at 14,300 V. Substations use very large, oilcooled transformers (a photograph of one is in the text) to accomplish this conversion. What is the ratio of primary to, secondary windings in a transformer in such a substation?, , 15.4 The electricity supplied through a city power, line is transmitted at 14,300 V. On each city, block there are one or more neighborhood, transformers on utility poles that transform the, potential difference to 120 V, as used in, homes and businesses (240 V outlets are, supplied by two 120 V lines). What is the ratio, of primary to secondary windings in a, neighborhood transformer like the one shown, in the photograph?, , 15.5 (a) The emf produced by a typical hydroelectric generator, at the generator, is 20,000 V. The electricity produced by the, generator is fed through a step-up transformer before being transmitted across the country on "high tension" lines like those, shown in the textbook. If the primary winding of the transformer has 1550 loops, what must be the multiplicity of the secondary, winding to produce a transmission emf of 100,000 V? Of 225,000 V? Of 345,000 V? (b) A current of 4500 A flows from the, generator to the step-up transformer. What current flows through the transmission lines at each one of the three operating, emfs stated in part a? (c) If the total resistance of the power lines between the generator and a city it supplies with electricity, is 112 ȍ, use the results of part b to calculate the power dissipated in the lines at each of the three operating emfs. (d) In, commercial or public power systems, why is electric power transmitted at extremely high emfs and (relatively) low currents,, rather than the other way around?, (a), , for 100,000 V, , (b), , A for 100,000 V, , for 225,000 V, A for 225,000 V, , (c), , W for 100,000 V, , W for 225,000 V, , for 345,000 V, A for 345,000 V, W for 345,000 V, , (d), , Section 17 - Interactive problem: configuring a transformer, 17.1 Use the information given in the interactive problem in this section to determine the maximum potential difference in the, primary coil. Test your answer using the simulation., V, , 556, , Copyright 2000-2007 Kinetic Books Co. Chapter 29 Problems
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30.0 - Introduction, Radio and television signals, x-rays, microwaves:, Each is a form of electromagnetic radiation. If steam, and internal combustion engines symbolize the, Industrial Revolution, and microprocessors and, memory chips now power the Information Revolution,, it almost seems that we have neglected to recognize, the “Electromagnetic Revolution.” Think about it: Can, you imagine life without television sets or cell, phones? You may long for such a life, or wonder how, people ever survived without these devices!, These examples are from the world of engineered, Electromagnetic radiation: Rainbows and radios. Sundazzled reflections., electromagnetic radiation. Even if you think we might, Shadowlamps and lampshadows. Red, white, and blue., all prosper without such technologies to entertain us,, do our cooking, carry our messages, and diagnose our illnesses, you would be hard-pressed to survive without light. This form of, electromagnetic radiation brings the Sun’s energy to the Earth, warming the planet and supplying energy to plants, and in turn to creatures like, us that depend on them. There are primitive forms of life that do not depend on the Sun’s energy, but without light there would be no seeing, no, room with a view, no sunsets, and no Rembrandts., Some of the electromagnetic radiation that reaches your eyes was created mere nanoseconds earlier, like the light from a lamp. Other, electromagnetic radiation is still propagating at its original speed through the cosmos, ten billion years or more after its birth. An example of this, is the microwave background radiation, a pervasive remnant of the creation of the universe that is widely studied by astrophysicists., Back here on Earth, this chapter covers the fundamental physical theory of electromagnetic radiation. Much of it builds on other topics,, particularly the studies of waves, electric fields and magnetic fields., , 30.1 - The electromagnetic spectrum, , Electromagnetic spectrum: Electromagnetic, radiation ordered by frequency or wavelength., Electromagnetic radiation is a traveling wave that consists of electric and magnetic, fields. Before delving into the details of such waves, we will discuss the electromagnetic, spectrum, a system by which the types of electromagnetic radiation are classified., The illustration of the electromagnetic spectrum above orders electromagnetic waves by, frequency and by wavelength. In the diagram, frequency increases and wavelength, decreases as you move from the left to the right. The chart’s scale is based on powers, of 10. Wavelengths range from more than 100 meters for AM radio signals to as small, as 10í16 meters for gamma rays., , Electromagnetic spectrum, Radio waves in AM, FM bands, , All electromagnetic waves travel at the same speed in a vacuum. This speed is, designated by the letter c and is called the speed of light. (The letter c comes from, celeritas, the Latin word for speed. It might be more accurate to refer to it as the speed of electromagnetic radiation.) The speed of light in a, vacuum is exactly 299,792,458 m/s, and it is only slightly less in air., The unvarying nature of this speed has an important implication: The wavelength of electromagnetic radiation is inversely proportional to its, frequency. As you may recall, the speed of a wave equals the product of its frequency and wavelength. This means that if you know the, wavelength of the wave, you can determine its frequency (and vice versa). For instance, an electromagnetic wave with a wavelength of 300, meters, in the middle of the AM radio band, has a frequency of 1×106 Hz. This equals 3×108 m/s, the speed of light, divided by 300 m. The, frequencies of electromagnetic waves range from less than one megahertz, or 106 Hz, for long radio waves to over 1024 Hz for gamma rays., We will now review some of the bands of electromagnetic radiation and their manifestations. The lowest frequencies are often utilized for radio, , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 557
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signals. AM and FM radio waves are typically produced by transmitters that incorporate, electric oscillator circuits attached to antennas. The AM and FM radio bands are shown, in Concept 1. Microwave radiation is at the upper end of the radio band, and is used for, cellular telephone transmissions as well as for heating food in microwave ovens., Infrared radiation has a higher frequency than microwaves and is associated with heat., It is generated by the thermal vibration or rotation of atoms and molecules., Visible light is next as we go up the frequency range. It is of paramount importance to, human beings, although it occupies only a small portion of the electromagnetic, spectrum. Like some other forms of electromagnetic radiation, it is created when atoms, emit radiation as their electrons drop from higher to lower energy levels. Light consists, of electromagnetic waves that oscillate more than 100 trillion (1014 ) times a second., The wavelengths of the various colors of light are in the hundreds of nanometers. Red, light has the lowest frequency and longest wavelength, while violet light has the highest, frequency and shortest wavelength. The visible light spectrum is shown in Concept 2., The Sun emits a broad spectrum of electromagnetic radiation, including ultraviolet (UV), waves, with frequencies higher than those of visible light. These waves are the main, cause of sunburn: Sunscreen lotion is designed to prevent them from reaching and, harming your skin. This radiation can also harm your eyes, especially if you wear plastic, sunglasses that diminish the amount of visible light reaching your eyes, but do not block, UV rays. Typically, you squint when your eyes are exposed to strong light, and this, helps protect them. If you wear sunglasses that do not stop ultraviolet light, your pupils, will dilate, allowing an extra dose of harmful ultraviolet waves to enter your eyes., The Earth’s atmosphere, specifically the layer that contains a molecular form of oxygen, called ozone, absorbs a great deal of the Sun’s ultraviolet radiation, protecting plants, and animals from its harmful effects. However, substances once commonly used in, refrigerators and aerosols catalyze ozone-destroying chemical reactions in the, atmosphere. Fortunately, the use of such substances has been restricted, but a “hole”, in the ozone, a region where the amount of ozone has been significantly depleted, has, been created above Antarctica. This hole varies in size from year to year, but on, average is approximately the size of North America., , Visible light, Consists of spectrum of colors, ·White = mixture of all colors, ·Black = lack of visible light, , X-rays, Used in, , X-rays, the next band of electromagnetic frequencies, are even more dangerous than, ·Medical diagnosis and research, UV, but they are also useful. Doctors can use them to “see” shadowy images of the, ·X-ray diffraction studies, inside of the human body because they travel more easily through some tissues than, ·X-ray astronomy, others. Scientists also use them to discern the detailed crystalline structure of materials, and to deduce the spatial configuration of complex molecules. NASA launched the Chandra X-ray Observatory in 1999 to capture the radiation, emitted from high-energy regions of the universe, such as the parts of space around exploded stars. The acceleration of high-energy electrons, is one source of x-rays. Because these rays can damage or destroy living cells and tissues, human exposure to them must be strictly limited., Finally, gamma rays are the highest frequency electromagnetic waves. They are emitted by atomic nuclei undergoing certain nuclear reactions,, as well as by high energy astronomical objects and events. These rays may enter the Earth’s atmosphere from space, but they usually collide, with air molecules and do not reach the ground. Those few that do reach the Earth can cause mutations in the DNA of living cells. Gamma, radiation is highly destructive, which is one reason for the thick shielding used to protect workers from nuclear materials. However, like x-rays,, gamma rays also have some beneficial uses. “Gamma knife” radiosurgery uses concentrated beams of gamma rays to kill cancer cells., The distinction between the different parts of the spectrum appears clearer in the diagram above than it is in reality; the labeled bands actually, blend into each other. For instance, there is no clear-cut frequency at which the shortest “radio waves” stop and the longest “far infrared” waves, start; the bands in the diagram simply provide a convenient way to classify frequency ranges. Frequency classifications do become precise, when business people enter the discussion: The rights to use certain frequency ranges, such as those for radio and television stations and cell, phones, are worth tens of billions of dollars., Despite the different names í AM radio signals, visible light, x-rays, gamma rays í the phenomena organized in the chart above and illustrated, to the right are all forms of electromagnetic radiation. In the same way that you think of both tiny puddle ripples and long, slow ocean swells as, being water waves, so you should think about the types of electromagnetic radiation. The frequency of the wave does not alter the fundamental, laws of physics that govern it., , 558, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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30.2 - Electromagnetic waves, , Electromagnetic wave: A wave consisting of, electric and magnetic fields oscillating, transversely to the direction of propagation., Physicist James Clerk Maxwell’s brilliant studies pioneered research into the nature of, electromagnetic waves. He correctly concluded that oscillating electric and magnetic, fields can constitute a self-propagating wave that he called electromagnetic radiation., His law of induction (a changing electric field causes a magnetic field) combined with, Faraday’s law (a changing magnetic field causes an electric field) supplies the basis for, understanding this kind of wave., As the diagrams to the right show, the electric and magnetic fields in an electromagnetic, wave are perpendicular to each other and to the direction of propagation of the wave., These illustrations also show the amplitudes of the fields varying sinusoidally as, functions of position and time. Electromagnetic waves are an example of transverse, waves. The fields can propagate outward from a source in all directions at the speed of, light; for the sake of visual clarity, we have chosen to show them moving only along the, x axis., , Electromagnetic waves, Consist of electric and magnetic fields, ·Perpendicular to each other, Propagate as transverse waves, ·Perpendicular to direction of travel, , The animated diagram in Concept 2 and the illustrations below are used to emphasize, three points. First, the depicted wave moves away from the source. For example, if you, push the “transmit” button on a walkie-talkie, a wave is initiated that travels away from, the walkie-talkie., Second, at any fixed location in the path of the wave, both fields change over time. The, wave below is drawn at intervals that are fractions T/4 of the period T. Look at the point, P below, on the light blue vertical plane. The vectors from point P represent the, direction and strength of the electric and magnetic fields at this point. As you can see,, the vectors, and the fields they represent, change over time at P. Concept 2 shows, them varying continuously with time at the point P., Third, the diagrams reflect an important fact: The electric and magnetic fields have the, same frequency and phase. That is, they reach their peaks and troughs simultaneously., , Electric, magnetic fields, Vary in strength over time at each point, Have same frequency and are in phase, Drive each other by changing strength, , A wave on a string provides a good starting point for understanding electromagnetic waves. Both electromagnetic radiation and a wave on a, string are transverse waves. The strengths of the two fields constituting the radiation can be described using sinusoidal functions, just as we, can use a sinusoidal function to calculate the transverse displacement of a particle in a string through which a wave is moving., There is a crucial difference, though: Electromagnetic radiation consists of electric and magnetic fields, and does not require a medium like a, string for its propagation. Electromagnetic waves can travel in a vacuum. If this is troubling to you, you are in good company. It took some, brilliant physicists a great deal of hard work to convince the world that light and other electromagnetic waves do not require a medium of, transmission., Furthermore, when electromagnetic waves radiate in all directions from a compact source like an antenna or a lamp, the radiation emitted at a, particular instant travels outward on the surface of an expanding sphere, and its strength diminishes with distance from the source. The waves, cannot be truly sinusoidal, since the amplitude of a sinusoidal function never diminishes., In the sections that follow we will analyze plane waves, which propagate through space, say in the positive x direction, in parallel planar wave, fronts rather than expanding spherical ones. They are good approximations to physical waves over small regions that are distant from the, source of the waves. Plane waves never diminish in strength; they can be accurately modeled using sinusoidal functions, and we will do so., , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 559
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30.3 - Proportionality of electric and magnetic fields, Although the electric and magnetic field vectors of an electromagnetic wave point in, perpendicular directions, their magnitudes are strictly proportional to each other at all, positions and at all times. We graphically display the magnitudes at a particular instant, on the same coordinate system in Equation 1. Their proportionality is expressed in the, equation to the right, using a constant c that depends on two other fundamental, physical constants., This proportionality turned out to have important implications in the study of, electromagnetic radiation. Why? Because when calculated, the value of c was very, close to the measured speed of light. This crucial discovery accelerated the, understanding of the relationship between electromagnetic radiation such as light or, radio waves, and electric and magnetic fields., , Proportionality of electric and, magnetic field strengths, In an electromagnetic wave,, , E = electric field strength, B = magnetic field strength, , 30.4 - Calculating the speed of light from fundamental constants, One of the major discoveries of 19th century physics, was that light is a form of electromagnetic radiation., By applying and extending their knowledge of electric, and magnetic fields, physicists were able both to, create electromagnetic radiation (initially radio waves), and to predict the speed of the waves., James Maxwell published his four laws governing, electromagnetic phenomena in 1864, and at the same, time he predicted the existence of self-propagating, electromagnetic waves. His work enabled him to, calculate what the speed of these waves would have, to be., , Heinrich Hertz transmitter, 1888. Transformer voltage causes a spark to, jump between the postionable brass spheres, generating a radio pulse., , The angular frequency of any wave is Ȧ = 2ʌf,, where f is its frequency in cycles per second. The angular wave number is k = 2ʌ /Ȝ,, where Ȝ is the wavelength. The speed of any wave is v = Ȝf. This means that the speed, in terms of the angular frequency and wave number is v = (2ʌ /k)(Ȧ/2ʌ), or Ȧ/k., Maxwell had already shown that the wave speed Ȧ/k is a constant for electromagnetic, waves, a constant he had designated as c and expressed in terms of μ0 and İ0. We, state the relationship of c to these several variables and constants in Equation 1., The fundamental constants μ0 and İ0 are used elsewhere in physics. For example, the, permeability constant is used in equations that describe the magnetic field created by, various electric current configurations. The permittivity constant is used to express one, form of Coulomb’s law. In other words, these constants determine the strengths of the, electric and magnetic forces in the physical universe., The example problem on the right asks you to repeat Maxwell’s calculation of the value, of c. The result is 2.998×108 m/s. In the year 1864, the speed of visible light in a, vacuum had been known with fair accuracy for well over a century. The English, physicist James Bradley estimated it in 1728 to be 3.1×108 m/s, based on his study of, “stellar aberration,” or the apparent change in the positions of stars as the Earth moves, around the Sun. Because the calculated and observed speeds were so close, Maxwell’s, results provided the first evidence that light is a kind of electromagnetic wave., , Speed of light, Theoretical speed of radiation, = the measured speed of light, Conclusion: Light is electromagnetic, radiation!, , Understanding that light is an electromagnetic wave, and knowing the general relationship between frequency and wavelength, sparked the, discovery of additional types of electromagnetic radiation. In 1888 Heinrich Hertz created what we would now call radio receivers and, transmitters, one of which you see in the illustration above. He proved the existence, and wavelike nature, of radiation having frequencies, around 100 MHz., , 560, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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30.5 - Creating electromagnetic waves: antennas, Radio antennas create electromagnetic waves. A, radio antenna is part of an overall system called a, radio transmitter that converts the information, contained in sound waves into electromagnetic, waves. A radio receiver then reverses the process,, converting the signals from electromagnetic waves, back to sound waves., The system depicted to the right shows the, fundamentals of a radio transmitter. In the, illustrations, the terminals of an AC generator are, connected to two rods of conducting material: an, Radio-wave transmitters. A cable carries a modulated electric signal, antenna. The AC generator produces an emf Ǜ that, to the dipole antenna rod assembly visible at the top of each tower., varies sinusoidally over time. The emf drives positive, and negative charges to opposite ends of the antenna. The separation of the charges, on the rods produces an electric field. The AC generator causes the amount and sign of, the charge on each rod to vary over time, so that the resulting electric field varies in, strength and orientation as well. (The flow of charge í that is, the current í also, produces a varying magnetic field close to the antenna, part of what is called the near, field, which we do not show here.), The electric field produced by the antenna at each instant in time propagates outward in, all directions at the speed of light. For simplicity’s sake, we only show it traveling in the, positive x direction in Concept 2. The electric field changes continuously with time and, that change induces a magnetic field. To be precise, it induces a magnetic field, proportional to the rate of change of the electric flux with respect to time, as described, by Maxwell’s law of induction. In turn, the changing magnetic field regenerates the, electric field as the wave travels., This coupling of changes in the magnetic and electric fields enables the electromagnetic, wave to cross vast gulfs of space over immense spans of time. Electromagnetic, radiation from distant stars, including light, reaches the Earth after billions of years of, travel., , Simple antenna components, A source of varying emf, Two conducting rods, , How does an antenna differ from other circuits you may have studied in which current, flows or charge is stored? Consider a battery-resistor circuit or a battery-capacitor, circuit in equilibrium; the current in the first creates a constant magnetic field, and the, stored charge in the second creates a constant electric field. Both fields rapidly diminish, as they extend outward in space. The crucial difference with the antenna is that not only, does charge accumulate at its ends, but the AC generator continually causes the, distribution of charge to change. The electric field varies sinusoidally over time, and a, constantly changing electric field is the crucial element required to create continuous,, self-propagating electromagnetic radiation., Electromagnetic waves are generated when charges move at nonconstant velocities, as, in an antenna. That is, they are generated by accelerating charges. In an antenna, the, acceleration is in a straight line. Charged particles moving in uniform circular motion, also emit electromagnetic radiation, called synchrotron radiation, due to their centripetal, acceleration., The AM and FM radio bands are located in different parts of the electromagnetic, spectrum, and they are used in different ways to broadcast program content. The, difference between them consists in how the information they convey is encoded. In, amplitude modulated or AM radio, sound waves are encoded by varying the amplitude, of a carrier radio wave around some reference value. Changes in amplitude convey the, signal. The frequency of the carrier wave is around 1 MHz for AM radio., , Simple antenna operation, Generates varying charge on rods, Charged rods create electric field, Electric field varies over time, Changing E-field induces B-field, , In frequency modulated or FM radio, sound is encoded by slightly varying the frequency, of the carrier wave around its base frequency. For FM radio, and television, the carrier, wave frequency extends upward from around 100 MHz., , AM and FM radio, Radio wave is “carrier”, AM varies amplitude of carrier, FM varies frequency of carrier, , 562, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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30.6 - Earth's seasons, Almost all the energy we use on Earth originates in the Sun and arrives in the form of, electromagnetic radiation. Roughly the same amount of solar power reaches the planet, throughout the year, yet many places on the globe experience significant seasonal, variations in the rate at which they receive this energy., The cause of seasonal changes in the Earth’s climate is the tilt of its axis, the line about, which it rotates. The illustration in Concept 1 shows the position of the Earth in its orbit, at different times of the year, as well as the direction in which the axis points. The axis, is tilted at a 23.5° angle away from a line perpendicular to the Earth’s orbital plane. It, always points towards the same direction in space (which is why Polaris remains the, North Star throughout the year)., March 21 and September 22 are known as the equinoxes. The name refers to the equal, lengths of night and day (12 hours each) for all locations on Earth on these dates., December 21 and June 21 are the solstices. As the season progresses from autumn to, winter, the Sun rises to a lower high point in the sky each day, and the days get shorter., On the winter solstice (meaning “sun stop”), the Sun stops getting lower and begins to, rise to a higher apex each day, as it does through the rest of the winter and spring. The, opposite happens after the summer solstice í the Sun once again peaks at a lower, point each day. (The dates of the equinoxes and solstices vary from year to year, but, are always around the 21st of the month.), , Cause of varying intensity, Earth’s axis of rotation is tilted, , While June 21 is the summer solstice in the Northern Hemisphere, it is the winter, solstice for the Southern Hemisphere. Concept 2 illustrates why this is true. It shows a, Northern Hemisphere city, Beijing, and a Southern Hemisphere city, Perth, at noon on, June 21., Imagine a solar collection plate of area A lying flat on the ground, tangent to the Earth’s, surface in either of these cities. Light rays from the Sun arrive approximately parallel to, the plane of the Earth’s orbit. On June 21, the sunlight intersects the collecting plate in, Beijing at a steeper angle than in Perth. Because Beijing is receiving sunlight more, vertically, the energy from that light is more concentrated í Beijing is receiving more, power over the area of its collecting plate., In Perth, a smaller amount of sunlight is being spread over the same collecting area, because of the oblique, slanting angle at which it hits the plate. The plate absorbs less, power. It is summer in Beijing, and winter in Perth., , The seasons, Greater intensity in summer, less in, winter, Northern, Southern Hemispheres have, opposite seasons, , Six months later, on December 21, the situation will be reversed: Perth will receive more direct sunlight than Beijing., Generally speaking, locations farther than Beijing or Perth from the equator experience a greater variation in the power they receive throughout, the year, and places closer to the equator experience less change. At the poles í dark six months of the year í this difference is extreme., Some people mistakenly believe that the seasons are due to the eccentricity of the Earth’s orbit í the fact that the Earth’s distance from the, Sun changes throughout the year. Your first clue that this belief is false is the observation that summer in the Southern Hemisphere occurs at, the same time as winter in the Northern Hemisphere (you merely have to make a long-distance phone call to confirm this). In fact, during winter, in the Northern Hemisphere, the Earth is actually closer to the Sun than in summer. The reason the eccentricity has only a slight effect is that, the Earth’s orbit is only slightly elliptical. The annual variation in insolation due to the eccentricity of the Earth’s orbit is about 7%, in contrast to, an approximately 110% increase from winter to summer (at the latitude of Beijing) due to axial tilt., , 30.7 - How electromagnetic waves travel through matter, Light and other forms of electromagnetic radiation can travel through a vacuum, and it is, often simplest to study them in that setting. However, radiation can also pass through, matter: If you look through a glass window, you are viewing light that has passed, through the Earth’s atmosphere and the glass. Other forms of radiation such as radio, waves pass through matter, as well., This section focuses on how such transmission occurs. It relies on a classical model of, electrons and atoms that predates quantum theory. In this model, electrons orbit an, atom. They have a resonant frequency that depends on the kind of atom. On a larger, scale, atoms themselves and the molecules composed of them also have resonant, thermal frequencies at which they can vibrate or rotate., We will use the example of light striking the glass in a window to discuss how, substances transmit (or do not transmit) electromagnetic radiation. When an, Glass opaque to infrared, electromagnetic wave encounters a window, it collides with the molecules that make up, Radiation is in resonance with, the glass. If the frequency of the wave is near the resonant thermal frequency of the, molecules, glass molecules, which is true for infrared radiation, the amplitude of the molecules’, Waves do not travel through, vibrations increases. They absorb the energy transported by the wave, and dissipate it, throughout the glass by colliding with other molecules and heating up the window., Because it absorbs so much infrared energy, the glass is opaque to radiation of this frequency, preventing its transmission., , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 563
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Scientists in the 19th century noted a phenomenon in greenhouses caused by the, opaqueness of glass to infrared radiation, which they called the greenhouse effect. The, glass in a greenhouse admits visible light from the Sun, which is then absorbed by the, soil and plants inside. They reradiate the solar energy as longer infrared waves, which, cannot pass back out through the glass and so help warm up the greenhouse. The, same phenomenon occurs on a vaster scale in the atmosphere as gases like methane, and carbon dioxide trap solar energy near the Earth’s surface., In contrast to infrared radiation, higher frequency radiation such as visible light does not, resonate thermally with atoms or molecules, but may resonate with the electrons of the, atoms of a substance. In glass, visible light experiences much less reduction in the, amplitude of its waves than infrared radiation does, and most of its energy passes, through the glass quite easily. Atoms with resonant electrons that do absorb energy, from a light wave quickly pass on that energy by re-emitting it as radiation of the same, frequency to other atoms, which in turn pass it on to their neighbors., This chain of absorptions and re-emissions, called forward scattering, follows a path, close to the light’s original direction of travel. A beam of light that strikes a pane of glass, will reach the “last atom” on the far side of the pane in an extremely short time. We see, the light after it emerges, and think of glass as transparent., , Glass transparent to visible light, Waves absorbed by electrons of atoms, Re-emitted from neighbor to neighbor, Waves pass through but are slowed, , This process does slow the transmission of the wave, which is why light travels slower in glass than it does in air or a vacuum (a fact captured, numerically by the index of refraction of glass). For instance, light travels through a typical piece of optical glass at about 2/3 of its speed in a, vacuum. Of course, 2/3 of the speed of light in a vacuum is still a rather rapid pace…., If atoms of certain substances, such as cobalt, are added to glass, they may absorb certain frequencies of light without re-emitting them. Cobalt, glass has a deep blue-violet color, which indicates that all the lower visible frequencies (from red through green) are absorbed and cannot pass, through it. Substances which absorb all frequencies of visible light are called opaque., , 30.8 - Polarization, , Polarized wave: A, transverse wave that, oscillates in a single plane., Polarized radiation: A form, ofradiation in which the, electric field of every wave, oscillates in the same plane., , Polaroid sunglasses block some light waves., , Polaroid sunglasses like the ones shown above reduce the amount of light that passes, through them. How do they do it? They only let through light waves whose electric fields, oscillate in a certain plane. When a source such as the Sun emits light, waves emerge, whose electric fields vibrate in every plane parallel to the direction of propagation. This, is shown in Concept 1. The electric field of each individual wave does oscillate, consistently in its own plane, which is called the wave's plane of polarization, but the, radiation as a whole does not have this property., The electric field of any electromagnetic wave is a vector quantity. A polarizing lens or, filter works by only letting through a certain component of the electric field of every light, wave that strikes it. An ideal polarizing filter can be visualized as a set of narrow parallel, slits whose direction is the filter’s transmission axis., , Unpolarized light, , When a wave passes through the filter, the component of its electric field parallel to the, Radiation can oscillate in many planes, transmission axis is what passes through. The component perpendicular to the axis is, absorbed. As a result, waves that oscillate parallel to the slits pass through unhindered,, while waves that oscillate perpendicularly to them are completely absorbed.You may want to consider an analogy: a rope passing through a, gap in a picket fence. If you shake the rope vertically, the wave you create passes through unhindered. If you shake the rope horizontally, the, wave collides with the pickets, transfers energy to them, and does not pass through. In the case of a wave passing through at an oblique angle,, the component of its transverse displacement along the “picket axis” passes through while the fence absorbs the other component., This basic model of how polarization works is shown with an ideal polarizing filter in Concept 2. In the diagram, the transmission axis happens, to be vertical. A “slit” allows waves oscillating in a vertical plane to pass through. For waves that oscillate in other planes, only the vertical, component of the electric field of the wave can pass through. Waves with a horizontal plane of polarization cannot get through the slit at all., Ordinary light consists of waves whose electric fields are randomly oriented in all lateral directions. This is called unpolarized radiation. Two, common sources of unpolarized light are the Sun and incandescent light bulbs. If the radiation is created or filtered so that it has only waves, oscillating in a single plane, then it is linearly polarized. In the illustration in Concept 2, the polarizing filter is exposed to unpolarized radiation,, and it transmits linearly polarized radiation., , 564, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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In Concept 3 you see an end-on “close up” of several light waves striking a polarizing, filter that has a vertical transmission axis. The light waves are coming toward you. In, each case, the vertical component of the electric field of the light is transmitted, and the, filter absorbs the horizontal component. The original field is shown as a solid, dimmed, vector; its components are hollowed out. The vertical component that passes through is, drawn with a bright color, and the horizontal component that is blocked is dimmed and, marked with a red ×. In each example illustrated, the electric field amplitude of the, transmitted light, and with it the light's intensity, is reduced., The final illustration, in Concept 4, displays an experiment with two polarizing filters., Unpolarized light is coming toward you from a distant source. It passes first through the, upper filter, which allows the passage of light that is polarized at the angle shown by the, “slit” lines. This polarized light continues toward you and passes through the lower filter., The left-hand part of the illustration shows the orientation of the filters, with parallel lines, indicating the transmission axis of each filter. The right-hand side shows you the, amount of light that passes through the area of overlap, and how that changes with the, angles between the two axes. When the axes are perpendicular, no light at all can pass, through., , Polarized light, The filter transmits polarized light, ·Incoming light is unpolarized, ·Transmitted light vertically polarized, ·Transmitted waves all in same plane, , Radiation also can be partially polarized, having a few waves oscillating in all planes,, but with most of its waves concentrated in a single plane. This is true of sunlight, scattered by the atmosphere. As the photo above shows, the sky in certain directions is, partially polarized in a vertical plane so that most of its light can pass through a pair of, sunglasses whose transmission axis is vertical. Less light (but still some) passes, through the rotated sunglasses. (Polarizing sunglasses are specifically intended to, reduce horizontally polarized glare reflected from roadways and water, not skylight.), Many forms of artificial electromagnetic radiation are polarized. A radio transmitter emits, polarized radiation. If the rods of its antenna are vertical, then so is the electric field of, every radio wave it creates. In this case, the most efficient receiving antenna is also, vertically oriented; a horizontal receiving antenna would absorb radio waves much less, efficiently. You may be familiar with this fact if you have ever tried to maneuver a radio, antenna wire or a set of television “rabbit ears” to get the best reception. (If you do not, know what “rabbit ears” are for television, well, before there was cable television, there, was….), , E-field component passes, through filter, If transmission axis is vertical, filter …, ·Transmits vertical E-field component, ·Absorbs horizontal E-field component, , Two Polaroid filters, Light passes through back (upper) filter, Then passes through front (lower) filter, Intensity depends on orientation of filters, , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 565
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30.9 - Scattering of light, , Scattering: Absorption and, re-emission of light by, electrons, resulting in, dispersion and some, polarization., The answer to a classic question í Why is the sky, blue? í rests in a phenomenon called scattering. In, this section, we give a classical (as opposed to, quantum mechanical) explanation of how scattering, occurs., , Scattered sunlight gives the sky its blue glow., , When light from the Sun strikes the electrons of various atoms in the Earth’s, atmosphere, the electrons can absorb the light’s energy, oscillating and increasing their, own energy. The electrons in turn re-emit this energy as light of the same wavelength., In effect, the oscillating electrons act like tiny antennas, emitting electromagnetic, radiation in the frequency range of light., An electron oscillates in a direction parallel to the electric field of the wave that, energizes it, as shown in Concept 1. The electron then emits light polarized in a plane, parallel to its vibration. We show a particular polarized wave that is re-emitted, downward toward the ground, since we are concerned with what an observer on the, surface of the Earth sees. Other light is scattered in other directions, including light, scattered upward and light scattered forward in its original direction of travel., Scattering explains why we see the sky: Light passing through the atmosphere is, redirected due to scattering toward the surface of the Earth. In contrast, for an astronaut, observer in the vacuum of space, sunlight is not scattered at all so there is no sky glow:, Except for the stars, the sky appears black. To the astronaut, the disk of the Sun, a, combination of all colors, looks white. We illustrate this below:, The full spectrum combines to form white light., , Scattering, Light from Sun hits electron, · Electron oscillates, · Re-emits polarized wave to ground, · Shortest wavelengths scattered most, , View from space, No scattering: sky is black, Sun appears white, , The question remains, why is our sky blue rather than some other color? Light at the blue end of the visible spectrum, which has the shortest, wavelength, is 10 times more resonant with the electrons of atmospheric atoms than red light. This means blue light is scattered more than red,, so that more of it is redirected toward the ground., Scattering also explains why we see the Sun as yellow rather than white. When you look up at the disk of the Sun from the Earth’s surface, the, bluest portion of its light has been scattered away to the sides. The remaining part of the Sun’s direct light appears somewhat yellowish., , Why the sky is blue (and the Sun, is yellow), Shortest waves are scattered: sky is, blue, With blue scattered, Sun appears, yellowish white, , 566, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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You may also have noted how the Sun appears to change color when it sets. As the Sun’s disk descends toward the horizon, its light must, pass through a greater and greater thickness of atmosphere in order to reach you. Since a certain amount of sunlight is scattered aside for, each kilometer of atmosphere it passes through, its position at sunset causes it to lose large amounts of light at the blue end and even toward, the middle of the visible light spectrum. At sunset, practically all the shorter wavelengths of light have been scattered out of it, leaving only light, at the red end of the spectrum to be viewed by you. The “missing” blue light is not really missing. People to the west of you perceive it as the, daytime sky., , Sunset, Longer wavelengths also scattered, Sun appears red, , Scattering also explains why skylight is partially polarized. When the Sun is low in the sky, as depicted in Concept 1, horizontally polarized light, that gets scattered down from the overhead sky is polarized in the plane of the downgoing wave you see in the illustration. Incoming sunlight, that is polarized in other planes also gets scattered, but not straight down towards the ground., You can experiment with skylight polarization yourself if you have a pair of polarizing sunglasses. In the early morning or late afternoon hold, your glasses against the northern or southern sky at arm’s length. Turn one of the lenses slowly, recalling that its transmission axis is vertical, when the sunglasses are worn normally. You will find that the skylight is partially polarized in a plane perpendicular to the direction to the Sun., , 30.10 - Physics at work: liquid crystal displays (LCDs), Substances that change the direction of polarization, of light passing through them are called optically, active. In other words, an optically active substance, rotates the plane of polarization of light passing, through it., The liquid crystal display (LCD) in the watch face, above demonstrates variable optical activity at work., LCDs are found in many common devices, including, calculators, cellular telephones and clocks. There are, two types of LCDs: backlit and reflective. The one, shown above is a reflective LCD, but we will explore, both types in this section. Backlit LCDs generate light, behind their displays; reflective LCDs like the one, above utilize ambient light., , This digital watch displays the time with a reflective LCD., , LCDs rely on polarization. The characters of a digital watch display consist of “digit”, segments: regions that can be made dark. These segments are filled with a substance, called liquid crystal that is optically active in its natural state but becomes inactive when, a potential difference is applied across it. You see this phenomenon illustrated for a, backlit LCD in Concepts 1 and 2. In Concept 1, the power is off (there is no potential, difference across the crystal), so the crystal is optically active and rotates the plane of, polarization of the light. The thickness of the crystal is designed to rotate the light by, 90°. In Concept 2 a potential difference is applied, the crystal becomes inactive, and, there is no rotation of the light’s plane of polarization., The liquid crystal is sandwiched between two polarizing filters with perpendicular, tranmission axes, as shown in Concept 3. In the backlit LCD shown there, a light behind Power OFF, the display shines through the left-hand filter. Following our usual practice we will call, Liquid crystal is optically ACTIVE, this filter that is closer to the light source the polarizer, and the one farther away the, Rotates polarized light 90°, analyzer. The polarizer allows light with a particular plane of polarization to pass, through. When the liquid crystal is “on” (optically inactive), it does not rotate the, polarized light, and the analyzer prevents any of the light from passing through, creating a dark area., When no potential difference is applied across the liquid crystal, it is “off” and reverts to its normal optically active state. The perpendicular, orientation of the analyzer now allows the polarized light to pass through. Instead of being black, the material of the segment looks the same as, the adjoining material. In order to make the transmitted light difficult to see, the display’s background is colored to resemble it. This is shown in, Concept 4., If you examine the watch in the illustration above or one on your wrist, or a friend’s wrist, you will see that the digits it displays are pieced, together from segments that appear black when they are “turned on.” When they are off, these segments are invisible to the casual glance, but, they can still be made out as faint shadows if you look very carefully., We have been describing a typical backlit LCD, which can be read in the dark. The backlighting consumes more power than any other part of, , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 567
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the display, since the amount of power required to turn on each digit segment is, negligible. Backlit LCDs are typically found in automobile dashboards, where their, power usage is not a particular concern and where readability at night is important., A reflective LCD, such as the one in the watch above, consumes less power. We show, how a reflective LCD works in Concept 5; it is a bit more complicated than a backlit, LCD. A mirror that can reflect ambient light coming from in front of the LCD replaces the, backlight behind the polarizer. As with the backlit display, the transmission axes of the, polarizer and analyzer are at right angles to each other., Concept 5 shows what happens when a liquid crystal digit segment in this type of LCD, is turned off. In this state, light passes through each of the analyzer, the liquid crystal, segment, and the polarizer twice. Unpolarized light enters the display from the right and, becomes horizontally polarized as it passes through the right-hand filter (the analyzer)., It rotates 90° to a vertical orientation as it passes through the optically active liquid, crystal component. In this orientation it can pass through the polarizer, reflect off the, mirror, and come back out through the polarizer without hindrance. (A keen-eyed, observer will note that the angular symmetry of reflection is apparently being violated, where the light strikes the mirror: We drew the incident ray at a different angle in order, to fit all the details into the diagram.), , Power ON, Liquid crystal is optically INACTIVE, Does not rotate polarized light, , After passing through the polarizer, the light retraces its path through the liquid crystal,, again rotating and again becoming horizontally polarized. It passes back out through, the analyzer without absorption, creating a region that blends into the background color, of the display., When the segment is turned on, it becomes optically inactive; optically, it can be treated, as if it were no longer there. Light entering from the right gets horizontally polarized by, the analyzer, and propagates to the left until it strikes the polarizer, where it is, absorbed. This light never even reaches the mirror, and the segment appears dark. You, see this happening in the watch above, where the dark segments spell out the time of, day., More elaborate LCDs of both types can be manufactured with “segments” having any, shape, not just the parts of digits. For example, the battery and signal strength icons on, a cellular telephone display, or the letters “H” and “M” on the watch above, use specially, shaped segments. Some flat panel display screens use an array of thousands or even, millions of tiny LCD dot-segments to produce virtually any image. With the appropriate, refinements, color images can be produced., , Backlit LCD í ON, Analyzer absorbs unrotated light, ·Segment appears black, , As with polarized light from the sky, you can use a pair of Polaroid sunglasses to, experiment with the polarized light emitted by a liquid crystal display. For example, you, will find that a watch, viewed through the glasses, looks quite different depending on its, orientation. At some angles the display looks more or less normal, while at others it, becomes completely unreadable., , Backlit LCD - OFF, Analyzer admits rotated light, ·Segment blends into background, , Reflective LCD, When power OFF, segment is not seen, When power ON, segment is black, , 568, , Copyright 2000-2007 Kinetic Books Co. Chapter 30
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30.11 - Gotchas, A light wave is a transverse wave. Yes. Both of its components, an electric and a magnetic field, oscillate perpendicularly to its direction of, travel., Radio signals and light waves are fundamentally different. Both are forms of electromagnetic radiation, so we lean toward “no” in response to, this statement. The wavelength and frequency of radio transmissions and light are significantly different, and humans can see light, but not, radio waves, so one could say “yes”. However, both are electromagnetic waves, and both move at the speed of light., , 30.12 - Summary, An electromagnetic wave is a traveling wave consisting of mutually perpendicular, electric and magnetic fields that oscillate transversely to the direction of, propagation. Electromagnetic radiation moves at the “speed of light,” or c, which is, 299,792,458 m/s in a vacuum. The value 3.00×108 m/s is often used., , Proportionality of fields, , Every electromagnetic wave has a characteristic frequency and wavelength. The, electromagnetic spectrum is an ordering of electromagnetic radiation in accordance, with these two properties and extends far beyond the tiny gamut called visible light, that we can detect with our eyes. Some other kinds of electromagnetic radiation are radio waves (AM and FM), television signals, microwaves,, infrared light, ultraviolet light, x-rays, and gamma rays., Maxwell proved that the speed of an electromagnetic wave equals the reciprocal of the square root of the product of two fundamental physical, constants, the electric permittivity of free space İ0, and the magnetic permeability of free space μ0. This value turned out to be equal to the, empirically well-measured speed of light, providing strong evidence that light is a form of electromagnetic radiation., Linearly polarized light consists of light waves whose electric fields all oscillate in the same plane. This type of light can be created by several, methods, such as by passing unpolarized light, whose electric fields oscillate in many directions, through a polarizing filter. The direction of, polarization that results is called the transmission axis of the filter. As a randomly polarized wave passes through such a filter, only the, component of its oscillating electric field that is aligned with the transmission axis passes through., The scattering of light passing through a transparent substance is the absorption and re-emission of light waves of characteristic frequencies, by atoms in the substance. Scattering in the atmosphere is responsible for the blue glow of the sky, the yellowish hue of the Sun, and the red, color of sunsets., Optical activity forms the technological basis of the liquid crystal displays (LCDs) used in many consumer electronic devices., , Copyright 2000-2007 Kinetic Books Co. Chapter 30, , 569
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Chapter 30 Problems, , Chapter Assumptions, Unless stated otherwise, electromagnetic waves are assumed to be propagating as plane waves through a vacuum., When converting light-years to meters, assume that a year has 365 days., The speed of light in a vacuum is 3.00×108 m/s., The Sun radiates energy at the rate of 3.91×1026 W., , Conceptual Problems, C.1, , In a phenomenon called diffraction, waves are able to bend around obstacles. For example, ocean waves can bend around, a piling so that it does not cast a down-wave "shadow" of calm water. The shorter the wavelength of a wave, the less able it is, to diffract around obstacles. You have experienced this yourself: You cannot see the people in a room down the hall, but you, can hear them talking, corresponding to the fact that a typical wavelength of light is about ten million times shorter than that of, a typical sound. If you live near the big city, but behind a rather large mountain, which of the urban radio broadcasts are you, likely to receive the best: AM or FM? Explain., , C.2, , Half of all the electromagnetic radiation reaching the Earth from the Sun lies in the visible spectrum that can be perceived by, the human eye: ranging from extremes of 390 nm to 780 nm. This is a tiny portion of the whole electromagnetic spectrum, and, it can't be just a coincidence. Why is so much of the Sun's radiation visible?, , C.3, , Clouds scatter light just as the atmosphere, does. Since clouds consist mainly of, microscopic water droplets of many sizes, they, scatter all the wavelengths of visible light, equally. The smallest droplets scatter the, shortest wavelengths at the violet end of the, spectrum, and larger droplets scatter light of, longer wavelengths. Since all wavelengths are, equally scattered by a cloud, it appears white., So does a jet's condensation trail, or fog, or, steam, or mist on a window., Well, not quite white. In fact if you look at a real (cumulus) cloud, you will see that parts of it appear bright white, while other, parts appear gray, and at the bottom of a thick cloud, almost black. What accounts for these variations?, , C.4, , It is early morning, and the Sun is rising (where else?) in the east. (a) What is the direction of polarization of the skylight, directly overhead? (b) What is the direction of polarization of the skylight as you look toward the northern horizon?, , AM, , (a), (b), C.5, , (b), , 570, , East-west, North-south, Vertical, Horizontal, , Now it is noon, and the Sun is directly overhead. (a) What is the direction of polarization of the skylight coming from the, eastern horizon? (b) What is the direction of polarization of the skylight as you look toward the northern horizon?, (a), , C.6, , i., ii., i., ii., , FM, , i., ii., i., ii., , Vertical, Horizontal, Vertical, Horizontal, , Sunlight which reflects off smooth surfaces like those of roadways or standing water becomes horizontally polarized in the, process. Polarizing sunglasses are designed to completely block this reflected glare, while at the same time they reduce the, intensity of unpolarized ambient light by one-half. (a) If a sunglasses lens is to block reflected glare, what should the, orientation of its transmission axis be? (b) LCD displays, such as flat panel computer display screens, and certainly, automobile instruments and digital writswatch faces, should be readable by people wearing polarized sunglasses. In order for, them to be readable, what should the transmission-axis orientation of their front "analyzer" filters be? (Note: The conceptual, diagrams of LCDs in the text may or may not have been drawn to reflect this requirement.), (a), , Vertical, , Horizontal, , (b), , Vertical, , Horizontal, , Copyright 2000-2007 Kinetic Books Co. Chapter 30 Problems
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Section Problems, Section 1 - The electromagnetic spectrum, 1.1, , The speed of light in a vacuum is exactly 299,792,458 m/s. This speed is sometimes used to provide a convenient yardstick, for large astronomical distances. (a) A light-second is the distance light travels in one second. If the Moon is 3.84×108 m from, the Earth (center to center), how many light-seconds away is it? (b) A light-minute is the distance light travels in one minute. If, the Earth is 1.50×1011 m from the Sun, how many light-minutes away is it? (Your answer also represents the number of, minutes it takes for light to reach the Earth from the Sun.) (c) A light-year is the distance light travels in one year. The Sun is, 28,000 light-years from the center of the Milky Way galaxy. What is this distance in meters?, (a), , 1.2, , 1.3, , light-seconds, , (b), , light-minutes, , (c), , m, , Nowadays many highway tunnels are equipped with AM "repeaters" so that motorists can listen to their AM radios while, inside. Otherwise, AM radio cannot be received in the tunnel. FM repeaters are not usually necessary, since FM reception in, tunnels is often acceptable without any special assistance. A typical tunnel is 5 meters high. (a) What is the wavelength of the, electromagnetic radiation from an FM station broadcasting at 94.1 MHz, which penetrates easily into the tunnel? (b) What is, the wavelength of the electromagnetic radiation from an AM station broadcasting at 940 kHz, which ordinarily cannot, penetrate into the tunnel?, (a), , m, , (b), , m, , A certain radio station broadcasts electromagnetic radiation with a wavelength of one quarter mile (0.25 mi). (a) What is the, frequency of the broadcast? (b) Is the station AM or FM?, (a), (b), , 1.4, , Hz, AM, , FM, , The color to which the human eye is most, sensitive is a yellowish green with a, wavelength of approximately 550 nm. (For this, "scientific" reason there was a brief vogue in, the 1980s for making fire trucks this color, a, fad that faltered when departments found that, firefighters took little pride in non-red fire, trucks and consequently maintained them, poorly. In fact, the photographer of the, accompanying picture was told that the truck, will be painted red "within a month í red is, traditional".) (a) What is the frequency of waves of this color? (b) How many wavelengths of this yellowish green light are, there in the thickness of a human hair that is 110 ȝm in diameter? (c) How many wavelengths of the light are there in the, width of a pencil, 6.00 mm?, (a), , Hz, , (b), (c), 1.5, , The rather unappetizing confection you see in, the photograph is a carrot strip that has been, cooked in a microwave oven from which the, rotating turntable was removed. When it is, turned on, the oven uses a "magnetron" tube, to generate standing microwaves in the cavity, of the oven. No cooking heat is generated at, the locations of the standing-wave nodes, and, maximal heat (microwave "hot spots") is, generated at the locations of the standingwave antinodes. The manufacturer's label on, the oven states that it operates at a frequency of 2450 MHz. (a) Based on this frequency, what is the wavelength of the, microwaves generated inside the oven? (b) Based on your answer to part a, what is the distance between two antinodes of, the standing microwaves? Give your answer to three significant figures. (c) Is the result in part b compatible with the, measurement shown in the photograph?, (a), , cm, , (b), , cm, , (c), , Yes, , No, , Copyright 2000-2007 Kinetic Books Co. Chapter 30 Problems, , 571
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1.6, , Apollo 11 astronauts left an array of "corner mirrors" on the Moon, each one consisting of three mirrors arranged at mutual, right angles like the corner of a box. A corner mirror has the property that light shined on it from any angle is reflected back, exactly in the direction from which it came. Scientists subsequently directed a brief pulse of light from a powerful laser at the, mirror assembly on the Moon, and measured the time required for it to travel to the array and return back to Earth: 2.56 s., How far away was the array of mirrors at that time?, , 1.7, , The human eye contains receptors for just, three colors of light: red, green, and blue. All, the colors we perceive in the world are mixed, in our brains from different proportions of, these three primary colors., Computer and video display screens, both, CRTs and flat panels, take advantage of this, fact by using color-generating elements of the, same three colors to create the appearance of, any imaginable color. An image on the screen, is built of tiny dots of color called picture, elements, or "pixels". (By default, the period at the end of this sentence consists of just one black pixel.) Each pixel has three, different color levels associated with it, one level for each of red, green, and blue, with integer values from 0 (representing no, color), to 255 (representing the brightest level of a color). For example, (R, G, B) = (255, 0, 0) represents the brightest pure, red, (170, 170, 0) represents an equal mixture of red and green, without any blue, which comes out a sort of dark yellow, and, (127, 127, 127) represents an equal mixture of all three primary colors, each one at half its maximum intensity: a color, sometimes called "50% gray"., (a) If black is the "color" corresponding to a lack of visible light, what are the RGB values of a black pixel? (b) If white is the, "color" corresponding to the brightest possible equal mixture of all colors, what are the RGB values of a white pixel? (c) How, many different colors can a pixel be? (The answer is large, but enter it as a whole number.), , m, , (a) (, , ,, , ,, , ), , (b) (, , ,, , ,, , ), , (c), 1.8, , The brightness of a computer screen color is, a percentage value calculated by dividing the, brightest of its three color levels by 255, the, maximum possible brightness. For example,, the brightness of the magenta hue (R, G, B) =, (127, 0, 135) is 135/255 or 53%. What is the, brightness of each of the following colors: (a), black (0, 0, 0); (b) brightest white, (255, 255, 255); (c) chartreuse (132, 251, 67);, (d) brown (114, 73, 32)., (a), , 1.9, , 572, , colors, , %, , (b), , %, , (c), , %, , (d), , %, , Equally balanced computer screen colors like, (R, G, B) = (0, 0, 0), (255, 255, 255), or, (120, 120, 120) have no predominant hue, and, they look black, white, or some shade of gray, in between: The grays with greater brightness, are closer to white, and the grays with less, brightness are closer to black. A color like, (104, 100, 100) will appear slightly red, but it is, mostly the gray color (100, 100, 100) with only, a small additional amount of red. Color, specialists say that this kind of color is, "unsaturated"., Colors that are "saturated" have a rich predominant color with little gray mixed in. For example the RGB color (246, 15, 18) is, a very saturated red. It only contains a slight constituent of gray, corresponding to its lowest-level component: (15, 15, 15)., The saturation of an RGB color equals the percentage of its brightest component color that does not form part of its gray, constituent. If the brightest component of an RGB color has value N, and the dimmest component has value n, then its gray, , Copyright 2000-2007 Kinetic Books Co. Chapter 30 Problems
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constituent is (n, n, n) and its saturation equals (N í n)/N. For example, the saturation of the vibrant red mentioned above is, (246 í 15)/246 = 94%. (a) What is the saturation of the red-tinged gray color (124, 100, 100)? (b) What is the saturation of the, dark green (33, 76, 45)? (c) What is the saturation of the light blue (213, 225, 255)?, (a), , %, , (b), , %, , (c), , %, , 1.10 A live outdoor radio broadcast is disrupted when someone fires a gun into the air right in front of the microphone. The sound, of the shot is broadcast via radio waves to a communications relay satellite in a geosynchronous orbit 35,790 km directly, overhead, and back down to the radio in a police car parked 85.0 m away from the microphone. The sound waves of the, gunshot also travel directly through the air to the car, at a speed of 343 m/s. Which sound does a police officer in the car hear, first, the broadcast sound or the direct sound?, Broadcast, , Direct, , Section 2 - Electromagnetic waves, 2.1, , The diagram below shows the electric and, magnetic field components of an, electromagnetic wave at a certain location and, at a certain instant in time. Is the wave, traveling toward you or away from you?, , Toward you, , Away from you, , Section 3 - Proportionality of electric and magnetic fields, 3.1, , An electromagnetic wave in a vacuum has an electric field amplitude of 417 V/m. What is the magnetic field amplitude of this, wave?, T, , 3.2, , An electromagnetic wave in a vacuum has an instantaneous magnetic field strength of 7.50eí7 T at a certain point. What is, the instantaneous electric field strength of the wave at this point?, V/m, , 3.3, , Here is a pair of sinusoidal functions describing an electric and a magnetic field that vary in space and time., , They cannot be the field equations of an electromagnetic wave. Why not?, , Copyright 2000-2007 Kinetic Books Co. Chapter 30 Problems, , 573
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31.0 - Introduction, “Image is everything,” one well-known advertisement blared for years. (An ad from, another company then trumpeted the contradictory message that “image is nothing.”, Go figure!), For physicists, image is as much a matter of mirrors and lenses as it is of, appearance and athleticism. By arranging mirrors and lenses, they can magnify or, shrink images and place them where they are needed. The design of a wide range, of devices, from reading glasses and cameras to telescopes, depends on a, thorough understanding of how to manipulate images created with light., This chapter begins with a discussion of light rays and mirrors, two essential, elements of your understanding of images. The simulation to the right features a, concave mirror. The mirror creates images of objects. In this case, the object is a, penguin. You can move the penguin left and right, and observe how its positioning, changes the location and size of the image created by the mirror. The image is, shown as a faded out version of the object itself. At some locations of the object, the, image will be off the screen, but it will reappear when you drag the object to another, location., As you move the object back and forth, consider the following questions. Where can you place the object so that the image and object are on, the same side of the mirror? On opposite sides? Where can you place the object so that the image is smaller than the object? Larger? The, variety of images produced by different types of mirrors is a major topic in this chapter., If you press the SHOW RAYS button in the simulation, you will see three light rays that emanate from the penguin, reflect off the mirror, and, converge to define the top of the image. These rays are used extensively in the study of mirrors and lenses, and this is your chance to begin to, experience their properties. (The rays do not always converge perfectly; this is a realistic depiction of the way curved mirrors work.) You can, turn off the rays by pressing the HIDE RAYS button., , 31.1 - Light and reflection, , Reflection: Light “bouncing, back” from a surface., When you look at yourself in a mirror, you are seeing, a reflection of yourself. When you look at the Moon at, night, you are seeing sunlight reflecting off that distant, body., Not all the light that reaches a surface reflects. In fact,, you see an object like a tree as having different colors, because its varied parts reflect some wavelengths of, To form a mirror image, light bounces back from the mirror's surface., light and absorb others. Light can pass through a, material, as it does with a glass window. It can also be absorbed by a material, as, evidenced by how a black rock warms up during a sunny day. All this can happen, simultaneously: Light will reflect off the surface of a lake (which is why you see the, lake), penetrate the water (otherwise, it would be completely dark below the surface),, and be absorbed by the water, warming it., To understand reflection, it is often useful to treat light as a stream of particles that, move in a straight line and change direction only when they encounter a surface. Each, light “particle” acts like a ball bouncing off of a surface, and like a ball, it reflects off the, surface at a rebound angle equal to its incoming angle. You see yourself in a mirror, because the light bounces back to your eyes from the mirror., The term “reflection” likely conjures up images of light and perhaps mirrors. Studying, mirrors is a good way to learn about reflection because they are designed to reflect light Reflection, in a way that creates a clear visual image. However, it is worth noting that reflection, Light bouncing back from a surface, does not apply only to light. Some creatures use the reflection of sound (echoes) to help, them perceive their surroundings and stalk their prey. For example, bats, seals and, dolphins emit high frequency sound and then listen for the reflected waves. By analyzing these reflections, they can “see” with great precision., Radar, used to track airplanes, is based on the reflection of radio waves. A sophisticated understanding of reflection can be used to design, “stealth” aircraft that are difficult to detect with radar. Stealth aircraft register on radar screens as being about as large as a BB, in part because, of their ability to reflect incoming waves in “random” directions., , 574, , Copyright 2007 Kinetic Books Co. Chapter 31
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31.2 - Light rays, , Light ray: A straight line used to represent the, path of light., Light rays are used to analyze devices such as mirrors and lenses. Light emanating, from a point in a particular direction is treated as moving in a straight line. You can think, of this as modeling light with a collection of laser beams, each ray moving in a straight, and narrow line until it intersects a surface., You can see an example of light rays depicted on the right, where a penguin views, himself in a mirror. Several rays are shown emanating from a point on the penguin’s, foot. We show the full path of a single ray to illustrate how the penguin can see his foot, reflected in the mirror. The light ray starts at the foot, reflects at the mirror and then, reaches his eye. The arrowheads indicate the direction of the light’s travel., The penguin’s feet are not the source of the light. That might be the Sun or an electric, lamp. But for the purposes of analyzing the mirror, we will proceed as though the light, rays originate from the object whose image we intend to analyze., , Light rays, Represent light with straight lines, Used to analyze mirrors, lenses, , Light is a complex phenomenon. Treating it as propagating in a straight line does not explain why a beam from a flashlight spreads out as it, travels; other concepts are needed there. But to understand mirrors and lenses, light rays prove very useful., , 31.3 - Mirror basics, At the right, you see a penguin looking at himself in a planar mirror. A planar mirror is a, flat, smooth surface that reflects light. A typical planar mirror is manufactured by coating, one side of a sheet of glass with a metallic film. It is designed to reflect as much light as, possible and to create distinct images. Either by looking at yourself in a planar mirror or, referring to the illustration in Concept 1, you can observe four essential properties of an, image produced by this kind of mirror:, 1., , Your image is the same size as you are. The variable h is used to represent its, height., , 2., , Your image appears as far behind the mirror as you are in front of it. Your, distance is called the object distance, and how far the image is behind the mirror, is called the image distance. The object distance, represented by do, is positive, by convention. When an image is on the opposite side of a mirror from the object,, its distance di from the mirror is negative by convention., , 3., , The image has front-back reversal. If you are facing north, then your mirror image, is facing south. If this reversal did not occur, you would see your back., , 4., , The image is not reversed either right-left or up-down. If you look in a mirror and, raise your left hand, its reflection will rise up on your left. Your head also still, appears on the top of the image. The image is not inverted., , Planar mirror object and image, Same height, Equally far from mirror, Image is right side up, , Mirrors do create the illusion of left-right reversal. For instance, if you raise your left hand, it will appear on your left but on the image’s “right,”, because your image is facing the opposite direction than you are. This is akin to “stage left” or “stage right” directions for actors, which are, based on the perception of what is left or right for the audience, not for the actors themselves as they face the audience. Their “left” and “right”, are reversed because they are facing in the opposite direction than the audience, just as your mirror image is., You cannot easily read normal writing in a mirror (unless you are as gifted as Leonardo da Vinci, who wrote in “mirror writing” so that others, could not easily decipher his work). Mirror writing does have its uses: An emergency vehicle may have “, ” written on its front so that, the text can be readily read when seen in the rearview mirror of a car., , Copyright 2007 Kinetic Books Co. Chapter 31, , 575
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31.4 - Virtual and real images, , Image: A reproduction of an, object by means of light., Virtual image: An image on, the opposite side of a mirror, than the object it is created, from. The image cannot be, projected onto paper at its, perceived location., , Kitten looking for its image behind a mirror., , Real image: An image on the same side of a, mirror as the object it is created from. The, image can be projected onto paper at its, perceived location., When you look into a planar mirror, your image appears to be on the opposite side of, the mirror, in a “looking glass” world that you cannot reach except in literature. The light, of the image registered by your eyes is not actually coming from behind the mirror, even, though your brain may interpret it this way., You may be so accustomed to virtual images such as those formed by planar mirrors, that you take them for granted and believe them to be as “real” as any other image. But, not as real as any other object: A true understanding of virtual images must be learned., A kitten, when first exposed to a mirror, believes its reflected image to be another kitten, and will search for the potential playmate behind the mirror. Over time, some animals, (including humans, chimpanzees and dolphins) can learn how to interpret the nature of, the images created by mirrors., , Image, A reproduction created by light, , It will now be helpful to be more formal about some basic terminology. Objects are “in, front” of a mirror: They are on its reflecting side. (In the diagrams in this book we will, almost always place an object to the left of the mirror, but that certainly is not a law of, physics, just a convention of sorts about which side is “in front.”) If the image is on the, same side of the mirror as the object, it also is in front of the mirror. If the image is on, the opposite side of the mirror from the object, we say it is “behind” or in back of the, mirror. A virtual image created by a mirror is behind the mirror. The distance between a, virtual image and the mirror is negative., Curved mirrors can create what is called a real image. In Concept 3, you see a diagram, of a concave mirror creating a real image: an image on the same side of the mirror as, the object. By convention, the distance between the mirror and the real image is, positive. We stress the conventions of sign because they will prove essential when, equations are used to calculate certain properties of an image, such as its position., , Virtual image, Opposite side of mirror, Negative distance, , The terms “real” and “virtual” can be confusing. One way to decide whether an image is, real or virtual is to imagine placing a piece of paper at the location of the image and, then observing the results. With real images, the light passes through the location of the, image; with virtual images, it does not. If you place a piece of paper “behind” a planar, mirror, where the virtual image is located, you will not find an image projected there., With virtual images the brain projects the location of the image as it interprets the light it, receives., On the other hand, if you place a piece of paper at the location of a real image, you will, see the image on the paper. For instance, images created by movie projectors are real,, which is why you can see them on movie screens., Concept 4 shows an optical illusion created by a popular toy. A special combination of, curved mirrors creates a real image that is so solid looking you are tempted to reach out, and grasp it., , 576, , Real image, Same side of mirror, Positive distance, , Copyright 2007 Kinetic Books Co. Chapter 31
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An optical illusion, Real image perceived as object, , 31.5 - The law of reflection, , Law of reflection: The angle, of incidence equals the angle, of reflection., At the right, you see a planar mirror. A light ray from, the penguin’s foot is reflecting off the mirror. The light, ray before reflection is called the incident ray; the light, ray that reflects off the mirror is known as the, reflected ray. In this illustration, the rays are in the, same vertical plane., , Diffuse and specular reflections from holiday ornaments., , The angle of incidence și is shown in the diagrams to the right. The angle și represents, the angle between a line perpendicular (normal) to the mirror’s surface and the incident, ray of light., The light leaves the surface at the angle of reflection șr . This is the angle between the, normal line and the reflected ray. As you can see in the diagrams, the angles of, incidence and reflection are the same. This is the law of reflection: The angle of, incidence equals the angle of reflection. This law is confirmed by experiments and, theory., The law of reflection applies to smooth surfaces, which exhibit specular reflection. The, right-hand ornament in the photograph above is extremely smooth and produces a, specular reflection. Rays that are parallel and close together when they strike the, ornament will all be moving in a new direction after they reflect, but they will still be, parallel. Specular reflection is the topic of this chapter., If the reflecting surface is rough, diffuse reflection results. The left-hand ornament, above provides an example of diffuse reflection. With a rough surface, neighboring, incident rays will reflect in a variety of directions. Rays that are parallel and close, together when they strike the surface will not be parallel after they reflect. Diffuse, reflection is sometimes desirable. For instance, matte (low gloss) wall paint is designed, to achieve diffuse reflection for surfaces where a shiny appearance is undesirable., , The law of reflection, Incidence angle equals reflection angle, Angles measured between light rays, and normal line, , și = șr, și = angle of incidence, șr = angle of reflection, , Copyright 2007 Kinetic Books Co. Chapter 31, , 577
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What is the angle of reflection?, Same as angle of incidence: 25°, 31.6 - Ray diagrams for planar mirrors, In this section, we use ray diagrams to explore some properties of an image produced, by a planar mirror. To review these: The image created by a planar mirror is virtual,, upright (right side up), the same distance from the mirror as the object, and the same, size as the object., Using the ray diagram in Concept 1, we first show why the image is virtual. You see two, rays emanating from a point on the penguin’s head. They reflect off the mirror and travel, to the eye of an observer viewing the penguin’s image. In this diagram, we are not, concerned with what the penguin sees. We are using him as an object whose image is, being viewed by an observer., Because the observer’s brain assumes that light travels in a straight line, it projects an, image behind the mirror. Dashed lines, called virtual rays, show the paths the brain, presumes are followed by rays emanating from the source of the light. The brain locates, the top of the image at the point where these virtual rays converge. Since this point is, behind the mirror, and no light actually comes from it, it forms part of a virtual image., , Ray diagram, Draw rays using law of reflection, Find image location with virtual rays, , Rays also emanate from a point on the penguin’s foot and reflect from a lower point on, the mirror back to the observer’s eye. The observer’s brain locates the bottom of the, image at a point where the virtual extensions of those rays intersect, a point behind the, mirror and below the location of the image’s head. This confirms that the image is not, only virtual, but upright as well., The diagram in Concept 2 shows that the image is the same distance behind the mirror, as the object is in front. The two colored triangles share a common side (the mirror) and, two pairs of equal angles, including a pair of right angles. This means the triangles are, congruent, so their bottom legs must have equal lengths, and di and do must have, equal magnitudes. A similar argument with two other congruent triangles demonstrates, that the image has the same height as the object., , To relate object, image distances, Draw a ray, Identify equal angles, Congruent triangles prove object, image, distances have same magnitude, , 31.7 - Spherical mirrors, , Spherical mirror: A mirror, that is a portion of a sphere., Concave mirror: A curved, mirror whose reflective side, is on the inside of the curve., Concave and convex spherical mirrors., , 578, , Copyright 2007 Kinetic Books Co. Chapter 31
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Convex mirror: A curved mirror whose, reflective side is on the outside of the curve., Mirrors can be curved as well as flat. One type of curved mirror is a spherical mirror,, which consists of a section of a sphere. To describe spherical mirrors it is common to, consider a sphere with silvered reflecting surfaces. In a concave mirror, reflection, occurs on the inside surface, as illustrated in Concept 2. A convex mirror is made by, using a section of the exterior of the sphere, as shown in Concept 3., If you like mnemonics, a good way to remember which type is which is to recall that a, concave mirror is like a cave whose walls curve around you. Though we concentrate on, spherical mirrors here, concave and convex are terms that can apply to curved mirrors, other than spherical mirrors., The photos above show two examples of spherical mirrors and the images they create., The left-hand photo shows a concave makeup mirror. These mirrors, often found in, bathrooms, are designed to magnify an image: You look in the mirror and see a larger, image of yourself. Lucky you!, , Spherical mirror, A portion of a sphere, , The photograph on the right above shows a convex safety mirror that allows drivers to, see around a corner of a twisty seaside road. Security mirrors are also often convex., They provide a larger field of view than a flat mirror, enabling store personnel, for, example, to survey a large area. Convex auxiliary mirrors are often affixed to the flat, rearview mirrors of trucks and RVs., , Concave mirrors, Reflecting surface is interior of sphere, , Convex mirrors, Reflecting surface is exterior of sphere, , 31.8 - Mirror terminology, Before we proceed to a further discussion of mirrors, we need to introduce some, terminology. The diagrams to the right are used to illustrate these new terms. Many of, the same terms are used to describe lenses, a topic you will study later., The mirrors discussed in this section are spherical. The sphere’s center of curvature is, shown as point C in the diagrams., A line passing through the center of curvature and the midpoint of a spherical mirror,, like the one shown in Concept 1, is called the principal axis of the mirror. This line is, perpendicular to the surface of the mirror where they intersect., While planar mirrors produce only virtual images, a curved mirror can produce either a, virtual or a real image, depending on the shape of the mirror and the location of the, object., The image point corresponding to an object on the principal axis is the location on the, axis of its image. When an object is infinitely far away from the mirror, its image point, defines the focal point of a mirror or lens. You see this point, F, in Concept 2. “Infinitely, far away” means the incident light rays from the object are effectively parallel. The focal, point is located on the principal axis, and it is an unchanging characteristic of a, particular mirror., , Mirror terminology, Center of curvature C is center of, sphere, Principal axis passes through center of, curvature and midpoint of mirror, , The distance between the focal point and the midpoint of a mirror is its focal length, represented with the symbol f. By convention, concave, mirrors have positive focal lengths, while convex mirrors have negative ones. You see an example of a negative focal length in Concept 3. This, , Copyright 2007 Kinetic Books Co. Chapter 31, , 579
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convention allows the use of a single equation to locate images and objects for both, types of mirrors., Our study of mirrors will focus on paraxial rays. Paraxial rays are incident rays that are, relatively close to the principal axis. Rays that are far from the principal axis and do not, converge at a single image point are called nonparaxial rays. Nonparaxial rays cause, blurry images. This effect is called spherical aberration., For our purposes, the base of an object is located on the principal axis. When an, object’s base is on the principal axis, then the base of its image will be too, though it, may be inverted, like the image shown in Concept 4. The height of such an object or, image is measured from the principal axis. Like most quantities associated with objects, and images in optics, height has a sign. A positive value means the image is upright. Its, top is still on top. A negative sign means an image is inverted, as you see in Concept 4., Its base still lies on the principal axis, but its top is now below the axis., , Focal point (F), Image point of infinitely distant object, , Focal length (f ), Distance from mirror to focal point, ·Positive on object side, ·Negative on far side, , Height (h), Image point and height h describe, image, , 31.9 - Spherical mirrors: focal length equation, In Concept 1, you see a diagram of a concave mirror with center of curvature C and, radius of curvature r. A distant object and its image are shown. (The object and image, distances as well as the relative heights are not drawn to scale.) Because the object is, far away, the image point is at the focal point. The focal length f is the distance from the, mirror to the focal point., How does the focal length change as the mirror radius changes? Imagine you are using, a mirror to create an image of a distant object. A concave mirror with a very slight, curvature would correspond to a sphere with a large radius. If you consider the law of, reflection and how rays would reflect from this mirror, you will realize that the focal point, (and image) would be relatively far away from it. In contrast, a sharply curving mirror, would create an image quite close to the surface., The equation shown in Equation 1 quantifies this general relationship. The magnitude of, the focal length f equals one-half the radius rof the sphere. It is positive for a concave, mirror, and negative for a convex mirror., , 580, , Focal length, Positive for concave mirror, Negative for convex mirror, , Copyright 2007 Kinetic Books Co. Chapter 31
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Concave spherical mirror:, f = r/2, Convex spherical mirror:, f = í r/2, f = focal length, r = distance to center of curvature, , What is the mirror’s focal, length?, f = ír/2, f = í (0.50 m)/2, f = í 0.25 m, 31.10 - Parabolic mirrors, Parabolic mirrors provide a solution to the problem of, spherical aberration. Spherical aberration occurs, because some reflected rays do not pass precisely, through the focal point of a spherical reflector. The, farther the incident rays are from the mirror’s principal, axis, the more their reflected rays will miss the focal, point. The result is a blurry image., Spherical aberration is a serious problem for, astronomers and others who need high quality, images. Astronomers use telescopes to create, images of distant objects. These telescopes often, include very large concave mirrors to collect as much, light as possible, allowing them to create images of, objects that are exceedingly dim. (And by “large” we, truly mean large: The diameter of a modern reflecting, research telescope is in the eight to ten meter range.), If spherical mirrors were the basis of such telescopes,, the images they created would be blurry due to, spherical aberration., , This telescope mirror is a hexagonal mosaic of smaller spherical, mirror segments. Note the workman sitting at the focal point., , To produce a sharp image, some telescopes instead use a mirror that has a cross section in the shape of a parabola rather than a section of a, circle. Why? Reflected rays from distant objects converge to form a sharp image, no matter how far the incident ray is from the principal axis., , Copyright 2007 Kinetic Books Co. Chapter 31, , 581
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The parabolic design is used not just for visible light but for other forms of, electromagnetic radiation, like radio waves, as well., For instance, the dish antennas used for satellite television are small parabolic, reflectors. These antennas do not have to be as smooth as light reflectors because the, wavelength of the television signals they receive is far greater than the wavelengths of, visible light. For electromagnetic radiation of even longer wavelengths, like the longest, radio waves, the receiving parabolic reflectors can be even less smooth, but they must, be very wide. You see an example of a large radio telescope in Concept 3., Mirrors having a precise parabolic shape are difficult to manufacture. Some modern, optical telescopes, called mosaic telescopes, combine the best features of both, spherical and parabolic mirrors. They consist of a number of spherical mirror segments, arranged in an overall parabolic shape. The mirrors are cut into hexagons so that they, fit together neatly with no gaps. In the photograph above you see such a mirror., This arrangement works well because the individual spherical segments are small, enough to have negligible aberrations of their own and to be easy to assemble, and like, all spherical mirrors they are easy to manufacture. Yet each small mirror can be, positioned as part of a large overall parabolic shape that optimizes the net image quality, produced by the reflector., , Spherical mirrors, Suffer from spherical aberration, , Parabolic mirrors, Do not exhibit spherical aberration, , Parabolic reflectors in use, Common in telescopes and satellite, dishes, , 31.11 - Ray-tracing fundamentals, , What will be the nature of the, image?, , Ray tracing enables you to determine the fundamental properties of an image: Is it upright or inverted? Smaller or larger than the object? Real, or virtual?, Ray tracing also provides an effective tool for verifying the computational results you get from applying the mirror and lens equations. It is easy, to drop a sign, forget a reciprocal, or make other errors when applying these equations; ray tracing allows you to subject your answers to a, “reality check.”, In this section, ray tracing will be used to determine the basic attributes of an image that is produced by an object located farther from a, concave mirror than the center of curvature. You see this configuration above., , 582, , Copyright 2007 Kinetic Books Co. Chapter 31
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To locate the image, we start with three rays emanating from the top of the object: in this case, the top of the penguin’s head. The rays reach, the mirror at three different points. They then reflect and converge. The point at which they converge is the location of the top of the image, penguin’s head., With hand-drawn ray-tracing diagrams, we use the working approximation that the incident rays are paraxial, so that they converge at one, point. In the simulations in this chapter, you can observe accurate ray-tracing diagrams that display the effects of spherical aberration., Below, we show each ray separately; all three rays and the image are shown in Concept 5., Ray 1 starts as an incident ray that is parallel to the principal axis. It reflects off the mirror and passes through the focal point after it reflects., , Ray 1, Incident parallel to axis, reflects through, focal point, , Ray 2 starts as an incident ray that passes through the focal point and then reflects parallel to the principal axis., , Ray 2, Incident through focal point, reflects, parallel to axis, , Ray 3 begins as an incident ray that passes through the center of curvature, strikes the mirror perpendicularly, and reflects back, moving along, the same line as the incident ray., , Ray 3, Incident through center of curvature,, reflects straight back, , Why do these rays reflect in the way they do? The first incident ray is parallel to the principal axis. By definition, the reflection of this ray passes, through the focal point., The second ray is parallel to the principal axis after it reflects. The definition of focal point applied “in reverse” explains why this should be so:, Since the ray passes through the focal point before it reflects, it is parallel to the axis after it reflects., The final ray reflects straight back because, like a radius of the sphere, it meets the spherical surface perpendicularly. Its angle of incidence is, zero degrees, so its angle of reflection will be the same., , Copyright 2007 Kinetic Books Co. Chapter 31, , 583
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Now that the three rays have been drawn separately, we draw them together in Concept 5. They converge at the head of the image. The image, is both inverted (upside down) and smaller than the original object. It is located between the focal point and the center of curvature. The image, is real, being located on the same side of the mirror as the object. Since it is real, if a piece of paper were placed at the point of intersection of, the reflected rays, a small inverted image would be projected onto it., , Image, Located at intersection of rays, Inverted, smaller, real, , 31.12 - Ray tracing: a second example, , What will be the nature of the, image?, , We will use ray tracing to determine the nature of the image created by an object that lies within the focal point of a concave mirror. As with the, rays in the prior example, the incident rays all start or pass through a point at the top of the object, the penguin. They also have the same basic, properties as the previous rays: One is parallel to the principal axis, one passes through the focal point, and the third passes through the center, of curvature., In contrast to the prior example, the reflected rays do not converge. We will need to use virtual rays to locate the image., Ray 1 starts parallel to the principal axis, reflects, and passes through the focal point. Note that we extend the reflected ray backward through, the mirror surface as a virtual ray (dashed line)., , Ray 1, Incident parallel to axis, , Ray 2 must pass through the focal point before reaching the mirror. We draw it as passing through the focal point before intersecting the, penguin’s head. It then strikes the mirror and reflects parallel to the principal axis., , Ray 2, Passes through focal point, , 584, , Copyright 2007 Kinetic Books Co. Chapter 31
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Ray 3 starts at C. It passes by the penguin’s head and reflects back through C., , Ray 3, Passes through center of curvature, , In Concept 5, you see all three rays in the same diagram. The real rays do not converge; to locate a point of convergence we must use the, virtual rays shown in the illustration., , Virtual rays, Extend rays backward through mirror, , The point where the virtual rays converge, behind the mirror, is where the top of the image is located. We show the image in Concept 6. It is, upright, larger than the object, and virtual., , Image, Upright, larger, virtual, , When you look at yourself in a concave makeup mirror, you put your face within its focal point to see a magnified image of yourself., , 31.13 - Ray tracing: a convex mirror, , What will be the nature of the, image?, , In this section, we construct a ray-tracing diagram for the image created by a convex mirror. The focal point and the center of curvature are, always behind the reflecting surface of a convex mirror. We use three rays with the same essential properties as before, modifying their, construction to compensate for the fact that the two points are on the opposite side of the mirror from the object., , Copyright 2007 Kinetic Books Co. Chapter 31, , 585
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Ray 1. Ray 1 is incident parallel to the principal axis. If we extend the reflected component of this ray backward through the mirror, the virtual, ray will pass through the focal point., , Ray 1, Incident parallel to axis, , Ray 2. Instead of passing through the focal point, the incident part of ray 2 is directed toward it. Before it can reach the focal point behind the, mirror, it reflects parallel to the principal axis. Its virtual extension behind the mirror is also parallel to the axis., , Ray 2, Incident in direction of focal point, , Ray 3. The incident component of Ray 3 is directed toward the center of curvature on the far side of the mirror and reflects back along the, same line. The virtual extension of the reflected ray passes through the center of curvature., , Ray 3, Incident towards center of curvature, , In the following illustration we draw all three rays together. There is no intersection of the actual reflected rays in front of the mirror, but their, backward virtual extensions intersect behind the mirror., , Virtual rays, Extend rays backward through mirror, , 586, , Copyright 2007 Kinetic Books Co. Chapter 31
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Finally, in Concept 6, we show the image as determined by the convergence point of the virtual rays. The image formed is upright, smaller than, the original object, and virtual. Note that this is the only possible result for a convex mirror., , Image, Upright, smaller, virtual, , This ray-tracing diagram can be used to explain why security and automobile passenger-side mirrors are often convex. The mirror forms, viewable images when the corresponding objects extend quite far above (or below) the principal axis, providing a very wide field of view. As, you know, mirrors like this, typically used on the passenger side of vehicles, are often labeled with safety warnings that state, “Objects in mirror, are closer than they appear.”, However, if you look at concept 6, you will note that the object is actually farther from the mirror than the virtual image is, and this is always true, for images formed by a convex mirror. So are the warnings wrong? No, the convex mirror’s virtual image is always smaller than the image that, would be produced by a planar mirror. Since the human brain relates size to distance, the image appears distant: We do interpret the object as, being farther away than it really is., , 31.14 - Interactive problem: image in a convex mirror, Here you use an interactive simulation to view the image produced by a convex, mirror. You drag the object farther from and closer to the mirror and observe its, image. You can turn on ray tracing with the SHOW RAYS button, and see how the, rays can be used to determine the location of the image., Some questions to consider include: Is the image ever inverted? Is it ever larger, than the object? Is it ever real? Is it ever farther from the mirror than the object is?, The object can be moved to positions that allow you to answer all these questions., , 31.15 - Mirror equations, , Variables in the mirror equations, Interpretation of signs, , In this section, we discuss several equations used to determine the nature of images produced by mirrors. Before doing so, we will review and, explain the mathematical sign conventions used in these equations., Some of the notation we have already introduced: d is used for distance, f for focal length, and h for height. More specifically, do represents, the distance of the object from the mirror and di the distance of the image from the mirror. Also, ho is the height of the object and hi is the, height of the image., With a single mirror, the object distance is always positive, although it can be negative in the more complex optical systems discussed below., The image distance is positive when the image is real, on the same side of the mirror as the object from which the image is created. The image, distance is negative when the image is virtual, on the opposite side of the mirror. Remember that the focal length is positive for concave mirrors, and negative for convex mirrors. The image height is positive for upright images and negative for inverted ones., The magnification, represented by m, equals the image height divided by the object height. When the magnification is positive, the image is, upright; when it is negative, the image is inverted relative to the object. Sometimes, the magnification as defined here is called lateral, magnification to make it clear that it represents the change in height. When m is greater than one, then the image is larger than the object:, , Copyright 2007 Kinetic Books Co. Chapter 31, , 587
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what people ordinarily think of as “magnified.”, Now, on to the equations. The mirror equations are very useful for designing equipment, such as cameras and telescopes because they enable engineers to correctly focus an, image at a required point (for example, at the surface of a photographic film, or in a, digital camera, at the surface of a “charge-coupled device”). The formula in Equation 1, shows the relationship of object and image distance to the focal length of a mirror. It is, valid when the incident rays from the object are paraxial., The next two equations, shown in Equation 2, are used to define and calculate, magnification. The first equation defines magnification: It is the ratio of the image and, object heights. The magnification can also be calculated as the negative of the ratio of, image distance to object distance., If you now refer back to the table in Concept 1 for signs, you may see one entry that, surprises you: an object with a negative distance. Such a “virtual object” can be created, by a configuration of two mirrors, or by a lens and a mirror, as shown in the illustration, below., , Mirror equation, , di = image distance, do = object distance, f = focal length, , Incident rays from the original real object first reflect off some mirror, or pass through, some lens, which is not shown in the illustration, being off screen to the left. As a result,, converging rays come in from the left and strike the convex mirror in the diagram. The, image they create would be a real image “behind” the location of the mirror if the mirror, were not there. Since the mirror is there, the convergence point of the virtual rays on the, right defines the location of a virtual object, whose distance from the mirror is negative., The incident rays reflect off the front of the mirror to create a real image, with a positive, image distance, as shown. Since optical instruments such as telescopes and, microscopes often rely on a combination of lenses and mirrors, negative object, distances are not rare., , Magnification equations, , m = magnification of image, hi = height of image, ho = height of object, , 588, , Copyright 2007 Kinetic Books Co. Chapter 31
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What is the distance to the, object?, di = í2.0 m, f = í6.0 m, , do = 3.0 m, , What is the magnification?, ho = +0.60 m, hi = í 0.40 m, m = hi /ho, m = í 0.40 m / 0.60 m, m = í 0.67, , Copyright 2007 Kinetic Books Co. Chapter 31, , 589
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31.16 - Interactive checkpoint: mirror equations, A frog sits in front of a convex, spherical mirror. The mirror produces, an image of the frog with a lateral, magnification of 1/4. The mirror’s, radius of curvature is 0.260 m. How, far away is the frog from the mirror?, , Answer:, , do =, , m, , 31.17 - Interactive problem: optical bench with a mirror, The simulations here challenge you to use the mirror equations. In the first, you are, asked to create an image 13.7 cm from the mirror’s surface as shown in the, illustration and in the simulation. The object is 11.2 cm from the mirror., You set the focal length of the mirror by dragging the focal point F, or by setting its, value in the control panel. As you change the focal length, you will also be changing, the curvature of the mirror. Once you believe you have set the focal length correctly,, press the CHECK button to test your answer. Be careful with signs!, Your mission in the second simulation to the right is to create a mirror that will have, a magnification of +0.470 for the same object at the same position. Again, you, control the focal length, and to test your answer, press the CHECK button in this, simulation. As a hint: First, determine where the image must be using one formula,, then use another formula to determine the required focal length., If you have trouble with either of these tasks, review the prior section on the mirror, equations., , 31.18 - Gotchas, Virtual images do not exist. This may be a philosophical question. They do exist in our minds. They cannot be touched, or projected onto a, piece of paper. In contrast, real images can be projected onto a screen or a piece of paper., A convex and a concave mirror have the same radius of curvature. This means their focal lengths are identical. The magnitudes of their focal, lengths are the same, but the signs differ. The focal length is positive for a concave mirror and negative for a convex mirror., A virtual image has a negative image distance. Yes. Conversely, real images have positive image distances., , 590, , Copyright 2007 Kinetic Books Co. Chapter 31
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31.19 - Summary, Reflection occurs when light bounces back from a surface, rather than passing, through it or being absorbed. Reflection causes objects to be visible., , Law of reflection, In optics, it is often useful to represent the path of light in the form of a light ray, a, straight line that can be used to determine how light is affected by mirrors and, lenses., Planar mirrors are the simplest and most common type of mirror. In a planar mirror,, your image has the same dimensions as you do and it appears to be the same, distance behind the mirror as you are in front of it. The image is reversed front to, back, but not right to left or top to bottom., When you see your image in a planar mirror, you are seeing a virtual image, so, called because you cannot project this image onto a screen at its perceived, location. A virtual image appears to be on the opposite side of a mirror from the, object it is created from, and its image distance is negative., In contrast, a real imagecan be projected onto a screen, and it occurs on the same, side of a mirror as the object it is created from. Its image distance is positive., Curved mirrors can create virtual or real images., , și = șr, Focal length of concave mirror, , f = r/2, Focal length of convex mirror, , f = –r/2, Mirror equation, , The law of reflection states that for a light ray reflecting off a mirror, the angle of, incidence equals the angle of reflection. These two angles are measured between, the light ray and a line normal to the surface of the mirror., A ray diagram can help you determine the location of the image an object casts in a, planar mirror. Draw two incident rays from any point on the object to the mirror., Draw the reflected rays using the law of reflection, and trace their virtual extensions, backward behind the mirror as dashed lines. The point of the image corresponding, to the point source of the incident rays on the object is located where these virtual, rays intersect., , Magnification equations, , Spherical mirrors have a constant curvature. A mirror is called concave if it curves toward (around) the object it reflects. A convex mirror curves, away from the object it reflects., All points on a spherical mirror are equidistant from its center of curvature. The principal axis is a line through the center of curvature and the, midpoint of the mirror. Paraxial rays are incident rays close to the principal axis of a mirror. The reflections of these rays converge at the focal, point if the rays are parallel to the principal axis. The height of an image is negative if the image is inverted., The focal length of a curved mirror is the distance between a mirror and its focal point. By convention, it is positive for a concave mirror where, the focal point is on the same side of the mirror as the object. It is negative for a convex mirror where the focal point is on the opposite side., Spherical mirrors exhibit spherical aberration because only paraxial incident rays converge at the focal point. With a parabolic mirror, all, incident rays that are parallel to the principal axis converge at the focal point., Ray tracing for concave and convex mirrors is more complicated than for planar mirrors, but it can help you determine the orientation, relative, size, and location (or type) of an image., The mirror equation relates the distances of an object and its image from a mirror to the focal length of the mirror. The magnification equations, define the magnification of an image created by a mirror. If the magnification is known, we may calculate the image height from the object, height or the image distance from the object distance, or vice versa., , Copyright 2007 Kinetic Books Co. Chapter 31, , 591
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Chapter 31 Problems, , Conceptual Problems, C.1, , An object is placed between the focal point and center of curvature of a concave mirror. Draw a ray diagram to determine the, nature of the image. Check all the boxes that describe the image., Real, Virtual, Smaller, Larger, Upright, Inverted, , C.2, , An object is placed between a concave mirror and its focal point. Draw a ray diagram to determine the nature of the image., Check all the boxes that describe the image., Real, Virtual, Smaller, Larger, Upright, Inverted, , C.3, , An object is placed in front of a convex mirror. Draw a ray diagram to determine the nature of the image. Check all the boxes, that describe the image., Real, Virtual, Smaller, Larger, Upright, Inverted, , C.4, , An object is placed 20.0 cm in front of a concave mirror with focal length 9.00 cm. Draw a ray diagram to determine the, nature of the image. Check all the boxes that describe the image., Real, Virtual, Smaller, Larger, Upright, Inverted, , C.5, , If you want to create a real image, which object location and mirror combinations could you use? Check all that apply., Object in front of convex mirror, Object at distance over twice the focal length from concave mirror, Object between focal point and center of curvature of concave mirror, Object at distance less than focal length from concave mirror, , C.6, , You hold up your index finger very close to the surface of a concave mirror (inside the focal point). (a) As you move your, finger farther from the mirror, does the image of your finger get larger or smaller? (b) What happens to the image when the, distance between your finger and the mirror exceeds the focal length of the mirror? Select all that apply., (a), , Larger, , (b), , Stays the same, , Smaller, , Inverts, Becomes virtual, Becomes real, , Section Problems, Section 0 - Introduction, 0.1, , 592, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Where can you place the, object so that the image and object are on the same side of the mirror? (b) Where can you place the object so that the image, is smaller than the object? (c) Where can you place the object so that the image is larger than the object? Check all that, , Copyright 2000-2007 Kinetic Books Co. Chapter 31 Problems
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apply., (a), , To the left of C, In between C and F, To the right of F, , (b), , To the left of C, In between C and F, To the right of F, , (c), , To the left of C, In between C and F, To the right of F, , Section 3 - Mirror basics, 3.1, , As you get ready for your early morning physics class, you comb your hair in front of a planar mirror. You know that you are, 3.75 m in front of the mirror. How far away from you is your image?, m, , 3.2, , Two parallel planar mirrors face each other. An object is placed between them at a distance x from one and 3x from the, other. If the images created of the object are "first-order" images, and those created of the first-order images are "secondorder" images, how far away are the second-order images from the object, in terms of x?, x, , Section 5 - The law of reflection, 5.1, , Two mirrors are placed at a 90° angle to each other. A light ray strikes one mirror 0.710 m from the intersection of the mirrors, with an incident angle of 36.5°. The ray then travels a distance d before reflecting from the second mirror. (a) What is the, distance d? (b) What is the angle with which the light ray is reflected from the second mirror? All angles are measured from, the normal belonging to the appropriate mirror. (c) What is the angle between the direction of the incident ray and that of the, final reflected ray? Express your answer as a positive number between 0° and 360°., (a), , 5.2, , m, , (b), , °, , (c), , °, , You and a friend are playing a game of squirtgun tag in a maze. Suddenly you see your, friend's image in a small planar mirror (see, illustration). You take a shot over the barrier in, front of you and find that your friend is just at, the end of the 7.0 m range of your squirt gun., If you are 4.0 m from the point of reflection of, the light ray, how far from the point of, reflection of the light ray is your friend?, m, , 5.3, , Two 3.0 meter-tall planar mirrors are placed facing and parallel to each other 0.75 m apart. If a ray of light just passes the, bottom edge of one mirror and strikes the other at an incident angle of 10°, how many reflections will the light make before, exiting this "hall"?, , 5.4, , A bartender keeps an eye on her customers while her back is turned by way of a mirror mounted behind the bar. The mirror is, 0.580 m wide and she is standing 0.750 m from it. How wide an expanse of the parallel wall opposite the mirror can she see, in the mirror if the opposite wall is 6.00 m away from the mirror?, m, , Copyright 2000-2007 Kinetic Books Co. Chapter 31 Problems, , 593
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5.5, , Carlota is standing 3.50 m from a wall. Her, friend Lily is standing 2.10 m from the same, wall, and they are 4.40 m from each other. If a, mirror is mounted on the wall with its edge, directly in front of Lily and extending toward, Carlota as in the illustration, how wide must, the mirror be for the friends to see each other, in the mirror?, m, , Section 9 - Spherical mirrors: focal length equation, 9.1, , A shiny metallic Christmas ball ornament 4.60 cm in diameter is used as a mirror. What is the mirror's focal length?, , 9.2, , You want to create a spherical mirror with a focal length of +15.0 cm. (a) Will the mirror be concave or convex? (b) What, must be the radius of curvature?, , cm, , (a), , Concave, , (b), 9.3, , Convex, , cm, , You are lost in the wilderness, carrying only a piece of aluminum foil. Thanks to diligent study in your biology class, you, remember which mushrooms are poisonous, and you are able to pick others that won't kill you. Your talents in basket weaving, also pay off magnificently, as you create a basket and devise a plan to line it with aluminum foil, making a solar cooker., Mmmm, roast mushrooms! The basket is a section of a sphere with radius 21.0 cm. How far from the bottom of the basket, should you place the mushrooms to cook them most efficiently using the Sun's rays?, cm, , Section 14 - Interactive problem: image in a convex mirror, 14.1 Use the simulation in the interactive problem in this section to answer the following questions. (a) Is the image ever inverted?, (b) Is the image ever larger than the object? (c) Is the image ever real? (d) Is the image ever farther from the mirror than the, object is?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , (d), , Yes, , No, , Section 15 - Mirror equations, 15.1 An object is 17.3 cm from a concave mirror of focal length 8.10 cm. (a) What is the image distance? State your answer with, the correct sign. (b) What is the magnification? (c) Is the image upright or inverted?, (a), , cm, , (b), (c), , i. Upright, ii. Inverted, , 15.2 Use the mirror equation to show that for a planar mirror, the magnitude of the image distance equals the magnitude of the, object distance. Hint: A planar mirror can be thought of as a spherical mirror with an infinite radius of curvature., 15.3 The focal point of a convex mirror is 22.4 cm from the mirror's surface. An image seen in the mirror is between the mirror and, the focal point, 12.9 cm from the focal point. How far from the mirror is the object?, cm, 15.4 A mirror creates an image of magnification í1/3. (a) Is the image upright or inverted? (b) If the object is 7.80 cm tall, what is, the height of the image? State your answer with the correct sign., (a), (b), , 594, , Upright, , Inverted, cm, , Copyright 2000-2007 Kinetic Books Co. Chapter 31 Problems
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15.5 A mirror produces an inverted image 2.80 cm tall when a 5.60 cm tall object is 25.0 cm away from it. (a) What is the image, distance? State your answer with the correct sign. (b) What is the focal length of the mirror? State your answer with the, correct sign., (a), , cm, , (b), , cm, , 15.6 You wish to use a concave mirror to produce a virtual image that is three times the size of an object. (a) Which will be closer, to the mirror, the image or the object? (b) If the object is 22.0 cm from the mirror at this magnification, what is the radius of, curvature of the mirror?, (a), , (b), , i. Image, ii. Object, iii. They are equidistant from the mirror, cm, , 15.7 A planar mirror is placed at the origin on a number line with the reflective side facing the positive direction. A convex mirror, with focal length í22.0 cm is placed at x = 35.0 cm with its reflective side facing the planar mirror. An object is placed at, x = 19.0 cm. What is the distance between the image of the object created by the planar mirror and the image of the object, created by the convex mirror?, cm, 15.8 A mirror of focal length 22.6 cm creates an image with magnification í0.350. What is the image distance?, cm, 15.9 An object and its image are both 12.4 cm from a spherical mirror, and on the same side of the mirror. (a) Is the mirror, concave or convex? (b) What is the mirror's focal length? State your answer with the correct sign., (a), , Concave, , (b), , Convex, , cm, , 15.10 You look into a shaving mirror and the upright image of your face is 40.0 cm away from you. The image has a magnification, of 1.50. (a) How far away are you from the mirror? (b) What is the radius of curvature?, (a), , cm, , (b), , cm, , 15.11 You place an object 34.0 cm in front of a convex mirror and the image produced is half the height of the object. What is the, focal length of the mirror? State your answer with the correct sign., cm, 15.12 A convex mirror has a radius of curvature of 36.0 cm. When a 5.60 cm tall object is placed in front of the mirror, the, magnification is 0.250. (a) What is the height of the image? (b) How far is the object from the mirror?, (a), , cm, , (b), , cm, , Section 17 - Interactive problem: optical bench with a mirror, 17.1 Use the information given in the first interactive problem in this section to calculate the focal length required for the image to, appear at the desired location. Test your answer using the simulation., cm, 17.2 Use the information given in the second interactive problem in this section to calculate the focal length required to produce, the desired magnification. Test your answer using the simulation., cm, , Copyright 2000-2007 Kinetic Books Co. Chapter 31 Problems, , 595
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32.0 - Introduction, Light can refract í change direction í as it moves from one medium to another. For, instance, if you stand at the edge of a pool and try to poke something underwater, with a stick, you may misjudge the object’s location. This is because the light from, the object changes direction as it passes from the water to the air. You perceive the, object to be closer to the surface than it actually is because you subconsciously, assume that light travels in a straight line., Although refraction can cause errors like this, it can also serve many useful, purposes. Optical microscopes, eyeglass lenses, and indeed the lenses in your, eyes all use refraction to bend and focus light, forming images and causing objects, to appear a different size or crisper than they otherwise would. Where a lens, focuses light, and whether it magnifies an object, is determined by both the, curvature of the lens and the material of which it is made. Scientists have developed, quantitative tools to determine the nature of the images created by a lens. We will, explore these tools thoroughly later, “focusing” first, so to speak, on the principle of, refraction underlying them., To begin your study of refraction, try the simulation to the right. Each of your, helicopters can fire a laser í a sharp beam of light í at any of three submarines lurking under the sea. The submarines have lasers, too, and, will shoot back at your craft. Your mission is to disable the submarines before they disarm your helicopters. When you make a hit, you can, shoot again. Otherwise, the submarines get their turn to shoot until they miss., You play by dragging the aiming arrow underneath any one of your helicopters. Press FIRE and the laser beam will follow the direction of this, arrow until it reaches the water, where refraction will cause the beam to change direction., In addition to hitting the submarines before they get you, you can conduct some basic experiments concerning the nature of refraction. As with, reflection, the angle of incidence is measured from a line normal (perpendicular) to a surface. In this case, the surface is the horizontal, boundary between the water and the air. Observe how the light bends at the boundary when you shoot straight down, at a zero angle of, incidence, or grazing the water, at a large angle of incidence. You can create a large angle of incidence by having the far right helicopter, for, example, aim at the submarine on the far left., You can also observe how refraction differs when a laser beam passes from air to water (your lasers) and from water to air (the submarines’, lasers). Observe the dashed normal line at each crossover point and answer the following question: Does the laser beam bend toward or away, from that line as it changes media? You should notice that the laser beams of the submarines behave differently than those of the helicopters, when they change media., As a final aside: You may see that some of the laser beams of the submarines never leave the water, but reflect back from the surface between, the water and the air. This is called total internal reflection., , 32.1 - Refraction, , Refraction: The change in the, direction of light as it passes, from one medium to another., A material through which light travels is called a, medium (plural: media). When light traveling in one, medium encounters another medium, its direction can, change. It can reflect back, as it would with a mirror. It, can also pass into the second medium and change, direction. This phenomenon, called refraction, is, shown to the right. In the photo, a beam of light from, a laser refracts (bends) as it passes from the air into the water., , Refraction bends water waves into line with the shore., , Light refracts when its speeds in the two media are different. Light travels faster through air than in water, and it changes direction as it moves, from air into water, or from water into air., Although we are primarily interested in the refraction of light, all waves, including water waves, refract. Above, you see a photograph of surf, wave fronts advancing parallel to a beach. Deep-ocean swells may approach a coastline from any angle, but they slow down as they encounter, the shallows near the shore. The parts of a wave that encounter the shallow water earliest slow down first, and this causes the wave to refract., Sound waves can also refract. During a medical ultrasound scan, an acoustic lens can be used to focus the sound waves. The lens is made of, a material in which sound travels faster than in water or body tissues., The surface between two media, such as air and water, is called an interface. As with mirrors, light rays are often used to depict how light, refracts when it meets an interface. Lasers are often used to demonstrate refraction because they can create thin beams of light that do not, , 596, , Copyright 2007 Kinetic Books Co. Chapter 32
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spread out. These light beams are physical analogues of light rays., , Refraction, Change in wave direction at interface, Caused by change in speed of wave, , 32.2 - Index of refraction, , Index of refraction of a material: The speed of, light in a vacuum divided by the speed of light, in the material., Light travels at different speeds in different materials. The index of refraction of a, material provides a measure of the speed of light in that material. The symbol n, represents the index of refraction. The equation used to calculate the index of refraction, for a material is shown to the right. The table to the right shows the indices for some, common materials., In the equation for the index of refraction, the speed of light in a vacuum is in the, numerator. Light travels at its maximum speed in a vacuum, so the index of refraction, for other materials is always greater than one., , Index of refraction, Speed in vacuum / speed in material, , The index of refraction depends on the wavelength of the light. The table in Equation 2, is for yellow light with a wavelength of 589 nm. The index depends on the wavelength, because the speed of light in a material depends on its wavelength. The speed of light, in any material is slower for shorter wavelengths, so the index of refraction is greater for, light of shorter wavelengths. For example, for a type of glass called crown glass,, n = 1.50 for red light, and n = 1.53 for violet light, which has a shorter wavelength, than red light., The index of refraction of a material also depends somewhat on its temperature. For, certain crystalline materials, it also depends on the angle at which the light travels, through the crystal lattice. This is why the index of refraction for corundum is indicated, as approximate in the table., The index of refraction for visible light in the Earth’s atmosphere is about 1.0003 at, standard temperature and pressure. Since we typically use two or three significant, figures, we treat that index as 1.00. At the same level of precision, the speed of light is, 3.00×108 m/s., , n = index of refraction, c = speed of light in vacuum, v = speed of light in material, , Copyright 2007 Kinetic Books Co. Chapter 32, , 597
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Indices of refraction, , Green light travels at 1.99×108, m/s in crown glass. What is the, index of refraction of the glass, for this light?, , n = 1.51, 32.3 - Snell’s law, Snell’s law is used to quantify refraction. In doing so, it uses some of the same, terminology as the law of reflection. As with reflection, angles are measured between a, ray and a line normal (perpendicular) to a surface. You see this illustrated in Concept 1,, with both the angle of incidence (și ) and the angle of refraction (șr ) shown., Snell’s law, shown in Equation 1, expresses the relationship between these angles. This, law was discovered empirically by Willebrord Snell and written in its current form by, René Descartes. It states that the product of the sine of the incident angle and the index, of refraction of the incident medium equals the product of the sine of the refraction, angle and the index of refraction of the refracting medium., We also show the same equation in an alternate formulation: The ratio of the sine of the, incident angle to the sine of the refracted angle is the reciprocal of the ratio of the first, to the second index of refraction., To put it more concretely, light bends toward the normal when it slows down, for, instance, when it passes from air to water. You see this in Concept 1 to the right. It, bends away from the normal when it speeds up, as from water to air., , Snell’s law, Quantifies refraction, Slower light bends toward normal, , If light passes through several media, Snell’s law can be applied at each interface. You see this occurring in Concept 2. The light’s direction, changes toward the normal at the first interface as it slows, and then away from the normal as it crosses the second interface. It bends away, because light moves faster in the third medium, which has a lesser index of refraction than the second., , 598, , Copyright 2007 Kinetic Books Co. Chapter 32
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32.4 - Interactive checkpoint: Snell’s law, What is the index of refraction of the, second material?, , Answer:, , n2 =, , 32.5 - Everyday effects of refraction, The refraction of light can cause interesting and sometimes confusing results as the, brain interprets the position of objects it sees via refracted light. For instance, in the, upper illustration to the right, you see an ancient Egyptian fisherman trying to spear a, fish. The solid line indicates the refracted path of the light traveling from the fish to his, eyes. Since his brain expects light to travel in a straight line, he projects the fish to be in, the position indicated at the end of the dashed line, which causes him to think the fish is, nearer to the surface than it is. Experienced spear fishermen know how to compensate, for this effect., The fisherman is not the only one to experience the effects of refraction. The fish does,, too. The light from the world above seen by the fish also refracts. Light coming straight, down will pass through the water’s surface unchanged, but light at any angle will refract,, in the process giving the fish a wider field of view of the world above than it would have, if there were no refraction. In essence, the fish sees a compressed wide-angle view of, the scene above. Certain camera lenses, appropriately called fisheye lenses, can, create the same effect, as illustrated in Concept 2., , Refraction confuses fisherman, He misjudges position of fish, , Another interesting consequence of Snell’s law concerns light passing through a, window. The light refracts as it travels from air to glass. After passing through the glass,, it refracts again at the second interface. The ratio of the indices of refraction is now, reversed, so the initial change in angle is cancelled out. The light ray that emerges is, parallel to the initial ray, but displaced a small amount. The amount of displacement is, small enough that we ordinarily do not notice “window shift.” However, if you place a, newspaper page on a tabletop and cover half of it with a flat pane of glass, you will be, able to observe the displacement effect., , A fisheye view, Wide-angle view compresses images, , Light through a windowpane, Light refracts twice, , 600, , Copyright 2007 Kinetic Books Co. Chapter 32
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Emerges parallel to initial ray, , 32.6 - Why refraction occurs, Why do waves change direction when they change speed? Here we offer a mechanical, analogy to explain this phenomenon., Consider the diagram in Concept 1 to the right. It uses the example of a car, encountering ice on the side of a road. If this has ever happened to you, you know that, the car can dangerously change direction when one (or more) of its wheels encounters, ice, as shown in the diagram. Why is this so? The wheels on the ice provide less, traction, so the right side of the car moves more slowly. The left side of the car, continues to move at the same speed as before, which causes the car to rotate and, veer off the road. When all the wheels supplying power are on the ice, the car will once, again move straight ahead because both sides will be moving at the same speed., Why light refracts can also be explained using Fermat’s principle of least time. This, principle, developed by French mathematician Pierre de Fermat (1601í1665), states, that light will travel the path between two points that requires the least amount of time., This may seem like it should be a straight line, but it is not when the speed varies along, the path between two points., , Car hits ice, Speed change on one side changes, direction of car, , This timesaving “technique” is similar to what you would intuitively do if you were standing on a beach and saw a swimmer floundering, desperately in the water some distance down the shore. Instead of taking a direct straight-line path to the rescue, you would run along the, beach for some distance, counting on your greater land speed, before taking to the water., , 32.7 - Interactive problem: helicopter and submarines, The simulation on the right is similar to the one in this chapter’s introduction. As, before, submarines lurk under the waves, but now you have only a single helicopter., As before, you have a laser you can aim in an attempt to disable two submarines, before they disable you. You keep shooting as long as you keep making hits. To, shoot your laser, aim it by dragging the aiming arrow and fire it by clicking on FIRE., The angle of incidence is shown in an output gauge., As soon as you miss, it is the submarines’ turn. Warning: The computer has been, set to be far more accurate in this game than in the introductory one. Unless you are, very precise with your shots, it is unlikely you will win., However, there is good news: Now you have more intellectual firepower because, you have the aid of Snell’s law. You are also given some assistance from an able, comrade; she has computed the angles of refraction required for your laser to reach, the submarines, as shown in the diagram. You should use 1.33 for the index of, refraction of water, and 1.00 for the index of refraction of air. If you correctly set the, angle of incidence when you aim each of your shots, you can make two straight hits, and disable the submarines before they disable you., , 32.8 - Total internal reflection, , Total internal reflection: Light reflects, completely at an interface, back into the, medium with the higher refractive index., Critical angle: The minimum angle of incidence, at which total internal reflection occurs., Total internal reflection means no light passes from one medium to another. On the, right, we show how this occurs using the example of an underwater flashlight shining a, beam at the interface between water and air. As the diagram shows, all the light reflects, back into the medium with the higher refractive index, the water. No light passes into, the air above the water., , Total internal reflection, No incident light leaves initial medium, Light is reflected at interface, , Why does this occur? Consider what happens when light is directed from water into air., Water has a greater index of refraction than air. At relatively small angles of incidence, light passes from the water into the air, and as it does, so, it refracts away from the normal. The angle of refraction is greater than the angle of incidence., As the angle of incidence increases, the angle of refraction will increase as well, and it will always be greater than the angle of incidence. At a, , Copyright 2007 Kinetic Books Co. Chapter 32, , 601
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sufficient angle of incidence, the angle of refraction reaches 90°, perpendicular to the, normal and parallel to the surface of the water. The light no longer crosses the interface, but remains in the water. The minimum angle of incidence at which this occurs is called, the critical angle., Even at angles of incidence less than the critical angle there is some internal reflection, along with refraction. But when the critical angle is exceeded, there is no refraction at, all: All of the light is internally reflected., Total internal reflection is shown in Concept 1. All the light is reflected back into the, water. Concept 2 shows the situation when the angle of incidence is equal to the critical, angle. The critical angle depends on the ratio of the indices of refraction of the two, media. You see this as the equation on the right, which is obtained by setting șr = 90º, (so sin șr = 1) in Snell’s law., Diamonds rely on total internal reflection to achieve their sparkle. They are cut so that, light entering them reflects internally and emerges only at certain points, giving the, effect of scintillating light., , Critical angle, Minimum incident angle for total internal, reflection, , If you happen to have a diamond ring handy, you can demonstrate this dependence on, the ratio of indices by placing the ring in water. The sparkle disappears; this is because, the indices of refraction of diamond and water do not differ as significantly as those of, diamond and air, so the geometry of the ring no longer causes the same internal, reflection., Engineers use total internal reflection in many different applications. For instance, right, isosceles prisms are used in binoculars to redirect light, as in the illustration below., , Fiber optic cable is used to transfer information (in glass or plastic) using total internal, reflection. Data (such as speech) is first encoded as modulations of a beam of laser, light. The light remains inside the transparent cable as it travels due to its total internal, reflection off the inner surfaces of the cable walls. In this way, light is transmitted, through the cable with little loss., , șc = critical angle, n1, n2 = indices of refraction (n1 > n2), , Using light instead of electricity to transfer information has several benefits. Light can be, used to encode much more information than electric oscillations because of its, extremely high frequency (more than 1014 Hz for red light), and fiber optic cable is, immune to interference problems from nearby electrical applications., , You are cutting a sapphire to, make it as brilliant as possible., Find the critical angle for the, sapphire in air. The index of, refraction of a sapphire is 1.77., , 602, , Copyright 2007 Kinetic Books Co. Chapter 32
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32.9 - Dispersion and prisms, , Dispersion: Refraction that, causes light to separate into, its various wavelengths., Humans perceive different wavelengths of light as, different colors. When we see light of a single, wavelength, we perceive it as a pure color such as, red, green or violet. The light that comes from the Sun, and from standard light bulbs, which we perceive as, white light, consists of a mixture of many different, wavelengths of electromagnetic radiation., A rainbow provides an example of dispersion., , Prisms, such as the one shown to the right, separate, white light into its many wavelengths. This is called dispersion, or to be more specific,, chromatic dispersion, since dispersion can also occur with non-visible electromagnetic, radiation and mechanical waves as well. Sir Isaac Newton famously used a pair of, prisms to disperse white sunlight into colors and recombine the colors into white light., Prisms disperse light because the index of refraction depends on the light’s wavelength., The index of refraction for a given material is greater for waves of shorter wavelength,, so blue and violet light refract more than red or orange light. When white light is incident, upon a prism, the different wavelengths that make it up are refracted at different angles,, resulting in a rainbow of colors., The rainbows we see in the sky are also caused by dispersion. Light disperses as it, enters a spherical raindrop. Total internal reflection occurs inside the raindrop, and the, light then refracts again when leaving the drop, dispersing even more. You see this, illustrated in a raindrop to the right., Why does this process cause rainbows? Raindrops reflect back nested “cones” of light, of different colors. In Concept 3 you see how red light reflected back from every drop in, the outermost portions of the rainbow reaches the observer, but the violet light from, those drops passes him by. You also see how the violet light from the innermost, portions of the rainbow reaches him, but not the red light. Other topics in rainbow theory, are still being researched. For instance, physicists, although vaguely alluding to, quantum effects and the promise of string theory, have to date offered no convincing, explanation for the pot of gold at the end of the rainbow., , Dispersion, Creation of a spectrum by refraction, Different wavelengths refract differently, , Rainbow: inside a raindrop, Refraction plus total internal reflection, Colors refract at different angles, , Observer views rainbow, Red light from outer drops reaches him, Violet light from inner drops reaches him, , Copyright 2007 Kinetic Books Co. Chapter 32, , 603
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32.10 - Gotchas, You calculated a material as having a refractive index of 0.76. Oops! This is not possible. The speed of light in a vacuum is greater than its, speed in any medium, meaning that the refractive index of the material must be greater than 1.00. Remember that when calculating the index, of refraction, you put the speed of light in a vacuum in the numerator, not the denominator., Light strikes water in a placid lake at an angle of 30° from the horizontal, so its incident angle is 30°. No, the incident angle is 60°. It is, measured between the incident ray and the interface’s normal line. This line is perpendicular to the interface., All wavelengths (colors) of light refract at the same angle when crossing the interface between different media. No, the angle of refraction, varies by wavelength., , 32.11 - Summary, Refraction is the changing of a wave’s direction due to a change in its speed. It, occurs when a wave, like light, passes from one medium into another., A material’s index of refraction is the ratio of the speed of light in a vacuum (c) to its, speed in the material. The index of refraction is represented by the letter n., , Index of refraction, , n = c/v, , Snell’s law quantifies how much light bends when it crosses an interface. Light, bends towards a line normal to the interface between the media as it slows down, (passes into a medium with a higher index of refraction). When light passes through, multiple media its path can be calculated by applying Snell’s law at each interface., , Snell’s law, , Total internal reflection occurs when no light is refracted at an interface; it is all, reflected. This is only possible as light encounters an interface with a medium, having a smaller index of refraction, at a large enough incident angle. The critical, angle is the minimum angle of incidence at which total internal reflection occurs., , Critical angle, , ni sin și = nr sin șr, , The index of refraction for a material varies somewhat by the wavelength of light, being refracted. Light composed of many wavelengths, such as white light, can be, separated into a spectrum by refraction. This effect is called dispersion. Prisms, exhibit dispersion., , 604, , Copyright 2007 Kinetic Books Co. Chapter 32
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Chapter 32 Problems, , Chapter Assumptions, Unless stated otherwise, the incident medium is air, with an index of refraction of 1.00., , Conceptual Problems, C.1, , Sparkle refers to percentage of light out to light in. Why do cut diamonds, whose index of refraction is 2.42, sparkle more than, identically cut glass with an index of refraction of 1.60? Diamonds are cut so that a maximum of the light entering the, diamond's top eventually leaves through the top., , C.2, , If a fish in a still river tries to look up out of the, water at an angle (measured from the vertical), that is greater than the critical angle, will he be, able to see the world of air? If not, what will he, see?, Yes, , No, , Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. Consider the smaller angles, between the laser's path and the dashed normal line. (a) Which is larger, the angle in the air, or the angle in the water? (b), When you shoot a laser from a helicopter toward a submarine, does the laser bend toward or away from the dashed normal, line as it changes media?, (a), , Air, , (b), , Toward, , Water, Away from, , Section 2 - Index of refraction, 2.1, , A laser pulse travels directly through a flint glass plate that is 0.150 mm thick, and then through a diamond sliver that is, 0.130 mm thick. In which substance does the pulse spend more time?, Flint glass, Diamond, , 2.2, , Yellow light travels 0.680 times as fast in a certain type of plastic than in a vacuum. What is the index of refraction for yellow, light in the plastic?, , 2.3, , At what speed does light of wavelength 589 nm travel in flint glass? Flint glass has an index of refraction of 1.61 for that, wavelength., m/s, , 2.4, , Red light travels at 2.07×108 m/s in a particular type of oil. What is the index of refraction for red light in the oil?, , Section 3 - Snell’s law, 3.1, , A beam of light travels from a medium with an index of refraction of 1.25 to a medium with an index of refraction of 1.46. If the, incoming beam makes an angle of 14.0° with the normal, at what angle from the normal will it refract?, °, , 3.2, , A beam of light travels from a medium with an index of refraction of 1.65 to a medium with an index of refraction of 1.30. If the, beam refracts at an angle of 23.5° from the normal, at what angle to the normal did it enter the second medium?, °, , Copyright 2000-2007 Kinetic Books Co. Chapter 32 Problems, , 605
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3.3, , A beam of light crosses the interface between two media with an angle of incidence of 34.0° and an angle of refraction of, 24.6°. The second medium has an index of refraction of 1.70. What is the index of refraction of the first medium?, , 3.4, , A double-paned window consists of two parallel panes of glass with an air gap between them. The outside pane has an index, of refraction of 1.61 while the inside layer has an index of refraction of 1.51. If light enters the window from outside at 21.0° to, the normal, at what angle to the normal does it exit the window on the inside?, °, , 3.5, , A beam of light travels from one medium to another, with the incident beam making an angle of 19.0° from the normal. The, beam then refracts at an angle of 32.0°. How many times faster does the light travel in the second medium than the first?, , 3.6, , A glass of water (n = 1.33) has a layer of vegetable oil (n = 1.47) floating on top. A beam of light enters the oil from above the, glass with an incident angle of 57.0°. What is the refraction angle of the light leaving the oil-water interface?, °, , 3.7, , A ring consists of a diamond (n = 2.42) and another jewel mounted side by side on a flat plate. Their top surfaces are also, parallel to the plate. When laser light is aimed at the ring at 80.0° to the normal, the refraction angle in the diamond is 11.1°, less (from the normal) than the refraction angle in the other jewel. What is the index of refraction of this other jewel?, , Section 7 - Interactive problem: helicopter and submarines, 7.1, , Use the simulation in the interactive problem in this section to calculate the angles of incidence required to hit the (a) left and, (b) right submarine., (a), , °, , (b), , °, , Section 8 - Total internal reflection, 8.1, , Find the critical angle for a ray of yellow light traveling from diamond (n = 2.42) into air (n = 1.00)., , 8.2, , The critical angle for light traveling from a certain plastic into air is 35.5°. What is the index of refraction of the plastic?, , 8.3, , You are constructing a fiber optic cable. You wish to surround the core of the cable with a cladding material such that the, critical angle to go from the core to the cladding is 73.0°. If the index of refraction of the core material is 1.61, what must the, index of refraction of the cladding material be?, , 8.4, , A point source of light is at the bottom of a koi pond, at a depth of 0.525 meters. What is the radius of the circle of light, formed on the water's surface? Take the index of refraction of water to be 1.33. Hint: Some of the light emitted experiences, total internal reflection inside the water., , °, , m, , Section 9 - Dispersion and prisms, 9.1, , A red ray and a blue ray each pass from air into silica at an incident angle of 29.00°. How large is the angle between the two, refracted rays? Red light has an index of refraction in silica of 1.459 and blue light has an index of 1.467., °, , 9.2, , A narrow ray of white light traveling in air enters a glass block at an incident angle of 40.00°. The white light consists of a, range of wavelengths. For these colors, the indices of refraction for the glass range from 1.512 to 1.531. Find the angular, width of the ray inside the block., °, , 606, , Copyright 2000-2007 Kinetic Books Co. Chapter 32 Problems
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33.0 - Introduction, Lenses may be the application of refraction with which you are most familiar., Lenses of glass or plastic, fashioned into eyeglasses and contact lenses, are worn, by about 60% of the population of the United States, and this percentage is, projected to increase as the population ages. Even people who do not require visual, correction are “wearing” lenses anyway, since the eye itself contains an organic, lens., The human eye has the ability to refract light and focus it on the retina at the back of, the eye, where signals are then transmitted to the brain. If the focusing is not, precise enough, glasses or contact lenses can improve the results. These days, the, eye itself can be tuned via laser surgery as well., Lenses augment vision in other fashions as well. Lenses in magnifying glasses,, telescopes, and microscopes allow you to see things that would otherwise be too, small or too far away to see in detail., At the right, you can begin your experimentation with lenses. The convex lens will, produce an image of the penguin. You can move the penguin left and right to see, how its distance from the lens changes the resulting image. Where must the object be relative to the focal point F to create an image on the, same side of the lens as the object? Where must it be to create an image on the far side? Can you make the image appear right-side up (rather, than inverted, as it is in the picture on the right)? A note: When the object is very near the focal point, the image will be too far away from the, lens to be shown in the simulation’s window., , 33.1 - Lenses, , Lens: A device that uses, refraction to redirect and, focus light., Light changes direction í refracts í as it passes from, air to a lens material and back to air. Lenses take, advantage of refraction to create images., In general, lenses are made of a transparent material, with a curved surface or surfaces. Although lenses, are typically curved, we will use two prisms, as shown, to the right, to introduce the fundamental functioning of lenses., , Lenses are used in eyeglasses to focus images., , In the upper diagram, you see two prisms placed so that their flat bases are touching., Three parallel rays of light strike the prisms. Because of the sloping surfaces of the, prisms, the rays are refracted in different directions., As you can see, after passing through the prisms, the three rays converge on the far, side. This first set of prisms serves as a model for a converging lens. Converging lenses, are thickest at their centers. With a two-prism lens, all the incoming rays would not, converge at one point like the three we have chosen to show, which is why prisms do, not make good lenses: The image they create will be blurry. A curved surface will focus, the rays more precisely., In the second diagram, we invert the prisms, so they touch tip to tip. The rays diverge, after passing through them; they will not intersect on the far side of the prisms. This, configuration provides a model for a diverging lens. The image created by a diverging, lens is a virtual one. By extending the paths of the rays that have passed through the, prisms backwards, we could locate the virtual image on the same side of the lens as the, object that creates it., , Converging lens, Two stacked prisms form crude lens, Prisms refract light, Thicker at center, , In this chapter, we focus on lenses that are both spherical and thin. A spherical lens has, two surfaces, each of which is a section of a sphere. The curvatures of the two sides of a lens can differ. Saying it is “thin” means the distance, across a lens is small relative to the distance to the object that is the source of the image. Thin lenses provide an accurate model for optical, equipment ranging from contact lenses to telescopes., Much of the terminology and many basic concepts coincide for mirrors and lenses. A line that passes through the center of a lens, perpendicular to its surface is called the principal axis. Parallel rays from an object infinitely far away refract through a lens and converge at the, lens’s focal point. The focal lengthf is the distance between the lens’s center and its focal point. In contrast to mirrors, lenses have two focal, points, one on each side of the lens, since incident rays can originate on either side of a lens. Rays that are close to the principal axis and, , Copyright 2007 Kinetic Books Co. Chapter 33, , 607
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converge at the focal point are called paraxial rays., Lenses produce real and virtual images, just as mirrors do. As with mirrors, a real image, can be projected onto a piece of paper, while a virtual image cannot. However, the, locations of virtual and real images are reversed for lenses and mirrors. With lenses,, virtual images occur on the same side of the lens as the object. Real images occur on, the opposite side of the lens as the object., The prisms in Concept 1 í the ones stacked base to base í produce a real image on, the opposite side from the object. The ones placed tip to tip create a virtual image, one, on the same side as the object., As with mirrors, incoming rays that are far from the principal axis of a lens do not, converge precisely at the focal point. This blurs the image and is called spherical, aberration., , Diverging lens, Thinner in center, light “spreads out”, , 33.2 - Converging lens: ray-tracing diagram, , Rays for a converging lens, What will be the nature of the image?, , As with mirrors, ray-tracing diagrams are used to determine the orientation, and approximate size and location, of an image produced by a, lens. The diagrams in this textbook employ a convention to make them easier to understand: Objects are shown fairly close to lenses that are, relatively thick compared to the distances shown. For the approximations in the theory of thin lenses to hold true, the objects should be quite, far from the lenses and the lenses should be much thinner than shown. Being literal would make for illustrations that would stretch far across, the page or computer screen, so the diagrams are not drawn to scale., Light that strikes the lens at an angle refracts twice, once upon entering the lens, and again while exiting. However, with a thin lens, we simplify, things by modeling the light ray as changing direction just once as it passes through the lens., Physicists use three specific rays to analyze the behavior of a lens; these rays differ somewhat from those used to analyze mirrors. A single, converging lens can create different image types based on the position of the object relative to the focal point. In this example, we discuss the, image created by an object more than twice the focal length away from a converging lens., Ray 1 begins as a horizontal line, parallel to the principal axis, starting at the top of the object. It refracts at the lens and passes through the, focal point on the far side of the lens. (We use “near” to describe the object’s side of the lens, and “far” to describe the other side. We will, typically place objects on the left side of the lens. Images can be created on the near or far side of the lens.), , Ray 1:, Parallel to axis, then passes through far, focal point, , Ray 2 starts at the top of the object and passes through the center of the lens. It does not change direction., , Ray 2:, Passes unchanged through center of, lens, , 608, , Copyright 2007 Kinetic Books Co. Chapter 33
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Ray 3 starts at the top of the object, passes through the focal point on the near side of the lens, is refracted by the lens, and continues parallel, to the axis on the far side of the lens., , Ray 3:, Passes through near focal point, then, parallel to axis, , The image created by the lens is shown in the diagram below, where all three rays are combined together. The point at which the rays, converge indicates the location of the top of the image. The image is inverted, smaller than the object and real., , Image, Inverted, smaller, real, , This example provides a model for a typical camera lens, which creates an inverted real image on the film or digital recording device on the far, side of the lens., , 33.3 - Diverging lens: ray-tracing diagram, , Rays for a diverging lens, What type of image will be produced?, , Like a convex mirror, a single diverging lens produces only one kind of image. The image is always upright, smaller than the object, and virtual., We use three rays to analyze the image., Ray 1 starts parallel to the principal axis. With a diverging lens, it refracts away from the principal axis. To have the ray pass through a focal, point, we must extend it backward using a virtual ray as shown in the diagram., , Ray 1, Parallel to axis, virtual ray passes, through near focal point, , Copyright 2007 Kinetic Books Co. Chapter 33, , 609
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Ray 2 passes through the lens without changing direction., , Ray 2, Straight through center of lens, , Ray 3 must refract to be horizontal on the far side of the lens. The forward extension of the incident portion of this ray is a virtual ray passing, through the focal point on the far side. The backward extension of the refracted portion of the ray is a virtual ray parallel to the principal axis on, the object side of the lens., , Ray 3, Virtual ray through far focal point, virtual, ray parallel to axis, , As the illustration below shows, the image is upright, smaller than the object, and virtual. This is the case for all images produced by a single, diverging lens., , Image, Upright, smaller, virtual, , The example to the right leads you through another ray tracing exercise for a diverging lens. The object in the example is closer to the lens, than the one above., , Determine the size and location, of the image., Draw ray 1, Draw ray 2, Draw ray 3, Image is upright, smaller, virtual, , 610, , Copyright 2007 Kinetic Books Co. Chapter 33
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33.4 - Interactive problem: image with a diverging lens, This simulation lets you experiment with a diverging lens. You can drag the object, toward and away from the lens, and note how the image changes. Also note which, properties of the image remain consistent., You can press “Show rays” to use the simulation to create a ray diagram. Pressing, “Hide rays” turns off the rays., , 33.5 - Sample problem: object outside the focal point of a converging lens, What kind of image will be produced?, , As shown above, the penguin is between the focal point and a distance twice the focal length. What kind of image will result? Once you know, the result, can you think of any application that relies on a configuration like this?, Strategy, Use ray tracing to establish the nature of the image., Diagram, , Step-by-step solution, 1., , Ray 1 starts parallel to the principal axis from the head of the penguin, refracts and passes through the focal point on the far side of the, lens., , 2., , Ray 2 passes through the center of the lens without changing direction., , 3., , Ray 3 passes through the focal point on the near side and refracts at the lens to be parallel to the principal axis on the far side., , The result is an inverted image that is larger than the initial object. It is a real image, since light rays pass through the position of the image,, and they would create a projected image on a screen placed there., Above, we also asked: What might be an application that relies on a configuration like this? A movie theater projector is one example. The, audience views enlarged, real images on the screen. Light is projected through the film, creating a luminous object from a tiny likeness, and, then passes through a projection lens to be magnified. The real image appears on the movie screen. The “object” is upside down on the film,, so when the lens inverts the image, it appears upright to the audience., , Copyright 2007 Kinetic Books Co. Chapter 33, , 611
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33.6 - Sample problem: object inside the focal point of a converging lens, What kind of image will be produced?, , The penguin is inside the focal point of a converging lens. What kind of image will result? Again, once you know the result, can you think of any, application that relies on a configuration like this?, Strategy, Use ray tracing to establish the nature of the image., Diagram, , Step-by-step solution, 1., , Ray 1 starts parallel to the principal axis. It refracts at the lens and passes through the focal point on the far side. To obtain a, convergence point for the image, we extend the ray backward, showing this extension with a dashed line., , 2., , Ray 2 passes through the center of the lens without changing direction. We again extend a virtual ray backward., , 3., , Ray 3 must be horizontal on the far side of the lens and it must also pass through a focal point. Here, it starts at the top of the object,, directed upward so that its backward extension passes through the focal point on the near side. It refracts at the lens, becoming, horizontal on the far side. By extending the horizontal portion backward, we create a virtual ray that passes through the convergence, point., , The result is an upright, virtual image that is larger than the object. This configuration is typical of objects being viewed in a magnifying glass., The magnifying glass enlarges objects, making it easier to view their fine detail. The image cannot be projected, but it is upright, a matter of, great convenience to physicists and textbook authors over 40 who may use a magnifying glass to see fine details!, , 33.7 - Lens equations, , Lens and magnification, equations, Interpretation of signs, , Ray diagrams provide a qualitative technique for establishing the nature of an image created by a thin lens. The image’s position and size can, be quantified with a set of three equations called the lens equations. These are the focus (we could not resist) of this section., The thin lens equation is shown in Equation 1. It states the relationship between object distance, image distance, and focal length: The sum of, the reciprocals of the object distance and the image distance equals the reciprocal of the focal length., For instance, suppose you want to calculate the location of an image created by placing an object 0.50 meters away from a lens with a focal, length of 0.33 meters. The thin lens equation and a little arithmetic reveal that the image distance is 1.0 meters. (We solve this problem in, , 612, , Copyright 2007 Kinetic Books Co. Chapter 33
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Example 1.) The positive value for the image distance places it on the far side of the, lens, which means the image is real. These values also confirm the answer to a, question posed in an earlier section, where the nature of the image of an object, between F and 2F was determined using ray tracing., The formulas in Equation 2 deal with lateral magnification, which is defined in the same, way for lenses as it is for mirrors: the ratio of the image height to object height. It can, also be computed using distances: The lateral magnification equals the negative of the, ratio of image distance to object distance. This is the second equation. For instance, in, the example just discussed above, this ratio is í2.0: the image is twice as tall as the, object. The negative sign means the image is inverted., The equation in Equation 3 is the lensmaker’s equation. It relates the focal length of the, lens to some of its physical properties, specifically its index of refraction and the radius, of curvature of each surface of the lens. The equation enables a lensmaker (or physics, student) to calculate the focal length for a lens that has differing curvatures on the two, sides of the lens, or to create a lens with a particular focal length., We show the version of the equation for a lens in air. If the lens is immersed in another, substance, say water, then the variable n has to be replaced by the ratio of the index of, refraction of the lens to the index of the surrounding material (for example,, n = nlens/nwater). The object side of a microscope objective lens is often immersed in a, special optical oil, as with the German “Oel Immersion” objective lens shown below., , Thin lens equation, , do = object distance, di = image distance, f = focal length, , The equations can be tricky to apply because various values like image distance can be, either positive or negative. You need to use the table in Concept 1 above, or perform, the more difficult task of memorizing the conventions., To review perhaps the trickiest sign convention, the radius of curvature R in the, lensmaker’s equation is negative when the center of curvature is on the near (the, object) side of the lens and positive when it is on the far side. If this sounds confusing,, consider each lens surface as part of a sphere. If the sphere’s center of curvature is on, the object side, so the sphere would enclose an object relatively near the lens, then R is, negative., With a converging lens that is convex on both sides, like the ones used in the, illustrations to the right, the radius of curvature for the near surface, R1, is positive, and, the radius of curvature for the far surface, R2, is negative. As the diagram illustrates,, with a lens that is convex on both sides, the center of curvature for each surface is on, the opposite side of the lens from the surface. With a diverging lens made up of two, concave surfaces, the signs are reversed and the centers are on the same side as the, lens surfaces. Remember, we said this was tricky!, , Magnification equation, , m = magnification, hi = height of image, ho = height of object, , There are additional conventions: Focal lengths are positive for converging lenses and, negative for diverging lenses. Object distances are positive when the object is real and negative when it is virtual. (Virtual objects can arise, when there are multiple lenses. A virtual object is on the side of a lens opposite to the source of the light.) Image distances are positive when, the image is real and negative when it is virtual. Magnification is positive for an image that is upright relative to the object and negative for one, that is inverted relative to the object., , Copyright 2007 Kinetic Books Co. Chapter 33, , 613
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Lensmaker’s equation, air, , n = index of refraction, R1 = radius of near surface, R2 = radius of far surface, , Find the distance to the image., , di = 1.0 m, , 614, , Copyright 2007 Kinetic Books Co. Chapter 33
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33.8 - Interactive problem: optical bench with a lens, Your mission here is to create the image at the location shown in the graphic at the, right. The image is 13.5 cm tall, and 15.0 cm from the lens. The object is 9.0 cm tall., (Note: We are deliberately being ambiguous about mathematical signs here!), You set both the object location and the lens’s focal length to the nearest 0.01 cm in, this simulation. Press CHECK to test your answer., A hint: Start this problem by determining the magnification required for the specified, image height, and then consider how that helps you to specify the object distance. If, you need help with this problem, consult the section on lens equations. You will, need to apply three equations from that section., , 33.9 - The human eye, The human eye contains a variable lens. This organ í a remarkable product of evolution, í employs the lens to focus images on the retina at its back surface. Special cells called, rods and cones line the retina. Light stimulates these cells, and they send signals in the, form of electric impulses to the brain via the optic nerve., Light passes through four different components of the eye on its way to the retina: the, cornea, the aqueous humor, the lens, and the vitreous humor. Each has a different, index of refraction, ranging from about 1.30 to 1.40. This means the greatest amount of, refraction occurs when the light first crosses from the air (n = 1.00) to the cornea,, where the indices differ the most., In order for an object to be perceived clearly, its image must be focused on the retina, at, a fixed distance from the lens. The real image projected on the retina is upside down,, but the brain automatically corrects for this., The eye must be able to create crisp images of objects that are located at varying, distances. It accomplishes this by changing the shape of the lens, a process called, accommodation. By causing the lens to contract or expand, the eye changes the radius, of curvature and the focal length of the lens. On the right, you see the lens in both a, relaxed state for viewing objects far away, and tensed as its curvature has been, changed to focus on a nearer object., , Human Eye, Contains variable lens, Image forms on retina, , The closest distance at which an object can be brought into focus is called the near, point. A young adult with a fully tensed lens can clearly see objects that in some cases, are as close as 15 cm, with an average near point for this age being 25 cm. By the time, a person reaches her early forties, the near-point distance increases to an average of, 40 cm. By age 65, the average near-point distance is 400 cm. Changes in the eye that, occur with aging explain this increase. In fact, this progression is so predictable that the, distance to the near point can pinpoint age to within a few years., The far point is the greatest distance at which an object can be seen clearly. For people, who do not need glasses for distance vision, this point is effectively infinitely far away., For those who are nearsighted, it can be much closer., If you have 20/20 vision, you are lucky: You perceive objects crisply without the use of, prescription lenses. If you have 20/60 vision, it means that you must be as close as 20, feet to see what a person with typical vision can see at 60 feet. Baseball players are, noted for having acute vision, such as 20/15 vision. This means they see at 20 feet, what most people need to be within 15 feet to see clearly., , Near point, Closest in-focus object distance, , Far point, Farthest in-focus object distance, , Copyright 2007 Kinetic Books Co. Chapter 33, , 615
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33.10 - Nearsightedness, , Nearsightedness, Distant objects out of focus, , Nearsightedness: Ability to focus on nearby, objects but not distant objects., To people with nearsightedness, or myopia, faraway objects are blurry and look like the, image above., To view an object that is far away, the eye’s ciliary muscle completely relaxes, causing, the lens to be as flat as possible. In a nearsighted eye, the eye is too long, causing, images of faraway objects to form in front of the retina, even when the lens is, completely relaxed. You see this in the conceptual diagram in Concept 2., To correct this problem in focusing, a diverging lens can be placed in front of the eye., As illustrated in Concept 3, the lens spreads out the parallel rays before they reach the, eye, so that the rays do not come to a focus quite so soon, but farther back upon the, retina where they should., , The problem, Image forms in front of retina, , More recently, laser surgery has become a popular method for correcting, nearsightedness. A laser directed at the eye reshapes the cornea, causing its outer, surface to be flatter so that an image can form more crisply on the retina. The angle of, incidence for incoming light rays is less, causing them to converge farther back in the, eye, at the retina., Contact lenses can also correct nearsightedness; they change the radius of curvature of, the front surface of the eye. Since the contact lenses are thicker at the edges than the, center, they work by flattening the eye’s surface., , A solution, Corrected with diverging lens, , 33.11 - Farsightedness, , Farsightedness, Close objects out of focus, , Farsightedness: Inability to clearly see objects that are relatively close., With farsightedness, or hyperopia, the image of a nearby object forms behind the retina. The eyeball is too short, and no matter how hard the, ciliary muscle strains, it cannot contract the lens enough to focus the image on the retina. You see this shown in Concept 2., People with hyperopia have difficulty reading things that are close and often use reading glasses as a result. A related condition called, presbyopia (“old vision”) occurs in many people starting in their forties. With presbyopia, the eyeball is not too short, but the lens itself has, become stiff with age and is unable to contract enough to focus images on the retina., A converging lens can compensate for this problem by creating a virtual image beyond the eye’s near point, so that the lens can focus an, image on the retina. This solution is shown in Concept 3., , 616, , Copyright 2007 Kinetic Books Co. Chapter 33
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The problem, Image forms behind retina, , A solution, Corrected with converging lens, , 33.12 - Physics at work: laser eye surgery, Laser eye surgery, a procedure now quite popular, provides a way to alter the shape of, the cornea so that glasses or contact lenses are not needed, or are required less, frequently. It has proven quite popular amongst both celebrities (such as Tiger Woods), and non-celebrities., To explain how the procedure works, we first review some fundamentals of how the eye, focuses an image. The eye has two components that are most responsible for the, location of the image it creates: the cornea (its outermost surface) and an internal, variable lens. Light refracts as it passes through both of these components., Although one might think most of the refraction occurs at the lens, in fact about twothirds of the refraction occurs at the cornea. It occurs there because the index of, refraction of the cornea differs substantially from that of air. This difference is much, larger than differences in the indices of refraction within the eye. The cornea and the, lens combine to focus the light at the retina, which is at the back of the eye. The eye, adjusts the curvature of the lens in order to alter its focal length, enabling it to focus, objects at various distances. When the eye can change the lens shape enough so that, light converges crisply at the retina for objects both near and far, a person has no need, for glasses., , The problem: nearsightedness, Image forms in front of retina, , However, many people need glasses or contact lenses in order to see clearly., Nearsightedness and farsightedness occur when the lens’s shape cannot be changed, enough to sharply focus images on the retina. Eyeglasses and contact lenses, compensate for this by providing a second lens that assists in the process of focusing, the image at the proper position., Laser eye surgery takes a different tack: It changes the eye itself. The surgery changes, the shape of the cornea in order to correct for the limitations of the lens. The procedure, does not alter the lens inside the eye; rather, it “tunes” the cornea., Currently, a type of laser surgery called LASIK (laser insitu keratomileusis) is the most, widely used. To begin the process, medical personnel first determine how the cornea’s, curvature must be modified in order to improve vision. Then, a surgeon temporarily, peels back the epithelium (the thin outermost layer of the eye) and trains a laser on the, eye to remove small amounts of the cornea. The epithelium is then put back to cover, the eye’s surface., , The solution, Cornea flattened to correct vision, After surgery í image forms on retina, , Laser surgery can be used to treat both nearsightedness and farsightedness. Nearsightedness occurs when the lens cannot become, sufficiently flat, resulting in an image formed in front of the retina. This is shown in Concept 1. To compensate for this, the surgeon flattens the, , Copyright 2007 Kinetic Books Co. Chapter 33, , 617
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cornea. This reduces the angle of incidence for light entering it, and the angle of, refraction as a result. Concept 2 shows the eye after laser eye surgery for, nearsightedness. The dashed line shows the original curvature of the eye., With farsightedness, the muscles of the eye cannot contract the lens enough. Close, objects are focused behind the retina, again causing blurring. This problem is shown in, Concept 3. To address it, the surgeon increases the steepness of the cornea. Since the, angle of incidence of incoming light rays is increased, they refract more. The bottom, illustration shows a farsighted eye after laser eye surgery. Again, the dashed line, represents the initial shape of the eye., , The problem: farsightedness, Image forms behind retina, , The solution, Cornea steepened to correct vision, After surgery í image forms on retina, , 33.13 - Angular size, , Angular size: How much of, the field of vision is filled by, an object or image., Imagine yourself in Seattle, standing two kilometers, from the Space Needle. You look at the structure, and, then raise your hand so that your thumb is a halfmeter from your eyes. Your thumb now appears to be, about the same size in your field of vision as the, Space Needle, although you know that the Space, Needle is much bigger., , Sizing up the Needle., , This effect occurs because of the relative angular sizes of your thumb and the Space, Needle. The angular size refers to the angle, measured in radians, subtended by an, object when viewed from a particular distance. The subtended angle is perhaps as well, explained with a diagram as in text. The subtended angle is labeled ș in Concept 1 to, the right. It measures the angle of your vision “blocked out” by the object., Trigonometry can be used to determine angular size when the height h of the object, and the distance d to the object are known. The angular size can be approximated as, the height divided by the distance to the object. This approximation uses a small-angle, approximation, ș § tanș for small angles, where the angle must be measured in, radians. This approximation for angular size is very good (within one percent) for, angular sizes less than 0.17 rad (about 10°). Approximations like this can be useful in, astronomy to approximate distance or size. For instance, it can be useful to know that, the Moon subtends about 0.5° (0.009 radians). If you know its distance, you can quickly, approximate its diameter using this fact and some trigonometry., , Angular size, Amount of visual field filled by object, Measured as angle, , To leave space and return to Seattle: The thumb in the photograph above is 5 cm or so, tall and 60 cm away. It has an angular size of 5 cm divided by 60 cm, which equals 0.08 rad. At a distance of 2 km, the Space Needle, which is, 184 meters tall, would subtend an angle of about 184 m/ 2000 m, or 0.09 rad. These values confirm that the thumb is about the same angular, size as the Space Needle., , 618, , Copyright 2007 Kinetic Books Co. Chapter 33
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ș = angular size (in radians), h = height of object, d = distance to object, , What is the angular size of this, person’s thumb at the given, distance?, șThumb § 0.05 m /0.5 m, șThumb § 0.1 rad, , How far is this person from the, Space Needle? She measures, the angular size of her thumb, and of the Needle as 0.11, radians., șNeedle § hNeedle/dNeedle, dNeedle § hNeedle/șNeedle, dNeedle § 184 m/0.11, dNeedle § 1700 m, , Copyright 2007 Kinetic Books Co. Chapter 33, , 619
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33.14 - Telescopes, , Refracting telescopes, Objective lens gathers light, Eyepiece magnifies image, , Simple refracting (lens-based) telescopes have two lenses. A clever configuration of, these lenses enables the telescope to serve three purposes: to gather as much light as, possible from the object, to magnify the object and to provide as relaxed viewing of the, image as possible., The first lens encountered by incoming light rays, the objective lens, is used to gather, as much light as possible. Light is at a premium in telescopes since faraway objects are, typically faint. You cannot see most of the stars in the sky because they are too dim, not, because they are too small (in angular size). Astronomers are willing to pay high prices, to obtain telescopes with large objective lenses. Amateur astronomers purchasing their, first telescope are often advised to pay more attention to its light gathering capability, than to its magnifying power., The eyepiece of a telescope performs two important tasks: It magnifies the real image, created by the objective lens and creates a virtual image whose image distance is, essentially at infinity. Creating a distant image is useful because the human eye is most, relaxed when viewing objects at a distance., At the right, you see the combination of lenses that constitutes a basic refracting, astronomical telescope. To analyze how this system functions, you must first determine, the location of the real image created by the objective lens. This image serves as the, object for the eyepiece., , How it works, Two converging lenses, Focal points at same location, Final image inverted, at infinity, virtual, , Telescopes are pointed at faraway objects, so the light rays entering the telescope are, essentially parallel. Both the objective lens and the eyepiece are converging lenses., The objective lens creates a real image of the distant object at its focal point. The, lenses are arranged so that this point is also at the focal point on the incident side of the, eyepiece. Putting the object here causes the virtual image created by the eyepiece to, be at infinity. The length of the telescope between the two lenses equals the sum of, their focal lengths. You see all this in the sketch to the right., As you can see in Concept 2, the image produced by the objective lens is inverted,, smaller than the object, and real. The image produced by the eyepiece from this real, image subtends a larger angle than the object, and it is virtual. The larger angle is an, example of angular magnification, provided by the “magnifying glass” of the eyepiece. In, this case it is being used to view the real image created by the objective lens, rather, than a concrete object. The eyepiece does not invert the real image created by the, objective lens, so the final image is inverted compared to the original object., , Reflecting telescope, Uses mirror in place of objective lens, , The telescope configuration we have been discussing works well for astronomy. The image it produces is inverted, but this typically poses no, problem for astronomers, since they are accustomed to seeing telescope-rendered views of the objects they study. However, telescopes used, to study terrestrial objects (like birds) use other configurations, including perhaps a diverging lens or a set of prisms, in order to avoid, presenting an inverted image to the eye., A different type of telescope designed for astronomy is called a Newtonian reflector. “Newtonian” refers to the developer of this type of, telescope, Sir Isaac Newton, and “reflector” refers to the use of mirrors. In a reflector, a relatively large concave mirror performs the “light, collecting” and first image creation duties that the objective lens performs in a refracting telescope. You see a diagram of a reflector to the right., Light passes down the tube, is collected by a curved mirror, redirected by a flat mirror and then passes through the eyepiece., Why mirrors? Consider the reflecting telescope at California’s Mount Palomar Observatory, which is famous for its 200 inch reflector. A glass, objective lens of the same diameter would have a mass of about 40,000 kg (and weigh about 50 tons), more than three times the mass/weight, of the mirror actually used. The weight of the mirror is still huge, but reducing the mass by more than two-thirds is a worthwhile, accomplishment. In addition, lenses are subject to chromatic aberration (different colors of light focusing at different distances), while mirrors, are not. A mirror is also easier to support (since it can be supported from behind), and only one side of the mirror needs to be manufactured, with optical precision, versus the two sides of a lens. These factors play out in the construction of large research telescopes: The largest, refractor in existence, at the Yerkes Observatory in Wisconsin, is only a little over one meter in diameter, while the largest reflectors, the Keck, telescopes, are nearly ten meters in diameter., , 620, , Copyright 2007 Kinetic Books Co. Chapter 33
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33.15 - Lens aberrations, , Lens aberration:, Imperfection in the images, formed by a lens., Spherical aberration: Image, blurring due to the spherical, contour of the lens., Chromatic aberration: Image, blurring due to differing, refraction of various, wavelengths (colors) of light., , Spherical aberration., , You may have noticed that even when an image created by a lens is focused as sharply, as possible, it is still somewhat blurry. The blurriness results from lens aberrations., Blurring means that the light rays from a single point on an object do not precisely, converge at one point in the image, as the examples to the right illustrate, using, horizontal rays from a point on a very distant object. A well-crafted lens causes little, blur; lower quality lenses can suffer from substantial blurriness., There are several types of aberration. Some aberrations may be caused by, imperfections in the manufacturing of the lens, but one type arises unavoidably from the, nature of a spherical lens. The lack of focus arising from spherical aberration most, affects rays that strike a lens near its periphery. Spherical aberration means the light, rays refract at an undesired angle; rather than converging to a single point, they, converge to a larger region. The spherical nature of the lens causes this problem., , Spherical aberration, Due to curvature of lens, , Instead of a single point of focus for incoming parallel rays, the lens creates a circle of, least confusion, a region in which the rays approximately converge and an image is, most satisfactorily viewed. This is diagrammed in Concept 1., A typical image created by a spherical lens is shown above. The paraxial rays focus, sharply in the center of the real image. Nonparaxial rays, having already passed, through their closer focal points, are beginning to spread out again. Their contribution to, the image shows up as a set of concentric rings. The outermost ring corresponds to, rays passing through the outermost portions of the lens which have the shortest focal, lengths., Spherical aberration can be counteracted in a variety of ways. For instance,, photographers are aware of this phenomenon, and when precise focus is important,, they “stop down” the aperture of the iris that lets light into the camera lens, restricting, the remaining light to the center portion of the lens. Mirrors are also subject to spherical, aberration: In large astronomical telescopes, the problem is avoided by using parabolic, or other non-spherical reflectors., , Chromatic aberration, Different wavelengths refract differently, , Chromatic aberration occurs because the refractive index of a material is a function of the wavelength of light. For instance, since blue light, refracts more than red light when it passes from air into glass, a ray of white light disperses into its component colors when refracted by glass., This notably occurs in prisms, although there it is often the desired effect. You see a diagram illustrating chromatic aberration in Concept 2., The photo below provides an example of chromatic aberration. There is a noticeable purple halo around the circular windows on the left side of, the image. The right side of the image is actually the center of a larger cropped image. The chromatic aberration is less noticeable in this area,, which corresponds to the center of the lens. Chromatic aberration worsens toward the edge of the lens., , Copyright 2007 Kinetic Books Co. Chapter 33, , 621
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A compound lens made with lenses of different indices of refraction can counteract this effect. This type of achromatic lens is found in high, quality optical equipment., , 33.16 - Gotchas, “Real” means the same thing with lenses as it does with mirrors. On the one hand, yes. Real images can be projected onto paper with both, lenses and mirrors. On the other hand, no. With lenses, a real image is on the opposite side of the lens from the object. With mirrors, a real, image is on the same side of the mirror as the object., Signs. You need to be careful with signs when using the lens equations, and we have provided a table to help you do so., The ray diagrams for lenses are entirely accurate and represent physical reality. The diagrams in this (and other textbooks) employ a few, conventions in order to make them easier to understand: Objects are shown fairly close to lenses that are relatively thick compared to the, distance to the objects. For the approximations inherent in the theory of thin lenses to hold true, the objects should be quite far from the lenses, and the lenses should be much thinner. Applying these changes to diagrams would make for illustrations that stretch far across the page or, computer screen., , 33.17 - Summary, Lenses redirect light by refraction. Converging lenses are thicker in the middle and, bring light rays together. Diverging lenses are thinner in the middle and spread light, out., Lens terminology is very similar to that of mirrors. An important difference is that, when we consider a single lens, virtual images appear on the same side of the lens, as the object, and real images appear on the opposite side. Another difference is, that lenses have a focal point on each side., , Thin lens equation, , Lateral magnification, , Ray diagrams for converging and diverging lenses are also similar to those for, mirrors., And like mirrors, lenses have equations that quantify the relative size, orientation, and distance of the images they produce. In addition, the lensmaker’s equation, determines the focal length of a lens with differing radii of curvature on its two sides., Sign conventions can be trickier for lenses, so pay special attention to them., , Lensmaker’s equation, air, , Several lenses can be used together to enhance their magnification properties, such, as in a refracting telescopes., The human eye contains a lens that can change shape in order to create a focused, image on the light-sensitive retina at the back of the eye. A person’s near point and, far point are the closest and farthest distances on which she can focus., Lenses can exhibit spherical aberration just as mirrors can. They also exhibit chromatic aberration due to the differing refractive indices of, different wavelengths of light. Both kinds of aberration can cause an image to look blurry., , 622, , Copyright 2007 Kinetic Books Co. Chapter 33
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Chapter 33 Problems, , Conceptual Problems, C.1, , An object is placed between the focal point of a converging lens and a distance twice the focal length from the lens. Draw a, ray diagram to determine the nature of the image and describe the image., i., ii., iii., iv., v., , C.2, , An object is placed between a converging lens and its focal point. Draw a ray diagram to determine the nature of the image, and describe the image., i., ii., iii., iv., v., , C.3, , Virtual; larger; upright, Real; smaller; inverted, Virtual; smaller; upright, Real; larger; inverted, Real; larger; upright, , An object is placed farther from a converging lens than twice the lens' focal length. Draw a ray diagram to determine the, nature of the image and describe it., i., ii., iii., iv., v., , C.5, , Virtual; larger; upright, Real; smaller; inverted, Virtual; smaller; upright, Real; larger; inverted, Real; larger; upright, , An object is placed in front of a diverging lens. Draw a ray diagram to determine the nature of the image., i., ii., iii., iv., v., , C.4, , Virtual; larger; upright, Real; smaller; inverted, Virtual; smaller; upright, Real; larger; inverted, Real; larger; upright, , Virtual; larger; upright, Real; smaller; inverted, Virtual; smaller; upright, Real; larger; inverted, Real; larger; upright, , A single lens is used to create a real image. Is the image upright or inverted? Explain., Upright, , C.6, , A converging lens is placed very far away from an object, so that an image is created very near the focal point of the lens., The image is real, smaller than the object, and inverted. As you move the lens closer to the object, (a) does the image get, farther away from the lens, or closer to it? (b) Does the image get larger or smaller? When the lens reaches a distance of, twice the focal length from the object, (c) where is the image located? (d) What is the nature of the image at this point?, (a), , Farther, , Closer, , (b), , Smaller, , Larger, , (c), , (d), , C.7, , Inverted, , i., ii., iii., iv., i., ii., iii., iv., , Infinity, At the focal point, At twice the focal length, At half the focal length, No image: rays refract parallel, Real; larger than the object; inverted, Real; equal size as the object; inverted, Virtual; equal size; upright, , Would you expect mirrors to exhibit chromatic aberration? Why?, Yes, , No, , Copyright 2000-2007 Kinetic Books Co. Chapter 33 Problems, , 623
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Section Problems, Section 0 - Introduction, 0.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) Where must the object be, relative to the focal point F to create an image on the same side of the lens as the object? (b) Where must the object be to, create an image on the opposite side of the lens? (c) Where must the object be to create an upright image?, (a), , (b), , (c), , i., ii., iii., iv., i., ii., iii., iv., i., ii., iii., iv., , On the opposite side of the focal point as the lens., Between the focal point and the lens., It is impossible., Anywhere., On the opposite side of the focal point as the lens., Between the focal point and the lens., It is impossible., Anywhere., On the opposite side of the focal point as the lens., Between the focal point and the lens., It is impossible., Anywhere., , Section 4 - Interactive problem: image with a diverging lens, 4.1, , Use the simulation in the interactive problem in this section to answer the following questions about a single object and a, diverging lens. (a) Is it possible to make a real image? (b) Is it possible to make an inverted image? (c) Is it possible to make, an image that is larger than the object?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , Section 7 - Lens equations, 7.1, , 7.2, , 7.3, , An object is placed 23.0 cm in front of a thin lens. The image created is 5.00 cm behind the lens. (a) Is the object distance, positive or negative? (b) Is the image distance positive or negative? (c) What is the focal length of the lens? (d) Is the lens, converging or diverging?, (a), , Positive, , Negative, , (b), , Positive, , Negative, , (c), , cm, , (d), , Converging, , Diverging, , An object is placed 23.0 cm in front of a thin lens. The image created is 5.00 cm in front of the lens. (a) Is the object distance, positive or negative? (b) Is the image distance positive or negative? (c) What is the focal length of the lens? (d) Is the lens, converging or diverging?, (a), , Positive, , Negative, , (b), , Positive, , Negative, , (c), , cm, , (d), , Converging, , Diverging, , The focal length of a thin lens is 0.110 m. You want to use it to produce a real image at 0.600 meters from the lens. How far, from the lens should the object be placed?, m, , 7.4, , The focal length of a thin lens is í9.00 cm. You place an object 15.0 cm from the lens. (a) Is the image produced real or, virtual? (b) What is the image distance (include the correct sign)?, (a), (b), , 624, , Real, , Virtual, cm, , Copyright 2000-2007 Kinetic Books Co. Chapter 33 Problems
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7.5, , 7.6, , An object is placed in front of a thin lens at a distance of 7.00 cm, and the lens produces an inverted image that is twice the, size of the object. (a) Is the image height positive or negative? (b) Is the image real or virtual? (c) How far is the image from, the lens? (d) What is the focal length of the lens?, (a), , Positive, , (b), , Real, , Negative, Virtual, , (c), , cm, , (d), , cm, , An object is placed in front of a thin lens at a distance of 12.0 cm, and the lens produces an upright image that is one-third, the size of the object. (a) Is the image height positive or negative? (b) Is the image real or virtual? (c) Is the image distance, positive or negative? (d) What is the image distance? (e) What is the focal length of the lens?, (a), , Positive, , (b), , Real, , (c), , Positive, , Negative, Virtual, Negative, , (d), , cm, , (e), , cm, , 7.7, , You are making a thin lens. The near surface (the surface on the object side) has a radius of curvature of +5.00 cm and the, far surface has a radius of curvature of +6.00 cm. What is the focal length of the lens if the index of refraction of the material is, 1.60?, , 7.8, , The thin lens equation given in this book is sometimes called the Gaussian form of the equation. Another form, called the, , cm, Newtonian form, uses the distance x between the object and the first focal point and the distance x' between the second focal, point and the image. Use the Gaussian form to derive the Newtonian form:, , 7.9, , A thin lens has a near surface with a radius of curvature of í5.00 cm and a far surface with a radius of curvature of +7.00 cm., (a) Is the lens converging or diverging? (b) What is the focal length of the lens if the index of refraction of the material is 1.74?, (a), , Converging, , (b), , cm, , Diverging, , 7.10 A lens has two convex surfaces (both surfaces bulge at the middle). The left surface has a radius of curvature whose, magnitude is 11.2 cm and the right surface has a radius of curvature whose magnitude is 13.0 cm. The index of refraction of, the lens is 1.41. (a) What is the focal length of the lens for light traveling from left to right? (b) What is the focal length of the, lens for light traveling from right to left?, (a), , cm, , (b), , cm, , 7.11 The cornea and lens of the eye each refract light as it passes into the eye. We consider the cornea and lens together as one, thin lens whose shape, and effective focal length, can be changed by the muscles around the eye. When incoming light is, parallel, such as for an infinitely far away object, the eye muscles shape the effective lens to have a focal length of about, 2.50 cm to produce a focused image on the retina. What focal length lens do the muscles cause to create a focused image of, an object 35.0 cm away?, m, 7.12 A camera has a lens of focal length 50.0 mm and is focused on an object 0.800 m from the lens. (a) To focus on an object, effectively at infinity, will the lens have to move toward or away from the film? (b) How far will the lens have to move? Give the, answer in millimeters., (a), , Toward, , (b), , Away, mm, , 7.13 A small light bulb is 1.50 meters from a screen. You have a converging lens with a focal length of 0.250 m. There are two, possible distances from the bulb at which you could place the lens to create a sharp image on the screen. What is the larger, of these distances?, m, , Section 8 - Interactive problem: optical bench with a lens, 8.1, , Use the information given in the interactive problem in this section to calcuate (a) the object distance and (b) the focal length, of the lens in order to create the desired image. Test your answer using the simulation., (a), , cm, , (b), , cm, , Copyright 2000-2007 Kinetic Books Co. Chapter 33 Problems, , 625
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Section 10 - Nearsightedness, 10.1 Nellie is nearsighted. She cannot focus on objects farther than 40.0 cm from her unaided eye. (a) Does she require a, converging or diverging lens to correct her vision? (b) What focal length must her corrective contact lens have to bring into, focus the most distant objects?, (a), , Converging, , (b), , cm, , Diverging, , Section 13 - Angular size, 13.1 A quarter is 2.3 cm across. What is its approximate angular size in radians when viewed from 3.5 meters away?, radians, 13.2 A penny is 1.8 cm across. What is its approximate angular size when viewed from 1.6 meters away? State your answer in, degrees., °, 13.3 You measure a crater on the moon to have an angular size of 3.64×10í4 radians. The distance to the moon is 3.84×108 m., What is the approximate diameter of the crater?, m, 13.4 The Hubble Space Telescope has a resolution of about 2.4×10í7 radians for visual wavelengths. (a) If you were to view a, quarter at a distance such that it subtended the same angle, how far away would it have to be? A quarter is 2.3 cm in, diameter. (b) The resolution of the Hubble at infrared wavelengths is about 9.7×10í7 radians. How far away would a quarter, have to be to subtend this angle?, , 626, , (a), , m, , (b), , m, , Copyright 2000-2007 Kinetic Books Co. Chapter 33 Problems
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34.0 - Introduction, Light is a particle. Many of the great scientists of the 17th and 18th centuries who made, fundamental contributions to the study of optics, including Isaac Newton, thought that, light consisted of a stream of “corpuscles,” or particles. In the 20th century, Albert, Einstein explained the photoelectric effect. His explanation, for which he was awarded, the 1921 Nobel Prize, depended on the fact that light acts like a particle. This property, of light led to the coining of the term “photon” for a single particle of light by the chemist, Gilbert Lewis., Light is a wave. Between the 18th and 20th centuries, physicists discovered many, wave-like properties of light. They found that a number of phenomena they routinely, observed with water waves they could also observe with light., For instance, the English scientist Thomas Young (1773-1829) showed that light could, produce the same kinds of interference patterns that water waves produce. At the right,, you see examples of interference patterns formed by light and by water waves. The, similarities are striking. In this chapter, you will apply to light some of what you have, studied about the interference of sound waves and traveling waves in strings., , Interference of light waves, Pattern of bright and dark on screen, , Let there be light. Is light a particle, a wave, or both? Perhaps an Early Authority had it, right. Light is light. It is a combination of electric and magnetic fields. Trying to classify, light as a particle or as a wave may be a fruitless effort í better to revel in its unique, properties. In this chapter, we will revel in its wave-like properties, and discuss the topic, of interference. Your prior study of electromagnetic radiation modeled as a wave, phenomenon will prove useful., , Interference of water waves, Expanding circular ripples, Pattern of disturbance and calm, , 34.1 - Interference, In Concept 1, you see an interference pattern created, by causing a beam of light to pass through two, parallel slits to illuminate a viewing screen., Constructive interference of light waves accounts for, the bright regions (called bright fringes) while, destructive interference causes the dark fringes., In this section, we review some of the fundamentals, of interference, and discuss the conditions necessary, for light to make the pattern you see to the right. You, may have already studied the interference of, mechanical waves; for instance, what occurs when, Grass is green because it selectively absorbs nongreen colors. In contrast, the, two waves on a string interact. In this chapter, you will, shimmering "color" of a peacock's feathers is due to interference effects., study what happens when electromagnetic waves, meet. Some of the same principles and terminology are used in discussing both kinds of, interference., When two light waves meet, the result can be constructive or destructive interference., In the following discussion, we assume that the waves have equal amplitude., Constructive interference creates a wave of greater amplitude and more intensity than, either source wave; destructive interference results in a wave of smaller amplitude and, less intensity than either source wave. At any point in a two-slit interference pattern, such as that to the right, light waves from the two sources meet and interfere, constructively, destructively, or partially (exhibiting a degree of interference somewhere, between complete constructive and destructive interference)., To create an interference pattern, a physicist needs light that is:, 1., , Monochromatic. This means light with a specific wavelength. For instance,, experimenters can produce the pattern you see in Concept 1 by using pure red, light., , Interference pattern, Bright and dark fringes, , Copyright 2000-2007 Kinetic Books Co. Chapter 34, , 627
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2., , Coherent. This means the phase difference between the light waves arriving at, any location remains constant over time., , The first condition, monochromatic light, means that all the light passing through the, slits must have the same wavelength. White light, which is a mixture of many, wavelengths (colors), does not produce distinct interference patterns. We often used a, similar condition when analyzing the interference of mechanical waves, restricting our, attention to what occurred when two waves of the same wavelength met., We also implicitly used the second of the two conditions stated above with waves on a, string. The phase difference between the waves sometimes remained constant, throughout the string, and over long time periods., For light waves, coherence can be achieved by causing light from a single point source, to pass through two narrow slits, separating an initial beam of light into two beams with, a constant phase difference. You see what happens when coherent light emerges from, two slits and falls on a surface in Concepts 2 and 3., , Constructive interference, Waves in phase, Result: bright region (fringe), , If the light were not coherent, we would not be able to see interference. If the phase, differences varied over time, then there would be constructive, partial, and destructive, interference occurring at every particular spot at different times, and no overall pattern, of interference could be readily observed. For example, if the coherent light from the, two slits were replaced with incoherent light emanating from two light bulbs, only a, uniform glow of illumination would be visible on the viewing screen., For complete constructive interference to occur, two light waves have to be in phase,, meaning there is zero phase difference (or else the phase difference is an integer, multiple of 2ʌ radians, or 360°). At the point shown in Concept 2 the waves have no, phase difference, peak meets peak and trough meets trough, and there is complete, constructive interference. This causes the maxima, or regions of maximum intensity in, the interference pattern., In Concept 3, you see the result at a different location on the viewing screen where light, waves from the two slits meet out of phase, with a phase difference of ʌ radians, or, 180° (the phase difference could also be an odd integer multiple of ʌ radians, such as, 3ʌ radians, í5ʌ radians and so on). With this phase difference, the result is complete, destructive interference. Peak meets trough, trough meets peak, and the waves cancel., Complete destructive interference causes the minima, the stripes of darkness in the, interference pattern., , Destructive interference, Waves out of phase, Result: dark region (fringe), , The regions in between the lightest and darkest points are the result of partial interference. Here, the waves are somewhat out of phase, so, they do not reinforce each other completely, nor do they completely cancel. The visual result is a progression of shades of intensity between, the bright fringes caused by complete constructive interference and the dark fringes caused by complete destructive interference., , 34.2 - Double-slit experiment: wavelike nature of light, The English scientist Thomas Young explored interference and diffraction in a series of, experiments that demonstrated the wavelike nature of light. Young performed his first,, crucial experiment in 1801. Here, we provide a summary of what he did. His actual, procedure was slightly more complex than the description below due to the pioneering, nature of his equipment., Young shined light at a pair of barriers containing slits to create the interference pattern, you see on the right. The first barrier had a single slit that acted as a coherent point, source of light. (This single-slit barrier is not shown in the diagrams.) The coherent light, then traveled through two parallel slits in the second barrier, each the same distance, from the single slit, and then reached a viewing screen. In the diagrams you see the, double-slit barrier and the pattern of interference he observed on the screen., The pattern was one of equally spaced bright and dark fringes. Young knew that water, waves passing through a pair of slits could cause a similar interference pattern. The fact, that both water waves and light produced exactly the same type of pattern supported, his hypothesis that light acted as a wave. Young further reinforced his position when he, demonstrated that he could use interference patterns to calculate the wavelength of the, light that he used in the experiment., , Young’s double-slit experiment, Interference pattern appears on screen, Shows wavelike nature of light, Path lengths of rays to screen differ, , Why do waves create the interference pattern you see here? To answer this question, let’s consider the diagram in Concept 2. Two rays of light, meet at a point that is the same distance from each slit. The drawings of the rays emphasize their wavelike nature., In this case, the two rays intersect the center of the screen in phase because both travel the same distance to this point. They were in phase, when they passed through the slits, and since they traveled the same distance, they remain in phase at their point of convergence. They, constructively interfere and produce a bright area of illumination. Other bright fringes occur to either side of the central fringe where the path, difference between the waves is not zero, but one full wavelength, two wavelengths, and so on., Now look at Concept 3. The rays of light reaching the viewing screen intersect at the first dark fringe to the left of center. They travel different, distances to the surface: The ray from S2 travels a half wavelength farther than the one from S1. Because they start in phase at their, , 628, , Copyright 2000-2007 Kinetic Books Co. Chapter 34
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respective slits, this means the waves are ʌ/2 radians (180°) out of phase when they, arrive at the screen. This causes destructive interference, and darkness., The pattern of bright and dark fringes extends to both the left and the right on the, screen. The light is interfering constructively at the bright fringes, and destructively at, the dark fringes, because of different path lengths to these regions and the resulting, phase differences., There are a few limitations to showing Young’s apparatus in a compact diagram. First,, the diagram is far from being drawn to scale. The screen should be much farther from, the double-slit barrier than we show here, and the slits should be narrower and closer, together. In actual interference experiments, the interfering rays from the two slits are, practically parallel. Second, we vastly exaggerate the wavelength of the light., You may have a question about what you would see if you conducted this experiment, yourself. What if, at some instant, two waves meet at the screen and are in phase, but, their electric and magnetic fields both happen to be zero at that point? Would you see, “flickering” as the two reinforcing waves moved from peak to trough and back again?, The answer is no: The frequency of light is so great that you only perceive the average, brightness of a region; the human eye does not perceive changes in intensity due to the, oscillation of a light wave., , Pattern of bright and dark, Constructive interference: bright fringes, , You do not even perceive flicker in systems oscillating at far lower frequencies, much, less than the frequency of visible light, which is on the order of 1014 Hz. For example, a, computer monitor refreshes its display 60 times a second, but you do not ordinarily, perceive any flicker when you look at it., , Pattern of bright and dark, Destructive interference: dark fringes, , 34.3 - Michelson interferometer, , Interferometer: A device, that uses the interference of, two beams of light to make, precise measurements of their, path difference. Historically, used to study the nature of, light., Michelson interferometer. In this view the telescope is on the left,, the adjustable mirror is on the right, and the fixed mirror is in front., , Albert A. Michelson, an American physicist, is famous, for conducting an experiment that helped Albert Einstein to develop his theory of special, relativity. Michelson’s experiment helped to disprove the existence of the “ether,” an, invisible medium that many scientists believed was required for the transmission of light, waves. In this section, we discuss the design of one piece of equipment he used, now, called the Michelson interferometer, and how it uses the interference patterns of known, wavelengths of light to measure minute differences in the path lengths of two beams of, light., , You see a photograph of an interferometer above, and a conceptual diagram to the, right. We use compass directions like “north” to explain the directions of the various, beams. In the center of the diagram is a beam splitter. It is a plate of glass whose back, is coated with a thin layer of silver. The beam splitter reflects half the light that falls on it,, and lets the rest pass through., The interferometer works in the following fashion:, , Michelson interferometer, Used for precise length measurements, Interference patterns change with path, difference, , 1., , Monochromatic, coherent light strikes the beam splitter from the west side., , 2., , The initial beam splits into two at the beam splitter. The silver coating reflects, beam N to the north, and the splitter refracts beam E, which passes to the east., , 3., , Beam N travels to an adjustable mirror, reflects back, and then passes back through the splitter to a viewing telescope., , 4., , Beam E goes through a compensating plate of glass, reflects off a fixed mirror, passes through the compensating plate again and then, reflects directly off the back of the splitter to reach the viewing telescope., , This system causes two mutually coherent beams of light to travel by different paths. An operator can control the length of the north-south path, , Copyright 2000-2007 Kinetic Books Co. Chapter 34, , 629
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by changing the location of the adjustable mirror. The compensating plate ensures that both beams pass through the same amount of glass,, which is important since light slows down in glass., The purpose of this equipment is to control and detect with extreme precision changes in the path length difference between the two beams of, light. Michelson knew the relationship between path length differences and interference patterns. A path difference of half a wavelength would, cause the beams to interfere destructively, and a path length difference of an entire wavelength would cause complete constructive, interference., To measure a tiny increment of path length, Michelson could place a bright fringe at the center of the image created in the viewing telescope by, setting the distance of the adjustable mirror. When he then moved the adjusting mirror a very small distance using a finely threaded screw, adjustment, the image in the telescope would shift so that a dark fringe would be at the center. He knew he had moved the mirror one-quarter, wavelength because the light traveled that additional distance to the adjusting mirror and then traveled back, making for one-half wavelength, difference in total. A half wavelength path difference is the difference between complete constructive and complete destructive interference. In, actual practice, Michelson’s experiment had a few more “tricks” to it than we have described here, but this section conceptually summarizes, how it worked., , 34.4 - Thin-film interference, , Thin-film interference: The, interference caused by light, waves reflecting off the two, different surfaces of a thin, film., In soap bubbles and in thin layers of gasoline or oil, floating on water, you sometimes see “rainbows,” light, Thin-film interference in soap bubbles., of all the colors of the spectrum. The causes of, rainbows in the sky and rainbows in soap bubbles are, quite different. Those celestial “Pot-of-gold” rainbows are caused by refraction and, reflection. Soap-bubble rainbows are caused by a phenomenon called thin-film, interference., To understand thin-film interference, we start with the fact that when light reflects off a, material with a higher index of refraction than the medium it is traveling in, it changes, phase by 180°. When it reflects off a material with a lower index of refraction than the, medium it is traveling in, there is no phase change., Specifically, when light reaches the front surface of a thin film surrounded by air, some, of it reflects, and changes phase by 180°. Some of the light passes through this first, surface, and then reflects off the far surface of the film. No phase change occurs here., As the light moves between the two media, it refracts as well., This combination of reflection and refraction is illustrated in Concept 1. You see a, downward ray striking the film from the upper left. Some of it reflects when it reaches, the film’s top surface, and we call that reflected ray 1. Some of the light passes into the, film, reflects at the bottom surface, and passes back through the film again. We call that, ray 2., , Thin-film interference, Interference “rainbow” caused by:, ·change of phase on first reflection, ·difference in travel distance of rays, ·variations in thickness of film, , The initial ray of light becomes two rays that have a somewhat complicated relationship in terms of their phase difference. Ray 1 changes, phase 180° as it reflects, and ray 2 does not change phase as it reflects but it travels an extra distance through the film that ray 1 does not., Depending on the path length difference, which equals roughly twice the thickness of the film, the two rays could end up completely in phase,, completely out of phase, or somewhere in between., Interference depends not just on path difference but on wavelength as well. The type of interference occurring at a specific point in a thin film, will differ by the local thickness of the film and the wavelength, or color, of the light. White light has components consisting of many colors, and, in a film of variable thickness these components will interfere differently at different points., What about the rainbow of colors you can observe on a soap bubble or oil sheen? The film of a soap bubble can be thicker or thinner at various, locations. The path length difference at a certain point on a soap bubble may cause destructive interference for red light, but constructive, interference for blue light. This means you perceive this region as blue. At another point, the bubble will have a different thickness, and the path, length difference may cause the opposite result, and you see red there. These patterns can change quickly. A slight breeze or the flow of soap, to the bottom of the bubble will cause parts of it to change thickness, resulting in a new pattern., A region of a bubble that is about to burst is often thin enough that all wavelengths of light destructively interfere there, and the bubble appears, dark. The path length difference through this part of the film is essentially zero, so the change of phase accompanying reflection at the front, surface, which affects all wavelengths equally, causes destructive interference., , 630, , Copyright 2000-2007 Kinetic Books Co. Chapter 34
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34.5 - Diffraction, , Diffraction: The expansion or spreading of a, wave front., In this section, we switch from light to sound waves to discuss the important, phenomenon of diffraction. We briefly switch to sound waves because they provide a, way to describe diffraction using an everyday experience., In Concept 1, we show the diffraction of wave fronts of a sound wave as they pass, through a doorway. Only a section of each wave front can pass through the doorway. If, this section of the wave front did not expand, it would move straight along the gray path., Instead, it expands spherically after passing through the doorway. The spherical, expansion of the sound wave allows the person in the picture to hear the sound., When you hear a friend down the hallway calling you, even though she is out of your, line of sight, you hear her in part because the sound waves diffract as they emerge from, the doorway. For a given size opening, waves of longer wavelength diffract more than, those of shorter wavelength. Because light has a much shorter wavelength than sound,, it would spread out far less after it passed through the doorway. This is why you can, hear the voice of your friend, even when you cannot see her. However, even though, you cannot observe light diffraction for doorway-sized openings, scientists have long, observed it for smaller openings., , Diffraction, Wave fronts “expanding”, , The debate over the cause of diffraction created a tale of considerable irony. The story, starts with a competition sponsored by the French Academy on the subject of, diffraction. The Frenchman Augustin Jean Fresnel (1788 - 1827) submitted a paper to, the judges in 1818 discussing diffraction that was premised on considering light as a, wave., A noted mathematician, Simeon-Denis Poisson, had earlier mocked Fresnel’s theories,, pointing out that Fresnel’s theory predicted that the shadow of a circular opaque object, subjected to a bright light would exhibit a bright spot in its center. Since the circular, object blocks the center portion of a screen from the light source, this result is highly, surprising., , Poisson spot, Bright spot caused by diffraction, , Noted mathematicians can be wrong (or should trust their mathematics). The, Fresnel/Poisson debate was resolved by experiment. In Concept 2, the reddish image, on the screen was created by shining a bright light at a small ball bearing. You can see, the expected bright light surrounding the circumference of the ball, but í surprise! í, there is also a bright spot at the center of the shadow caused by light diffracting around, the ball’s perimeter., Fresnel was right, and Poisson unintentionally helped to confirm his theory. The irony is, that though Fresnel did all the work, and Poisson initially ridiculed the theory, the bright, center today is often called the Poisson spot. Others call it the Fresnel spot, perhaps a, more appropriate name, with Arago spot yet another name (François Arago was the, judge of the competition)., , Straight-edge diffraction, , Waves also expand around sharp edges into regions that would otherwise be in shadow Causes interference pattern, if the wave traveled only in a straight line. This type of diffraction causes an interference, pattern. The photograph in Concept 3 shows straight-edge diffraction. Light from a point source passes by a sharp, well defined edge, and the, resulting diffraction causes a pattern on a screen behind the object., , 34.6 - Huygens’ principle, The Dutch scientist Christian Huygens (1629 - 1695) concluded that light was a wave,, made up of tiny points that emit spherical wavelets like the ones you see in Concept 1, to the right. Although his model has its limitations, it provided a basis for explaining, many phenomena that scientists observed, and his model was usefully employed and, expanded. Here we explain his fundamental principle, and show how his model can be, used to explain diffraction., As Huygens wrote:, …each particle of matter in which a wave spreads, ought not to communicate its, motion only to the next particle which is in the straight line drawn from the, luminous point, but that it also imparts some of it necessarily to all the others, which touch it and which oppose themselves to its movement. So it arises that, around each particle there is made a wave of which that particle is the center., , Huygens’ principle and, diffraction, , Copyright 2000-2007 Kinetic Books Co. Chapter 34, , 631
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In Concept 2, you see the wave front created by the wavelets moving toward a barrier, with a gap in it. Before the wavelets reach the barrier, each wavelet has another, wavelet adjacent to it., , Wave front made up of spherical, wavelets, , Huygens’ principle states that the position of the wave front at any time can be found by, drawing a line tangent to the leading surface of each wavelet. As each wavelet moves, forward, the wave front moves forward as well., Let’s consider what happens when the wave front meets the barrier. The barrier, prevents most of the wave front from moving forward (for simplicity’s sake, we ignore, the reflection of the wave). One section of the wave front (and the wavelets that make it, up) passes through the opening. They continue to move forward after passing through., As Huygens mentions in the quote above, you can think of the sources of the wavelets, as a set of vibrating particles. After the wave front passes through the slit, the oscillating, particles that now make up the wave front are free to interact with other particles above, and below them. The interaction of the particles with their formerly stationary neighbors, and the spherical form of the wavelets account for the spherical expansion of the wave, front. This is illustrated in Concepts 3 and 4., If you like, you can think of the wavelets as a group of subway passengers exiting a, subway car. Their “wave” is constrained by the width of the subway doors, but when, they exit the subway and enter the subway station, their wave can expand amongst the, people awaiting the subway., , Huygens' principle, Wave front tangent to wavelets, same, speed, , Diffraction, Portion of wave passes through slit, After the slit, particles at edges interact, with nearby particles, , Diffraction, Result is a spherical wave front, , 34.7 - Single-slit diffraction, , Single-slit diffraction pattern, Light and dark fringes, , Waves, including light waves, diffract as they pass through a single narrow slit. When a screen is placed on the other side of the slit, a, , 632, , Copyright 2000-2007 Kinetic Books Co. Chapter 34
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diffraction pattern will be created on the screen. This pattern arises from interference, among light waves coming through different portions of the slit. We will assume that the, screen is far enough away from the slit that the rays that pass through are, approximately parallel to each other. This is called Fraunhofer diffraction. (Fresnel, diffraction occurs when the rays cannot be treated as parallel, and is discussed in more, advanced texts.), The result of Fraunhofer diffraction is a pattern of light and dark bands (often called, fringes) on the screen, as shown above. The black-and-white image emphasizes the, light and dark pattern., This pattern of light and dark can be explained using the concept of interfering waves., As with double-slit interference, bright fringes result from constructive interference of, waves and dark fringes from destructive interference. The intensity of the bright fringes, diminishes the farther they are from the midpoint. The first bright fringe is located, straight across from the single slit. The other fringes are located in a symmetric pattern, on both sides of the center., , Use Huygens' principle, Model light as emanating from point, sources, , In the illustrations, we simplify the configuration necessary to produce the diffraction, pattern shown. A lens is typically used to focus the light, making the diffraction pattern, clearer., To explain the source of the interference, we use Huygens’ principle and treat the light, passing through the slit as though it were made up of individual waves (wavelets), emanating from a series of point sources, as shown in Concept 2. We focus on the, waves emanating from just two of those point sources to simplify the drawings and, explanations. As with double-slit interference, the difference in the waves’ path lengths, to the screen (and any resulting phase difference) determines whether they interfere, constructively, and create a bright fringe, or destructively, to create a dark fringe., First, we show constructive interference. In Concept 3, we consider two waves from the, edges of the slit. They meet at the center of the screen. Since they travel the same, distance, there is no path length difference, which means they arrive in phase. Any, point source within the slit can be matched to a corresponding “mirror point” that is an, equal distance from the midpoint, but on the other side. The interference is completely, constructive, and the result is the central bright fringe, which is the brightest in the entire, diffraction pattern., , Constructive interference, Causes bright fringes, , In Concept 4, we show how completely destructive interference creates a dark fringe. In, this case, we consider a wave on the left edge of the slit and a wave from the center of, the slit. The wave on the left travels one-half wavelength less to reach the screen than, the wave on its right. This means they will be completely out of phase when they meet, at the screen. In fact, every wave has a corresponding wave exactly half a slit away that, will cancel it at that screen location., The regions between the lightest and darkest points are the result of intermediate, interference. In these regions, the overall interference is neither completely destructive, nor completely constructive, and the brightness at these points is between that of the, points just discussed., , Destructive interference, Causes dark fringes, , 34.8 - Resolving power, , Resolving power: Ability to distinguish, between two objects., Resolving power expresses the capability of an optical system to show the separation of, objects that are close together. This correlates to the instrument’s ability to show fine, detail. A microscope is one example of an instrument that needs a good deal of, resolving power. To be effective, a microscope must enable you to distinguish very, small, close details. Telescopes also must supply great resolving power to allow the, viewer to separate distant objects., In concept 1 we show the importance of resolving power. The image of galaxy M100 on, the left was taken by the Hubble Space Telescope before a defect in its mirror was, corrected; the image on the right was taken after the defect was fixed. The right-hand, image shows much more detail, and many more distinct objects are visible due to the, greater resolving power achieved., , Resolving power, Measures ability to distinguish between, two objects, , The Hubble telescope can resolve light sources that are less than 0.00003° (§ 0.0000005 rad) apart. What does that mean on a human, scale? It could resolve a pair of headlights roughly 3000 km away, approximately the distance from Denver to Miami. A great conceptual, design, some repairs and ongoing maintenance, and the fact that the telescope is above the blurring effects of the Earth’s atmosphere, all, , Copyright 2000-2007 Kinetic Books Co. Chapter 34, , 633
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contribute to this mind boggling capability., You do not need a telescope to understand what is meant by resolving power. You can, experience it simply by looking at a car’s headlights at night. At a large distance, the, car’s lights look as if they come from a single source, in one blur of light. As the car, comes closer, the blur resolves into two separate sources of light., Diffraction causes the stars and the headlights to “blur” together. Concept 2 shows the, diffraction pattern created by two sources that are far enough apart that their diffraction, patterns do not significantly overlap. Concept 3 shows what occurs when the diffraction, patterns overlap. One “blurry” central image is the result., , Two resolved images, Diffraction patterns remain separate, , Blurry image, Significant overlap of diffraction patterns, , 34.9 - Resolving power of the eye, Human eyes, and those of other animals, resolve objects. From ten meters away, a, person with good eyesight can resolve two objects separated by only three millimeters., Some animals can do far better. At the same distance, an eagle can resolve two objects, that are only 0.8 mm apart. This ability allows eagles in flight to pick out potential prey, on the ground with amazing accuracy. For example, an eagle can distinguish a rabbit, from its surroundings from as far as a mile away., Painters and other artists sometimes exploit the limitations of the human eye to create, their effects. The style of painting known as Pointillism was famous for this. To the right,, we have created our own Pointillist art. The “zoom-in” shows that the picture is, composed of many small “dabs” of color. As you step back, your eye can no longer, resolve the individual dabs, and you see the collective image., If your school lacks a first rate collection of Post-Impressionistic paintings, the same, effect is demonstrated in your daily newspaper. Look closely enough at a picture and, you will see the dots that compose it., , Lack of resolution leads to art, , 34.10 - Physics at play: CDs and DVDs, Most computers today contain a CD-ROM drive and many contain a DVD drive. A CD, or DVD can hold a great quantity of information. The drive relies on the principles of, interference to “read” the disc., A CD or DVD contains a long spiral track with pits in it. These pits are formed in a disc, by an injection molding process, and represent some of the smallest mechanically, manufactured objects. A thin layer of metal such as silver or aluminum covers the pits., This layer in turn is covered by a thin layer of plastic., The pits are created on the top of the CD or DVD but the disc is read using a laser that, is projected up from the bottom. From the bottom of the disc, the pits appear to be, raised areas. Non-pitted areas of the disc are called land. (An incidental fact: The pits, are nearer the top of the disc than the bottom, so scratches on its top are more likely to, damage the CD than scratches the bottom, or “reading” side!), CDs or DVDs created by burners do not create pits in the fashion described above, but, rather change the color of a layer within the disc., The CD or DVD reader contains a laser diode that emits a beam of light that reflects off, the disc. The intensity of the reflected light varies as the disc rotates and the light, , 634, , CDs and DVDs, Laser shines light onto track of CD, DVD, Discs have “pits” in surface, Sensor reads signal from laser, , Copyright 2000-2007 Kinetic Books Co. Chapter 34
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reflects off pits and land. The intensity is measured and interpreted as a series of ones, and zeros (digital information) by photodetectors. This information is then relayed to, other systems that interpret it., How does interference factor in? The laser beam reflects off of the CD. If all of the, beam hits a land or a pit, then the path length difference back to the photodetector is, essentially the same, and the result is constructive interference: bright light. You see, this case in Concept 1., On the other hand, when the disc moves and laser light is half on a pit, and half on the, land, the path length difference is significant. The two parts of the laser beam have a, total path length difference of one-half a wavelength, and the result is destructive, interference: darkness., You see this in Concept 2, where we emphasize “sides” of the same laser beam, and, how one side reflects off of a pit and the other off a land., DVDs contain more data than CDs and employ a variety of strategies to do so. For, instance, DVD drives use lasers with shorter wavelengths. A shorter wavelength means, smaller pits are possible, and these smaller pits can be placed more closely together,, allowing more data to be stored., , The signal, Interference pattern depends on where, light strikes, Sensor receives brighter or dimmer light, , 34.11 - Gotchas, In a string, complete destructive interference occurs when the peaks of one wave meet the troughs of another wave with the same amplitude., The same is true with light. This is correct. With a string, the result is zero displacement. With light, the result is darkness., A double-slit interference pattern is caused by differences in the path lengths traveled by the light emanating from each slit. This is true., If the path length difference in a double-slit interference pattern equals one wavelength, the result is complete destructive interference. No, the, result is complete constructive interference. When the path difference is one wavelength, the waves are in phase: Peak meets peak, and, trough meets trough. A half wavelength path difference will cause destructive interference., , 34.12 - Summary, Light exhibits the properties of both a particle and a wave. The interference patterns created by light can be explained by treating it as a wave., Interference patterns consist of alternating bright and dark bands called fringes. To create an interference pattern, you need at least two, sources providing light that is both monochromatic (having only one wavelength) and coherent (light from the different sources has a phase, relationship that does not change over time). You also need a screen on which to view the pattern., The bright fringes in a two-slit interference pattern are the result of completely constructive interference of the light from the two sources, while, the dark fringes are created by completely destructive interference., Whether a given point on a viewing screen will be a point of constructive, destructive, or intermediate interference depends on the difference in, the path lengths from each of two slits to that point., An interferometer is an instrument that takes advantage of interference to make precise measurements of length. It relies on the fact that a, specific difference in the path lengths of two monochromatic coherent beams of light causes a specific interference pattern at a point where the, beams meet, and a microscopic change in the path difference causes an easily visible change in the interference pattern., When a light wave reflects from a material with a higher index of refraction than the one in which it is traveling, it experiences a 180° phase, change. When a light wave reflects from a material with a lower index of refraction, there is no phase change. This affects the nature of thinfilm interference, where waves that reflect from the front surface of a thin film interfere with other waves that refract through the front surface, and then reflect from the back surface., Diffraction is the expansion or spreading of a wave front as it passes through an opening or past a sharp edge., Huygens’ principle is a model to explain diffraction. It says that a wave front is made up of a series of spherical wavelets., Diffraction is most often discussed in terms of light waves. When light passes through a slit in a mask, it diffracts. We can model the light, passing through the slit as coming from a series of point sources which each emit a spherical wave front. The light from each point source, interferes with light from the others. This creates an interference pattern of dark and light fringes., Diffraction limits the ability of optical systems to distinguish between two objects, to resolve them., , Copyright 2000-2007 Kinetic Books Co. Chapter 34, , 635
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Chapter 34 Problems, , Chapter Assumptions, Unless stated otherwise, use the following indices of refraction, , nair = 1.00, nwater = 1.33, nglass = 1.50, , Conceptual Problems, C.1, , Two beams of light, initially in phase, follow, the paths shown. In terms of the wavelength Ȝ,, what is the minimum length d that will result in, complete constructive interference between, the two beams?, , Ȝ/6, , C.2, , Ȝ/2, , 3Ȝ/4, , In a single-slit diffraction pattern, when destructive interference occurs at a dark fringe, does this mean energy has been lost, from the light wave? Explain your answer., Yes, , C.3, , Ȝ/3, , No, , Two beams of light, initially in phase, follow, the paths shown. In terms of the wavelength Ȝ,, what is the minimum length d that will result in, complete destructive interference between the, two beams?, 4Ȝ, , Ȝ/8, , Ȝ/4, , 3Ȝ/2, , C.4, , In the old days before cable television, everyone received broadcast television signals by using an antenna perched on the, television or on the roof. You might think that if you had a clear line of sight to a nearby broadcasting station, you would, always receive a strong signal, but this is not true. Explain why the presence of surrounding buildings, which do not block the, signal but do reflect some of it, might reduce the signal received at a television antenna with a clear line of sight., , C.5, , In recent years, cancer researchers have proposed a new way of treating previously inoperable brain tumors. Some, malignant tumors can be destroyed by focusing a narrow, high-energy x-ray beam on them. The beam must have a high, power, but at high intensities, a single beam will destroy healthy tissue in the path to the tumor. At lower powers, a single, beam will keep the healthy cells alive, but it will not kill the cancer cells. In order for the treatment to be effective, a tumor must, receive a full concentration of energy at one instant in time, but healthy cells must never receive the full concentration. How, can this be done?, , Section Problems, Section 3 - Michelson interferometer, 3.1, , You are using a Michelson interferometer to measure the wavelength of the orange-red emission line in the spectrum of the, krypton-86 atom. When the movable mirror is shifted by a distance of 303 nm, the interference between the two beams moves, from constructive to destructive back to constructive. To three significant digits, what is the wavelength of this emission line?, m, , 636, , Copyright 2000-2007 Kinetic Books Co. Chapter 34 Problems
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3.2, , The adjustable mirror in a Michelson interferometer is moved a distance of 0.223 mm and 769 bright fringes cross the field of, view of the telescope. What is the wavelength of light that is being used?, m, , 3.3, , A standard helium-neon laser has a wavelength of 633 nm, and is used in a Michelson interferometer experiment. If the, adjustable mirror is moved a distance of 0.0216 mm, how many bright fringes cross the field of view of the telescope?, , 3.4, , The textbook speaks of moving the adjustable mirror in a Michelson interferometer by as little as a quarter wavelength of, light, using a "finely threaded screw adjustment." This is an interesting claim: Just how finely threaded does such an, adjustment screw have to be? Assume that the screw can be turned by as little as one-quarter of a degree with fair accuracy., How many screw threads must there be in each millimeter of the screw's length in order for a turn of one-quarter degree to, advance the screw by 125 nm, a quarter of a wavelength of blue light? State your answer to the nearest hundredth. Hint: The, pitch of any screw is defined as the distance between adjacent threads, or equivalently, the distance the screw advances after, one full turn, 360°., , fringes, , threads/mm, , Copyright 2000-2007 Kinetic Books Co. Chapter 34 Problems, , 637
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35.0 - Introduction, With a pair of brilliant papers published in 1905 and 1915, Albert Einstein inaugurated a revolution in physics. Scientists still are happily, grappling with the implications of his work in these and other papers. His special theory of relativity and his later work predicted a range of, phenomena from the amount light is “bent” by gravity, to time passing at a different rate for a passenger in a moving airplane than for an, observer on the ground. To their great delight, when scientists went looking for these effects, they found them., This revolution was all the more surprising since a distinguished scientist, Lord Kelvin, had not long before remarked: "There is nothing new to, be discovered in physics now. All that remains is more and more precise measurement." Today’s physicists are more realistic, which certainly, makes their research more interesting!, Strange and wonderful ideas emanate from Einstein’s work. His special theory of relativity predicts that an identical twin could leave her sister,, fly off into space at nearly the speed of light and, upon her return, have aged 20 years less than her sibling. Other parts of his work, outside the, realm of this chapter, have led to the discovery of black holes, objects with mass so concentrated and possessing such strong gravitational, fields that even light cannot escape them., Einstein himself found some of the implications of his research and that of his peers too incredible to believe. For instance, his work predicted, that the universe is endlessly expanding. He found this idea troubling enough that he added a “cosmological constant” so the equations would, predict a universe of a constant size. Subsequently, when the astronomer Edwin Hubble introduced evidence that the universe is indeed, expanding, Einstein gladly removed the constant from his work, saying it was the biggest mistake he had ever made. Today, the debate about, the merits of the cosmological constant continues as physicists continue their research., Although at times seemingly “incredible,” Einstein’s essential theories have been tested and proven by scientific experiments. The results have, confirmed his work to a level of precision of about one part in 1015. Physicists believe that with advances in equipment and approach, they can, confirm that the data conforms to his theories to even higher levels of precision., This chapter focuses on Einstein’s special theory of relativity, as opposed to his later and more complex general theory. Einstein’s special, theory of relativity is based on two postulates., The first states that the laws of physics are the same in any inertial reference frame. What constitutes an inertial reference frame merits more, discussion; briefly, you can consider any system moving at a constant velocity to be an inertial reference frame. In such a reference frame, the, “classical” or Newtonian laws of physics hold true. Let’s say you are either in a train moving at a constant 125 km/h, or standing on the ground., In either situation, you can throw a ball up in the air, and predict where it will land. (Note: the Earth itself is not truly an inertial reference frame, because its rotational motion means that supposedly “fixed” objects are actually accelerating. However, we typically ignore this because of its, minor impact.), In sum, the laws of physics hold true in the train and on the ground, in France as well as in Germany. This is a postulate you likely assumed:, that the physics you study do not vary by location. They hold true for any inertial reference frame. You could not use Newtonian mechanics in a, bumpy truck as it drove along a winding country road, but this would be a quite atypical location for you to conduct lab experiments., Einstein’s second postulate states that the speed of light in a vacuum is the same in all inertial reference frames: It does not change due to the, motion of the source of the light or the motion of the person observing the light. This insight is surprising, and does not accord with, observations of everyday events., For instance, if someone in a train moving toward you at a high speed throws a tennis ball out a window, you expect the tennis ball to be, moving toward you. You intuitively add the velocity of the train to the velocity with which the ball was thrown to determine its overall velocity., You combine the velocities., Einstein correctly stated that this is not true with light: Its speed does not change. If someone flashes a light at you from the train, whether the, train moves toward you or away from you, the speed of light you measure remains the same., With these two postulates, Einstein started a revolution. This chapter covers these two postulates in more depth, and then summarizes many of, their amazing implications., , 35.1 - Reference frames, , Reference frame: A coordinate system used to, make observations., We discussed reference frames many chapters earlier, in the context of motion in, multiple dimensions. Since that was many chapters ago, we reprise the section here,, though we have changed the examples and some of the discussion in order to make, this section more appropriate for this chapter. Reference frames are a crucial element, of special relativity., A reference frame is a coordinate system used to make observations. If you stand next, to a lab table and hold out a meter stick, you have established a reference frame for, making observations. Your choice of a reference frame determines your perception of, motion., , 638, , Reference frame, Point of view for observing motion, , Copyright 2007 Kinetic Books Co. Chapter 35
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We use the classroom shown on the right to discuss reference frames. The classroom, shows a professor on a skateboard and Katherine, a student in the professor’s class., The professor conducts a demonstration in the class: He throws an eraser up, and, catches it in the same hand. The professor does all this while moving across the, classroom on a skateboard. Both the professor and Katherine have stopwatches to, measure the time interval between “the throw” and “the catch”., In Concept 1, we show what Katherine observes. She sees the professor moving by at, a speed v. The blue arc shows the path of the eraser from her reference frame. It, moves both vertically and horizontally in projectile motion. A grid is shown in the, illustration. Each side of the grid is 1.0 meter. Katherine sets the position of the “throw”, at (x = 0, y = 0) m, and the catch at (3.00, 0) m. The eraser reaches its peak at (1.50,, 1.50) m in her reference frame., Katherine can also establish the time coordinates of these events. She starts her, stopwatch when the professor throws the ball, so the throw is at t = 0 s. He catches the, ball at t = 1.11 seconds., , Reference frame, Motion is perceived relative to a, reference frame, , In Concept 2, we show the exact same series of events as observed in the professor’s, reference frame. He considers himself as stationary and views the class as moving by. It is the same experience as looking out the window of, an airplane: You consider yourself stationary as you sit in a seat, and the ground is passing by., In his reference frame, the eraser travels solely vertically. Its initial and final positions are (0, 0) m. Its peak position is (0, 1.50) m. He also has, a stopwatch, which he also starts when he tosses the eraser. His observations of the time (to the precision of this stopwatch) are identical. He, throws the ball at t = 0.00 s, and catches it at t = 1.11 s. To describe the catch, he could describe its space time coordinates as (0 m, 0 m,, 0 m, 1.11 s). The first three coordinates state its x, y and z position, and the last states the time., The conclusion of all this: Reference frames determine the observations made by observers. Katherine observes the eraser moving, horizontally as well as vertically; the professor sees it move only vertically., You may object: But does the professor not know he is moving? Should he not factor in his motion? Consider throwing an eraser up and down, and catching it. Did you catch it at (roughly) the same position? In your reference frame, perhaps your classroom, the answer is “yes”. But to an, observer watching from the Moon, the answer is no, since the Earth is moving relative to the Moon., There is no correct inertial reference frame. Katherine cannot say her reference frame is better than the reference frame used by the professor., Measurements of position, time and other values made by either observer are equally valid., In your earlier studies, you were asked to assume that the time intervals in each reference frame were identical. At the speed the professor is, moving, and to the hundredth of the second, the two observers measure the same time interval. However, if the professor were moving at, say,, 75% of the speed of light, the time intervals would be quite distinct. Discussing how measured time intervals change as the observers’ relative, speeds approach the speed of light is a major topic in this chapter., , 35.2 - Events and observers, , Event: Something that can be pinpointed using, position coordinates and time., Observer: Person who records where and when, an event occurs in a particular reference frame., The definition of an event may accord with your own sense of the word í it is something, that occurs at a specific place and time. A bat striking a ball is an event; the ball striking, the glove of a fan in the bleachers is another event. In physics, a “Saturday night dance”, is not an event, since it does not occur at a specific time or a specific enough location., In the illustrations to the right, we consider a single event: The professor on the train, kicks a soccer ball. He is standing on a train that is moving at a constant velocity on the, track. Another observer, Sara, stands on the ground and observes the same event, the, professor kicking the soccer ball. An observer is someone who records when and where, something occurs in a reference frame., , An event, “Something that happens at a specific, time and location”, Example: Professor’s foot meets ball, , Both agree that the professor kicked the soccer ball, and that it rolls along the train. But they use different reference frames to describe where, and when the soccer ball was kicked., In Concept 2, you see the position and time of the event described using Sara’s reference frame. She uses a set of position markers on the, ground. (You can see markers like this, though at a scale greater than meters, on railroad tracks and highways). In her reference frame, the, ball was kicked at the position 2.8 meters, and her watch tells her it was kicked at 5:00 PM Pacific Standard Time., In Concept 3, we show the same event, but now the professor describes it using his reference frame. He measures position using a scale on, the train. Using that reference frame, he states that he kicked the ball at the position 0.7 meters. He also has not changed his watch since he, set out on his journey, so his watch tells him that he kicked the ball at 8:00 PM Eastern Standard Time., These two sets of observations can be reconciled, but they illustrate how the reference frames of the two observers determine the coordinates, , Copyright 2007 Kinetic Books Co. Chapter 35, , 639
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of the observations they make. Both sets of observations are equally valid, given the, reference frames of the observers, but they do differ. They provide a starting point in, special relativity: Reference frames play an essential role in the observations made by, observers., , An observer, Person who records where, when event, occurs, Uses reference frame to make, measurements, Sara: Foot met ball at 2.8 meter marker, at 5:00 P.M. PST, , Observer in different reference, frame, Observers in different reference frames, make different observations, Professor: Foot met ball at position 0.7, meters at 8:00 P.M. EST, , 35.3 - Light can travel through a vacuum, As the 20th century dawned, the topic of how light travels generated much debate in the, physics community. The evidence brought to bear on this debate helped Einstein to, confirm one of his central insights: The speed of light in a vacuum is constant. The, speed of light does not depend on the motion of its source or of the observer. This, conclusion is one of Einstein’s two postulates; the experiments that enabled him to, deduce this are an interesting story in the history of science., The primary experiment that propelled Einstein toward his conclusion was not intended, to lead to conclusions about the speed of light. Rather, it was an experiment about the, existence and nature of the medium through which light travels. Earlier observations, had convinced scientists that light acted as a wave. Experiments had shown that a, beam of light spreads out (diffracts) in the same manner as a wave of water, and that, light produces interference patterns that are consistent with the patterns caused by, waves as well., , Light travels through vacuum, If light traveled through ether, , Since physicists knew light acted as a wave, they went in search of the medium in, ·It would travel faster when moving with, which it traveled. They reasoned that waves always move through some form of, the ether, ®, medium í water, the wire of a Slinky , the strings of a violin. However, the medium for, light was mysterious. Scientists were puzzled by the fact that light travels through the, near vacuum of space, where there apparently is no medium. Physicists assumed there must be a medium, and called this elusive medium the, ether. They assumed the ether permeated the universe, including the Earth’s atmosphere., Some physicists were skeptical about the existence of the ether. Nonetheless, they looked for ways to measure its attributes. Two clever, American physicists, Albert Michelson and Edward Morley, conducted an experiment that played a key role in proving that ether did not exist., Michelson and Morley reasoned: If light travels in a medium, it should move faster when traveling in the same direction as that medium and, slower when traveling against it. In other words, light should act like a swimmer in the ocean: An observer on the ground would see the, swimmer moving faster when swimming with the current and slower against it., , 640, , Copyright 2007 Kinetic Books Co. Chapter 35
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The most prevalent theory of ether stated that it was stationary and the Earth moved, through it. An object stationary on the Earth would be moving through the ether., Michelson and Morley decided to measure the speed of light when the light was moving, in different directions relative to the Earth’s motion around the Sun. If ether existed, the, speed of light would be different, and would be slowest when it moved in the same, direction as the Earth’s motion through the ether., Here is an analogy that may make this clearer. Imagine two birds flying at the same, speed in opposite directions, while you observe them from a slow-moving train, with the, train’s motion causing you to feel air in your face. If the train is moving in the same, direction as one of the birds, that bird will appear to fly slower than the other bird. The, birds are like light, the air is like the ether, and the train is like the Earth., The two physicists searched for changes in the speed of light using an apparatus now, known as a Michelson interferometer. Michelson estimated that the Earth’s motion, through the ether would cause a change in the speed of light on the order of, 30,000 m/s. The interferometer was capable of detecting this 0.01% change in the, speed of light., , Light travels through vacuum, No difference in speed measured, ·There is no ether. The speed of light is, constant, , Michelson made multiple observations, measuring the speed of light with different, orientations relative to the Earth’s rotation and motion around the Sun. He observed no changes in its speed. As he tersely summarized: “The, result of the hypothesis of a stationary ether [with respect to the Earth] is thus shown to be incorrect.” (Michelson also considered the case that, the ether was dragged along with the Earth; this possibility was later shown to be false as well.), Michelson’s conclusion was correct. His experiment indicated that the effects expected when waves propagate through a moving medium did, not occur. Despite some efforts to revive the theory of ether, the clarity and simplicity of Michelson’s experiment was a major and conclusive, blow to the theory of ether., If this experiment had proven only that light could travel in a vacuum, disproving the existence of the ether, it would have been a landmark in, the history of science. But Einstein realized that something even more interesting was being shown by the experiment: The motion of the, equipment was not affecting the observed speed of light in the way classical physics predicted it should. As the light moved in the, interferometer, the equipment itself was moving, which according to the physics as Einstein had learned it should alter the effects of the, experiment. In other words, even without an ether, the equipment itself was moving, and that should have affected the measurement of the, light., In his 1905 paper on special relativity, Einstein wrote, “…the unsuccessful attempts to discover any motion of the Earth relative to the ‘light, medium,’ suggest that the phenomena of electrodynamics as well as of mechanics possess no properties corresponding to the idea of absolute, rest.” Michelson’s experiment helped Einstein to formulate his second postulate: The speed of light is absolute; it does not depend on the, reference frame. It also supported Einstein’s first postulate. If the ether frame existed, the laws of light propagation would have been different in, that frame, which would have contradicted the requirement that the forms of the laws of physics be the same for all observers in inertial, reference frames., , 35.4 - Simultaneity, or the lack thereof, , Simultaneity, Events perceived as simultaneous in, one reference frame, Not simultaneous in another, , Einstein’s simultaneity thought experiment: The relative motion of two, observers determines whether they perceive two events as simultaneous., One of the consequences of Einstein’s postulates is that the perception of “simultaneous” events is relative to an observer’s reference frame., Einstein demonstrated that his postulates proved that two observers would disagree on how much time elapsed between two events, or on, whether two events occurred simultaneously. In this, he disagreed with Sir Isaac Newton. Newton had stated: “Absolute, true and mathematical, time, of itself, and from its own nature, flows equably without relation to anything external.” Einstein was to prove this wrong., To make his point, he used a specific means to define simultaneity, noting that if an observer sees two events that are the same distance away, occurring at the same moment in time, she thinks the two events occurred at the same moment in time. It seems a commonsense conclusion, that another observer, also equidistant from the events, would see them occurring simultaneously. Einstein’s genius lay in challenging this, conclusion. He created a scenario showing that the relative motion of two observers influenced whether they perceived two events as occurring, simultaneously, or as occurring one after the other., Einstein made his case in part with a famous “gedanken experiment,” or “thought experiment.” A thought experiment is one that is conducted in, the mind, as opposed to in a laboratory. Einstein used his thought experiment to make the following revolutionary point: The measured length, , Copyright 2007 Kinetic Books Co. Chapter 35, , 641
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of time intervals differs when observers are in motion relative to one another., To state Einstein’s thought experiment, we use a train and a platform beside the train. A, professor stands still on an open railcar at the middle of the train. Another observer,, Katherine, stands still at the middle of the platform. Two lightning rods, shown as gray, towers in Concept 2 and Concept 3, are equidistant from Katherine and the midpoint of, the platform., We start this experiment with the train not moving, and the professor directly across, from Katherine, so he is also equidistant from the lightning rods. A lightning bolt strikes, each lightning rod. These lightning strikes cause flashes of light that will be visible to the, observers. Photons í packets of light í move from each lightning rod toward them., Each lightning strike counts as one of the two spatially separated events we are, analyzing. We draw the photons from one rod in blue, and the photons from the other, rod in red, so that you can more easily distinguish them., When will an observer conclude that the two events are simultaneous? The professor, concludes that the bolts struck simultaneously only if two photons reach him at exactly, the same moment. Why? Since both travel the same distance, the events are, equidistant from him. The professor reasons that if the photons reach him, simultaneously after traveling the same distance at the same speed, they must have, started their journey simultaneously., , Train stationary relative to, observers, Light reaches each observer at an, instant in time, Both think the strikes occurred, simultaneously, , Katherine also concludes that the bolts struck simultaneously only if two photons reach, her at a single instant in time. She too believes the events occurred simultaneously if, two photons, traveling the same distance and the same speed, reach her, simultaneously., All this is “as expected”. Two lightning bolts strike rods, photons from the strike reach, each observer simultaneously, and they conclude that the two bolts struck, simultaneously because the photons traveled the same distance to each observer at the, same speed. This situation is shown in Concept 2, as the photons reach Katherine and, the professor., Now, Einstein changed the scenario: He put the train in motion. Einstein asked again:, would they both say that the lightning bolts struck simultaneously? That the time interval, between the events was zero?, The answer, as illustrated by Concept 3 is: No. In the scenario we show, Katherine still, states that the lightning bolts struck simultaneously because both photons reach her, simultaneously. However, as you can see, the photons from the strike on the right have, reached the professor, but those on the left have not. His conclusion is that the strike on, the right occurred before the strike on the left., Although they cannot be readily seen in the illustration, this time we have the strikes, make slight scorch marks on the train. The professor can later walk up and down the, train and reassure himself that the distance between each strike and his position as an, observer is the same., , Train moving relative to one, observer, Professor on train: Strikes NOT, simultaneous, Katherine on platform: Strikes, simultaneous, Observers disagree on whether events, are simultaneous, , Einstein’s thought experiment with the train showed that the perception of the time interval between events depends on the observer’s frame of, reference. Specifically, he showed how two events that are perceived as occurring simultaneously in one reference frame would be perceived, as occurring at two different moments in time (non-simultaneously) in another reference frame that is in motion with respect to the first., Both observers were correct, based on observations in their own reference frames. This phenomenon is called the relativity of simultaneity. It, means that time intervals between events í something that Newton assumed was absolute í are actually relative, that is, they vary depending, on an observer’s state of motion., As Einstein wrote: “Are two events (e.g. the two strokes of lightning A and B) which are simultaneous with reference to the railway embankment, also simultaneous relatively to the train? We shall show directly that the answer must be negative.”, An observer’s measurement of time intervals depends on his or her frame of reference. Time is relative. Motion alters time. This is a strange, conclusion, but it is true, and it changes our view of the universe. Few thought about it before Einstein, because in his day, the effect was, inaccessible to measurements at the low speeds humans were accustomed to observing. For instance, for a jogger moving at several meters, per second past a stationary observer, the difference is negligible (about one part in 1016), but the effect becomes much more significant as the, relative speed between two observers increases., Since Einstein’s time, technological advances have made it necessary to factor this effect into the design of certain systems. You may have, seen or used a Global Positioning System (GPS) unit, which receives signals from a network of orbiting satellites to pinpoint its location on the, Earth’s surface. The lightning strikes we considered are analogous to the signals that are emitted by satellites. The GPS unit works by, interpreting the time delays in the electromagnetic signals that are received from these fast-moving satellites whose positions are precisely, known. Relativistic considerations play a central role in making the system work., , 642, , Copyright 2007 Kinetic Books Co. Chapter 35
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35.5 - Interactive problem: Conduct Einstein's simultaneity experiment, There are two simulations to the right. In the first, the professor, Katherine and the, train are all stationary. In the second, the professor and the train move., As discussed before, lightning bolts strike each lightning rod. These lightning strikes, cause flashes of light that will be visible to the observers. Photons í packets of light, í move from the position of each lightning rod toward them. Each lightning strike, counts as one of the two spatially separated events we are analyzing. We draw the, photons from one rod in blue, and the photons from the other rod in red, so that you, can easily distinguish them., The lightning bolts also scorch the sides of the train as they hit the rods. The, professor can use these scorch marks to confirm that the two events are equidistant, from the train’s center where he is standing., Press GO to launch the lightning strikes. The simulations run in extreme slow, motion. We have slowed time down by a factor of more than fifty million to clearly, show what is happening., The first simulation is in Katherine’s reference frame. The second simulation starts, in Katherine’s reference frame. You can view the same events in the professor’s, reference frame by pressing the “Professor’s reference frame” tab., Try each simulation. Do Katherine and the professor observe the lightning bolts as, striking the lightning rods simultaneously when the train is stationary? When the, train is moving?, In order to let you experiment further with this thought experiment, both simulations, contain a feature on the control panel that allows you to adjust the interval of time, that Katherine observes between the lightning strikes. You can use this controller to, see how she and the professor observe the events as this interval changes. For, instance, you can cause there to be a five-nanosecond interval of time between, when one photon reaches her and the next photon reaches her. This lets you go, beyond Einstein’s thought experiment, where the time interval in Katherine’s, reference frame was always zero and she observed the lightning strikes as, occurring simultaneously., In the second simulation, can you adjust the time interval in Katherine’s reference, frame to cause the photons to arrive at the same instant at the professor’s location?, If you cause this to happen, what does Katherine now observe? Can you cause them to both believe the strikes occurred simultaneously when, the train is moving? (Hint: Do not spend too much time trying!)., , 35.6 - Time dilation, , Proper time, Clock is at same location as events, , Time dilation: A clock moving relative to an, observer runs more slowly, measuring longer, time intervals, than clocks at rest relative to the, observer., Einstein pondered: If the speed of light is constant, what does this imply? He concluded, that it means that things we think of as absolute í like the time interval between events, í must be “relative.”, Consider the professor and the student, Katherine, in the diagrams above and on the, right. The professor moves rapidly through the classroom on his skateboard. He throws, and catches the ball, using his stopwatch to measure the time interval between the, throw and the catch. You see this illustrated in Concept 1., , Time dilation, Observers differ on how much time, passes between events, , Copyright 2007 Kinetic Books Co. Chapter 35, , 643
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Katherine watches all this and measures the time between the throw and the catch, using her stopwatch. This is shown in Concept 2., If they compare their observations afterwards and had incredibly precise equipment,, Katherine will discover that she measured more time passing between the two events, than the professor does with his stopwatch. If the professor moves at everyday speeds, as we show above, the effect will be very small, far too small to be measured on any, typical stopwatch. But if the professor were moving past at 87% of the speed of the, light, Katherine would measure over twice as much time between the two events as the, professor does., , “Moving” observer/clock measure less, time, , The professor’s time is said to be “dilated”, which means it is passing more slowly. This, is not due to some malfunctioning nor change in functioning of his clock. Any clock in, his frame, including the professor’s “biological clock”, will record less time passing., Time dilation can be quantified, as the equation at the right shows. To understand the, equation you must understand another term: proper time. The proper time between two, events is the time measured by a clock in the reference frame where the two events, occur in the same place. “Proper” comes from the German for the events’ “own time”., The proper time is measured with a single clock, which is at rest in the reference frame, in which the events occur at the same xyz coordinates. The professor’s stopwatch, measures the proper time., Another observer, the student Katherine in this case, watches the professor (and his, stopwatch) pass by at a velocity v. On the right side of the equation is the proper time, interval, t0. Katherine measures an interval of time t between the events., We also present the equation in another way. The term that multiplies the proper time,, that is, the reciprocal of the square root term, is frequently used in relativity. It is, represented by the Greek letter Ȗ (gamma) and is called the Lorentz factor., Physicists have confirmed the principle of time dilation by many methods, such as, studying muons. Muons are subatomic particles that have an average half life of about, 2.2 microseconds. That is, in a sample of muons that are stationary relative to an, observer, half of them will decay (change) into other particles during this interval of time., , t = “stationary” observer time, t0 = proper time in “moving” frame, v = speed of reference frame, c = speed of light, t = Ȗt0, , Fast-moving muons are produced when cosmic rays enter the Earth’s atmosphere and, collide at high speeds with atoms there. The muons travel toward the ground at about, 0.999c. Scientists observe that these moving muons decay more slowly than stationary, muons., Why? The fraction of muons that decay is a function of time, and less time elapses in, the reference frame of the moving muons. In fact, time passes about 22 times more, slowly in the moving muons’ reference frame. If 2.2 μs have passed according to the, scientists’ clock, then only 0.1 μs have passed in the moving muon reference frame,, which is a time interval much shorter than the half life of the muons. The scientists see, that far less than one-half of the muons have decayed in 2.2 μs, and conclude that the, moving muons decay more slowly. In the reference frame of the muons, the decay rate, is unchanged. Time dilation explains the discrepancy., You can confirm the “22 times” ratio using the time dilation equation. The proper time is, being measured by the muons’ “clock”, which is observed via their decay rate., Experiments with highly precise clocks have also confirmed Einstein’s conclusions, concerning the effect of motion on time. For instance, scientists have measured a, difference in the time interval measured by a clock in a plane to that measured by a, clock on the ground. Einstein’s general theory of relativity and its predictions figure, prominently into the results, but his two theories account for the discrepancies between, the time intervals measured by the clocks., Do you experience time dilation? Yes, but very small amounts of it, given how slowly, you move compared to the speed of light. Using the equation to the right, you could, determine that if you travel for an hour in an airplane flying at 1000 km/h, you would, have aged about 0.0000000015 (that is, 1.5×10í9) seconds less than a person who, remained stationary on the ground. To provide another sense of the magnitudes, if you, moved at a speed of 90 km/hr away from a twin for a 100-year lifespan, you would have, lived 11 microseconds less. On the other hand, if you could move at one-tenth the, speed of light during a century of his life, you would be six months younger, and if you, moved at 90 percent of the speed of light, 270,000,000 meters every second, you would, be 44 years old when he was 100., , 644, , 20 seconds have passed, between two events on the, rocket. How much time has, passed on Earth?, , t = 25 s, , Copyright 2007 Kinetic Books Co. Chapter 35
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35.7 - Exploring and deriving time dilation, , Light clock, Light flash bounces up and down, , We will use the experiment portrayed on the right to derive the equation for time, dilation. Following the steps of the derivation may also help you to understand how, Einstein’s postulates lead inescapably to the conclusion that time dilation occurs., The experiment uses a light-based clock mounted on a high-speed skateboard., Ordinary clocks use periodic mechanical or electric processes, such as the oscillation of, a pendulum or a timing circuit, to establish a unit of time. A light clock uses the amount, of time it takes a pulse of light to travel a particular distance. The light clock is, convenient to use in this scenario, but any clock would record the same result., A light pulse is emitted from the base of the clock. The pulse reflects off the top of the, clock and returns to the bottom. The clock measures time by using the relationship of, time to distance and speed. The elapsed time for the up-and-down journey equals the, distance the light pulse travels, divided by the speed of light., In our experiment, let’s consider one “tick” of the clock. The light rises from the bottom, of the clock, reflects off the top, and returns to the bottom. The professor, who is also on, the skateboard, sees the light pulse moving straight up and down, and he measures the, distance it travels as being twice the height of the clock., , Professor’s reference frame, Light moves strictly up and down, , Now, let’s consider what another observer sees. Katherine is standing on the ground, and watches the professor and the clock pass by. She also watches the light pulse as it, moves. However, she measures a different value for the distance traveled by the light, pulse. She sees it not only moving up and down, but also forward. She measures the, light pulse moving through the distance indicated by the two lines labeled s in Concept, 3., If this seems confusing, just think of a friend riding a train, throwing a ball straight up, and down. From your friend’s perspective, the ball just travels up and down. From a, vantage point on the ground outside, you would see the ball moving horizontally at the, same time it is moving up and down., The clock uses light, which according to Einstein’s second postulate has a constant, speed independent of any frame of reference. Einstein’s first postulate states that the, laws of physics are the same in any inertial reference frame. That is, in our scenario,, both observers can use the same equation: Time equals the distance traveled by the, pulse, divided by the speed of light., , Katherine’s reference frame, Light moves up, down, horizontally, , Having explained the experiment, we will now analyze it algebraically, calculating the, distance traveled by the clock and the time interval required for one tick of the clock in, each frame of reference. This will enable us to derive the equation for time dilation, shown in Equation 1., Variables, We will use the triangle shown to the right in Equation 1 to relate displacement, speed, and a time interval. The triangle reflects half of one tick of the light pulse, as it moves, from the bottom to the top. The clock moves L horizontally during half of one tick, and, Katherine observes the light moving a distance s., measured by, Katherine, , measured by, professor, , clock’s horizontal displacement, half tick, , L, , 0m, , distance light pulse moves, half tick, , s, , h, , clock’s speed, , v, , 0 m/s, , height of clock, , h, , h, , elapsed time, , t, , speed of light, , c=, , Time measurements differ, Time = distance / speed of light, Katherine, professor measure different, distances, Speed of light is constant, Katherine, professor measure different, time, , t0, 3.00×108, , m/s, , Copyright 2007 Kinetic Books Co. Chapter 35, , 645
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Strategy, 1., , Use the Pythagorean theorem to determine the distance the light pulse moves in, one tick as measured by Katherine., , 2., , Use the fact that distance equals the product of speed and time to replace the, distances in the diagram., , 3., , Simplify the equation., , Mathematics principle, We will use the Pythagorean theorem, , Equation for time dilation, , Step-by-step derivation, In Katherine’s reference frame, the light traces out the hypotenuses of two right, triangles. We use the Pythagorean theorem to calculate the hypotenuse. The triangle’s, height is calculated using the speed of light and the time measured by the professor. Its, base is calculated using Katherine’s measurement of the clock’s speed., , Step, , Reason, , 1., , Pythagorean theorem, , t = “stationary” observer time, t0 = proper time in “moving” frame, , 2., , light travels two hypotenuses, , v = speed of reference frame, c = speed of light, , 3., , 2h = ct0, , definition of speed, , 4., , 2L = vt, , definition of speed, substitute equations 3 and 4 into, , 5., , equation 2, We have one last distance left, and again we substitute for it using the speed equation. Then a series of algebraic steps yield the equation for, time dilation., , Step, , 6., , 2s = ct, , Reason, , definition of speed, , 7., , substitute equation 6 into equation 5; square both sides, , 8., , expand and re-arrange, , 9., , divide both sides by c2 – v2; simplify, , 10., , take square root, , 646, , Copyright 2007 Kinetic Books Co. Chapter 35
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35.8 - Interactive problem: Experiment with the light clock, In this simulation, you experiment with a light clock. The professor rolls across a, basketball court with a light clock on his skateboard, while Katherine watches from, the side. You will view the light clock in operation from the professor’s reference, frame, and from the student’s reference frame. You are asked to calculate the time, measured by each observer for the professor to cross the basketball court., The simulation launches in the professor’s reference frame, where the light clock is, stationary. Press GO and watch the basketball court and the background pass by. A, counter will record the number of light clock cycles, the number of round trip, journeys made by the light pulse. The clock is 3.0 m tall., Then press the tab labeled “Student’s reference frame” and press GO again. You, will see the same series of events from Katherine’s reference frame. The simulation, displays the path of the light and indicates some key distances in a fashion similar, to the derivation of the prior section., If you asked the professor how long it took him to cross the basketball court, what, would he say? What if you asked Katherine the same question? Check your, answers by entering them in the simulation., Now a thought question for you, foreshadowing a future topic. Katherine measures the length of this court as 24.0 m (alas, she does not have, an NBA standard court). You can use this length and the time she measures to determine the professor’s speed. This is the same speed the, professor would measure of the ground moving beneath him., Let’s say the professor decided to use the time he measures and this speed to determine how long the court is. How long does he think the, court is? Does he think it is shorter or longer than 24.0 m?, , 35.9 - Length contraction, , Length contraction: The length of an object is, less when it is moving relative to an observer, than when it is stationary., Einstein’s postulates require that time dilates í observers measure different time, intervals between two events when their reference frames are in relative motion. The, postulates also require that length “contract”. The length of an object will be less when it, is moving relative to an observer than when it is stationary. (You will see that the effect, is far too small to measure for everyday speeds, which is why you do not notice this, effect.), Consider the train scenario on the right. The train is moving rapidly past an observer at, 0.8c. This is not an everyday speed for even a bullet train. The professor is onboard the, train, at rest relative to it. He measures the length of his car as 10 meters. This is the, car’s proper length, the length measured by an observer stationary relative to the length, being measured. The use of proper here is analogous to the use of proper in “proper, time”., , Length contraction, Observer stationary with respect to, object measures one length, , Sara, standing on the ground, measures a different value. She measures the train car, as being just six meters long. The relative motion of the two observers causes the, differing measurement., The equation on the right quantifies length contraction. This equation converts the, proper length measured by an observer who is stationary relative to the length being, measured, to the length observed by an observer who views the object as moving., As with time dilation, you face two challenges: First, believing it to be true, since it, defies your intuition and experience, and second, understanding the notation for the, equation. We are not sure how much we can change your intuition, but at the least we, can work on the notation. To stress the notation once more: The proper length (L0) on, the right side of the equation is the length measured in a reference frame stationary, relative to the object being measured. In the illustration, it is the professor measuring, the length of the car., , Length contraction, Observer viewing moving object, measures shorter (contracted) length, , The quantity L on the left-hand side of the equation represents the length that will be measured by a person who views the object (and its, reference frame) as moving. This person observes the object as moving at velocity v., Length contraction occurs only along the direction of the relative motion. In the scenario in this section, only the car’s length changes, not its, height nor depth. If the train were carrying a light clock, the vertical dimension of the clock, used by onboard travelers to measure time, intervals, would not be affected by length contraction., Does the moving object “really” contract? Well, to pose an analogous question from everyday life, do faraway objects really get smaller, as they, , Copyright 2007 Kinetic Books Co. Chapter 35, , 647
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appear to our eyes? The observations and measurements made by different observers, do differ. In discussions of relativity, much focus is placed on observers and, measurements, since experiments show effects such as time dilation and length, contraction are real., On the other hand, we are taking liberties in showing the object’s contraction in these, illustrations. If the train car were photographed from a station as it passed by, it would, appear rotated. The rear of the car is farther away from the stationary observer, and, light from the rear will take longer to reach her than light from the front of the vehicle., The light that arrives simultaneously for the observer is emitted at different instants in, time from the car, which moves during that interval of time. Using geometrical, arguments, it can be shown that this effect creates the perception of rotation., , L = length measured by observer, watching object pass by, L0 = proper length (measured by, observer stationary relative to object), v = speed of reference frame, c = speed of light, , How long is the box on the train, from Sara’s perspective? The, professor measures its length as, 4.0 meters., , L = 4.0(0.6) = 2.4 m, 35.10 - Doppler shift for light, , Doppler shift: light, Motion changes light frequency (color), Shift depends on velocity between, source, observer, , The Doppler effect is frequently thought of in terms of sound. The siren on a police car heading toward you will sound different than when it is, , 648, , Copyright 2007 Kinetic Books Co. Chapter 35
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moving away from you. This effect occurs because the motion of the siren changes the, frequency of the sound waves that reach you., The change in the frequency depends on whether you move toward the source of a, sound, or the source moves toward you. There is an asymmetry because sound travels, through a medium, and it matters which reference frame í yours or the source’s í is, stationary with respect to the medium., In contrast, light requires no medium. With light (and other electromagnetic radiation),, the Doppler effect depends solely on the relative motion of the source of the light and, the observer of the light. In some ways, this makes the Doppler shift “easier” to, calculate with light., On the right, we show equations for calculating the change in perceived frequency due, to the Doppler shift. The symbol f0 represents the frequency of the light as it is emitted, from the source (or as it would be perceived by an observer stationary relative to the, source). This is called the proper frequency. The symbol f represents the frequency, perceived by an observer in a reference frame in which the source is moving at speed, v. The speed is always positive. The two equations are the same except for the signs;, the signs differ for the reasons discussed below., , Light source closing in, Blue shift, , When the observer and the source are moving toward one other, the frequency of the, light seen by the observer will increase. The speed of the wave í the speed of light í, must remain constant. Since the wavelength equals the speed of light divided by the, frequency, then as the frequency increases, the wavelength decreases. When the, source is moving away from the observer, the opposite effect occurs: The wave, frequency decreases and the wavelength increases., With light, decreasing wavelength due to the approach of the light source is called a, blue shift since shorter wavelengths occur near the blue end of the visible spectrum., Increasing wavelength due to the recession of the light source is called a red shift., Almost all distant stars and galaxies, in whatever direction you look, exhibit red shift. In, fact, the farther away they are, the greater the red shift. This is used as evidence that, the universe is expanding., , Light source moving away, Red shift, , The Doppler effect proved to be crucial to astronomers trying to understand the nature, of the universe. Einstein’s work provided grounds for believing that the universe was, expanding, an implication that proved quite troubling to Einstein himself, who, “corrected” his equations to provide for a static universe. But several years after, Einstein performed this correction, the American astronomer Edwin Hubble, (1889í1953) showed that distant stars and galaxies exhibit a red shift., Astronomers like Hubble were trained to expect certain patterns in the light emanating, from stars. Elements like calcium in the atmospheres of stars absorb certain, wavelengths of light; these show up as “gaps” í spectral lines í in the light emitted from, stars. Hubble expected those spectral lines to occur at certain wavelengths. Instead, he, found them “red shifted” to different wavelengths. Although Hubble knew that the, Doppler shift could account for these changes, he and his colleagues were astonished, that stars in every direction were moving away., The debate about the nature and fate of the universe continues; but the red shift of, starlight provided crucial data to scientists that caused them to further examine the, possibility that the universe is expanding. The second sample problem to the right, challenges you to compute the speed at which a galaxy is moving away from the Earth, based on the shift in its observed wavelength., , Doppler shift, source/observer closing:, , source/observer separating:, , f = observed frequency, f0 = proper frequency of source, v = source/observer relative speed, , Copyright 2007 Kinetic Books Co. Chapter 35, , 649
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What frequency of light is, observed on Earth?, , f = 5.0×1014 Hz, , How fast is the galaxy receding?, Wavelength Ȝ0 of H K lines is 394 nm, Redshifted wavelength Ȝ is 458 nm, f = c/Ȝ, , v = 0.149c = 4.47×107 m/s, , 650, , Copyright 2007 Kinetic Books Co. Chapter 35
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35.11 - Relativistic linear momentum, In discussing relativistic linear momentum, and the equations that describe it, we start, with three premises., First, the same laws of physics hold true in any inertial reference frame. This is, Einstein’s first postulate. If a property like momentum is conserved in one inertial, reference frame, it must be conserved in any other inertial reference frame., Second, Einstein’s second postulate, that the speed of light is constant. These two, postulates were used to deduce equations for time dilation, length contraction and so, on., Third, we expect that at much lower speeds, the equations in this section will essentially, reduce to the “classical” ones that work perfectly well at low speeds. For instance, the, momentum of an object moving at say 30 m/s should very, very closely equal mv. This, has held true with prior equations in this chapter. At speeds much less than the speed, of light, the equations predict results that accord with “classic” mechanics equations., These three premises are satisfied by the relativistic equation for momentum at the, right. This extended equation for momentum is required because it can be shown that if, Newtonian, or classical, momentum is conserved in one reference frame, then when the, velocities are converted to the values that would be observed in another frame, the total, momentum is never conserved in that second reference frame., , Relativistic momentum, New equation for momentum required, for high velocities, , Specifically, if a collision occurs in a moving reference frame S´, and the velocities are, converted to the values that would be observed in S, the quantity mv before and after, the collision is changed by the collision. Using the equation shown to the right ensures, this will not occur, and that momentum is conserved when the collision is observed from, any reference frame., The larger the relative speed of the frames, the greater the discrepancy between the, classical momentum and the relativistic momentum will be. It turns out that the Lorentz, factor Ȗ appearing in the definition of momentum shown in Equation 1 provides exactly, the right corrective factor for momentum to be conserved in both S and S´, no matter, how fast the frames move with respect to each other. (The variable u in the formula, refers to the speed of an object in its frame, not the speed of one frame with respect to, another.), , Equation for relativistic, momentum, , The extended momentum equation obeys the first two premises stated above. It also, obeys the third. At velocities much less than the speed of light, momentum approaches, mv, since dividing the square of a velocity like 300 m/s by the square of speed of light, means the relativistic effect is near nil., As the example on the right shows, at half the speed of light, relativistic effects increase, momentum by 15% over its classical value. The effect increases significantly as an, object moves at speeds closer to the speed of light: At 99% of the speed of light,, momentum is about seven times its classical value., , p = momentum, , This phenomenon is of great importance in the modern physics research done with, large particle accelerators, such as the Fermilab accelerator in Illinois, or the CERN, accelerator in France and Switzerland. Objects move near the speed of light in these, accelerators, and calculating their momentum requires the use of the equation shown in, this section., , c = speed of light, , m = mass, u = velocity of object, Ȗ = Lorentz factor, , Copyright 2007 Kinetic Books Co. Chapter 35, , 651
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What is the momentum of the, proton? Its mass is 1.67×10í27, kg., , p = 2.89×10í19 kg·m/s, (Relativistic effects add 15%), 35.12 - Mass and energy, The world’s “most famous equation” is shown in Equation 1: Rest energy (also known, as mass energy) equals mass times the speed of light squared. By “rest energy”, we, mean the energy equivalent of its mass., If a mass is moving, its total relativistic energy can be calculated as the sum of this, value and its kinetic energy. The equation for calculating this total energy is shown in, Equation 2. The equation factors in relativistic effects when calculating KE., It is interesting to consider Equation 1 in the context of the Sun. The Sun becomes less, massive as its mass is transformed by fusion reactions into the energy it radiates. It, radiates 3.91×1026 joules of energy per second, and as a consequence, it “loses” more, than four billion kilograms per second! (But do not worry! Its total mass is 1.99×1030 kg)., Earlier scientists had proposed the conservation of mass in chemical reactions. This, textbook has discussed the conservation of energy. At non-relativistic speeds, these, principles hold true to a high degree of accuracy. Einstein showed both mass and, energy must be considered when applying conservation principles., Thermonuclear processes inside the sun convert a very large quantity of mass into a, stupendous quantity of energy every second. Much more comprehensible amounts of, matter yield tremendous amounts of energy as well. For example, if one gram of matter, were converted entirely to energy, it would produce the energy equivalent of, combusting fifteen thousand barrels of oil. On the dark side of things, converting into, energy a tiny fraction of the 2 kg of matter in a nuclear bomb can produce a blast, equivalent to the chemical energy released from the explosion of 50 billion kilograms of, dynamite., , Energy at rest, E0 = mc2, E0 = rest energy, m = mass, c = speed of light, , Atomic particles are often subject to relativistic effects because of their high speeds. For convenience, the energy and mass of such particles is, often measured in units other than joules and kilograms. An electron volt (eV) equals the change in the potential energy of an elementary, charge (1.60×10í19 C) when it moves through a potential difference of one volt. Electron volts are convenient measures of particle energy., Particle mass is often measured in units of eV/c2, or energy divided by the speed of light squared. The fact that these are mass units follows, from the equation E = mc2., , 652, , Copyright 2007 Kinetic Books Co. Chapter 35
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Total energy, E = Ȗmc2, E = total energy, Ȗ = Lorentz factor, m = rest mass, c = speed of light, , When a proton and antiproton, collide, they annihilate each, other and their mass energy is, converted into a pair of gamma, ray photons. Calculate the total, energy of the gamma rays., , E = Ȗmc2, E = (70.7)(938 MeV/c2)c2, E = 6.63×104 MeV, 2E = 1.33×105 MeV, , Copyright 2007 Kinetic Books Co. Chapter 35, , 653
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35.13 - Gotchas, If you measure the speed of light from a star that is moving toward you, you will get a higher value than if you measure the speed of light from a, star that is moving away from you. No! You will measure the same speed. The speed of light does not change based on motion of source or, observer. This is one of Einstein’s postulates., You are flying from Earth to Alpha Centauri at a constant speed of 150,000,000 m/s. Given your incredible speed, you should expect some of, the laws of physics to change. No, Einstein’s other postulate is that the laws of physics are the same in any inertial reference frame. You are, moving at a constant speed, so you are in an inertial reference frame., I’m watching an airplane flying overhead at 400 km/hr. The pilot throws and catches a ball, measuring the time with a stopwatch. Her stopwatch, measures the proper time. Yes, that is correct., Since you are standing still as you watch the plane pass by, you measure its proper length. No. You have to be at rest relative to an object to, measure its proper length. However, in the case of an airplane, the difference is negligible í not so for relativistic speeds!, Einstein was very smart. Yes., , 35.14 - Summary, Einstein radically changed our perception of space and time with his two postulates, of special relativity. The first postulate states that the laws of physics are the same, for observers in any inertial reference frame. The second postulate states that the, speed of light in a vacuum is the same in all inertial reference frames., An event is specified by giving its space and time coordinates. Observers who are in, different inertial reference frames will assign different coordinates to the same, event., Observers in a reference frame where two events occur at the same place measure, a time interval between those events called the proper time. Observers moving, relative to that frame will always measure a longer time interval. This effect is known, as time dilation., The inability to agree on time intervals led to Einstein’s thought experiment about, the concept of simultaneity. In this thought experiment, he showed that observers in, reference frames moving relative to one another would not agree that there was, zero time interval between separated events í in short, they would disagree about, whether events occurred simultaneously or not., Special relativity also correctly predicts that different observers may not agree on, the spatial interval between two events. The length of an object at rest in an inertial, reference name is known as its proper length. Observers who are moving relative to, that frame will always measure a shorter length than the proper length. This effect is, known as length contraction., , Time dilation, , t = Ȗt0, Lorentz factor, , Length contraction, , Doppler shift, , Source/observer closing:, , The Doppler effect applies to light. The astronomer Hubble used this effect to argue, that the universe is expanding., The intertwining of space and time means that a relativistic definition of linear, momentum must be accepted if consistent relations among them are to be retained, at both low and high speeds, and for observers in all inertial reference frames., Special relativity reveals a previously unseen connection between two fundamental, concepts: mass and energy. A mass has a tremendous amount of rest energy, locked up, ready to be released either in a terrifying chain reaction or in a controlled,, useful manner. Meanwhile, the converse is also true: Energy has a mass equivalent, and may be transformed into matter., , Source/observer separating:, , Relativistic momentum, , Rest energy, , E0 = mc2, Total energy, , E = Ȗmc2, , 654, , Copyright 2007 Kinetic Books Co. Chapter 35
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Chapter 35 Problems, , Chapter Assumptions, Assume the speed of light in a vacuum is 3.00×108 m/s for all observers., , Conceptual Problems, C.1, , Would a windowless spaceship, rotating at constant speed in deep space, qualify as an inertial reference frame?, , C.2, , A scheme for a modern-day Fountain of Youth is introduced by a marketing company. Their idea is to launch their clients on, a round trip at relativistic speeds. Their ads proclaim: "After 10 years have elapsed on Earth, you will only have aged 5 years!", Will the passengers enjoy a longer life according to their own biological clocks?, , C.3, , A spaceship races by the Earth at a constant velocity of 0.999c. An astronaut in the ship holds a mirror up and looks at her, image in it. (a) Does she see any unusual effects? (b) Describe the effects or explain why she doesn't see any., , Yes, , Yes, , (a), , No, , No, , Yes, , No, , (b), C.4, , Consider a box on a relativistic train car, moving at 0.866c along the positive x axis as viewed from the Earth. The length of, the box along the x axis, as measured by Earth observers, will be only half of its proper length. If V is the volume of the box, measured by an observer on the train, what is volume of the box as measured by an observer on Earth?, V, , 0.866V, , V/2, , V/8, , V/ʌ, , Section Problems, Section 3 - Light can travel through a vacuum, 3.1, , Here is an analogy to the Michelson-Morley, experiment. Suppose a swimmer can swim at, 3.0 m/s in still water (that is the "medium" of, travel for the swimmer). Consider three fixed, rocks A, B, and C in a flowing river, as shown,, such that distance AC is 12 meters and AB is, also 12 meters. The river flows at 1.0 m/s. (a), How much time does the swimmer take for, round trip ACA, parallel to the river flow? (b), How much time for round trip ABA,, perpendicular to the flow?, Explanation of the analogy: By rotating their apparatus, and swapping the role of B and C, Michelson and Morley hoped to, measure the speed of the earth with respect to the ether. (The analogy is to try and measure the water speed by making, round trip time measurements.) What they found was that there was no difference in the times, no matter what time of year, they did their experiments, and no matter what the orientation of the apparatus., (a), , s, , (b), , s, , Section 5 - Interactive problem: Conduct Einstein's simultaneity experiment, 5.1, , Use the simulations in both interactive problems in this section to answer the following questions. In all cases, use the default, setting for the time interval between lightning strikes, 0 ns. (a) Does Katherine observe the lightning bolts as striking the, lightning rods simultaneously when the train is stationary? (b) Does the professor? (c) Does Katherine observe the lightning, bolts as striking the lightning rods simultaneously when the train is moving? (d) Does the professor?, (a), , Yes, , No, , (b), , Yes, , No, , (c), , Yes, , No, , (d), , Yes, , No, , Section 6 - Time dilation, 6.1, , A rocket ship is 45.0 m long according to measurements made in its rest frame, and moves at 0.385c with respect to the, , Copyright 2000-2007 Kinetic Books Co. Chapter 35 Problems, , 655
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Earth. Astronaut Naomi sends a laser pulse from the tail of the rocket to the nose, where it reflects off a mirror, and returns to, Naomi. (a) How long does the round trip take, according to Naomi? (b) How long does the round trip take according to Tariq,, an observer on Earth?, , 6.2, , (a), , s, , (b), , s, , An interstellar spacecraft races by the Earth at a velocity of 1.85×108 m/s, as measured by observers on the Earth. The, spacecraft's galley cook places a pot of Minute® rice on the stove for exactly 60.0 s, according to his watch. How long will the, pot have been on the stove according to the earthbound observers?, s, , 6.3, , An interstellar cruise ship races by a Martian spaceport. According to galactic regulations, the safety strobe lights on the, ship's bridge are supposed to blink every 1.50 seconds. A rookie policemartian, based on the planet, issues a ticket, claiming, he measured the time interval between blinks as 1.75 seconds. You are called upon by the indignant captain to explain the, discrepancy, who maintains that her lights are fully compliant with code. What spacecraft speed will show both the captain, and policemartian to be correct?, m/s, , 6.4, , You have probably never noticed time dilation effects in everyday life. Do some calculations to see why this is. (a) Calculate, the speed that a friend would have to move with respect to you so that your measured time intervals are 1.00% larger than, hers. That is, for each 100 elapsed seconds on her wristwatch, 101 seconds elapse on yours. (b) How many times larger is, the speed you just calculated than the speed of an orbiting space shuttle, 8.00×103 m/s?, , 6.5, , (a), , m/s, , (b), , times the speed of the shuttle, , A pi meson, or pion, is an elementary particle that exists for a brief time. Charged pions at rest have an average lifetime of, 26.0 ns. A group of freshly created pions travel, each at a constant speed, an average distance of 28.0 m in the laboratory, before decaying. What was the average pion speed as measured in the lab?, m/s, , Section 7 - Exploring and deriving time dilation, 7.1, , Consider a very tall light clock that is 555 meters high. It is mounted vertically on a spacecraft that is moving along the, positive x axis at 2.00×108 m/s, as measured from the Earth. During a certain time interval, an earthbound observer measures, the clock moving a distance of 497 meters along the x axis, while the light pulse has made a trip from the bottom of the clock, to the top. (a) How far has the light pulse traveled, according to an astronaut on the spacecraft? (b) How far has the light, pulse traveled, according to earthbound observers? (Your answers to parts a and b should differ.) (c) What is the time interval, required for the light pulse to travel from the bottom to the top of the clock, according to the astronaut? (d) What is the time, interval required for the light pulse to travel from the bottom to the top of the clock, according to earthbound observers? (e), Calculate the Lorentz factor for the spacecraft. (f) Use the Lorentz factor to calculate the dilated time interval corresponding to, the astronaut's measurement in part c. (Your answers to parts d and f should be the same.), (a), , m, , (b), , m, , (c), , s, , (d), , s, , (e), (f), , s, , Section 8 - Interactive problem: Experiment with the light clock, 8.1, , Use the simulation in the interactive problem in this section to answer the following questions. (a) How long does the, professor think it took him to cross the basketball court? (b) How long does Kathering think it took the professor to cross the, basketball court? (c) Suppose Katherine measures the length of the court as 24.0 m, and then uses this to calculate the, professor's speed. If the professor uses that value for his speed and the time he measured, is the professor's measurement of, the basketball court's length going to be longer, shorter or the same as Katherine's measurement?, (a), , 656, , s, , (b), , s, , (c), , i. Longer, ii. Shorter, iii. The same length, , Copyright 2000-2007 Kinetic Books Co. Chapter 35 Problems
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Section 9 - Length contraction, 9.1, , 9.2, , A newly-constructed spaceliner boasts a 105 m long and 65.0 m wide soccer field for the enjoyment of its international crew, during extended missions. The long dimension of the soccer field is aligned along the ship's direction of flight. As the ship, cruises by Earth, FIFA (Fédération Internationale de Football Association) officials on Earth measure the length of the field as, 99.0 m and declare the field ineligible since it is shorter than the required 100 m. The captain protests that the discrepancy is, due to length contraction. (a) How fast was the ship traveling according to the Earth-bound observers (as a decimal fraction of, c)? (b) What is the width of the field according to the FIFA officials?, (a), , c, , (b), , m, , You want to visit your friend Ford Prefect, who lives in the vicinity of Betelgeuse, 425 light-years from Earth (that is, light, takes 425 years to arrive from Betelgeuse). You hitch a ride on a passing starcruiser, and as measured by Earth observers,, you travel so fast that the trip only takes 426 years. (a) How long was the trip according to you? (b) What is the distance, between Betelgeuse and Earth, according to you?, (a), , years, , (b), , light-years, , Section 10 - Doppler shift for light, 10.1 A spaceship is flying directly away from Earth at a speed of 0.725c, and the captain's family at home on Earth broadcasts, their well wishes on the Family Radio System frequency of 462.5625 Mhz. To what frequency must the homesick captain tune, the ship's radio to receive the broadcast? (Round your answer to three significant figures.), MHz, 10.2 Starting in Toronto, Canada, a spaceship flies down Yonge Street (reportedly the longest street in the world) at a velocity of, 0.330c. An astronaut on board measures her pulse rate at 50.0 beats per minute. (a) What is the time interval between, heartbeats according to her? (b) What is the time interval between beats according to observers who are standing by the side, of the road? Now, the astronaut is attached to a machine that sends a brief radio pulse for each beat of her heart back toward, a monitoring device at her starting point. (c) What is the time interval measured by the monitor between consecutive radio, pulses?, (a), , s, , (b), , s, , (c), , s, , 10.3 Albert protests the $150 ticket he received for piloting his spaceship through a red light (wavelength Ȝ = 630 nm) on, Intergalactic-5. His argument is that he was moving so quickly toward the light that it appeared to have a wavelength of, , Ȝ = 546 nm, which is green. The judge agrees to dismiss the ticket, but instead fines him for speeding. The speed limit is, 5.0% of c, with a charge of $20 for each percentage point in excess of that (for example, a speed of 0.085c = $70 ticket). (a), How fast was Albert traveling down the highway? Express your answer as a decimal fraction of c. (b) How much does he, have to pay?, (a), , c, , (b) $, 10.4 A police sergeant has set up a speed trap on the popular Rocketbahn route. While she is idling next to a signpost, she fires, her 12.0 GHz radar gun at an oncoming sports rocket that is heading directly toward her. The radar wave hits the rocket,, reflects off it, and returns directly to her detector. She detects a reflected frequency of 15.0 GHz. At what speed is the sports, rocket traveling? Express your answer as a decimal fraction of c., c, , Section 12 - Mass and energy, 12.1 Find the (a) speed (as a decimal fraction of c) and (b) momentum of a proton that has a kinetic energy of 1000 MeV. The, proton mass is 1.673×10í27 kg, or 938 MeV/c2., (a), , c, , (b), , MeV/c, , 12.2 On the television show Star Trek®, a handheld device called a phaser vaporizes humanoid-sized bad guys, presumably, turning their mass into energy, with no ill effects to the operator or to nearby objects. On the show, the energy is represented, by a brief, glowing outline. In the real world, if a 50 kg alien were turned completely into energy, the energy released would be, equal to how many tons of exploding TNT? The explosion of one ton of TNT releases approximately 4.0×109 joules of energy., tons of TNT, , Copyright 2000-2007 Kinetic Books Co. Chapter 35 Problems, , 657
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12.3 In an antimatter warp drive engine, an electron traveling at a speed of 0.410c meets an antielectron (i.e., a positron) head on,, traveling at a speed of 0.740c. They annihilate each other, and produce two gamma rays. What is the total energy of the, resulting gamma rays? The electron and the positron have identical mass, of 9.11×10í31 kg, or 0.511 MeV/c2., MeV, 12.4 A neutral K meson (with mass 498 MeV/c2), or kaon, decays at rest into two pi mesons, or pions (with mass 140 MeV/c2, each) (a) The two pions must fly out in exactly opposite directions. Why? (b) Find the kinetic energy of each pion., (a), , (b), , i. Conservation of momentum, ii. Conservation of energy, iii. The speed of light is always constant, MeV, , 12.5 As stated in the text, the Sun radiates 3.91×1026 joules per second, and thus its mass diminishes by about 4.34×109 kg each, second. (a) If you had a cube of water whose mass was equal to 4.34×109 kg, what would be the length of one side of the, cube, in km? Assume a density of 1000 kg/m3. (b) The mass of the Sun is 1.99×1030 kg. Calculate how long the sun would, last, in years, if it were to keep radiating at its present power level until all its mass is gone. (This is much longer than its, lifetime as predicted by more sophisticated astronomical models.), (a), , km, , (b), , years, , 12.6 In a high energy accelerator, the Tevatron at the Fermi National Accelerator Laboratory, a proton and an antiproton moving at, equal speeds make head-on collisions. Suppose that a new theory predicts that this collision can create a single particle of, mass equal to 80.0 GeV/c2. You are in charge of running the initial experiment. The proton and antiproton each have equal, rest masses of 938.38 MeV/c2. What minimum proton/antiproton speed is necessary for the new particle to be created?, Express your answer as a decimal fraction of c to five significant figures (or else you will just find that v = c)., c, 12.7 The Stanford Linear Accelerator is capable of producing beams containing 30 GeV electrons. (That is, the total energy of, each electron is 30 billion electron volts.) What is the Lorentz factor for these electrons?, , 658, , Copyright 2000-2007 Kinetic Books Co. Chapter 35 Problems
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36.0 - Introduction, Quantum physics is the branch of science required to fully explain the behavior of, light, its interaction with matter, and the behavior of exceedingly small particles such, as atoms and electrons. As scientists began to discover in the late 19th and early, 20th centuries, certain principles and techniques of classical physics fail utterly when, applied to light and to atomic-scale systems., Although “quantum physics” often connotes mystery and difficulty, its applications, are very real. You may be pleasantly surprised at how much you can understand, when you are equipped with just a few fundamental concepts from this science., In particular, you can learn the principles governing the functioning of two of the, most pivotal technologies of the last half-century: semiconductors and lasers., Why are these two technologies so important? Without semiconductors, there would, be neither transistors, nor the microprocessors built from them. Semiconductorbased microprocessors serve as the “brains” of computers and are found in digital, cameras, cell phones, and automobiles: wherever engineers want “smart” behavior., Semiconductors are also used in various types of computer memory, such as, random access memory (RAM). Semiconductor chips not only “think,” they also “remember.”, In recent years, connecting all these semiconductor devices has become the central thrust of the information processing industry. The, Internet, cell-phone “fixed-rate calling plans”, video on demand, downloadable music, and even the Web-based version of the textbook you are, now reading all rely on the cheap and rapid transmission of information over wired or wireless networks. The two technologies most, responsible for creating this networking revolution of rapidly decreasing costs and dramatically increasing bandwidth have been the, microprocessor and the laser., How do these two technologies enable networking? Communication networks use devices like routers and high-speed switches to transmit, data. These devices rely on microprocessors to determine where to send their information and they form parts of extended physical systems, that use lasers to move the information at light speed over fiber-optic cables., Lasers give you access to data from sources both distant and nearby. In addition to sending data around the world, they are used on your, desktop or in your home to read the data stored on CDs and DVDs (not to mention their use in stores to read data codes on your purchases)., Without lasers, vinyl records and “floppy disks” might still be the primary means of storing audio and digital data. Believe us: If you have never, used a floppy disk, you haven’t missed much., How does quantum physics relate to the working of these devices?, Explaining what is meant by “quantum” is the place to start. A key tenet of quantum physics is that particles in some systems, like the electrons, in hydrogen atoms, exist only at certain energy levels. Physicists say the energy levels of the electrons in an atom are quantized., It may be easiest to explain a quantum property by first considering its opposite, a property that is continuous. Consider the potential energy of, a bucket that is raised or lowered by a rope. You can raise it 1.000 meters off the ground, or 1.001 m, or 1.002 m, or however much you like., By controlling its height, you can make its potential energy whatever you like. The range of possible energy values is continuous: say 10.00, joules, 10.000017 J, 10.027 J and so forth., Electrons prove not to be as flexible. The electrons around a hydrogen (or other) atom exist only in states with certain discrete energy values;, for instance, two possible energy levels for the hydrogen electron are í1.51 eV (electron volts) or í3.40 eV (í2.42×10í19 J or í5.44×10í19 J, respectively). Between these two values lies a forbidden gap, and a hydrogen atom’s electron is never observed with energies in that range., Physicists say an electron’s energy is quantized, that it only exists at certain levels. In the example of the hydrogen electron mentioned above,, you will never observe an energy of í1.6 eV or í2.9 eV, since those are forbidden., You may ask: How does an electron change between energy levels? How can it “move” across a “forbidden gap” to a higher or lower energy, state, if intermediate energy values are forbidden? You may not be satisfied with the answer, but it is most straightforward to say: Those are, simply the only values that have ever been measured. Any time a scientist measures a property of an atomic electron (such as its energy, or its, angular momentum), she only observes results from a particular set of values that can be predicted with extreme accuracy by quantum, physics. It is impossible to “catch” an electron in any in-between state., The simulation at the right reproduces one of the key experiments that led to the widespread acceptance of the ideas of quantum physics., Einstein explained data from a more sophisticated version of the same experiment, known as the photoelectric effect, to earn his Nobel Prize., Describing the experiment is simple. Scientists had noticed that when they shined a beam of light on a metal, electrons were released from the, illuminated surface. From their perspective, this was not particularly surprising: The energy of the light was transferred to the electrons,, allowing some of them to escape their bonds to the atoms of the metal., The emission of electrons could be explained by classical physics. Light was a wave with energy, and that energy could provide a “kick” to, electrons as the metal absorbed the light., Although the experiment and the expected outcome are simple in concept, the detailed results were quite surprising and could not be, explained by classical physics. Some colors of light, such as red, could not eject electrons from the surface, no matter how bright the beam, was. On the other hand, other colors, such as violet, were effective at ejecting electrons from the metal surface, even when the beam intensity, was very low. It was the frequency of the light, and not its intensity, that determined whether or not electrons were ejected., , Copyright 2007 Kinetic Books Co. Chapter 36, , 659
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What the scientists observed does not make sense if light is conceived of solely as a wave. Let’s compare their observations in terms of water, waves crashing against a wooden dock. It is as if low-frequency waves (with their crests arriving, say, every five seconds) could never rattle, the dock enough to knock free the timbers that make it up, even if they were giant waves 50 meters tall., Now imagine centimeter-high waves arriving more frequently, say every second. Imagine that these small but frequent waves could knock, loose pieces of the wood from the dock. Water waves with these effects would be confusing to observe, and you might be as confused as the, scientists who observed dim but high frequency light freeing electrons from samples of metal., Einstein successfully explained the photoelectric effect. He argued that light has both a wave nature and a particle nature. Electromagnetic, radiation, he said, consists of small packets called photons. Photons are small “chunks” of light energy. More energetic light consists of more, photons, not larger, more energetic waves. In addition, he stated that the higher the frequency of light, the more energetic the photons that, make up the light. Dim but high frequency light can eject electrons because of the interaction between the energetic photons that make it up, and the atoms of the metal. It is the energy of the individual photons that matters, not the overall energy of the light., Conceiving of light as consisting of photons could explain another mysterious result: More intense light of a certain color caused more electrons, to be emitted, but their maximum kinetic energy was exactly the same. The classical physicist would expect the “larger wave” of the more, intense light to cause higher-energy electrons to be released, but this did not happen. Einstein’s theory explains why: more intense light, consists of more photons, each with the same energy as before. Again, it is the interaction between an individual photon and an individual atom, that matters., You just read a brief summary of some crucial points in quantum physics. You will become familiar with the photoelectric effect by using the, simulation on the right. In your experiment with this effect, the flashlight can shine red or violet light. It can be set to emit either low or high, intensity light., When you press GO, you will see photons moving in slow motion from the flashlight toward the metal. When appropriate, we show electrons, escaping from the metal., Start the simulation with the light set to LOW. One color of light will cause electrons to escape the metal being used in our simulation; another, will not. Red light has a longer wavelength but a lower frequency than violet light. Which of the two colors of light do you think will cause, electrons to be emitted?, Now set the intensity of the light to HIGH, and try both colors again. What do you expect will change when you make this change? What do you, think will stay the same?, , 36.1 - Quantum, , Quantum: The smallest amount of something, that can exist independently., Quantum refers to the fundamental or least amount of something. For instance, the, quantum of money in the U.S. is the penny. Your net worth will be a multiple of that, quantum. You can be a pauper worth one penny, a millionaire with a worth of, 100,000,000 pennies, or a starving college student with a net worth of í5,012 pennies., However, you cannot legally use three-quarters of a penny, or 1.45 pennies, or 3ʌ/4, pennies. There are many similar examples of things that come in discrete amounts: the, number of siblings you have, the number of eggs you can purchase at a store, and so, on., A physicist would say that things like money or siblings or eggs are quantized. Calling, something quantized means that it is grainy; it is the opposite of continuous. Using the, example mentioned in this chapter’s introduction, one would say that the height and, energy of a bucket being raised by a rope are continuous quantities, as are the height, and energy of an elevator car. In contrast, the height and energy of elevator stops are, quantized; they occur solely at discrete points. You only see buttons for the first, second, and third floors, not for the 1.75th floor., , Quantized things, Exist with discrete values, , A mathematical example of something that is continuous is shown on the right: real, numbers. Examples of real numbers are 3, or 3.1, 3.01, 3.001, 3.002 and so forth. The, set of real numbers is continuous, not quantized., Although the idea of quantization may seem intuitive for money, it is much less obvious, in some areas of physics: Scientists now know that many things once thought to be, continuous are in fact quantized. Albert Einstein, for instance, showed that the energy of, any precise color of light is quantized., , Opposite of quantized, , Prior to Einstein scientists expected properties of light, such as its energy, to be, Continuous, continuous. Why? In the 18th and 19th centuries, a series of discoveries had led most, ·Real numbers are an example, scientists to conclude that light was a wave. They knew very well that the energy of a, mechanical wave is continuous, not quantized. An ocean wave, for instance, can have a, height (amplitude) of 1.01 meters, or 1.04 meters, or any value in between, and its energy will depend on that amplitude. Since light was, believed to be a wave, scientists concluded that its energy would be continuous as well, and that for example, they could create a beam of a, certain frequency of blue light with any desired energy simply by making the light brighter or darker., However, as the next sections discuss, in the early 20th century it became increasingly clear that light is quantized: It consists of small chunks, , 660, , Copyright 2007 Kinetic Books Co. Chapter 36
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or packets of energy. The energy of a beam of a particular color of light must be a multiple of the energy of the packets that make it up. This, realization had profound implications for the understanding of both light itself and the atoms that emit the light., , 36.2 - Balmer series, “The most important result of the application of, quantum mechanics to the description of electrons in, a solid is that the allowed energy levels of electrons, will be grouped into bands.” So wrote Andy Grove,, then an employee of the Intel Corporation and a, faculty member of the University of California at, Berkeley, in his text on the physics and technology of, semiconductor devices. Grove became the chairman, of Intel during its rise to power, prestige and, profitability., Grove cites two quantum principles. First, the concept, that there are “energy levels” for electrons, and, second, that these are grouped into bands (the, emphasis in the quote is his)., , The electrically excited neon gas in this sign emits light, at several sharply defined red-orange wavelengths., , For now, we will simplify our discussion by focusing on energy levels. Bands refer to the, fact that certain electrons exist at energy levels that are close to one another. In, practical applications, the distinction between a band and an energy level is often, dropped., Grove’s words convey how important the ideas of quantum theory are for, semiconductor science and technology. We use these words to motivate the next few, sections of this book. One might wonder: How did physicists discover that atomic, electrons had discrete energy levels? In other words, how did they first learn that the, energy levels of electrons are quantized, not continuous?, Physicists advanced the theory as their observations forced them to. The story begins, in 1666 when Newton showed that a prism could disperse sunlight into a spectrum of, colors. Today, one would say Newton showed that sunlight comprises light of many, wavelengths: Light perceived as white is in fact made up of a rainbow of components of, various hues. When it was first discovered, the spectrum of sunlight seemed to be a, continuous gamut of colors., , Hydrogen emission spectrum, Is not continuous, , A series of later experiments convinced scientists that light had a wavelike nature. They, could create interference patterns with light that were conceptually identical to patterns, created by water waves. Physicists even found that they could measure the, wavelengths of various colors of light. One mystery of science seemed to be solved:, Light was a wave., However, in 1814, the German physicist Joseph von Fraunhofer made careful, observations using a thin slit, and discovered that the spectrum of sunlight contained, many narrow dark lines, or gaps. In other words, certain wavelengths of light were not, present in the spectrum he was observing. He discovered that the spectrum of sunlight, was not continuous., Throughout the 1800s, scientists studied the light emitted and absorbed by various, gases. They discovered that a gas like hydrogen only emits or absorbs light of specific, wavelengths. By 1880, the wavelengths of the spectral lines of various elements,, including most famously hydrogen, were well known. In Concept 1 you see an, illustration of the spectral lines in the emission spectrum of excited hydrogen gas., The distinct colors and wavelengths of light you see are characteristic of light emitted by, this element. Similar lines, but with different colors í wavelengths í can be found when, the light of a neon sign, or the glow of a fluorescing ruby, is analyzed. (Each element, has a corresponding absorbtion spectrum, consisting of dark lines at exactly the same, wavelengths, against a rainbow background.), , Balmer series, , Ȝ = wavelength, RH = Rydberg constant, n = integer, Constant RH = 1.097 37×107 mí1, , As you can see, the spectral lines of hydrogen are sharp and distinct, not blurred. For, instance, the red light you see has a wavelength of 656.3 nanometers, the blue-green light has a wavelength of 486.1 nm, and the violet light, has a wavelength of 434.1 nm. Hydrogen atoms emit this light after being “excited” by an electrical discharge through the gas, caused by a, potential difference of 5000 volts applied between two electrodes., The discrete nature of the hydrogen spectrum puzzled and intrigued physicists. Why did the light emitted by hydrogen only exist at certain, wavelengths, rather than being continuous like a rainbow? And why at these particular wavelengths? Was there any way to predict the, wavelengths?, A Swiss high school teacher, J. J. Balmer, analyzed the pattern. He determined that the wavelengths were not random, but could be, determined using the formula in Equation 1. The constant RH that appears in Balmer’s formula is called the Rydberg constant., Intriguing mathematical patterns did not stop with the set of spectral lines known as the Balmer series. There are also wavelengths emitted by, , Copyright 2007 Kinetic Books Co. Chapter 36, , 661
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hydrogen that lie outside of the visible spectrum, which are predicted by formulas very, similar to the one in Equation 1., Later scientists determined that similar relationships existed for the spectral lines of, other elements, as stated by the Rydberg-Ritz combination principle. Although the, hydrogen atom is often used to discuss this principle in order to keep things simple, it, applies to all atoms. For instance, the neon sign you see in the photograph above is, displaying red light at the wavelengths of several of its spectral lines., The physicists who determined these mathematical relationships did not know why the, wavelengths of the spectral lines followed the patterns they did. These data were just, too far ahead of the theory of atomic structure. Scientists could observe the discrete, spectral lines of the emitted and absorbed light, and note the mathematical relations, that predicted their wavelengths, but could only speculate as to the cause., However, the work was underway. The light emitted by hydrogen was found to have a, discrete spectrum, one that could be predicted by a formula. It was a tantalizing clue, about the quantized nature of atoms., , What is the lowest visible, frequency of light emitted by, hydrogen?, , f = 4.57×1014 Hz, 36.3 - Photons, , Photon: A packet of light. The fundamental, unit or quantum of light., Max Planck showed how the radiation emitted by an ideal object in thermal equilibrium, with its surroundings could be explained if the radiation emitted or absorbed by the body, was quantized. His theory had a somewhat marginal existence for four years until Albert, Einstein, then employed as a Swiss patent clerk, began to consider it in depth., Planck’s theory stated that the energy (radiation) absorbed or emitted by an object, had to be taken up or released in discrete chunks, as quanta. However, he still, conceived of the radiation itself as a wave. His theory simply stated that matter, absorbed or emitted the radiation in discrete amounts., Einstein took the next bold step: He stated that the radiation itself was quantized. He, saw that what Planck had been studying was not just how matter absorbed and emitted, radiation, but the basic nature of the radiation itself., , Pre-Einstein conception of light, Light is a wave, ·Energy can vary continuously, ·Einstein demonstrated flaws in the, wave model of light, , Einstein stated that light of any frequency is quantized in units now called photons. The, energy of a photon is proportional to its frequency f., He wrote, “The energy in a beam of light is not distributed continuously through space, but consists of a finite number of energy quanta, which, are localized at points, which cannot be subdivided, and which are absorbed or emitted only as whole units.” (This was in the same year that, he published his special theory of relativity; not a bad year.), This new model challenged the previous concept that light behaved solely as a wave. Instead, it stated that light could also be conceived of as, a stream of packets of energy, almost as particles., The energy of each photon could be calculated with the equation E = hf. In other words, red light of frequency 4.60×1014 Hz cannot have just, any energy level; its energy is always an integer multiple of hf. One photon of red light of frequency 4.60×1014 cycles per second has, 3.05×10í19 J of energy, two photons of this frequency red light have 6.10×10í19 J of energy, and there is no such thing as 1.5 photons of red, light, any more than there can be 1.5 electrons., , 662, , Copyright 2007 Kinetic Books Co. Chapter 36
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It was not easy for scientists to accept Einstein’s new theory. When four elite scientists, í including Planck í nominated Einstein to the Prussian Academy of Science they, wrote, “That he may have missed the target in his speculations, as, for example, in his, hypothesis of light quanta, cannot really be held too much against him, for it is not, possible to introduce fundamentally new ideas, even in the most exact sciences, without, occasionally taking a risk.”, One year later, the American physicist Robert A. Millikan reported a precise, confirmation of Einstein’s equation for the energy of the photon, E = hf., , Photon, Packet of energy, ·Light is quantized, ·Brighter light = more photons, , Energy of a photon, E = hf, E = energy, h = Planck’s constant, f = frequency, Constant h = 6.63×10í34 J·s, , What is the energy of this photon, of red light?, E = hf, E = (6.63×10í34 J·s)( 4.41×1014 sí1), E = 2.92×10í19 J, , Copyright 2007 Kinetic Books Co. Chapter 36, , 663
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36.4 - Sample problem: solar radiance, Take the “typical” wavelength of light, to be 550 nm. What is the energy of a, photon of that light? If you were to, stare into the Sun (a very bad idea!),, how many photons per second would, enter one of your eyes? Use, 7.85×10í7 m2 for the surface area of, the pupil and assume that the, intensity of sunlight on the Earth’s, surface at your location is 1000 W/m2., , Variables, total energy, , E, , energy of a single photon, , Ep, , frequency of light, , f, , wavelength of light, , λ = 550×10í9 m, , Planck’s constant, , h = 6.63×10í34 J·s, , intensity of light, , I = 1000 W/m2, , surface area of one pupil, , A = 7.85×10í7 m2, , speed of light, , c = 3.00×108 m/s, , number of photons per second, , N, , What is the strategy?, 1., , Determine the energy of each photon by using Einstein’s equation for the energy of a photon. To determine the photon’s frequency, use, the relationship between the speed of light, its wavelength and its frequency., , 2., , Calculate the power of the light entering your eye by using the relationship of intensity, power and surface area., , 3., , Compare the power of sunlight hitting your eye to the energy of a single photon and calculate how many photons are hitting your eye, each second., , Physics principles and equations, The energy of a photon is, , The relationship between wave speed (in this case, c), wavelength and frequency is, , Intensity can be calculated as, , Power equals, , 664, , Copyright 2007 Kinetic Books Co. Chapter 36
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Step-by-step solution, We first compute the energy of a single incident photon, answering the first part of the question above., , Step, , Reason, , 1., , Ep = hf, , energy of a photon, , 2., , c = Ȝf, , wave speed, wavelength and frequency, , 3., , solve for frequency, , 4., , substitute step 3 into step 1, , 5., , evaluate, , Next we calculate the power of the sunlight entering your pupil. To do so, we use the intensity equation above, the given intensity of the light, and the area of the pupil., , Step, , 6., 7., , Reason, , intensity, P = IA, , 8., , solve for power, evaluate, , We have determined the power of the light entering your pupil and the energy of a single photon. Now we will relate these two values to find, the number of photons per second that enter your eye., , Step, , Reason, , 9., , definition of power, , 10. P = NEp, , power and photon rate, , 11., , solve equation 10 for N, , 12., , evaluate, , Note that N is a very large number of photons per second. Sensors in the eye’s retina can actually respond to just a single photon. A single, molecule in one “rod” cell in your eye can absorb one photon, triggering a chemical reaction that sends a signal to the optic nerve. However,, neural filters only let a signal go to the brain if approximately five to nine photons arrive every 100 ms. This limit prevents the visual fuzziness or, “noise” that would exist under conditions of low intensity light if the eyes were too sensitive., , Copyright 2007 Kinetic Books Co. Chapter 36, , 665
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36.5 - Photoelectric effect, , Photoelectric effect: The ejection of electrons, from a material due to light striking it. Aspects, of this effect were used by Einstein to, demonstrate the quantization of light., What caused Einstein to believe that light was quantized? In the year 1905 he used a, quantum model of light to explain the results of an experiment that could not be, explained using classical electromagnetic theory. In fact, Einstein won the 1921 Nobel, Prize in Physics “for his services to Theoretical Physics, and especially for his discovery, of the law of the photoelectric effect.”, In 1887, Heinrich Hertz had shown that shining light on metal could cause electrons to, be ejected from the metal. You can think of this process as analogous to evaporation., When light shines on water, it can cause some of the water molecules to escape as a, gas. When light shines on metal, some of the electrons can escape the metal., , Light and electrons, Shining light on metal causes it to emit, electrons, , We illustrate this phenomenon in Concept 1, with electrons escaping from the metal, electrode on the right. The “ejected” electrons could be readily explained by the, classical model of light as a wave composed of electric and magnetic fields. These, waves transported energy, and it made sense that some of the electrons in the metal, could absorb enough of this energy to escape the attractive force binding them to atoms, in the metal., In Concept 1, you also see parts of an apparatus used to conduct experiments whose, results were not so readily explained using classical theory. It consists of two electrodes, enclosed in a vacuum. The left electrode has a lower electric potential than the right, electrode, which means it will repel electrons. You may also think of the apparatus in, this way: an electric field is established between the electrodes that points from the right, to the left. This field “pushes” the electrons to the right, back toward the electrode they, have escaped from., , Incorrect expectation, Speed/energy of an electron, corresponds to light’s intensity??, , Shining light on the right-hand electrode causes electrons to be ejected from it and to, move to the left. The faster they are moving, the more negative the electric potential of, the left electrode has to be to keep them from reaching it. By adjusting the electric, potential on the left electrode so that the electrons “just fail” to reach it, an experimenter, can determine the maximum kinetic energy of the electrons., An expected result of this experiment would be that, the more intense the light shining, on the right electrode, the more energy its electrons would absorb, and the faster they, would move when ejected from the metal. It would be like chopping wood with an ax. As, you chop the wood, the harder you strike, the faster you expect some of the chips to fly, off., However, to the surprise of the experimenters, this proved not to be the case. There is, no correlation between the intensity of the light and the maximum kinetic energy of the, electrons. Rather, when the intensity of the light is increased, more electrons are, emitted. Instead of the kinetic energy of the escaping electrons increasing, only their, number increases. The incorrect “classical” expectation is shown in Concept 2, and the, actual observed behavior in the experiment is shown in Concept 3 (refresh your, browser to restart these animations). In terms of the ax analogy: It is as if hitting the log, harder results not in more energetic chips, but in more chips of the same energy flying, off., , What was actually observed, Number of electrons increases with, light’s intensity, , Another surprising result was that when the frequency of the light was below a certain, value, known as the cutoff frequency, the light could be of great intensity, but no, electrons at all would be ejected. Classical physics cannot explain these phenomena., To use the ax example one more time: If you strike a log very infrequently, but with, great force (energy), you would expect chips to fly off. However, if the photoelectric, effect applied to wood chopping, then when you chop at a slow rate, no chips will fly off,, no matter how hard you strike the log. Very odd!, Quantum theory, however, explains both these results quite neatly. Considering light as, packets of energy means that with more intense light more packets are striking the, metal each second. The energy of each packet does not change with intensity; the, number of packets does. More photons (each having the same energy) striking the right, electrode each second increases the rate at which electrons are ejected from the metal,, but not their maximum kinetic energy., This can also be likened to knocking over bottles with bullets from a rifle. In classical, theory, more “intensity” means switching to a bigger, more powerful rifle, so that each, , 666, , Relationship of frequency and, electrons, No electrons emitted, ·For colors below a certain frequency, ·No matter how intense the light, , Copyright 2007 Kinetic Books Co. Chapter 36
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bullet will hit a bottle with more energy. The same number of bottles will still be hit; they, just go flying off the target faster., In quantum theory, more “intensity” does not mean switching to a larger caliber, more, powerful rifle. More intensity means firing more bullets per second from the same rifle,, which means more bottles get knocked over. The increased intensity does not involve, changing the energy of the bullets, so the bottles always get struck with the same, amount of energy., Quantum theory also explains why low-frequency light cannot cause any electrons at all, to be emitted. The energy of each photon equals hf. Using intense light of low, frequency means that many low-energy packets of light are striking the metal, but no, individual packet has enough energy to raise an electron to an energy level high, enough for it to escape., The minimum energy an electron needs to escape the metal is called the work function,, and low-frequency photons have less energy than this. To use the rifle analogy:, Employing low-frequency light is like changing from bullets to spitballs. No spitball alone, has enough energy to knock over a target. In quantum theory, increasing the intensity of, light means the number of spitballs increases, not the energy of a spitball. Just as no, spitball is energetic enough to knock over a bottle, no low-frequency photon in the, experimental apparatus is energetic enough to free an electron., , Quantum theory (and Einstein) to, the rescue, More intense light ĺ more photons, More photons ĺ more ejected electrons, , Frequency/energy of light and, electrons, Lower frequency light ĺ photons less, energetic, Less energetic photons cannot free, electrons, , 36.6 - Sample problem: photoelectric effect, Ultraviolet light with a frequency of, 1.03×1015 Hz is incident on a metallic, sodium surface. The work function,, the minimum amount of energy, required for an electron to be ejected, from this material, is ij = 2.36 eV., What is the maximum kinetic energy, an escaping electron can have?, , The electrons in a particular metal have a range of different energies and require various amounts of additional energy in order to be “freed”, when the metal is struck by a photon., The work function of a material is the least amount of energy required to release any electron from it. In other words, it is the amount of energy, required to free the “least attached” electron., Variables, energy of photon, , E, , maximum kinetic energy of electron, , KEmax, , work function for sodium, , ij = 2.36 eV, , Planck’s constant, , h= 6.63×10í34 J·s, , frequency of ultraviolet light, , f =1.03×1015 Hz, , What is the strategy?, 1., , Use the conservation of energy. This means that the energy of the incoming photon must equal the energy required to free the electron, plus the kinetic energy of the electron as it flies away from the surface. (Ultraviolet light does have enough energy to free an electron in, , Copyright 2007 Kinetic Books Co. Chapter 36, , 667
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this case.), 2., , Observe that the electrons that are least bound to the metal will be the electrons that leave it with the maximum kinetic energy. The less, energy that is “spent” by a photon to free an electron, the more energy there is “left” that can go to increasing the electron’s kinetic, energy., , Physics principles and equations, Energy is conserved., The energy of a photon is, , E = hf, Step-by-step solution, , Step, , Reason, , 1., , E = KEmax + ij, , conservation of energy, , 2., , KEmax = E – ij, , solve for KEmax, , 3., , E = hf, , energy of a photon, , 4., , KEmax = hf – ij, , substitute equation 3 into equation 2, , 5., , photon energy, , 6., , convert units, , 7., , evaluate, , We converted to electron volts in step six because in quantum physics, energies are commonly stated in electron volts rather than joules. An, electron with kinetic energy KEmax = 1.89 eV would be moving at a speed of 8.19×105 m/s, which is approximately 0.27% of the speed of, light. At this speed, relativistic effects do not play a significant role and are ignored in the solution above., The equation derived above, KEmax = hf í ij, was an important result that was used to confirm the particle-like nature of light. Einstein first, stated it in 1905, and Robert A. Millikan experimentally verified it shortly thereafter., , 36.7 - Interactive problem: photoelectric effect, Laser engineers use the principles of quantum physics. In the simulation on the, right, you will experiment with one of the fundamental phenomena that underlie the, working of a ruby laser. Rubies are crystals of aluminum oxide with chromium, impurities that give them their distinctive red color. More importantly for engineers,, chromium is primarily responsible for ruby’s lasing properties., The simulation focuses on one of the outermost electrons of a chromium atom. The, initial step to make a laser work is to excite the electron so that it is in a high energy, state. The electron starts at a low energy state called E1. Your first goal is to, increase this electron’s energy, and cause it to jump to a higher energy state called, E3., How can you boost the energy of the electron from E1 to E3? Here, you may fire at, the atom either photons of red light with a frequency of 4.32×1014 Hz, or photons of, green light with a frequency of 5.45×1014 Hz. One photon of the red light used in the, simulation has an energy of 1.79 eV, and a photon of the green light has an energy, of 2.26 eV. These photon energies are calculated using the formula E = hf, where, h is Planck’s constant and f is the frequency of the light., Conduct a few experiments by selecting photons of different colors (energies) and then pressing GO. Do photons of both frequencies have an, effect on the outermost electron of the chromium atom, or just one frequency? After testing the photons, can you deduce what the energy, difference is between E1 and E3?, After you fire the appropriate photon, the electron will have energy E3, but this energy level is highly unstable. The electron will rapidly and, spontaneously fall to an intermediate energy level E2. This new state is relatively stable compared to the higher state, and we will pause time in, the simulation when it reaches this “metastable” state. (A photon is emitted when the electron drops from E3 to E2 but that photon is irrelevant, to the operation of the laser.), , 668, , Copyright 2007 Kinetic Books Co. Chapter 36
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Your final goal is to stimulate the chromium atom to emit another photon, which it will do when the electron drops from energy E2 to E1. You do, this by firing a new photon at the chromium. Which color do you think will cause the emission of an additional photon?, If you see two photons moving to the right and the electron in the chromium atom returns to energy level E1, you have successfully simulated, the workings of a laser. Congratulations! The second photon you fired by pressing GO has been “amplified” since it results in two photons, moving through the ruby. It has taken advantage of the energy stored in the atom to do this., As you conduct your experiments, you may note that the details of the laser process differ from material to material. With neon, the crucial, transition is from the highest metastable level to a middle level. With chromium, the material used for this transition, the crucial transition is from, a middle metastable level back to the lowest energy state., , 36.8 - Bohr atom, , Bohr atom: The atom consists of a nucleus, surrounded by electrons orbiting it at specific, radii and energy levels., Einstein showed that light was quantized. The Danish physicist Niels Bohr proposed a, model of the atom in which the energy of electrons was quantized, and could only exist, at certain values. Together, these theories explained the frequencies of spectral, emission and absorption lines., Before discussing Bohr’s theory, we will briefly explain some work that preceded his,, and then explain his crucial hypotheses., In 1897, the scientist J. J. Thomson showed that the “rays” often observed flowing, between charged electrodes in a vacuum were streams of negatively charged particles., In other words, he discovered the electron. His discovery caused scientists to update, their model of the atom. Some theorized that atoms consisted of a “mix” of negative, particles and positive regions, like negatively charged chocolate chips embedded in, positively charged cookie dough. (In fact, Thomson called it the plum pudding model,, after the plums scattered throughout a pudding.), , Bohr atom, Electron moves in circular orbit around, nucleus, Change in energy alters orbital size, Orbits are quantized, , However, in 1910 Ernest Rutherford conducted experiments that led scientists to reject, the plum pudding model. He fired alpha particles at a thin gold foil, expecting them to, sail through. (An alpha particle consists of two protons and two neutrons, and is, positively charged.), Rutherford discovered that although most of the particles passed through the foil with, minimal deflection, a few had violent collisions. These particles were deflected at, extreme angles, or even rebounded straight back., Only a dense, positive nucleus could explain this result. Rutherford’s subsequent, analysis led him to conclude that this tightly packed nucleus must be surrounded by, electrons that were orbiting the nucleus at relatively great distances, a model eerily, similar to the solar system. In short, he developed the basis of a model that is still, commonly used today to describe the atom., , Energy levels of hydrogen atoms, , During the period of these discoveries, the physicist Robert Millikan measured the, magnitude of the elementary charge, which is the amount of charge of an electron or a, proton. In a little over a decade, the basic model of a positively charged nucleus orbited, by negatively charged electrons had been established, as had the charge of an, electron., The model of the atom was radically advanced by the work of Thomson, Rutherford and, Millikan. However, physicists remained puzzled by a paradox stemming from their, understanding of electromagnetic theory and orbital mechanics., Electromagnetic theory predicts that accelerating charges emit electromagnetic, radiation. Electrons circling around the nucleus of an atom are constantly accelerating, because they are constantly changing direction; this is similar to how the electrons in an, antenna repeatedly accelerate back and forth as they oscillate over its length in simple, harmonic motion. If orbital electrons were emitting radiation due to their acceleration,, they would be losing energy, and they should eventually crash into the nucleus. The, analogous effect is witnessed with a satellite orbiting the Earth: If it continually loses, energy due to atmospheric resistance, its orbital radius decreases, and it eventually, crashes., , ǻE = Ef í Ei, ǻE = change in energy of atom, , However, since most atoms are stable (phew!), electrons are not “crashing” into nuclei,, E = energy of an electron’s orbit, but rather are maintaining orbits of a constant radius. Bohr could not explain why the, electrons acted as they did, but he formulated a theory that was consistent with what physicists were observing. He stated that electrons in, atoms could only exist in certain orbits called stationary orbits or stationary states. Bohr postulated that in these states the size and energy of, , Copyright 2007 Kinetic Books Co. Chapter 36, , 669
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an electron’s orbit is stable and constant. (By using the term stationary, Bohr did not, mean that the electrons stood still, but rather that their orbital radii and energies, remained constant.), In Bohr’s model, both the orbital radius of an electron and the total energy of its orbit are, quantized. Electrons can only exist at certain distances from the nucleus that, correspond to certain energy levels. (Bohr used the concept of angular momentum to, determine the sizes of the orbits.) His model also led to the conclusion that an electron, in such an orbit does not radiate energy continuously and that its orbital radius cannot, gradually decay, but remains constant unless it is disturbed., In Concept 2 we show a table of energy levels for a hydrogen atom. The lowest energy, level is called the ground-state energy level. At this level, the electron is at its closest to, the nucleus, and this distance is called the Bohr radius. This is the smallest possible, orbit of the electron. The ground-state energy of a hydrogen atom is í13.6 electron, volts., Note that the value is negative. Physicists liken this to an electron being placed in a, well. It takes 13.6 eV to remove the electron from the atom so that it is free, no longer, bound to the proton that is the nucleus of the hydrogen atom. The closer it is to the, nucleus, the more negative its energy. This is akin to measuring the gravitational, potential energy of, say, a rock at the bottom of a well. Its gravitational potential energy, is stated to be negative there, and it becomes less negative as it approaches the top of, the well at the surface of the Earth., , When an electron moves from N3, to N2, what happens to its, energy? Calculate the change in, energy of the atom., ǻE = Ef íEi, ǻE = (í3.40 eV) í (í1.51 eV), ǻE = í1.89 eV, , Today, electrons are not considered to be particles moving like satellites in orbits, around the nucleus, and a quantum-physics electron cloud model of the atom has, replaced the Bohr model. Nevertheless, many of Bohr’s ideas have proven to be very, useful, and his model greatly advanced the understanding of the atom., How does Bohr’s work relate to spectral lines? His model provided the first steps toward, a quantized view of the atom. As an electron moves between specific energy levels, it, emits or absorbs a quantized amount of energy in the form of a single photon of a, specific frequency., The spectral lines that result when a gas emits or absorbs energy are thus also, quantized. Bohr’s work provided a model on the atomic side of why this should be the, case., , By how much does the atom’s, energy change when the photon, strikes, moving the electron from, N1 to N3?, ǻE = Ef íEi, ǻE = (í1.51 eV) í (í13.6 eV), ǻE = 12.1 eV, 36.9 - Energy levels, photons and spectral lines, The Bohr model, combined with Einstein’s and, Planck’s work, explained the discrete spectral lines, that physicists were observing., Why does excited hydrogen gas only emit light at, certain frequencies? When it is excited, say by heat,, or by an electric current, its atoms absorb energy., When a hydrogen atom absorbs energy, its electron, jumps from one orbit to another, from a lower energy, level to a higher one. The change in energy is, quantized because electrons can only exist at the, specific energy levels prescribed in Bohr’s model., Hot, excited metal atoms emit light at characteristic frequencies. For example,, , When the atom loses energy, it does so by releasing, fireworks packed with copper salts radiate blue light when they explode., a single photon. The energy of the photon, corresponds to the amount of energy the electron loses as it returns to a lower-energy orbit., , The energy of the photons emitted by excited gas atoms must be quantized because the electron energy levels are quantized. The frequency, of a photon is proportional to its energy, f = E/h. We perceive a given frequency (or wavelength) of light as a specific color., For example, consider the red-colored line having wavelength 656 nm and frequency 4.57×1014 Hz in the emission spectrum of hydrogen. A, photon of this color must have energy 3.03×10í19 J, or 1.89 eV., It is possible to calculate the orbital change of the electron in a hydrogen atom that creates light of this frequency. An electron in the n = 3 orbit, has í1.51 eV of energy. An electron in the n = 2 orbit has í3.40 eV of energy. An electron that drops from N3 to N2 gives up 1.89 eV of, energy: exactly the energy of the red-color photon. A mystery solved! Some Nobels won!, , 670, , Copyright 2007 Kinetic Books Co. Chapter 36
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In sum, in the first decades of the 20th century, scientists had already discovered the, physics that Andy Grove and his peers would need in order to create semiconductors, and lasers. In fact, the first patents for basic semiconductor transistors were granted in, Germany and the United States in the 1920s and 1930s. The practical manufacture of, these devices required progress in the material sciences, but transistors were, manufactured in the late 1940s, and the first modern-day field effect device, a type of, transistor, was proposed in 1952 and built in 1953., , Bohr, photons and spectral lines, Orbits (and energy levels) are quantized, ·Only certain energy changes possible, Photons of light are quantized, ·Energy = Planck’s constant times, frequency, ·Observed spectral lines match energy, differences, , 36.10 - Interactive checkpoint: photons and electron energy levels, A helium atom emits light of, wavelength 471 nm when an electron, makes a transition from the n = 4 state, to the n = 3 state. If the energy of the, n = 3 state is -6.04 eV, what is the, energy of the n = 4 state?, , Answer:, , E=, , eV, , 36.11 - Conduction in solids, This chapter started off by noting that the development of semiconductor-based devices, such as transistors, has been one of the most, important technological advances of the past 50 years. We also promised that understanding their functioning would require only a basic, understanding of some principles of quantum physics. With your introduction to quantized energy levels and photons essentially complete for, the purposes of this chapter, we can now start the discussion of semiconductors, and shortly, lasers., To explain semiconductors, we need to discuss the conduction of electrical currents in solids. In terms of conducting electricity, semiconductors, lie between the extreme cases of conductors and insulators (hence the “semi” in their name). The ability of engineers to influence how readily, semiconductors conduct a current is a key to their utility., To review some terminology: Some substances (like copper or aluminum) are considered conductors; current flows relatively easily in a, conductor. Others (like silicon laced with impurities) are considered semiconductors. Others, like silicon dioxide, are insulators, where it is very, difficult to cause a current to flow., In this section, we discuss why current flows more í or less í easily in conductors, semiconductors and insulators. We start with an energy, diagram of two electrons in a single atom, as you see below. The electrons exist at distinct energy levels., , Energy levels of a single atom, , Copyright 2007 Kinetic Books Co. Chapter 36, , 671
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To understand the relative ease or difficulty of causing a current to flow, we need the concept of energy bands. Below we show what happens, when many atoms are brought together, say atoms that have crystallized into a solid. When atoms are brought very close together, their, energy levels interact and merge into a set of broader energy bands. Instead of the electrons of an individual atom being restricted to particular, energy levels that are so sharp and distinct they are represented with lines, the electrons now can exist with their energies in ranges of values,, or bands., , Energy bands in a solid, Presence of many atoms broadens, energy levels into bands, , In fact, we can get more specific about these bands. We will use the diagram below to do so. As you can see, the lower energy band is the, valence band. If you have studied chemistry, you may recall that valence electrons are those most likely to participate in chemical bonds., Although atomic electrons in the valence band can be shared with neighboring atoms in chemical reactions, they are too strongly bound to the, nucleus of their own atoms to be able to flow freely in an electric current. Electrons at lower energy levels are even more closely tied to the, nucleus, cannot flow in current, and are irrelevant to our discussion., The highest-energy electrons are in the conduction band. It is electrons in this band that can flow in a current. They are crucial to our story:, Electrons in the valence band cannot flow in a current; electrons in the conduction band can., , Two outer bands of insulator, Valence band, Conduction band, Large energy gap between bands, ·“Forbidden gap”, , It takes energy to move from one band to another, just as it takes energy for an electron to move from one energy level to another in an, isolated atom. When energy is added to an atom, its electrons can be promoted from the valence band to the conduction band, where they are, free to move and become part of a current. How much energy it takes to accomplish this determines whether the material is classified as a, conductor, a semiconductor or an insulator., Let’s now discuss this concept with specific materials. The diagram above shows the energy band diagram of an insulator, say silicon dioxide,, for the valence and conduction bands. As in the prior illustration, the valence electrons occupy states in the lower energy band shown in the, diagram. The conduction band is the upper band. Between these two bands lies the forbidden gap, or band gap. This is the Mojave Desert of, electrons. They are not allowed to exist there. Electrons can only exist with energies in ranges like the valence or conduction bands., As you see, the forbidden gap is comparatively large in an insulator. It takes a relatively large amount of energy (8 eV for silicon dioxide, if you, like specifics) to cause an electron to move from the valence band to the conduction band. With few electrons in the conduction band, it is, extremely difficult to produce a current., , Conductor, Energy bands, ·Bands overlap, , Now we consider the other extreme: a conductor. Its energy diagram is shown above. Note how the energy bands overlap; in a conductor, this, overlapping region is only partially filled with electrons. It is “easy” for an electron to move from the valence to the conduction band when a, potential difference is introduced. This means a conductor like copper can provide a ready supply of conduction electrons, ready to rumble, when the slightest electric field from a source like a battery is applied., , 672, , Copyright 2007 Kinetic Books Co. Chapter 36
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Semiconductor, Energy bands, ·Medium gap between bands, , Above, you see the energy level diagram for the third type of material, a semiconductor. (We use silicon; purists may rightly complain that pure, silicon is not truly a semiconductor, but we hope they will let us slide for a moment.) Its properties lie between those of insulators and, conductors. As the diagram suggests, it takes less energy than with an insulator for an electron to move from the valence to the conducting, band. At room temperature, it takes 1.12 eV., To contrast the differences, the diagram below shows energy band diagrams for the three types of material side by side by side. You can see, how the energy gaps differ by the type of material., , Three types of material, Contrast between energy gaps, , 36.12 - Mobile electrons and holes, , A hole: is caused by the departure of an, electron and is positive., In a physics topic like direct current circuits, an electric current is described as a flow of, electrons, since it is moving electrons that are the charge carriers in a copper wire., However, semiconductor engineers think about the flow of current slightly differently., We will use a silicon atom or two to explain., Here is a traditional discussion of covalent bonds as taught in first-year chemistry: A, single silicon atom has four electrons in its valence band, but it “wants” to have eight, there. When it is part of a solid piece of silicon, it forms covalent bonds with neighboring, atoms. It “shares” electrons with neighboring silicon atoms, and by sharing, it fills its, valence band with eight electrons. You see a symmetrical, “satisfied” silicon atom in the, diagram of Concept 1., , Silicon in equilibrium, Valence band is filled, , Now let us consider what happens when an electron makes a jump from the valence, band to the conduction band. It might do so as the silicon increases in temperature and, the internal energy of the material increases., When an electron makes this jump, a silicon atom has lost its valence electron. The, valence band now has an opening, called a hole. The diagram in Concept 2 shows the, “missing” electron as a hole. A hole is positive since it is caused by the departure of an, electron. The number of protons in the region now exceeds the number of electrons by, one., Holes are crucial in semiconductors. Why? Because they provide a place for electrons, to flow to. They provide natural “landing spots” for mobile electrons., However, there is more to it than that. When considering semiconductors, the flow of, electrons is crucial, as electrons constitute the electrical current in a conductor., However, equally important and real in the eyes of semiconductor engineers is the flow, of holes. In the animation in Concept 3, we illustrate some moving electrons and the, resultant motion of a hole. Refresh your browser screen if you did not see this animation, yet and wish to do so., , Silicon with an electron in the, conduction band, Valence band is “missing” an electron, A hole is created, ·Holes are positive, , You may consider hole movement as being akin to a bubble moving through a fluid. A, , Copyright 2007 Kinetic Books Co. Chapter 36, , 673
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bubble is the absence of fluid. The motion of the bubble can be described more, concretely as a movement of the fluid around it. When the bubble moves one way, there, is a flow of fluid in the other direction. However, the “movement” of the bubble is much, more noticeable than the motion of the fluid itself. Semiconductor engineers and, physicists treat holes as though they were particles as real as electrons., , Holes resemble bubbles, Holes move, , 36.13 - Doping, , Doping: Increasing the availability of, conducting electrons or holes by adding, impurities to a material like silicon., Commercial semiconductors are doped. This is thoroughly frowned upon in the, Olympics, but quite a desirable thing to do in a semiconductor foundry. Adding, impurities to a substance like silicon allows engineers to tailor its electrical properties, and makes it more useful in building devices like diodes and transistors., We start with the diagram in Concept 1 showing silicon atoms in an ideal or equilibrium, state, perhaps at a temperature near absolute zero. The atoms fill their valence bands, by sharing electrons., , Pure silicon, , Now let’s consider the same pure silicon, but at room temperature. The average thermal, energy of the atoms has increased. This increase in energy means some electrons will, have enough energy to spontaneously make the energy jump from the valence to the, conducting band., In fact, in absolute terms, a large number of electrons will make the jump. A cubic, centimeter of silicon contains 5×1022 atoms, and 2×1023valence electrons. At room, temperature (293 K), you will find about 1×1010 electrons in the conducting band in this, volume of silicon. Since they have left the valence band, you will find an identical, number of holes. On the one hand, this is a vast number (10,000,000,000 electrons, and an equal number of holes). On the other hand, it is extremely small compared to, the total number of electrons: About one electron out of every 1013 valence electrons, has become a conducting electron., One out of 1013 is a small fraction, less than semiconductor engineers desire. To, increase the number of conducting electrons and holes, the silicon is doped. In doping,, impurities are added to the silicon to increase its ability to conduct current. Arsenic and, phosphorus are common elements that are added to increase the number of available, conducting electrons, while hole-increasing elements include gallium and boron., Arsenic and phosphorus atoms both have five valence electrons, one more than silicon, has. When an arsenic atom takes the place of a silicon atom in the atomic lattice, four of, its valence electrons fit easily into the covalent bonds and in essence become members, of the valence band of the adjacent silicon atoms. The fifth electron, however, enters, the conduction band since there is no room for it in the valence band., , Doping, Adding a different material to silicon, n-type: inserts extra electron, “Extra” electron readily moves to, conduction band, , In Concept 2, you see an arsenic (As) atom and its “fifth” electron. When it is doped with, an element like arsenic, the semiconductor is called an n-type semiconductor. The “n”, stands for negative, since the charge carriers supplied by the dopant are negative. The, “extra” electrons are called donor electrons., To facilitate the flow of current, an engineer may want a material with holes: places for, those electrons to flow to. She would use an element like gallium or boron whose atoms, have just three valence electrons. When such an element replaces silicon in the atomic, lattice, it is one electron short. The result is a hole, as shown in Concept 3., When it is doped with an element like gallium (Ga), the semiconductor is called a p-type, semiconductor. The “p” stands for positive, representing the fact that holes act like, positive charges when they move., Arsenic is used to supply carrier electrons and gallium is used to supply holes. In, quantitative terms, the band gap separating the “fifth” arsenic valence electron from the, , 674, , Creating holes with doping, p-type: an electron “short”, Provides destination for mobile, electrons, , Copyright 2007 Kinetic Books Co. Chapter 36
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conduction band is about 1/20th the size of the band gap for the silicon valence electrons., In sum, an n-type semiconductor has a set of electrons ready to move, while in contrast a p-type semiconductor has a place ready for valence, electrons to go, freeing up holes that can then move. When an external electric field is applied to a doped semiconductor, current will flow, much more readily than in a pure semiconductor., , 36.14 - p-n junction, , p-n junction: p-type and n-type semiconducting, materials placed adjacent to one another. This, type of junction is the basis of devices like, diodes and transistors., Diode: A component that readily allows the, flow of current in one direction, and is highly, resistant to current in the other., In this section, we discuss what happens when p- and n-type materials are placed in, contact with one another. One result is a useful device, the diode., , p-n junction, p-type material has excess holes, n-type material has excess electrons, , Consider the p-n junction shown in Concept 1. A junction refers to the region or a, device where the two types of semiconducting material are touching. Remember that, the n section has excess electrons that can flow fairly readily, while the p section has, excess holes that could accept those electrons., When these two types of material are placed next to one another, some holes flow from, the p to the n material, and some electrons flow from the n to the p material., This is called a diffusion current. The electrons diffuse from the n-type material, where, there is a higher concentration of them, to the p-type, where they are relatively scarce., The holes move in the opposite direction, from the p-region where they are abundant to, the n-region where they are scarce. They diffuse, just as perfume molecules diffuse, across a room., The p-n junction is the essential element of a diode. When it is connected in a circuit in, the fashion shown in Concept 2, the net flow of current is relatively great at low, voltages. This is called a forward-bias connection., , Diode, Allows current to flow in one direction, , Why does the current flow easily? Electrons flow from the n-region of the semiconductor, rather readily to fill holes on the p side. The negative terminal of the battery acts to, replenish the supply of electrons in the n-region and the positive terminal replenishes, the holes in the p-region., Now we reverse the orientation of the battery, as shown in Concept 3. The diode will, block nearly all the current for low applied voltage: It acts as a resistor of great, resistance. This is called a back-bias connection., Why does this orientation prove so resistant to current? As Concept 3 shows, the, battery causes a significant depletion zone. Holes in the p-region are moved away from, the junction as they move toward the negative terminal of the battery, and free electrons, in the n-region move away from the junction toward the positive terminal. The region, around the junction on both sides loses its mobile charge carriers; it becomes depleted., The battery can “pull” harder and harder, but in effect, all it does is expand the depletion, zone, instead of causing a continuing current., , Battery reversed, Creates depletion zone, Prevents flow of current, , The two graphs on the right show current versus voltage curves for forward bias and, reverse bias connections. In the forward bias case, the current increases with potential, difference. The diode acts roughly like a resistor. In the reverse bias case, the diode acts almost like a break in the circuit, and even relatively, large potential differences cause negligible currents., , Copyright 2007 Kinetic Books Co. Chapter 36, , 675
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Forward bias, Current increases with ǻV, , Reverse bias, Current does NOT increase with |ǻV|, , 36.15 - Physics at work: MOSFET transistors, , Transistor: A three-terminal, semiconductor device that, forms the basis of random, access memory and, microprocessors., The physicist William Shockley proposed the first, modern-day transistors in 1952, although earlier, The first transistor, built at Bell Labs in 1947. The semiconductor substrate is, scientists had devised conceptual prototypes. In a, a germanium crystal. The three electrical leads are the source, gate and drain., circuit, transistors can be controlled so that they work, either as strong resistors or effective conductors, which is what makes them important, in electronics applications, such as electronic on/off switches (in computers), or as part, of signal amplifiers (in your stereo)., The transistors discussed in this section are called field effect transistors, since the, conductivity (or resistance if you like) of such transistors is regulated by an electric field., The first transistors, developed earlier, were of a type called junction transistors or, junction field effect transistors, but MOSFET transistors dominate many applications, today., MOS stands for metal-oxide-semiconductor and FET stands for field-effect transistor., MOSFET transistors are crucial in microelectronics, forming the basis of random access, memory (RAM) and of charge-coupled devices (CCDs). A CCD is used to record, images in digital cameras and digital video cameras., Enough acronyms: There are many more! (Other acronyms were suggested for, MOSFET, including MISFET…) Let’s move to the design of an n channel MOSFET, transistor. The basic design is shown in Concept 1., Silicon is lightly doped to form a p-type semiconductor, which is the bottom layer you, see in the diagrams on the right. This layer, the substrate, is deliberately very lightly, doped so that it will be a poor conductor, but the holes it contains are necessary for a, reason you will soon learn. Then two much more heavily doped islands of n-type, semiconductor, which are shown in green in the diagram, are formed on top of the, substrate. One island is the source S and the other is the drain D., , 676, , Structure of a MOSFET transistor, Substrate is lightly doped p-type, semiconductor, Two islands of n-type material sit atop, substrate, ·One island is source S, one is drain D, n channel connects islands, Insulator covers substrate, islands, , Copyright 2007 Kinetic Books Co. Chapter 36
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A thin channel of n-type material, called appropriately enough, the n channel, connects, the two islands. A layer of insulating material (such as silicon dioxide, whence the “O” in, MOSFET), mere nanometers thick, is deposited on top and penetrated by the two metal, leads (whence the M) shown in the diagram. On top of the insulating material lies a third, deposit of metal, called the gate G. Because of the insulating layer, no charge can flow, from the gate to the rest of the transistor., , Metal gate G is above insulator, , This is a reasonably complex configuration. It is all the more impressive when you, consider that semiconductor manufacturers are now building transistors where the, gates measure 65 nanometers across, and that a single microprocessor chip can, contain 500 million or more transistors. (It is a safe bet that humans have manufactured, more transistors than any other device.), The diagrams on the right show a transistor as part of a circuit. There is always a, potential difference between the source and the drain. The key to how the transistor, functions resides in whether there is a potential difference between the source and the, gate. We will call this difference the gate voltage., Let’s consider what happens when there is no potential difference at the gate: In this, case it has no effect on the other parts of the device. The illustration in Concept 2, shows this state. Electrons flow from the negative lead to the source island, then across, the n channel to the drain island and the positive lead, because of the potential, difference between the source and the drain. In short, a current flows through the, transistor when the gate voltage is zero., , How a transistor functions (ON), When gate voltage = 0 with, Potential difference between source and, drain, ·Electrons flow from source to drain, , Now we assume that there is a negative gate voltage. This is illustrated in Concept 3. (A, signal from another circuit might “turn on” this potential difference.) When it is turned on,, we show the gate as negatively charged. Electrons repel each other, so the field caused, by the electrons in the gate drives the n channel’s electrons into the p-doped substrate, where they find holes to occupy., The channel is said to be depleted: The n channel has less n, electrons. Another way to, put it is that the channel becomes narrower. This increases its resistance, and with a, strong enough electric field from the gate, no electrons can flow from the source to the, drain at all. This is where the “field effect” in MOSFET comes into play. (The substrate, is too lightly doped and too poor a native conductor to allow current to flow there.) The, entire process can be likened to stepping on a flexible garden hose to cut off the flow of, water., With a variation of the electric field of the gate, the transistor can be turned from ON to, OFF, from allowing current to flow to preventing it. This simple idea, enabling circuits to, be set to “ON” or “OFF”, to represent “1” or “0”, “true” or “false”, underlies the design of, computer memories and microprocessors., , How a transistor functions (OFF), When gate has negative voltage, Electrons driven out of n channel, ·n channel is depleted of electrons, Current cannot flow, , After all this, you may think: It just turns on and off. Indeed. Just repeat that 500 million, times or so and you have a microprocessor! The sophistication of the manufacturing, process allows many transistors to be packed into a very small region., The transistor also can function as part of an amplifier. Imagine that the potential, difference between the source and the drain is very large: This is the “power” part of the, amplifier. We show this in Concept 4., When there is no sound, the microphone creates a negative gate voltage, depleting the, n channel and preventing current from flowing through the transistor. Although the, voltage from the “large emf” source is much greater than the voltage of the microphone,, the microphone is preventing any current from flowing., Now imagine that a singer starts her song. As the sound of her voice becomes louder,, the microphone creates a smaller negative gate voltage, or even a positive one, and, current is allowed to flow from the drain to the source, unleashing the power of the large, emf and driving the loudspeaker. The transistor functions as a variable resistor that is, controlled by the microphone signal., The strong current flowing through the channel from the source to the drain is in perfect, synchronization with the amount of charge on the gate, which depends on the voltage, applied by the microphone. The large signal mimics the small one., , A transistor at work, Source/drain emf is large, Gate emf is small (signal), Signal determines output of amplifier, , Copyright 2007 Kinetic Books Co. Chapter 36, , 677
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36.16 - Physics at work: photovoltaic cells, , Photovoltaic effect: Electron, flow caused by photons., Solar cells are semiconductor devices that absorb, light and convert its energy into electricity. Since, sunlight is free and the operation of the cells has no, environmentally hazardous side effects, there is an, obvious appeal to their use. The challenge for, researchers and manufacturers is to lower their cost, and to raise their efficiency so that they may, effectively compete with oil, coal and other traditional, energy sources., Banks of photovoltaic cells populate a “solar energy farm”, , How do solar cells work? In Concept 1 we show a, that can produce hundreds of kilowatts of electric power., typical n-on-p junction solar cell, the most common, type of cell. A wafer of p-type silicon has an element such as phosphorous diffused into, its upper surface. This results in an n-type material, one with mobile electrons, being, located above a p-type material, whose charge carriers are mobile holes. In short, this, is a p-n junction, or diode., With solar cells, the concept of diffusion current is important. Near the junction, mobile, electrons from the n-region diffuse into the p-region, leaving positively charged donor, ions (holes) in their wake. These holes remain there and form a positive region., Mobile holes also diffuse from the p-region in the other direction, into the n-region. This, flow of charge, a diffusion current, occurs for the same reason perfume diffuses across, a room, from where the concentration is higher to where the concentration is lower. The, result is a depletion zone near the junction. That is, the n-region is relatively depleted of, mobile electrons and the p-region is depleted of the same number of holes., The effects of this diffusion are shown in Concept 2. After enough charges move,, equilibrium is reached: The result in the depletion zone is a built-in electric field that, points from the n-region toward the p-region. This field opposes any further motion of, charge, and the diffusion current quickly stops., , Structure of photovoltaic cell, p-substrate with n-type material diffused, into surface, ·A p-n junction, , In an earlier section, we discussed how placing a p-n junction in a circuit with a battery, could create a depletion zone. The point here is that a depletion zone also forms, spontaneously when the materials are simply placed adjacent to one another., Now let’s consider what happens when a photon strikes the solar cell. The only photons, relevant to our story are those that are energetic enough to promote an electron from, the valence to the conduction band. When such a photon strikes a valence electron in, the semiconductor, it increases the energy of the electron, promoting it to the higher, band., The electron will flow toward the n-type semiconductor, since that side of the depletion, zone is positively charged, and the hole will flow to the negatively charged p-type, semiconductor. To put it another way, the electric field caused by the diffusion current, “pushes and pulls” the holes and electrons freed by photons striking the semiconducting, material. Press the refresh button in your browser to see this occur in Concept 3., Since many photons will strike the material, many electrons and holes will flow: A, current is born. By placing metallic contacts on either side of the junction, this current, can be used to power a load. In Concept 4, you see the solar cell in a circuit powering a, home. There is a flow of negatively charged electrons out of the n-region into the circuit,, and a flow of positively charged holes out of the p-region into the circuit., , When materials placed together, Electrons spontaneously diffuse, n to p, Holes spontaneously diffuse, p to n, , The price of photovoltaic cells has been steadily decreasing over the past 30 years., However, the electrical power produced by such cells still costs more than power from, fossil fuels (coal and oil), or wind-generated power. To cite some approximate numbers, (since energy prices fluctuate), a kilowatt-hour produced by burning a fossil fuel costs, from 3.5 to 4.5 cents. Wind power costs just 4.5 to 5.5 cents per kilowatt-hour, although, there are issues with it: What do you do when the wind is not blowing?, In contrast, the cost of solar power is approximately 25 to 45 cents per kilowatt-hour., Solar power costs more money, although it could be cost effective for supplying power, to remote locations, since power lines would not have to be run from distant power, plants., Analysts in the environmental community do raise issues about the “true” costs of, different energy sources, such as the costs of environmental side effects, including, health issues, pollution and global warming. Fair enough! With those costs factored in,, , 678, , When a photon strikes, A mobile electron/hole pair is formed, The electron moves to n-region, , Copyright 2007 Kinetic Books Co. Chapter 36
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different conclusions can be drawn about which power source really is the cheapest., However, the numbers above reflect the costs that companies and consumers pay in, the short term for power from different sources, and they help to explain the current, dominance of fossil-fuel energy., , The hole moves to p-region, , Current flows, As photons cause electron/hole pairs, , 36.17 - Lasers, , Laser: Light Amplification, by the Stimulated Emission, of Radiation., Semiconductor transistors rely on physics that was, pioneered in the early 20th century, and so do lasers., Lasers amplify electromagnetic radiation. They can, be used to amplify radiation at frequencies ranging, from infrared to x-rays. To simplify the discussion, we, will simply say that they amplify light, not worrying, about whether or not the radiation is visible to humans., , Laser emitting coherent red light., , The light that emanates from a laser has three crucial properties. It is (1) coherent,, (2) monochromatic, having one frequency, and (3) highly directional. Coherent means, that the emitted waves are all in phase with one another: The light can be considered as, a single wave. In contrast, the light waves that emanate from a light bulb are out of, phase, or incoherent. Such light is often described as consisting of a collection of finite, wave trains, and the trains are not synchronized., In Concepts 1 and 2, you see light emanating from a flashlight and from a laser. The, light from the flashlight consists of many wave trains, which as you can see have, different wavelengths and frequencies, and are traveling in different directions. The, wave trains from the flashlight are not in phase: The locations of peaks and troughs, vary by wave train. The contrast with laser light is clear: its waves are in phase, have, one frequency, and travel in a single direction., , Light from flashlight, Incoherent, multi-colored, divergent, , A working laser has three essential parts: a laser medium, a pumping process, and a, feedback mechanism., The laser medium can be manufactured from a wide variety of materials. The first, operational laser used a ruby crystal. Today, the laser medium can be a gas (such as, helium-neon), a liquid, or a solid, as is the case with diode lasers, the type you would, find in a DVD player., The laser has a pumping process, during which the atomic electrons of the medium are, excited to high energy levels. This can be done by means such as electric discharges,, flash lamps, or even light from other lasers., The pumping process increases the energy of the atoms of the medium. Once these, atoms are excited, photons injected into the medium cause it to emit other photons:, Light shined into the laser medium generates additional light., The container has silvered reflective walls, so that photons in the medium are reflected, back into it. The process repeats as these photons cause even more in-phase photons, to be emitted, and the original light is amplified., , Laser light, Coherent, monochromatic, highly, directional, , This basic system is illustrated in Concept 3. (Press the refresh button in your browser if, you want to see the animation.) The laser medium is inside the container. All the photons reflect off of the mirror on the left, back into the, medium, and most of them reflect off the mirror on the right, while some are allowed to pass through. The mirrors form an optical feedback, mechanism., , Copyright 2007 Kinetic Books Co. Chapter 36, , 679
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This feedback mechanism can be likened to the audio feedback that may occur with a, sound amplifier. If you have ever winced at the loud squeal as someone experiments, with an amplifier, you have heard the unintended consequences of feedback: Too much, of the amplified sound is feeding back into the microphone, and is amplified again, and, again, until a high-amplification runaway reaction occurs. In a laser, the mirrors, reflecting light back into the laser medium provide the feedback. When the system is, properly configured, coherent oscillation occurs, and a highly monochromatic, highly, directional output beam is created., Above, we have described the fundamentals of a laser. There remains a basic question,, however: Why should the light emanating from the laser be of one color, or frequency?, In a laser, you pump energy into the medium, and then use that energy to create an, intense beam of light. But why is that light all of the same wavelength, say 633 nm? To, explain why, we have to turn to quantum theory., , Fundamental laser components, Laser medium: accepts energy, emits, light, Pumping process: how electrons get, excited, Feedback mechanism: enhances and, focuses the signal, , 36.18 - Laser pumping and stimulated emission, , Pumping: Exciting atoms, into higher energy levels., Stimulated emission: An, “excited” atom emits a, photon when a photon of the, right frequency passes by., Illumination with ultraviolet light makes these minerals fluoresce with, characteristic colors. A sample of calcite, in the middle, glows orange., The core of a laser’s functioning is its pumping, process, followed by the stimulated emission of, radiation. In this section, we describe three types of interaction between radiation and matter. The first, where matter absorbs radiation, is, relevant to pumping. The second and third are processes in which matter emits radiation., , Below, we show absorption. An atom in its ground state absorbs a photon and one of its electrons changes states, moving to a higher energy, level. These levels are typically described with subscripts; so, for example, the atom might go from energy level E0 to level E1., , Absorption, Atom absorbs photon, raising energy, state, , After an atom is excited, how does it return to its ground state? One answer is shown in the illustration below, which illustrates spontaneous, emission: An atom can spontaneously, with no outside influence, lower its energy state by emitting a photon in a random direction. (This may, not seem spontaneous since the atom first had to be “excited,” but this is the terminology, and it contrasts with what will be described below.), , Spontaneous emission, Atom releases photon, lowering energy, state, , 680, , Copyright 2007 Kinetic Books Co. Chapter 36
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If you shine ultraviolet light upon certain minerals, such as calcite, they fluoresce, that is, they glow as long as the energizing ultraviolet, radiation continues. A mineral absorbs high-energy photons of radiation and then spontaneously emits photons in a series of steps. These, minerals only absorb and emit certain wavelengths of light, due to the quantized nature of their energy levels, so they glow a particular color., Spontaneous emission is an interesting topic but not the controlled process desired in lasers. If all the laser medium did was to absorb light and, then spontaneously re-emit it in random directions, there would be no amplification of the light., Stimulated emission is the key to lasers. Below, you see an atom that has already been excited to a higher energy state., , Stimulated emission, Photon passes near excited atom, Photon triggers emission of 2nd photon, Two photons leave atom, , A photon of the appropriate frequency passes close to the excited atom. The result: The atom emits its own photon of the same frequency and, returns to its initial energy state. Here is the “gain” produced by lasers: One photon of light causes a net result of two: the process of light, amplification by the stimulated emission of radiation. The presence of one photon causes another to be emitted., We use words like “passes close” and “presence” because photons do not “collide” with atoms during stimulated emission, but they do cause, the atom to emit a second photon when the energy of the stimulating photon corresponds to a difference in energy levels allowed in the atom., In sum, with stimulated emission, you start with one photon, and end with two., To explain further what is occurring in a laser medium, we will use two analogies. The simpler one is mechanistic. Imagine a pool ball that has, been raised off the ground and placed on the surface of a flat kitchen table. This increases its gravitational potential energy and corresponds to, an atom that has been excited. A second pool ball is rolled at the first and both of them fly off the table. The combined kinetic energy of the two, balls as they reach the ground is greater than the original kinetic energy of the ball that was rolled. The collision has “unleashed” the potential, energy of the pool ball that was resting on the table., There is a limitation here, perhaps an issue that concerned you as you considered this metaphor and applied it to a large number of pool balls,, in order to make it more similar to the many atoms in the laser medium. The process would not produce a coherent stream of pool balls. If you, did this experiment with many pool balls, they would fly off the table in many directions., It is better to think of the light as a wave and to consider the phenomenon of resonance, as exhibited by mechanical or electromagnetic waves., Quantum physics states that atoms act like electromagnetic oscillators with particular resonant frequencies. The passing light wave causes the, atoms in the medium to begin vibrating in a resonant, coherent relationship. The oscillations are driven by, and in phase with, the stimulating, light wave. The atoms respond to the incoming light and reradiate like tiny antennae. The reradiated waves reinforce the waves that cause, them., , 36.19 - Population inversion, , Population inversion: The number of excited atoms is greater than the number, of lower-energy atoms., Stimulated emission is required for lasers to function. For there to be stimulated emission, photons must be passing by and interacting with, excited atoms in a medium, from which they can cause coherent photons to be emitted, rather than just passing by atoms whose electrons are, at lower energy levels. Not only do the low-energy atoms fail to participate in stimulated emission, but they may also sabotage the process by, absorbing photons., Under normal circumstances, more atoms in a medium will be in lower energy states than in higher energy states. A population inversion is, required: The majority of the atoms must be pumped to a higher energy level. When photons strike a medium with an inverted population,, stimulated emission is more likely to occur., , An “unexcited” population, Lower-energy atoms outnumber higherenergy atoms, When light shines into medium, ·More photons absorbed than emitted, ·Light is not amplified, , To illustrate the need for an inverted population, we start by showing you the opposite in the diagram above: photons striking a material with a, , Copyright 2007 Kinetic Books Co. Chapter 36, , 681
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normal population. Some of the atoms are at higher energy levels, but most are not. This is typical of a material that is in thermal equilibrium, with its surroundings, such as one at room temperature., When photons strike the material, most of them are absorbed. This raises the energy levels of some of its atoms, which then spontaneously, emit photons in random directions. Occasionally, there is stimulated emission when a photon encounters an atom that already contains an, excited electron, but this is rare. This is no way to run a laser!, , Population inversion, Higher-energy atoms outnumber lowerenergy atoms, When light shines into medium, ·More photons emitted than “fired in”, ·Light is amplified, , In contrast, the medium above has a population inversion. A photon is far more likely to strike an excited atom, and stimulated emission, becomes commonplace. Placing the medium in a reflecting container can further aid this process. Photons will reflect off the container walls, and strike other excited atoms, causing a “chain reaction” of stimulated emissions. Energy can be continually added by pumping the medium,, ensuring a ready supply of excited atoms., , 36.20 - Physics at work: operating a laser, Today, there is a variety of ways to create a, population inversion that enables light amplification., To discuss one, let’s consider a type of laser that was, developed early on, in 1960 and 1961, and remains, common, the helium-neon (He-Ne) laser. A glass tube, is filled with helium and neon gases. An electric, current passes through the tube, and its electrons, collide with helium atoms, raising their energy levels, from their ground state E0 to an elevated state EA., In this state, the helium atoms are said to be at a, metastable level. They will stay there for a relatively, He-Ne laser in a laboratory., long period of time, as opposed to quickly and, spontaneously emitting photons and falling back to a, lower energy level. (Relatively long, in this context, means on the order of a thousandth, or a ten-thousandth of a second.), The helium atoms, including those at the metastable level EA, are continually colliding, with neon atoms in the gas mixture. The energy level of a metastable helium atom is, very close to that of a high-energy neon atom. In a collision with a ground-state neon, atom, a metastable helium atom causes the neon to rise to energy level EB while the, helium atom itself returns to the ground state. This is shown in Concept 1. Neon atoms, at EB are also metastable; they will persist for a relatively long time at this elevated, energy level., Why not just let the electric discharge excite the neon atoms, and skip the helium?, Helium is required since neon atoms do not respond readily to electron bombardment, by the current that excites the helium atoms., , Laser process: part 1, , So far, we remain in the dark, so to speak. Many neon atoms have had their energy, raised to EB. The population has been inverted: There are more neon atoms at EB than, at E0. The next goal is to stimulate the neon atoms to emit photons, dropping to the, , Energy of helium atoms raised, Energy transferred to neon atoms, Population inversion results, , intermediate energy level EC in the process. To accomplish this, light is beamed into the laser medium consisting of photons having an energy, corresponding precisely to the difference between EB and EC. These photons interact with neon atoms whose electrons are at energy level, EB, causing stimulated emissions: The EB neon atoms each emit a photon of the same frequency, and drop to energy level EC. The atoms, undergo what is called a laser transition., It is important to note that the energy of each stimulating photon must equal the energy difference between EB and EC in order for the laser to, function. Here is where quantum physics comes into play. A precise understanding of the energy levels of helium and neon atoms is required, to design a successful He-Ne laser, as is knowledge of the relationship between the energy and frequency of a photon., The additional photons that are emitted interact with other energized neon atoms, causing further stimulated emissions. Mirrors on both ends of, the laser reflect these photons back into the laser medium, along the axis of propagation. The distance between the mirrors is crucial since it, ensures the constructive interference of the light. They require extremely precise fabrication, and reflect more than 99.5% of the photons that, hit them., One last detail: How about the neon atoms that are now at energy level EC? They rapidly and spontaneously decay to their ground state,, , 682, , Copyright 2007 Kinetic Books Co. Chapter 36
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emitting photons in all directions in the process. Those photons are in essence “noise”, while the stimulated emission photons are “the signal.” The spontaneous emission, occurs after about 10í8 seconds, ten thousand or so times faster than it occurs with, metastable states., This rapid transition is important. It means that not many neon atoms linger in the EC, state. They quickly return to the E0 state where collisions with helium atoms can excite, them back to EB, and the lasing process can begin anew. And there is another reason, why it is important that few atoms have energy EC: If there were too many of these, around, then they would absorb photons and return to energy EB, which would thwart, the lasing process. (Neon atoms at E0 cannot absorb photons and rise to either EB or, EC, since the required energy differences do not match the energy of the photons being, used.), He-Ne lasers are inexpensive and common. The laser tubes can be purchased for less, than $100. These may yield a power output of 1.0 mW when connected to a DC power, input of 10 W. This means they are about 0.01% efficient. Such lasers have been used, for purposes ranging from supermarket scanners to laser printers, but semiconductor, lasers, which are cheaper to fabricate, are rapidly superceding them., , Laser process: part 2, Photons enter laser medium, Stimulated emission occurs, ·light is amplified, Neon returns to initial state, , 36.21 - Sample problem: ruby laser, The light emitted by a ruby laser has, wavelength 694.3 nm, and ruby emits, this light when electrons drop from, energy level E2 to E1 in the ruby’s, chromium atoms., What is the energy level difference, between E2 and E1 in a chromium, atom?, , Theodore Maimann built the first operational laser in 1960, a flash-pumped ruby laser. Maimann placed a cylindrical ruby rod inside a flash, pump. The flash pump excited the chromium atoms in the ruby to a metastable state using an intense burst of light. Then the ions fluoresced, back to the ground state, releasing light of wavelength 694.3 nm. Carefully aligned mirrors were placed at either end of the ruby to produce a, coherent laser oscillation., Variables, energy, , E, , wavelength, , λ =694.3 nm, , What is the strategy?, 1., , Calculate the energy of a photon of the light emitted from a ruby laser., , 2., , Use the energy of the light to calculate the energy level difference between E2 and E1., , Physics principles and equations, Energy of a photon, , E = hf, Energy level difference, , ǻE = Ef í Ei, , Copyright 2007 Kinetic Books Co. Chapter 36, , 683
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Step-by-step solution, , Step, , 1., , E = hf, , Reason, , energy of photon, , 2., , frequency, wave speed and wavelength, , 3., , substitute equation 2 into equation 1, , 4., , energy difference, , 5., , substitute equation 4 into equation 3, , 6., , evaluate, , 36.22 - Gotchas, An atom’s electrons can exist at any energy level. No, this statement contradicts one of the key tenets of quantum physics. The electrons can, only exist at specific energy states, and will not be found with energies between those levels., A friend says: an electron falls from an energy level of í4.5 eV to í7.2 eV. It emits a photon with 2.7 eV of energy. Has he learned his quantum, physics? Yes, he has. The energy of the photon equals the amount of energy given up by the atom., All photons have the same energy. No, all photons of electromagnetic radiation (for example, light) of a particular frequency have the same, amount of energy. The energy of a photon increases with the frequency of the radiation., , 36.23 - Summary, Something is quantized if it has a smallest, indivisible unit. The opposite of, quantized is continuous. The eggs you buy in a carton at the grocery store are, quantized; the amount of milk you pour into a glass is effectively continuous., , Balmer series, , The Balmer series formula predicts the wavelengths of visible light that are present, in the emission spectrum of excited hydrogen gas. When discovered, it revealed a, mathematical pattern in the spacing of the spectral emission (or absorption) lines of, hydrogen that strongly hinted at an underlying order., The physicists Max Planck and Albert Einstein showed that radiation is quantized., The quantum of light is called a photon. The energy of a photon of electromagnetic, radiation equals Planck’s constant times the frequency of the radiation. Einstein, cleared up the mystery behind the photoelectric effect by making the assumption, that light is quantized., , Energy of electromagnetic radiation, , E = hf, , (one photon), , E = nhf, , (multiple photons), , Niels Bohr developed the basis for the modern-day quantum view of the atom. He, stated that the orbits of electrons around the nucleus of an atom are quantized: Electrons can only exist at orbits of specific radii and energy, levels. When its electrons jump between levels, an atom emits or absorbs photons whose energy corresponds to the change in electron, energy., Quantum theory is used in the design of semiconductors. Semiconductors are doped í mixed with impurities í to alter the nature of the, resistance they offer to currents. Doping creates more mobile electrons and holes than exist in pure semiconductor material., Key components of transistors include p- and n-type semiconductors. A p-type material has mobile holes that act as charge carriers; an n-type, material has mobile electrons. An external potential difference can increase the supply of these charge carrriers near a p-n junction in a, transistor, allowing current to flow more readily. However, if the potential difference is changed in magnitude or reversed, it can cause mobile, charge carriers to move away from the junction, reducing the flow of current or stopping it altogether., Laser designers also rely on insights from quantum theory. Some agent such as an electrical discharge excites the laser medium, for example, a helium-neon gas mixture. This elevates the medium’s electrons to higher energy states. When photons of light are injected into the medium,, the excited atoms are stimulated to emit identical, in-phase photons, increasing the flux of photons in the medium and amplifying the light., For a laser to work, there must be a population inversion: there must be more excited atoms than unexcited ones in the laser medium. If there, is no population inversion, then any photons injected into the medium will be absorbed, increasing the energy of the atoms, rather than causing, additional photons to be emitted., , 684, , Copyright 2007 Kinetic Books Co. Chapter 36
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Chapter 36 Problems, , Chapter Assumptions, When converting between electron-volts and joules, assume that, 1 eV = 1.60×10–19 J, , Conceptual Problems, C.1, , One estimate for the number of grains of sand on all the beaches on Earth is 5×1021 or five thousand billion billion grains., Making the simplifying assumption that sand grains are uniform, give examples where the weight of an amount of sand can be, measured without considering the quantized nature of the sand, and cases where this cannot be ignored., , C.2, , Suppose two different hydrogen atoms, labeled A and B, have their electrons in state n = 4. Atom A emits light when its, electron transitions directly from the n = 4 state to the n = 2 state. Atom B emits light twice, as its electron first transitions from, the n = 4 state to the n = 3 state, then transitions from the n = 3 state to the n = 2 state. (a) Are the three wavelengths of light, the same or different? (b) How does the energy of the photon from atom A compare to the total energy of the two photons, from atom B?, (a), , i., ii., iii., i., ii., iii., , (b), , C.3, , The wavelengths are the same, The wavelengths are different, Not enough information to tell, The energies are equal, The energies are different, Not enough information to tell, , Suppose that for a particular metal, the photoelectric effect occurs with light of wavelength Ȝ0. With this light incident on a, small region of metal, measurements show that electrons leaving the surface have kinetic energies as large as E0. If the, intensity of light is increased without changing the wavelength, (a) how does this affect the number of electrons leaving the, surface? (b) What is the effect on the maximum kinetic energy of the ejected electrons?, (a), , There are fewer electrons leaving the surface., The number of electrons leaving the surface is unchanged., There are more electrons leaving the surface., , (b), , The electrons' maximum kinetic energy is less than E0., The electrons' maximum kinetic energy is still equal to E0., The electrons' maximum kinetic energy is greater than E0., , C.4, , In a p-type semiconductor, a current can flow when an electric field is applied and the following happens:, i., ii., iii., iv., , There is movement of holes in the valence band., There is movement of electrons in the valence band., There is movement of holes in the conduction band., There is movement of electrons in the conduction band., , C.5, , In a p-n junction, can a hole cross from the p side to the n side without an electron crossing the junction from the n side to the, p side? Explain your answer., , C.6, , In your own words, explain why current does not readily flow in a circuit loop with a diode that is back-biased., , C.7, , Two doped semiconductors, one n-type and the other p-type, are brought together to form a p-n junction. Before being placed, in contact, they are each electrically neutral. They are brought into contact, and reach equilibrium (that is, the diffusion current, stops). (a) Is the net charge on the n-type semiconductor positive or negative? (b) Is the net charge on the p-type, semiconductor positive or negative?, , Yes, , C.8, , No, , (a), , Positive, , Negative, , (a), , Positive, , Negative, , Light is amplified in a laser. Consider photons with characteristics that are specific to the laser medium, that is, the laser will, emit light at the frequency of the photons. What property of these photons initially increases during the amplification process?, i., ii., iii., iv., , Wavelength, Number of photons, Energy of each photon, Frequency, , Copyright 2000-2007 Kinetic Books Co. Chapter 36 Problems, , 685
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C.9, , Does energy need to be continually supplied to maintain a population inversion? Explain your answer., Yes, , No, , Section Problems, Section 2 - Balmer series, 2.1, , What wavelength light is emitted from a hydrogen atom when the electron transitions from a state where n = 4 to one where, n = 2?, nm, , 2.2, , If a visible photon with wavelength 434 nm is emitted from an atom in a hydrogen gas lamp, what energy state was the atom, in before the photon was emitted? That is, what was the number n for that state?, , Section 3 - Photons, 3.1, , Visible light consists of photons having energies that range from 2.8×10í19 J to 5.0×10í19 J. What is the range of photon, frequencies in visible light?, Hz up to, , 3.2, , Hz, , Given the information in the previous problem, what is the range of photon wavelengths in visible light?, lowest wavelength:, , m, m, , highest wavelength:, 3.3, , Gamma ray bursts come from tremendous explosions in faraway galaxies. These bursts are thought to be produced during, the end stages of the lives of massive stars, and give off more energy than any other observed astrophysical phenomenon. A, gamma photon in such a burst could have a frequency of 1021 Hz. A typical visible photon has a wavelength of 550 nm., Roughly how many times more energetic is this gamma photon than a typical visible photon?, i., ii., iii., iv., , 10, 1000, 1000000, 1000000000, , Section 5 - Photoelectric effect, 5.1, , Electrons can be ejected from a clean silver surface using light with wavelengths as large as 262 nm. What is the work, function for silver?, , 5.2, , Potassium has a work function of 2.21 eV. What is the maximum kinetic energy of electrons escaping from a sample of, , eV, potassium if photons of frequency 6.00×1014 Hz strike it?, J, 5.3, , In a demonstration of the photoelectric effect, electrons leave a calcium surface with velocities as great as 2.0% of the speed, of light. If the work function for calcium is 3.2 eV, what is the wavelength of the incident radiation? Relativistic effects may be, neglected., nm, , 5.4, , Platinum has a work function of 6.4 eV. Radiation with wavelength 150 nm and intensity 1.3 W/m2 is directed at a platinum, surface. (a) Find the maximum kinetic energy of the ejected electrons. (b) If 1.0% of the photons striking a 1.0 cm2 area of, platinum eject electrons, how many electrons are emitted per second from that part of the surface?, (a), , J, , (b), , electrons per second, , Section 8 - Bohr atom, 8.1, , If a 12.1 eV photon was absorbed by a hydrogen atom initially in its ground state, which then emits 2 photons, what are the, emitted photon wavelengths?, shorter wavelength:, longer wavelength:, , 686, , nm, nm, , Copyright 2000-2007 Kinetic Books Co. Chapter 36 Problems
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Section 11 - Conduction in solids, 11.1 What is the longest wavelength that a photon can have to promote an electron from the top of the valence band to the bottom, of the conduction band in silicon? Assume a band gap of 1.12eV for silicon., m, 11.2 What is the longest wavelength that a photon can have to promote an electron from the top of the valence band to the bottom, of the conduction band in silicon dioxide? Assume an energy gap of 8.0 eV between the valence and conduction bands., m, , Section 14 - p-n junction, 14.1 Diodes in forward-biased circuits emit, electromagnetic radiation. Specifically, when, an electron from the conduction band in the ntype semiconductor of the diode fills a hole in, the valence band of the p-type semiconductor, of the diode, it emits a photon whose energy is, equal to the difference in energy between the, two bands. The energy required for this, emission is provided by whatever is driving the, electric current in the circuit., The accompanying photograph shows a circuit, containing green light emitting diodes (LEDs). LEDs actually look a lot like computer chips, but they are enclosed in protective, plastic lenses like the ones you see here so that they will project a bright light. Light emitting diodes are made from non-silicon, semiconductors such as gallium, and the energy difference between their conduction and valence bands is such that the, photons they emit have wavelengths in the visible light spectrum. LEDs are used for applications ranging from jumbo TV, screens in sports stadiums to the digits of glow-in-the-dark clocks., The LEDs in the photograph emit green light with a wavelength of 520 nm. What is the energy difference between their, conduction and valence bands?, eV, , Copyright 2000-2007 Kinetic Books Co. Chapter 36 Problems, , 687
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37.0 - Introduction, Richard Feynman (1918 í 1988) was an American physicist best known to the, general public for his leading role in the commission that investigated the, destruction of the space shuttle Challenger in 1986. His bestselling memoirs, delighted millions with his irreverence and tales of bongo drum-playing., In the physics community, Feynman was known for his insight into quantum, mechanics and the brilliance of his lectures. The “Feynman Lectures in Physics,”, still in print today, show his timeless ability to clearly and memorably explain, fundamental aspects of physics., We start this chapter by giving you a chance to conduct a thought experiment that, Feynman described as “impossible, absolutely impossible, to explain in any, classical way, and which has in it the heart of quantum mechanics.” In this thought, experiment, Feynman emphasized the shortcomings of considering matter solely as, composed of particles, and the need to adopt a vision where matter has both a, particle and a wave nature. This is known as the wave-particle duality., Feynman first asked the reader to consider classical particles, that is, those for, which the physics of Newton completely predicts the motion. Feynman used bullets in his experiment; we will be slightly more pacific and ask, you to imagine throwing baseballs at a picket fence. The picket fence is missing two of its slats. The missing slats provide gaps that the, baseballs can pass through. The balls will stop when they strike a wall behind the fence., If you conducted this experiment yourself, you would find that most of the baseballs hit the wall directly behind the missing slats. A few, baseballs would hit slightly to the sides of those areas, corresponding to the balls that went through the gaps at an angle. You would see two, piles of baseballs that accumulated behind the gaps in the fence., Feynman used this first thought experiment to remind his audience how they expect the world to “work” based on everyday experience. This, experiment provided the contrast with another experiment, one you can conduct for yourself using the simulation to the right., In the simulation, you are conducting a similar experiment, but with particles of far smaller scale. Instead of baseballs being fired, electrons are, fired one at a time toward a barrier with two slits. The slits are very narrow and very close together., The electrons pass through the slits and reach a screen, which is a photographic material that records where they land. The electrons cannot, be observed as they move from the source to the barrier, or from the barrier to the screen. Only the final position is marked, by using black, dots., Press FIRE to launch a single electron. Then, hold the FIRE button down to fire a stream of electrons so you can see the pattern of where they, land., Look at the pattern of where the electrons accumulate. Instead of accumulating in two piles, like the baseballs did in Feynman’s first thought, experiment, you see regions where many electrons accumulate, alternating with regions where very few electrons land. The pattern should, remind you of the dark and light fringes that are created on a screen when light shines through a pair of slits., In the theory of optics, physicists explain the pattern of light and dark fringes by modeling light as a wave, with wave-like properties such as, frequency and wavelength. This simulation shows that something which you have solely considered to be a particle, an electron, also displays, wave-like properties in a similar experiment. This is the essential point of Feynman’s second thought experiment: Particles such as electrons, have wave-like properties. A single particle can travel from a source to a screen and demonstrate interference effects due to the presence of, the two slits. The wave that is associated with a moving particle is called a matter wave., Now you can use the simulation again to see another fundamental aspect of quantum mechanics. Reset the simulation. Press the FIRE button, a few times and note the locations of the first three or four electrons. Then press RESET again, fire a few more electrons, and note their, locations. In your two experiments, did the first three or four electrons show up at the same location each time?, The answer to the question is “no”. In both this simulation, and in the real experiments that this simulation is recreating, the location of a single, electron cannot be predicted. Although the overall pattern of light and dark fringes as you fire hundreds of electrons can be predicted, where a, particular electron will strike the screen cannot be determined in advance., This is the second point of the animation: The locations of the electrons can be stated in terms of likelihoods, like the probability of drawing an, ace from a deck of cards. The pattern of light and dark fringes provides a “map” of where an electron is likely to strike the photographic paper, behind the slits. However, you cannot predict in advance where any one electron will land, any more than you can state with certainty that one, particular card will be an ace., You are witnessing a major point of quantum mechanics in this simulation. Scientists have performed the experiment with electrons, and the, results have been exactly as depicted. Electron after electron can be fired through slits, and the interference of their matter waves will create a, pattern like the one you see here., The simulation shows the wave-like properties of particles. Scientists like Einstein had also postulated the particle-like nature of light, which, had been considered a wave. This chapter will start with that topic, the particle nature of electromagnetic radiation, before returning to the topic, that the experiment in this section illustrates., , 688, , Copyright 2000-2007 Kinetic Books Co. Chapter 37
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37.1 - The Compton effect, , Compton effect: Considering x-rays as, composed of particles explains why their, frequency diminishes as they are scattered by a, material., Einstein stated that light was composed of indivisible packets of energy. His theory was, rooted in experiment. By treating light as a particle he could explain the photoelectric, effect and make successful predictions: Calculating the energy of the photons that, make up light of a particular color allowed Einstein to predict whether shining that color, light on a given material would cause it to emit electrons., This opened up another question: Did light have other particle properties, such as, momentum? In 1916, Einstein derived an equation quantifying a photon’s momentum., As shown in Equation 1, the momentum equals Planck’s constant divided by the, photon’s wavelength. However, it remained to be shown experimentally that photons, actually had momentum. Some scientists believed that the photoelectric effect might be, explained by updating the wave theory of light, but if light could be shown to have, momentum then it would be hard to deny its particle-like nature., , Electromagnetic radiation as, wave, When an x-ray met a crystal:, ·Scientists anticipated its frequency, would stay the same, ·They were wrong, , Physicists such as W.H. and W.L. Bragg í a famous British father-and-son pair í had, been using x-rays to analyze the structure of matter, particularly crystals. The small, wavelengths of x-rays, compared to the atomic spacing in the crystals, meant that the, scattered radiation would exhibit diffraction patterns, that is, the radiation would be, strong in particular directions. The diffraction pattern could be used to analyze the, regular, periodic layout of atoms within crystals, and this technique was later used to, deduce the double-helix structure of DNA. Since diffraction is a property only of waves,, you might think that x-ray diffraction experiments could not provide any support for a, particle view of electromagnetic radiation í but as it turned out, they did., Scientists experimenting with x-rays scattering from target materials started to observe, some disturbing data. To explain their discomfort, we first have to explain what they, expected., Their expectations were based on the view of light as an electromagnetic wave,, consisting of oscillating electric and magnetic fields. In this classical picture, when an, electromagnetic wave encounters an atom, it causes the atom’s electrons to oscillate at, the same frequency as the wave. Accelerating charges emit radiation, so these, oscillating charges then radiate their own electromagnetic waves. This process is called, scattering; some waves would be re-emitted in the same direction as the initial wave,, while others would be re-emitted in other directions., The critical point to focus on is the frequency of the re-radiated waves. Classical, electromagnetic theory, used by scientists like the Braggs, predicted that the outgoing, waves should have the same frequency as the incoming waves. Scattering would, change the direction of the incoming radiation, but scientists were confident that it, should not change its frequency., Alas, observations did not accord with theory. Scientists observed that atoms being, subjected to x-rays were re-emitting radiation of lower frequency than the initial, radiation, and the effect depended on the angle at which the radiation was scattered., Something was diminishing the frequency (and the energy) of the scattered radiation., At first, the data were not taken seriously. However, in 1923 the American physicist A., H. Compton published two papers that argued conclusively that if radiation were, quantized, one should expect the frequency to be reduced. The problem was not with, the experimental data; the problem was with a theory that regarded light solely as a, wave., , Compton effect, Consider x-ray as a photon (particle), A photon “collides” with an electron in, the crystal, Photon loses energy in collision, Less energy means lower frequency (E, = hf), , Momentum of a photon, , p = momentum, Ȝ = wavelength, h = Planck’s constant, 6.63×10í34 J·s, , How did Compton explain the discrepancy? He assumed that the electromagnetic radiation was made up of photons that had momentum. He, proposed that instead of picturing a light wave shaking electrons up and down, scientists ought to picture the interaction as akin to a collision, between two particles, a photon and an electron. Like Einstein, he was asking his peers to expand their conception of radiation to include, properties usually associated with particles, such as momentum., He then used classical mechanics, applying the laws of conservation of energy and momentum to the collision. He applied the same principles, that would be applied to a collision of two billiard balls. (His analysis had to be more complex than that for two balls, because he had to relate, the quantized photon energy to wavelength and frequency, and relate its energy and momentum.), Compton stated that when the photon in question collides with an electron belonging to an atom in the target, the electron gains some kinetic, energy from the collision. Energy must be conserved, and the photon loses that same amount of energy. The energy of a photon equals hf, the, product of Planck’s constant and its frequency. When its energy is diminished in the scattering process, so too is its frequency., , Copyright 2000-2007 Kinetic Books Co. 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Not only did Compton’s work explain the reduction in frequency, but his analysis of the collision also correctly relates the direction of the, scattered x-rays to their change in frequency. When a moving particle strikes a stationary target, its change in momentum and the angle at, which it scatters are related. Consider the photon-electron collision. If the direction at which the photon comes out is virtually unchanged from, its original direction, then its change in momentum is small, and very little momentum (and energy) will be given to the electron. In other words,, the two particles just suffered a glancing blow., When Compton observed x-ray photons whose direction was barely changed, he saw that their frequency was also practically unchanged., Again, the basics of collisions held true., As the angle between the emitted and incident radiation increased, the change in momentum increased. More momentum (and energy) is, transferred to the electron. Radiation that was backward scattered suffered the largest reduction in frequency., Compton’s work with x-rays confirmed that electromagnetic radiation possesses momentum, a property that had been classically associated, with particles. It was increasingly hard to argue that light should only be considered a wave when it could be demonstrated that it had, momentum, and its interaction with matter could be modeled using a classical explanation of collisions. It became necessary to admit that, under certain conditions the wave nature of light is observed while in different experiments the particle nature is needed to explain the results., , 37.2 - Matter waves, After Compton’s observations, physicists were forced to confront the fact that light,, which had been thought of as solely an electromagnetic wave, also has properties of a, particle. This new viewpoint enabled them to understand the experimental data they, were confronting., However, there still remained many enigmas. For instance, Bohr had constructed his, model of the hydrogen atom, which successfully predicted observed emission and, absorption spectra. Bohr proposed quantized energy states for the electron, starting, with a classical view of the electron as a negatively charged, point-sized particle circling, the positive nucleus, and using Planck's work. However, Bohr was not able to, demonstrate why his model was correct., In 1923, a French doctoral student named Louis de Broglie proposed a simple idea to, help to rescue the physicists from their intellectual tar pit. Recognizing that light has, both wave and particle properties, he reasoned that nature is symmetrical, and that the, same is true for matter. De Broglie asserted that particles such as electrons have both, particle and wave properties., , De Broglie, Matter is particle with wave-like, properties, , To quote de Broglie: “…I had a sudden inspiration. Einstein’s wave-particle dualism was, an absolutely general phenomenon extending to all physical nature…”, Although it was a simple idea, it propelled the next revolution in physics., De Broglie conceived of the electron in an atom as a standing “matter wave” vibrating, around the nucleus. In other words, the electron is “smeared out” instead of being a, single point-sized particle. Let’s consider the implications of representing an electron, with such a wave, as shown in Concept 2., As you may recall, a standing wave results from waves that interfere with one another., For there to be constructive interference, peak must meet peak, and trough must meet, trough. In contrast, if at a given location, the peak of one wave meets the trough of, another, the result is destructive interference í a flat line, in essence., Here, the string is looped into a circle, and the length of the string is the circumference, of the circle. A wave begins its circular path around the string, and when it makes a, loop, it meets up with itself., If as it makes a second journey around, peak meets peak, the result will be a standing, wave. On the other hand, if peak meets trough, the wave will cancel. The relationship, between the circumference of the circle and the wavelength determines whether there, is constructive interference. In Concept 2, we show an example where there is, constructive interference., , De Broglie and the electron, orbits, Wave-like properties explain quantized, orbits, , Wavelength of a matter particle, , The wave shown to the right does not represent the actual path of the electron through, space. It is a matter wave í a way to visualize the likelihood of finding the electron at a, given location. If the amplitude of the wave is zero, then the likelihood is zero, and there, will be no electron., This provides one piece of the puzzle. Only certain wavelengths are possible for a given, orbit, but that doesn’t prevent any orbit being possible í it just dictates possible, wavelengths., The other piece of the puzzle comes from considering the angular momentum of the, electron. De Broglie showed why Bohr’s quantization argument could be justified by, considering the angular momentum of the electron and the relationship between, momentum and wavelength., , Ȝ = wavelength, p = momentum, h = Planck’s constant, 6.63×10í34 J·s, , To quantify the wavelength of a matter particle such as an electron, de Broglie proposed that the same equation that describes the momentum, , 690, , Copyright 2000-2007 Kinetic Books Co. Chapter 37
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of a photon, p = h/Ȝ, could also be applied to matter. A particle’s wavelength, sometimes called the de Broglie wavelength, is related to its, momentum by Ȝ = h/p, as shown in Equation 1., De Broglie’s insight provided a crucial step in the understanding of the atom. He and other physicists used the idea to write down equations for, the standing waves corresponding to any particle confined to a small space. The form of the matter waves (also known as the wavefunction of, the particle) leads to predictions about the behavior of the particle. This is the principle behind quantum mechanics, also called wave, mechanics., , 37.3 - Observing matter waves, When light is shone through a pair of small slits, an interference pattern results. We, show a light interference pattern in Concept 1., If electrons can act like waves, they should also display interference patterns, and they, do. The slits used should be of a width comparable to the wavelengths of the electrons, in the experiment, which are moving at a speed such that their wavelength is on the, order of 10í10 meters. You see the interference pattern caused by sending electrons, through slits in Concept 2., At the time that de Broglie proposed his theory of matter waves, it was not possible to, make slits small enough to demonstrate electron diffraction. However, in 1927 two, physicists named Clinton Davisson and Lester Germer inadvertently produced electron, diffraction using a crystal of nickel. The spacing between atoms in the crystal happened, to be on the order of the electron wavelength, causing the electrons’ matter waves to, interfere., After witnessing this and the diffraction of other particles such as neutrons and whole, hydrogen atoms, scientists began to take the wave-like nature of particles for granted,, or perhaps better put, to marvel at it as a fact of nature. They called it the wave-particle, duality., , Finding the wave in light, Light’s wave-like properties visible in, interference patterns, , Once they finished marveling, they also concluded that they could take advantage of, the wave properties of matter. Wave diffraction imposes a limit on how small of an, object can be resolved when it is probed with radiation of a certain wavelength. For, example, an ordinary microscope uses visible light and glass lenses, and cannot, resolve objects much smaller than 10í6 m, which is on the order of the wavelength of, visible light., To gain increased resolution, a transmission electron microscope (TEM) employs, electrons instead of light. The wavelengths of those electrons are about ten thousand, times smaller than that of light, which allows the TEM to resolve objects down to a size, of about 10í10 m. Just as the light in an ordinary microscope passes through a sample, that is fixed on a glass slide, the electron beam (think of it as a wave) passes through, the thin sample on the way to a detector. An ordinary microscope uses a glass lens to, focus light rays; the TEM uses magnetic fields to focus the charged electron beam., In Concept 3 is an artificially-colored image captured by a transmission electron, microscope. It shows a salivary gland of a mosquito infected by the Eastern equine, encephalitis virus (red dots). The individual viruses are about 60 nanometers in, diameter, and even smaller details than this are visible in the image. In comparison, the, best optical microscopes can only resolve details as small as 200 nanometers., , Matter waves, Electron interference pattern, Pattern also created by projecting, electrons at crystal, , Transmission electron, microscope, Takes advantage of small wavelengths, of electrons, Resolves details at a scale of 10í10, meters, , Copyright 2000-2007 Kinetic Books Co. Chapter 37, , 691
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37.4 - Matter waves are probabilistic, De Broglie correctly asserted that electrons and other particles can be described as, matter waves. Physicists, being quantitative types, wanted to know: How could they, mathematically describe the location of a wave-particle like an electron?, In classical mechanics, when the net force exerted on a particle of a certain mass is, known, the equation F = ma is used to calculate the acceleration of the particle. If the, initial position and velocity are also known, then other equations can determine the, particle’s location and velocity at any time., Quantum physicists found that applying classical equations to atomic-sized particles led, to paradoxes. For instance, the equations could be interpreted to predict that a quarter, photon, or half an electron, should be present at a given location. This was in, contradiction not only to the tenets of quantum physics, but to experiments, which, showed that photon and electrons were indivisible., Another conundrum can be discussed using the double-slit experiment. Electrons are, fired through a pair of slits toward a barrier, as shown in Concept 1. A photographic, plate records where the electrons strike the barrier. The pattern recorded on the plate, (and shown in Concept 1) looks the same as that created by shining light through the, slits., , Electron fired through slits, Create interference pattern, Where a single electron will land can, only be stated as likelihood, , Although physicists can state in advance what the overall pattern will look like based on, factors such as how wide the slits are, they cannot predict in advance where any given, electron will land on the photographic plate. They can only state the probability that it, will land near a given location í for instance, the probability is much higher in the areas, of the plate that contain the most dots., You can liken this to a game of cards. You know that on average, one out of four cards, will be a spade. However, given a shuffled deck and asked to state for sure whether a, spade will be the first card you draw, or the second, you cannot. You can only say that, there is a one in four chance that it will be a spade., If 12 cards are dealt from a shuffled deck of cards, the probability is greatest that 3, cards will be spades, but from observing the repetition of many such trials, you could, conclude that it is fairly likely that the actual number will be 2 or 4, and it is possible that, there will be 1 or 5 spades, or even 0 or 12. All you can do is describe the probability of, a given number of spades that will emerge when you deal 12 cards., , Position, Can be stated as a probability for, “classical” objects, , The section started with a question: How can the location of a particle like an electron, be described? Quantum physicists describe such a particle with a wavefunction. The, value of a wavefunction at a particular point and time is related to the probability of, finding the particle near that point, at that time. To be more precise, the absolute square, of the wavefunction is a probability density, a concept we discuss next., Quantum physicists use the idea of probability density to describe the likelihood of, finding a particle in a given region of space. Probability density is the probability per unit, volume and is the absolute square of the wavefunction. (In one-dimensional problems,, probability density is the probability per unit length, and in two-dimensional problems, it, is the probability per unit area.), The term “absolute square” covers the case when the wavefunction has complex values, (containing the imaginary number i). The absolute square equals the wavefunction, multiplied by its complex conjugate. If the wavefunction contains only real values, so, that its complex conjugate is the wavefunction itself, then the absolute square of the, wavefunction is the same as the “ordinary” square of the wavefunction., , Particle in a box, Probability density describes likelihood, that particle will be found in a given, region, , What is meant by probability density? It is a concept that can be applied to objects, typically described with classical mechanics. For instance, someone could create a, graph of your probability density at midnight. It is relatively likely that if someone is looking for you at that time, they would find you in bed. This, means that the probability density function has a relatively high value there. The probability density function at midnight would also have, smaller peaks at other locations in space, such as the chair in front of your computer, or in front of the refrigerator. Since the probability is zero, to find you on Mars at midnight (or at any other time), the probability density function would be zero there., The same idea can be applied to the position of a mass on a spring that is moving back and forth in simple harmonic motion with a certain total, energy (or, physicists say that the system is in a particular state). You could graph the probability density function for the location of the mass., In fact, we did so, and that graph is shown in Concept 2., How do you interpret this graph? Again, the locations where the graph is higher are the locations where it is more likely you would observe the, mass. If you were only allowed to take photographs of the mass at random times, you would create a graph like this. (This is the situation in, which quantum physicists find themselves.) The graph is higher near the endpoints of the mass’s motion because it is moving more slowly, there, which means it spends more time there and you are more likely to observe it there. Taking a snapshot at a random time, you are least, likely to find it near the center because it moves the most quickly through that region., In Concept 3, we move more to the realm of quantum mechanics. One intellectual construct that quantum physicists use to describe particles is, the concept of a particle moving in a one-dimensional rigid box. We use the concept of waves and a wavefunction to describe that particle,, , 692, , Copyright 2000-2007 Kinetic Books Co. Chapter 37
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which in this case is in its lowest energy state. We then construct a probability density, graph. It shows that the particle is more likely to be found in the center of the box. The, walls of the box are at 0 and A, and the particle never escapes the box. This means the, particle is always present within the box, that is, the probability of finding it outside this, region is zero., The probability density that describes the position of an actual particle is rarely as, simple as the one shown in Concept 3. In Concept 4, you see the graph of the, probability density function for the electron in the ground-state in a hydrogen atom (the, electron has the lowest energy that it can have). You can see that the electron is most, likely to be found at the Bohr radius, but that it is possible for the electron to be found at, radii other than this value. For large values of the radius, the probability density, approaches zero. The very low probability density at large radii means it is possible, but, highly, highly unlikely, that a bound electron can be found, say, one meter from the, nucleus. Quantum physicists can use the Schrödinger equation to create a, wavefunction that yields the graph you see in Concept 4. This is one of the fundamental, equations in quantum mechanics., , Position of bound hydrogen, electron, Height of graph reflects likelihood of, finding particle in given region, , The graph in Concept 4 tells us about the behavior of an electron in the ground-state of, a hydrogen atom. If you measured the distance of the electron from the nucleus, and then did so again and again, each time charting its, location, you would create a graph (a radial probability density) similar to that. You would note that the largest number of your observations, have the electron at a distance equal to the Bohr radius, but you would find it at other locations as well., Along with Einstein’s theories of relativity, quantum mechanics has changed the world’s conception of reality. Einstein showed that time and, length cannot be treated as absolutes, but that the motion of the observer affects these properties. Physicists have now shown that concepts, as common as the location of a particle must be stated in terms of probabilities. The work of these physicists has set the direction of physics for, the last 100 years or so., , 37.5 - Interactive exercise: observing the probabilities of a particle, The simulation on the right is used to show how an interference pattern emerges as, more and more electron waves pass through a pair of slits., You are asked to observe two things., First, press the FIRE button a few times and observe the location of the first five or, six electrons. Then press RESET, and do this again. Do you observe each electron, at the same location every time you run the simulation?, At the risk of ruining the punch line, the answer is no. If this is not clear, just fire one, electron, press RESET, and fire one electron again., You cannot predict in advance where a given electron will land on the screen. You, can predict the probability of it landing near a certain spot, but not where it will land, for sure., Although you cannot predict the location of any given electron, after enough, electrons pass through, the overall pattern is visible as predicted by the intensity, function., After you fire 200 or so electrons, look at the graph. (Depress and hold down the FIRE button.) It will very closely resemble the graph of the, intensity function for the interference of two electromagnetic waves passing through a pair of slits. Where the function is greater, you will see, more photons accumulate over time. The smaller the function at a point, the fewer the number of particles (be they photons or electrons) that, will accumulate nearby. Try firing 500 electrons, and you will see that the graph resembles the intensity function even more as the number of, particles increases., What you are observing here is in contrast to what one would expect using classical physics. If you toss a ball with a speed of exactly 25.0 m/s, at an angle of 30.0 degrees from a height of exactly 1.50 m in a vacuum, classical physics states that you can exactly predict its trajectory and, where it will land. If you again throw it with the same velocity from the same height, it will always land at the same location. Classical physics, would claim the same certainty and reproducibility for an electron launched in the same way., No such luck with quantum physics í or maybe one should say “it’s all luck” with quantum physics? If you fire two consecutive electrons at the, double-slit system, under identical conditions, you do not know for sure that the particles will land even remotely near the same spot. You can, only observe where they land this time., The more electrons that are fired, the more accurately you can predict the overall pattern. But you can no more predict the outcome of a single, spin of a roulette wheel than the landing point of any one photon or electron., , Copyright 2000-2007 Kinetic Books Co. Chapter 37, , 693
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37.6 - Heisenberg uncertainty principle, In prior sections, we discussed the interpretation of a matter wavefunction as relating to, the likelihood that a particle will be found in a particular location., Now we will turn to what happens when you try to measure the location of a particle., We will keep things simpler by assuming that the particle moves along a line, namely, the x axis., In general, if you want to know where a particle is, you make a measurement. All, measurements are imperfect and have some uncertainty. ǻx represents the uncertainty, in position. If you were making a measurement of the momentum of the particle in the xdirection, there would also be some uncertainty of momentum ǻpx., A classical physicist would agree there is some uncertainty in all measurements, just as, you may have learned while doing lab assignments. Better procedures and instruments, can reduce the uncertainty. For example, if you were trying to measure an electron’s, position and momentum 5.00 seconds after it had been propelled by a given electric, field, you might set up 1000 identical trials, use the best instruments available, and, average the results for momentum and for position., According to classical physics, these uncertainties can in principle be reduced to zero., There are no limits to knowledge about a classical particle. A classical physicist, considering the theoretical uncertainty of position and momentum could write an, equation like:, (classical physics), , Heisenberg uncertainty principle, , ǻx = uncertainty in position, ǻpx = uncertainty in momentum, = Planck’s constant/2ʌ, , The German physicist Werner Heisenberg, in contrast, made the bold statement that knowledge is limited. He related the uncertainties in the, particle’s position and momentum along the same axis:, , This inequality is a statement of the Heisenberg uncertainty principle. It states that there is a tradeoff between reducing the uncertainty in, position and trying to do the same for momentum. One can find situations where ǻx is relatively small, but this low uncertainty in position, comes at the price of a higher uncertainty in the particle’s momentum. Or if the momentum can be well determined, then the particle’s location, will be less precisely known., Heisenberg’s principle does not describe an “equipment problem”. The problem cannot be solved with a better microscope, or more expensive, lab equipment, or any other technique. Physicists hold that it is a fundamental property of reality: A particle’s position and momentum must, reflect a certain minimum amount of uncertainty because the particle simply cannot have both a definite position and a definite momentum. The, wave/particle is spread out in space., One important implication of this principle is that a particle can never have zero kinetic energy. If it did, it would be stationary (and have zero, momentum), and then both its position and momentum could be determined. Consistency with the uncertainty relation requires that the particle, must have at least some amount of kinetic energy, even at zero temperature., Again, as with much quantum physics, the departure from classical physics becomes apparent only when the mass of the particle is very small., , 694, , Copyright 2000-2007 Kinetic Books Co. Chapter 37
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37.7 - Gotchas, More massive particles always have shorter de Broglie wavelengths than less massive particles. Not necessarily. A particle’s wavelength, depends on its momentum, which equals its mass multiplied by its velocity. A more massive particle may have a longer wavelength than a, lighter particle, if its velocity is small enough., Only particles like protons and electrons have a de Broglie wavelength associated with them. “Matter waves” do not apply to larger things like, baseballs. No. The wavelengths of objects with relatively large momenta, such as moving baseballs, are so small that experiments do not, reveal their wave-like properties. But these objects are still subject to the laws of quantum physics. For large objects moving at ordinary, speeds, the predictions of quantum physics and those of Newtonian mechanics are identical for all intents and purposes, much as special, relativity essentially agrees with Newtonian mechanics at slow enough speeds., , 37.8 - Summary, Further evidence that light has properties of particles í in addition to properties of, waves í is provided by the Compton effect. When a photon and an electron collide,, they behave much like two colliding billiard balls. The photon’s frequency is, reduced, because it has transferred some energy to the electron. The Compton, effect provides evidence that though photons do not have mass, they do have, momentum., One of the basic tenets of quantum physics is that matter, like light, also exhibits, wave-particle duality. The wave nature of matter explains why the energies of, electrons in an atom are quantized., , Momentum of a photon, , Wavelength of a matter particle, , The equation Ȝ = h/p relates a particle’s wavelength to its momentum, and applies, to particles of matter or of light., Matter wave interference can be observed using an experiment similar to the, double-slit experiment for light. However, in the case of matter waves, the, wavelengths of even the smallest particles (generally those with the least, momentum), such as electrons, are very small and so require very small slits to, observe. Such small, closely-spaced slits are difficult to manufacture, but certain, crystals naturally have atomic spacing similar to the required slit spacing. Matter, waves were first observed using such crystals., , Heisenberg uncertainty principle, , Matter waves are probabilistic. The wave- and particle-like properties of matter are reconciled in the form of the particle’s wavefunction. The, value of the wavefunction is related to the probability that the particle will be found at a given location at a given time. Mathematically speaking,, a particle is described by a probability density function that tells the probability of observing the particle in any region., One consequence of the wave-particle duality is that it is impossible to know both the position and momentum of a particle to infinite precision, at the same time. In fact the product of the uncertainties in position and momentum has a precise lower limit: half of ʄ. This is the Heisenberg, uncertainty principle., , Copyright 2000-2007 Kinetic Books Co. Chapter 37, , 695
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Chapter 37 Problems, , Chapter Assumptions, 1 eV = 1.60×10–19 J, h = 6.63×10–34 J·s, ƫ = 1.05×10–34 J·s, mec2 = 0.511 MeV, , Conceptual Problems, C.1, , Electron A is accelerated from rest by a potential difference of 500 volts, and electron B undergoes a similar treatment by a, potential difference of 1000 volts. Afterward, which electron has the larger de Broglie wavelength? Explain your answer., , C.2, , An electron and a proton are each accelerated from rest by a potential difference whose magnitude is 500 volts. (The, electron is accelerated from a potential of 0 V to 500 V, whereas the proton is accelerated from 500 volts to 0 volts.), Afterward, which has the larger de Broglie wavelength? Explain your answer., , C.3, , A bowling ball is dropped from a tall building. While it is speeding up, does its de Broglie wavelength become smaller,, become larger, or stay the same? Explain your answer., , Electron A, , Electron, , Electron B, , Proton, , i. Becomes smaller, ii. Becomes larger, iii. Stays the same, , Section Problems, Section 1 - The Compton effect, 1.1, , What is the momentum of a photon of visible light with wavelength 555 nm?, kg · m/s, , 1.2, , What is the momentum of an x-ray photon with a frequency of 1.10e+18 Hz?, kg · m/s, , Section 2 - Matter waves, 2.1, , What is the wavelength of a 1.20×103 kg car that is traveling along the highway at 99.0 km/hr?, , 2.2, , The wavelength of a nonrelativistic electron is 2.71eí6 m. How fast is it moving? Express your answer as a fraction of the, speed of light, c., , m, , c, 2.3, , The de Broglie wavelength of a particle is 7.01×10í13 m and it is moving at 0.189% of the speed of light. What is its mass?, kg, , 2.4, , The kinetic energy of an electron is 29.3 eV. (a) How fast is it moving, expressed as a fraction of the speed of light c? (b), What is its wavelength?, , 2.5, , An electron, starting from rest, is accelerated through a potential difference and ends up with a de Broglie wavelength of, , (a), , c (b), , m, , 1.12×10í10 m. What potential difference is needed to accomplish this?, V, 2.6, , An electron is moving such that its de Broglie wavelength equals the Compton wavelength. How fast is it moving, expressed, as a fraction of the speed of light c? Hint: You must use a relativistic expression for momentum., c, , 696, , Copyright 2000-2007 Kinetic Books Co. Chapter 37 Problems
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Section 3 - Observing matter waves, 3.1, , A beam of neutrons, all moving at the same speed, is directed through a pair of slits whose centers are 2.00×10í6 m apart., An array of detectors is located 49.9 m from the slits. The intensity pattern of received neutrons is measured, and the first, zero-intensity point is 1.76 mm off-axis. The mass of a neutron is 1.68×10í27 kg. (a) What is the de Broglie wavelength of a, neutron in the beam? (b) What is the speed of the neutrons in the beam?, (a), , m, , (b), , m/s, , Section 6 - Heisenberg uncertainty principle, 6.1, , 6.2, , A barrier contains a vertical slit of width 5.0×10í6 m. An electron approaches the barrier and passes through the slit. (a) What, is the uncertainty in the horizontal position of this electron (measured parallel to the barrier, not in the direction of the, electron's motion) as it emerges on the other side? (b) What is the minimum uncertainty in its corresponding horizontal, momentum?, (a), , m, , (b), , kg · m/s, , The position of a 900 kg boulder's center of mass has been determined to within an uncertainty of 1.0 nm. (a) What is the, minimum uncertainty in the boulder's velocity? (b) Repeat the calculation, but for a proton with the same uncertainty in, position. (c) Repeat the calculation, but for an electron with the same uncertainty in position., m/s, , (a), , 6.3, , (b), , m/s, , (c), , m/s, , The y-component of a dust particle's velocity is measured with an uncertainty of 1.0×10í6 m/s. The particle has a mass of, 1.6eí9 kg. (a) What is the limit of the accuracy to which we can locate the particle along the y-axis? That is, what is the, minimum uncertainty in the y-position? (b) Does this place any limitation on the accuracy with which we can locate its position, along the x or z axes?, (a), (b), , m, Yes, , No, , Copyright 2000-2007 Kinetic Books Co. Chapter 37 Problems, , 697
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38.0 - Introduction, “Turn off the lights when you leave the room.” “Don’t, buy that gas guzzler.” “Turn the thermostat down a, few degrees in winter.” These phrases are all about, saving energy, a worthwhile topic, but this chapter is, about where that energy comes from., The energy powering your computer and lights is, electrical, but that electricity was likely generated from, the burning of oil or coal. In turn, these fossil fuels, came from the remains of ancient animals and plants,, which derived their energy from sunlight. If your, power was generated from hydroelectric sources í, Nuclear power plants convert the energy stored, letting water stored behind a dam spin a turbine as it, in atoms into energy humans can use., falls to a lower height í it was sunlight that, evaporated the water, which later turned into rainwater which was stored behind the dam., Ultimately, the source of our energy here on Earth is the Sun, and nuclear fusion is what powers the Sun. The fusion process takes lighter, chemical elements such as hydrogen, and forms helium and heavier elements, in the process releasing energy., In that respect, we owe our very existence to nuclear energy. However, to some, the term “nuclear physics” brings darker images to mind. For, better or worse, the world learned what nuclear fission was in 1945 at Hiroshima, although the exact same physics principles behind atomic, bombs are used peacefully every day., In this chapter, you will learn about the nature of atoms, which is a field of study called nuclear physics. You will become more familiar with, protons, neutrons and alpha particles. You will learn what occurs when atoms are split apart (fission), or when they are forced to merge, (fusion)., You will also learn about many peaceful applications in the field of atomic physics such as radioactive dating (no, this does not mean that two, radioactives go out for dinner and a movie) and energy production in nuclear reactors., , 38.1 - The atom and the electron, Nobel Laureate Richard Feynman posed a hypothetical situation in which our entire, base of scientific knowledge was destroyed, and we could pass on just one sentence to, our descendants. What would we choose to tell them, in order to convey the most, information in the fewest words?, In the famous “Feynman Lectures”, his choice was to tell the post-apocalyptic, population that all matter is composed of atoms, which are tiny particles that continually, move around, attracting each other when they are fairly close together, but strongly, repelling when they are pressed even closer together., How did scientists arrive at this modern picture of the atom? In fact, what is an atom?, Humans have spent many millennia asking the question “What is the world made of?”, We have come a long way from the ancient Greeks, who believed that the cosmos was, composed of four elements: water, earth, air, and fire. Now, we know of over 100, chemical elements, which are the fundamental building blocks of matter. You have, heard of many of the elements: hydrogen, oxygen, carbon, gold, lead, and so on., , Elements, Fundamental building blocks of matter, Atoms are smallest distinct component, of elements, , The modern definition of an element is a substance that cannot be divided or changed, into another substance using ordinary chemical methods. Each element has different, physical and chemical properties such as density, specific heat, and the way in which it bonds with other elements. You can see an example of, a common element, carbon, in Concept 1., An atom is the smallest piece of an element that still has its chemical and physical properties. Atoms consist of electrons circling a nucleus, composed of protons and neutrons. This is illustrated in Concept 2. The next few sections will discuss the inner structure of the atom., The modern theory of the atom began with the discovery of the first subatomic particle, the electron, in experiments performed over the course, of about 40 years in the late 1800s. Physicists found that when they applied a high voltage across a low-pressure gas, an electric discharge, was produced. They used a device called a cathode ray tube to achieve this effect. Today, you can witness a similar discharge in a neon sign, or a fluorescent light bulb., The physicists determined that the gas in the tube was conducting electric charge. The tube’s negative electrode, also called a cathode,, emitted a type of “invisible ray” that could cause a glow in the treated glass wall of the tube. Later experiments showed that the rays could be, deflected by electric and magnetic fields. This suggested that the rays were charged particles, and not a form of electromagnetic radiation., Today, scientists would say that an electric current is flowing through the gas, and the current consists of electrons. The discovery of the, electron is credited to J.J. Thomson in 1897. He showed that these rays were small, negatively charged particles. Thomson also made the first, measurement of the ratio of their charge to their mass., , 698, , Copyright 2007 Kinetic Books Co. Chapter 38
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As with many scientific discoveries, this discovery raised more questions. Scientists had, long known about electric charge. Since most matter is neutral, they knew there must, be positive charges to balance the negative charge of the electrons. To put it at the, microscopic level: Atoms must contain positive charges to balance the negative, electrons. But how were these positive and negative charges arranged in an atom?, Answering that question is the topic of the next section., , Atoms, Contain a nucleus and one or more, electrons, , 38.2 - Rutherford’s discovery of the nucleus, , Nucleus: A relatively small region in the center, of an atom where the positive charge í and, most of the mass í of an atom is located., The modern model of the atom describes it as a small positive nucleus, surrounded by, orbiting electrons. The radii of the orbits are far larger than the size of the nucleus. How, did scientists create this model of the atom? For instance, what led them to believe that, the nucleus was small compared to the size of the orbital radii?, Scientists in the early 20th century struggled to understand the nature of atoms. They, could not directly see the structure of atoms, but knew that they contained negativelycharged electrons. They reasoned that atoms must contain equal amounts of positive, and negative charges since matter tends to be electrically neutral., Without other data, most scientists thought that the positive charges that make up, matter were evenly distributed throughout the atom í why not? Since electrons have so, little mass, scientists knew that the positive charges carried almost all the mass of the, atom. Their mental picture of the atom looked like a blob of positively-charged cookie, dough with small chocolate chips (electrons) embedded in it. The model actually bore, the name of a more popular dessert at the time: The plum pudding model consisted of, negatively-charged electrons (plums) scattered throughout a massive cloud of positive, charge (pudding) that was distributed uniformly throughout the volume of the atom. This, model is shown in Concept 1., Lord Rutherford made a major breakthrough in this area, winning the 1908 Nobel Prize, in Chemistry for his experiments. As a good scientist, he wanted data to support (or, contradict) the plum pudding model of the atom. At the time, Rutherford was studying, alpha particles, which are massive, positively charged particles that are emitted at high, speed from some radioactive substances (such as radon). He realized that a beam of, alpha particles might serve as a tool to probe the atomic interior., In his experiment, he aimed a beam of these particles at a thin gold foil, and measured, the distribution of the outgoing alpha particles. This is known as a scattering experiment, í observing how the particles scatter í and it is a now-common technique to probe the, details of atomic-sized systems., , Old “plum pudding” model, Positive charge distributed evenly, with, embedded electrons, ·Wrong!, , Rutherford’s experiment, Fired alpha particles at gold foil, Some particles scattered by a large, amount, Implied compact, massive nucleus, , A mechanical analogy to what Rutherford did would be to probe the interior of a box by, shooting a high-powered BB gun at it. If the box were filled with sponge cake, the BBs, would pass through. If it had a metal plate inside, the BBs would rebound back. Or if, there were a small metal sphere inside, a small fraction would rebound or scatter, sideways, with a distribution of angles., Rutherford initially assumed that the alpha particles would be passing through a, “pudding” of positive charges spread uniformly throughout the foil. Relying on this, model, Rutherford predicted that most of the alpha particles should just pass straight, through or be only minimally deflected., This was not what he observed. Much to Rutherford’s surprise, a small fraction of the, particles were scattered by 90° or more. Occasionally, an alpha particle even, rebounded from the foil, straight back at the source. He concluded that the atom must, not have a uniform distribution of positive charge inside. Instead, the large force, necessary to cause such scattering of the positively charged alpha particle could be, , Nucleus consists of positive charge, , Copyright 2007 Kinetic Books Co. Chapter 38, , 699
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provided only if the atom’s positive charge (and mass) were highly concentrated within, the atom, in a region called the nucleus., , Nucleus is small compared to orbits of, , electrons, After analyzing the data and seeing how few of the alpha particles actually scattered,, Rutherford concluded that the nuclear radius must be about 10,000 times smaller than, the atomic radius, a figure that is still accepted today. Rutherford’s groundbreaking experiment proved that the atom is mostly empty space. His, atomic model is shown in Concept 3. Note that the diagram is not even close to being drawn to scale; the atomic diameter is far too small, compared to the nuclear size. If the diagram were drawn so that the nucleus were 1 cm wide, roughly the width of your pinky, the atom would, have to be drawn about 100 meters wide, about 10% longer than the length of an American football field., , 38.3 - Components of the nucleus, Rutherford proved that atoms consist of a compact, very dense positively-charged, nucleus surrounded by negatively-charged electrons. In this section, we take a deeper, look at the parts of the nucleus, and introduce some common terminology and notation, that scientists use when talking about elements and nuclei., The positive charge of the nucleus comes from particles called protons, which are about, 1800 times more massive than electrons. The simplest nucleus consists of a single, proton, and the simplest atom is hydrogen, which consists of a proton and an orbiting, electron., What distinguishes an atom of hydrogen from an atom of gold? A chemical element is, defined by the number of protons in its nucleus. For an atom of a particular element, the, nucleus consists of Z protons, where Z is called the atomic number of the element., Hydrogen has a single proton, so for hydrogen, Z = 1. Gold has 79 protons, so its, atomic number is 79. Protons and electrons have equal but opposite charges. This, means that an electrically neutral atom has the same number of electrons as protons,, so for example a gold atom has 79 protons and 79 electrons., There can be more to a nucleus than just protons; there may also be other particles, present, called neutrons. These are uncharged particles that are just a bit more massive, as the proton (about 0.1% more massive). Protons and neutrons are known as, nucleons because they make up the nucleus., , Parts of the nucleus, Protons, ·Z = number of protons, Neutrons, ·N = number of neutrons, , The neutron number, N, states the number of neutrons. Any atom of an element always, has the same number of protons, but it can have different numbers of neutrons. Two, forms of an element with different numbers of neutrons are known as isotopes., Hydrogen, for instance, always has a single proton, but it can have either zero neutrons, (“common hydrogen”), one neutron (an isotope called deuterium), or two neutrons, (called tritium). Deuterium and tritium are shown in Concept 2., Different isotopes of an element will have different atomic masses because of the, differing numbers of neutrons. The table in Concept 3 summarizes the charge and mass, properties for protons, neutrons, and electrons., Atomic masses can be measured using instruments such as the mass spectrometer., Masses are commonly given in terms of atomic mass units, u, defined such that the, mass of the most abundant kind of carbon atom, carbon-12, has a mass of exactly 12 u., The value of an atomic mass unit in kilograms is given in Equation 1., , Isotopes, Same number of protons, different, number of neutrons, , The sum of Z, the number of protons, and N, the number of neutrons, is called A, the, mass number of the atom. Since electrons have little mass compared to protons and, neutrons, the mass number is very close to the entire mass of the atom when it is, expressed in atomic mass units., An atom with a particular combination of Z and N is called a nuclide. A particular, nuclide always has the same type of nucleus. A widely adopted notation to identify a, nuclide is to write the chemical symbol of the element, with its atomic number, subscripted to the left and its mass number superscripted to the left. This is shown in, Equation 2, as applied to the most common carbon nuclide which has six protons and, six neutrons., , Neutrons and protons have similar mass, Protons far more massive than electrons, , 700, , Copyright 2007 Kinetic Books Co. Chapter 38
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Atomic mass unit, u = 1.66054 × 10í27 kg, u = atomic mass unit (amu), , Mass number, A=Z+N, A = mass number, Z = atomic number, N = neutron number, 38.4 - Interactive checkpoint: using Z, N and A, A neutral atom of sodium-23 has 11, protons and a mass number of 23., How many electrons does it have?, How many neutrons does it have?, , Answer:, , Number of electrons =, Number of neutrons =, , Copyright 2007 Kinetic Books Co. Chapter 38, , 701
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38.5 - The strong nuclear force, Atoms contain positively-charged nuclei that attract the negative electrons, and the, nuclei contain closely packed protons and neutrons. These conclusions bring up a, major question about the nucleus. Since protons are positively charged, and positive, charges repel, why do the protons in the nucleus not fly away from one another? In this, section, we will explore the nature of the force that holds nucleons together., One good hypothesis would be that the force of gravity attracts them. This is a good, hypothesis, but incorrect: It turns out that gravity is far too weak. (For two protons, by, using Newton’s law of gravity and Coulomb’s law, you can calculate that the attractive, force of gravity is about 1036 times weaker than the electrostatic repulsive force.), One hypothesis down. Since gravity cannot explain the stability of the nucleus, the only, alternative is that there must be another attractive force. This fundamental force is, called the strong force., The strong force has several important properties. It is “strong;” it manages to hold, together the protons in a nucleus despite their electrostatic repulsion. It always binds, particles together, even if their electric charges are the same, or if they are uncharged., , Strong force, Holds particles in nucleus together, Is very strong!, Always attractive, regardless of charge, Acts only over a very short range, , The strong force causes protons to attract protons, protons to attract neutrons, and, neutrons to attract other neutrons. The strong force acts only over a very short range., For example, once two protons are separated by more than about 10í15 m (roughly their, own diameter), there is hardly any attraction due to the strong force, though the electrostatic repulsion is still substantial., , Although much has been learned about the properties of the strong force from experiments, there is no simple formula to relate its strength to, distance. With the electrostatic and gravitational forces, the amount of force is inversely proportional to the square of the distance between the, particles. In contrast, no simple formula can be stated for the dependence of the strong force on distance, though there are complicated, numerical approximations., How have physicists studied the strong force? They experiment by bombarding target nuclei with high energy particles, which are influenced by, the nuclei via the strong force or the electrostatic force. These forces can change the paths of the incoming particles. By observing the, distribution of the outgoing particles, and comparing it to the predictions of theoretical models, scientists can test these models., , 38.6 - Nuclear properties, Are the nucleons rigid objects, or soft, compressible ones? In other words, do they, behave like hard marbles that are clumped together, or is there some flexibility to them,, like cotton balls being crammed into a bag? These questions can be answered if we, can determine how the size of the nucleus depends on the number of nucleons in the, nucleus., It turns out that nucleons are nearly incompressible. This conclusion can be drawn by, looking at how the radius of the nucleus relates to the number of nucleons inside. The, same experiments that physicists perform to study the strong force, where they fire, particles at nuclei, have also allowed them to measure other properties of the nucleus,, such as its size., They have determined that the radius of an atom’s nucleus is proportional to the cube, root of its mass number. As shown in Equation 1, the radius equals the cube root of the, number of neutrons and protons, multiplied by 1.2×10í15 m., There are some striking implications of this simple-looking formula. It holds the answer, to our question about the rigidity of nucleons, and implies other facts about the nucleus., Consider how the radius of a sphere relates to its volume. The volume of the spherical, nucleus is equal to 4ʌ/3 times the radius cubed. If you cube the radius, using the, equation to the right, you are cubing A1/3, which equals A. In other words, the volume is, proportional to A, the number of neutrons and protons: Each time you add a nucleon,, you are adding roughly the same amount of volume to the nucleus., , Nuclear properties, Nucleons are nearly incompressible,, tightly packed, Nuclear density is roughly the same for, all atoms, , The equation also allows one to conclude that the neutrons and protons must be tightly, packed. If there were large spaces between nucleons, then as their number increased,, the volume would increase at an even faster rate; for example, the volume would more, than double when you doubled A. (If this is not obvious to you, consider the change in, the volume of adding a tenth planet beyond Pluto, versus adding another marble to a, bag of marbles. Adding another planet would increase the volume of the Solar System, by more than the volume of the planet itself; but as you add hard marbles to a cluster of, marbles, the volume of the cluster increases by about the volume of a single marble, each time.), The equation tells scientists that the density of all nuclear material is constant. How do, we know this? Density equals mass divided by volume, and since the mass and volume, increase at the ratio of 1:1, the density does not change., , 702, , Nuclear radius increases with, mass number, A, , Copyright 2007 Kinetic Books Co. Chapter 38
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Dependence of radius on A, R = (1.2×10í15 m)A1/3, R = nuclear radius, A = mass number, 38.7 - Sample problem: nuclear density, Calculate the density of the hydrogen, nucleus (A = 1, mass § 1.0 u) and of, an aluminum-27 nucleus (A = 27,, mass § 27 u). Express the answer in, kg/m3, to two significant figures., , Variables, mass of hydrogen nucleus, , mH, , radius of hydrogen nucleus, , RH, , volume of hydrogen nucleus, , VH, , mass of aluminum nucleus, , mAl, , radius of aluminum nucleus, , RAl, , volume of aluminum nucleus, , VAl, , What is the strategy?, 1., , Convert the mass of each nucleus from atomic units to kilograms using the conversion factor stated below., , 2., , Find the radius of each nucleus using the relationship between radius and mass number., , 3., , Calculate the volume of each nucleus using the radius just calculated., , 4., , Divide mass by volume to find the density., , Physics principles and equations, The nuclear radius grows as the cube root of the mass number., , R = (1.2×10í15 m)A1/3, The nuclear shape may be modeled as a sphere. The volume of a sphere in terms of its radius is, , The definition of an atomic mass unit is, u = 1.66 × 10í27 kg, Mass density, , ȡ = m/V, , Copyright 2007 Kinetic Books Co. Chapter 38, , 703
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Step-by-step solution, We begin by converting the mass of the hydrogen nucleus into kilograms, then we calculate the radius of the nucleus. Using the radius, we, calculate the nuclear volume. Finally, we divide the mass by the volume to determine the density., , Step, , Reason, , 1.66×10í27, , 1., , mH = 1u =, , 2., , R = (1.2×10í15 m)A1/3, , kg, , hydrogen nuclear mass, definition of u, apply nuclear radius equation, , 3., , calculate volume of hydrogen nucleus, , 4., , definition of density, , Now perform the same calculations for aluminum., , Step, , Reason, , 27(1.66×10í27, , 5., , mAl = 27 u =, , 6., , R = (1.2×10í15 m)A1/3, , kg) aluminum nuclear mass, definition of u, apply nuclear radius equation, , 7., , calculate volume of aluminum nucleus, , 8., , definition of density, , The densities are nearly identical. This is further confirmation that nucleons are tightly packed and incompressible., The nuclear density is far beyond the density of materials in our experience. For example, consider gold, which is 19.3 times denser than water, and almost 1.7 times as dense as lead. A nucleus is 12,000,000,000,000 times denser than gold. (In case you were wondering, no, the zero, key did not get stuck down while we typed that number.) Recall that Rutherford found that the atomic radius was on the order of 10,000 times, the nuclear radius. The volume of a sphere scales as the cube of the radius, so the ratio of atomic to nuclear volume is approximately (10,000), 3. Since the mass of an atom mostly resides in the nucleus, and so much of an atom is empty, this explains the incredible density of a nucleus., , 38.8 - Stable nuclei, What are the rules for the number of allowed protons and neutrons in a nucleus? Is any combination of protons and neutrons possible? Could, there be a hydrogen atom whose nucleus has 1 proton and 7 neutrons? What about a silver atom with 47 protons and no neutrons? Could, there be an element with 150 protons and any number of neutrons?, The nuclei described above do not exist. You could not even momentarily create nuclei with such extreme imbalances of protons and neutrons,, or in the last case, a nucleus with so many protons. Only certain combinations of protons and neutrons can form nuclei, and even fewer, combinations can form stable nuclei., By stable, we mean elements that will not spontaneously decay. Gold-197 is the only stable isotope of gold, while there are several stable, oxygen isotopes: oxygen-16, oxygen-17 and oxygen-18., , 704, , Copyright 2007 Kinetic Books Co. Chapter 38
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Some nuclides are unstable, meaning they have a limited lifespan. An unstable nucleus, is called a radionuclide, and it will spontaneously and rapidly decay or split up into more, stable pieces. Such materials are radioactive, and the details of the decay process are, the subject of another section., In this section, we discuss what makes for a stable nucleus. The question of stability is, of fundamental importance. If all elements were unstable, then life would as we know it, would not exist í the carbon, oxygen and other elements that make up your body would, be constantly changing into other elements. Life is complicated enough!, On the other hand, if all elements were equally stable, then the nuclear fusion process, that powers stars (including the Sun) would never happen, and the universe would be, almost all hydrogen, with none of the heavier elements that make life possible., To explain why some atoms are stable and others are not, it helps to consider a, diagram of stable and unstable nuclides where Z is plotted against N. For example,, even though isotopes of silver with its 47 protons (Z = 47) have been created with, mass numbers as low as 96 and as high as 124, just two of these nuclides are stable (, A = 107 and 109). In other words, there must be 60 or 62 neutrons along with the 47, protons., , Stable nuclei, Stable only for certain combinations of, neutrons and protons, , Unstable nuclei, Other combinations do not exist or are, unstable, , Graph of stable and unstable, nuclei, Values clustered near “band of stability”, Ratio of neutrons to protons increases, with nuclear size, Larger nuclei: more neutrons required to, dilute protons, , We would like you to observe three important features of this diagram. First, note that the stable nuclides are clustered around a band running, through the diagram. The unstable nuclides exist on either side of this band. Second, you can see how the stable nuclei are distributed., Roughly speaking, for less massive nuclei, the number of protons and neutrons is approximately equal: They cluster around the line N = Z. In, contrast, for more massive nuclei, the number of neutrons exceeds the number of protons, N > Z. Third, observe that there are no stable, nuclei beyond bismuth (Z = 83)., Since neutrons are effective at diluting the repulsive electric force between protons (by spacing them out more), and the strong force binds, neutrons effectively to protons and other neutrons, it seems like having more neutrons can only bind the nucleus more tightly. You may be, wondering why an atom cannot have an extremely high ratio of neutrons to protons. For instance, why are there no hydrogen isotopes with, eight neutrons, or even just seven neutrons? Quantum mechanical principles dictate why., In contrast, explaining why very large nuclei are unstable, even if N is closer to Z, only requires considering the nature of the strong and, electrostatic forces, not quantum mechanics., , Copyright 2007 Kinetic Books Co. Chapter 38, , 705
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Heavy elements like uranium have a large number of protons, which all repel one another. As the number of protons increases, the repulsive, force keeps growing and growing, making the nucleus more and more unstable., Neutrons counteract this growing instability by increasing the distance between protons, which decreases the electrostatic forces, and by, attracting each other and the protons with the strong nuclear force., However, there comes a point when the nucleus gets too large, and will be unstable no matter how many neutrons are present. This can be, understood by considering the relative ranges of the strong and electrostatic forces. The strong force only acts between very close neighbors,, while the repulsive electrostatic force acts between all protons regardless of their position within a nucleus. When there are lots of protons, already present, and one more proton is added, it will be subject to repulsive electrostatic forces from every proton that is already there, while, the attractive strong force will only be exerted by a few very close neutrons or protons. Eventually, it becomes impossible for the nucleus to, “hold in” an additional proton because the strong force cannot overcome the electrostatic force., , 38.9 - Nuclear binding energy, , Binding energy: The energy that must be added, to disassemble, or unbind, a nucleus into the, protons and neutrons that make it up., You may have heard of radioactivity, and know that uranium atoms will spontaneously, decay into other elements, while other elements such as common iron (iron-56) are, stable. Stable nuclei such as iron all have one thing in common: Their nucleons are, tightly bound. Unstable atoms are not as tightly bound. What does it mean to be “tightly, bound”?, Note that the term “stability” in nuclear physics is not making a statement about the, tendency of an atom to enter into chemical reactions. For example, we say that iron is, “stable” in the nuclear sense, even though it rusts. When iron combines with oxygen to, form iron oxide, it is a chemical reaction, not a nuclear reaction. The iron remains iron, when it becomes iron oxide; it shares electrons with oxygen but the element’s nucleus, remains unchanged., , Binding energy, Energy that must be added to, disassemble nucleus completely, Increased energy of separate particles, reflected in increased mass, , What makes the protons and neutrons in a radioactive uranium atom less tightly bound, than the nucleons in an extremely stable iron atom? Can we quantify and compare the, stability of nuclei?, As it turns out, we can. One way to measure the stability of a nucleus is to try and rip, the nucleons apart, overcoming the strong force. When physicists conduct such, experiments, they find it takes energy to break the nuclear bonds (which the strong, force is responsible for), that is, to take apart a nucleus into separate protons and, neutrons. (This makes sense, because if no energy was required to separate them, they, would fall apart on their own.), The particles that emerge when the nucleus is forced apart can be analyzed. Careful, measurements show that the sum of the masses of the separate nucleons is always, greater than the mass of the nucleus when it is whole., Concept 1 shows this, using an isotope of hydrogen, deuterium, as an example., Why should the mass of the nucleus increase when it is broken up? To a classical, physicist, unacquainted with Einstein’s theory of special relativity, this would be a, surprise since mass is assumed to be conserved., , Binding energy becomes mass, Separate particles have more mass than, assembled nucleus, , However, we know that another conservation principle applies here: the total of mass, and energy (or mass-energy) must remain the same, though the individual terms may, vary. Einstein’s principle of mass-energy equivalence, summed up by the equation, E = mc2, applies. This is shown in Concept 2. It takes energy to separate the particles,, and the energy added to the nucleus to fragment it into nucleons shows up as the, “extra” mass. (We will assume for the sake of simplicity that the kinetic energies of the, particles and of the nucleus are negligible.), The energy that must be added to completely disassemble the nucleus is known as the, binding energy. This works in both directions. The binding energy is released when the, protons and neutrons come together to form a bound nucleus. This is illustrated in, Concept 3., The terminology could be a little confusing. You can think of it like gravity: You must, “add” energy to pull apart two particles, or lift a rock farther from the surface of the, Earth. That is analogous to the binding energy., Because of the equivalence of energy and mass, the binding energy may also be, related to mass. When energy is added to a nucleus to disassemble it, the mass of the, parts increases. On the other hand, when a nucleus is assembled, the release of, , 706, , Assembling the nucleus, Binding energy released when nucleus, is assembled, Decreased energy of nucleus reflected, in decreased mass, Mass becomes binding energy, , Copyright 2007 Kinetic Books Co. Chapter 38
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binding energy from the system shows itself as a corresponding reduction of mass in, the assembled nucleus. This is sometimes called the “missing mass”., , 38.10 - Interactive checkpoint: calculating the binding energy, For a carbon-12 nucleus with 6, protons and 6 neutrons, and a mass, of 11.9967 u, how much more mass, do the individual nucleons have than, the assembled carbon nucleus? This, is called the mass excess. What is, the binding energy of carbon-12’s, nucleus?, A neutron has a mass of 1.0087 u, and a proton has a mass of 1.0073 u., 1 u = 1.66×10í27 kg., Answer:, , mass excess =, binding energy =, , u, J, , 38.11 - Binding energy curve, The binding energy is a measure of stability for a nucleus, since the binding energy is, how much energy it takes to completely disassemble a nucleus. Roughly speaking, the, higher the binding energy, the harder it is to pull all the nucleons apart., However, determining which nuclides are stable is not as simple as calculating the, binding energy. In this section, we discuss how the stability of a nuclide can be, determined., We will use two nuclides as examples:, , (iron-56), and, , (uranium-235)., , Using a table of nuclear data, one may find that the binding energy of an iron-56 atom,, which has 26 protons and 30 neutrons, is 492 MeV. The binding energy of a uranium235 atom (Z = 92, N = 143) is 1784 MeV., Does this mean that the uranium nucleus is more stable since it has a greater binding, energy? Not necessarily. To compare the stability of different nuclei, a useful number to, consider is the binding energy per nucleon. To calculate this ratio, divide the binding, energy of the nucleus by the mass number, the total number of protons and neutrons., This provides a metric to compare the binding energy per nucleon in the iron isotope, with the binding energy per nucleon in uranium., , Stability and binding energy, Stability determined by binding energy, per nucleon, , The binding energy per nucleon in uranium-235 is 7.59 MeV and the binding energy per, nucleon in iron-56 is 8.79 MeV. The binding energy per nucleon is a good measure for, stability; the fact that the binder energy per nucleon is higher for iron than uranium, correctly predicts that iron-56 is more stable than uranium-235. (In fact, the binding, energy per nucleon for iron-56 is among the highest for all nuclides.), The graph in Concept 2 shows the binding energy per nucleon for naturally occurring, isotopes, plotted against the mass number A. This is called the binding energy curve., , Graph of binding energy per, nucleon versus mass number, Graph peaks near iron-56, , Copyright 2007 Kinetic Books Co. Chapter 38, , 707
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What is the total binding energy, for a uranium-235 nucleus?, , 38.12 - Shape of the binding energy curve, The graph of binding energy per nucleon versus mass number has a distinct shape that, proves to be very important. The higher on the graph an element is (indicating more, binding energy per nucleon), the more stable it is., The most stable elements are at the highest points, with iron-56 in this region, as you, can see in Concept 1., Very light nuclei (on the left of iron-56 in the binding energy curve) can become more, stable if they combine to form larger nuclei through a process called fusion. By this, process, the binding energy per nucleon is raised, which means that energy is released., This is the process by which stars, like the Sun, continually transform their mass into, energy. In a multistep process within the Sun, hydrogen nuclei fuse together to become, helium-4 nuclei., As mentioned, the most stable locations on the curve represent elements such as iron, and nickel. Heavier, radioactive nuclei to their right can increase their binding energy, per nucleon and become more stable by “moving to the left and up” on the curve. For, instance, you can see in Concept 3 that uranium is less stable than iron. A heavy, nucleus could become more stable by emitting particles and becoming slightly smaller, (the process of radioactive decay) or, in extreme cases, by splitting into two mediumsized nuclei (a process called fission). This is the principle behind radioactivity and, nuclear power., , Binding energy per nucleon, versus mass number, Highest in the middle, , The shape of the graph also illustrates the relative distances at which the strong and, electrostatic forces effectively operate. The argument that follows is reasonably, complex but provides a good example of how graphical data can be analyzed. The, difference between these two forces can be used to explain why the graph first shows a, rapid increase of binding energy per nucleon, then levels off, and finally declines., Earlier, we discussed the short-range nature of the strong force. It is so short-range that, it acts only between a nucleon and its nearest neighbors. The graph supports this, hypothesis. Why? When there are only a few nucleons, they are all very close and, Lighter elements can undergo, every nucleon interacts with every other. For instance, when there are two nucleons,, they are next to each other, and exert a strong force on one another. As a third nucleon fusion and release energy, is added, it has two neighbors to exert a force on, so the force increases faster than the, number of nucleons. This means the binding energy per nucleon increases, so the line, has a positive slope. (Mathematically, the binding energy of the smaller nuclei increases as the square of the number of nucleons.), As more nucleons are added, at some point they are too far apart to all be “neighbors”. For instance, when there are 100 nucleons, and, another is added, it can only interact with its close neighbors. A nucleon on one side of the nucleus is too far away to exert a significant strong, force on one on the far side. Adding a nucleon does not increase the binding energy per nucleon. This means with larger nuclei, the additional, binding energy per nucleon becomes constant. The sharp increase in binding energy per nucleon ceases., The strong force needs to be contrasted with the electrostatic force, which acts to push apart the protons, and which acts at a greater distance, than the strong force. When a new proton is added, it is attracted only to its nearest neighbors via the strong force, but is repelled by every, other proton that is already present because the electrostatic force acts at a greater range. This, in turn, makes it easier to disassemble the, nucleus when more of the protons want to be separated., In sum, at first the strong force dominates, causing the increase in binding energy per nucleon. But as the nucleus grows in size, the, electrostatic force plays a larger role, causing an eventual decrease in binding energy per nucleon., , 708, , Copyright 2007 Kinetic Books Co. Chapter 38
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Heavier elements can divide and, release energy, , 38.13 - Fission, , Fission: A heavy nucleus breaks up into two, smaller ones, releasing energy., You may have heard the term “splitting the atom” as something that humans first, accomplished in the 20th century. In this section, you will learn what it means to “split”, an atom. Fission is the process used both in nuclear reactors to produce electrical, power and also in the first atomic bombs., When a nucleus breaks up into smaller, more stable pieces, this is known as fission., Some unstable nuclei, such as uranium-236, do this spontaneously. When an atom, undergoes fission, it changes identity, as the new nuclei it breaks into have different, numbers of protons. There are many ways that the nucleus can break up. For example,, the uranium-236 nucleus can break into Xe-140 and Sr-94, in the process releasing two, neutrons. To see this fission process, press the refresh button in your browser and look, at Concept 1., However, being the impatient race that we are, humans learned to induce the process, to happen at a greater rate. Induced fission, also known informally as “splitting the, atom,” was first performed by Otto Hahn and Fritz Strassmann. They bombarded, uranium with neutrons and found that lighter elements (such as barium) were produced., , Fission, Nucleus breaks up into smaller, elements, ·And releases neutrons, ·And releases energy, , Why are neutrons effective at inducing fission? Was it not stated earlier that the strong, force they supply is crucial to holding a nucleus together? Yes, but it is possible to have, too much of a good thing. For a given number of protons in a nucleus í for a certain, element í the band of stability is quite narrow. Too few neutrons or too many neutrons, make the nucleus unstable., In fact, neutrons are a natural choice to induce fission. Since they are electrically, neutral, they can easily approach and hit the nucleus without being repelled by, electrostatic forces. If the incoming neutron has the right speed, the nucleus captures it, and becomes even more unstable. The nucleus has one neutron too many for the, number of protons, and that immediately causes the new compound nucleus to fission, into various elements., Energy is released during fission; in the above process, roughly 200 MeV. You can, understand why this happens by returning to the concept of binding energy per nucleon,, which is lower for very heavy nuclei than it is for intermediate-size nuclei., , Fission and chain reactions, Neutrons can be used to cause fission, Fission can be self-sustaining, Releases energy, , The binding energy is the amount of energy that is required to separate a nucleus into its separate nucleons, or it is the energy released when, separate nucleons are brought together to form a nucleus. The more tightly bound a nucleus is, the higher the binding energy, that is, the more, energy is released. Since the intermediate-size nuclei have the highest binding energy per nucleon, this means that heavier nuclei will release, energy as they split apart and become medium-sized., We have discussed fissioning a single atom with the release of energy. This is a crucial scientific achievement, but in order for this to be useful, (for example, in a nuclear reactor), the process needs to be self-sustaining., What makes fission a practical process is that in a fission reaction, one or more neutrons may be released, which can then induce more fission, reactions in nearby atoms, which produce more neutrons, and so on. Why are there “extra” neutrons? Recall that for heavier nuclei, the, number of neutrons exceeds the number of protons, while for lighter nuclei, the number of neutrons and protons tends to be nearly equal. This, means there are usually some neutrons left over after the nuclear re-arrangement., When there are enough uranium atoms so that at least one neutron, on average, is captured by another uranium atom, the critical mass has, been reached. The process is self-sustaining, and it is called a chain reaction. Press the refresh button to see the fission process occur in, Concept 2., , Copyright 2007 Kinetic Books Co. Chapter 38, , 709
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Since humans learned to split the atom in the 20th century, the process has been put to great use. The most well-known application was the, use of nuclear fission in the “atomic” bomb. A runaway chain reaction happens very fast, releasing a lot of energy in a burst which can be used, to devastating effect., On the positive side of the ledger, nuclear power has been used as an energy source. A slowly progressing chain reaction produces a steady, flow of heat that can be used to boil water, which then drives steam turbines to generate electricity. The poster-child for nuclear power is, France, which supplies about three-quarters of its electrical needs with nuclear reactors., However, nuclear power is not without its risks or costs. Some of the byproducts of the fission process are highly radioactive and remain, dangerous for tens of thousands of years. A typical way to dispose of these wastes is to bury them deep in the Earth. If they leak, they, contaminate water sources., , 38.14 - Fusion, , Fusion: Two light nuclei fuse into a heavier one,, releasing energy., Fusion is a process in which nuclei join together to become a single, larger nucleus., This process also releases energy. Because positively charged nuclei repel one, another, fusion does not occur spontaneously under normal conditions on Earth., However, fusion is commonplace in the Sun and other stars where hydrogen atoms, fuse into helium atoms. Fusion provides the energy to keep the star going, which Earth, ultimately experiences as light and heat. Fusion occurs in the Sun because of the high, temperature within the star; its interior is at about 100 million Kelvin., At this high temperature, the atoms are in an ionized state of matter called plasma., While normal matter consists of distinct neutral atoms, plasma is a “soup” of positive, nuclei and negative electrons. In the interior of the Sun, nuclei are hot enough and, moving quickly enough to overcome the electrostatic repulsion of their positive nuclei., They move close enough to be bound by the attractive strong force between them, then, fuse into a single nucleus., , Fusion, Light nuclei fuse into larger, more stable, pieces, Releases energy, , Why is energy released in the process? For light elements (atoms to the left of the peak on the binding energy curve), the binding energy per, nucleon increases with atomic number. As smaller nuclei are fused into larger ones, the result is a more efficient arrangement of nucleons; one, that is harder to break apart. Since the new nucleus is more efficient í more tightly bound í than the previous ones, there is energy to spare., It may be tempting to think that fusion and fission are opposite processes, since one combines nuclei and the other splits them. It may seem, confusing as to how two “opposite” processes can release energy. The key is that fission with energy release occurs only when very heavy, nuclei break apart into medium-size nuclei, and fusion with energy release occurs when very light nuclei fuse into heavier ones. The answer, again resides on the curve of binding energy; “mid-sized” elements have the highest binding energy per nucleon. As predicted by the curve, it, requires energy to combine two medium-sized nuclei into a single large nucleus, just as it requires energy to split a medium-size nucleus into, smaller nuclei., Fusion is critical to the universe we observe. Without it, there would be no stars, and in fact no elements heavier than Lithium (Z = 3)., However, there is also a practical reason for scientific interest in fusion. If we could create and control fusion, we could use it as an energy, source. Using fusion as an energy source has two huge appeals: the fuel (hydrogen isotopes) is present in the oceans in essentially unlimited, quantities (compared to the relative scarcity of, say, uranium) and fusion creates no radioactive byproducts., However, before fusion reactors become commonplace, some daunting engineering challenges will have to be solved. Recreating the, conditions inside the Sun, with its enormous temperatures, is no easy feat. Furthermore, simply getting the atoms hot is not sufficient; there is a, minimum plasma density that must also be achieved so that collisions between nuclei will occur frequently enough to release energy, which, keeps the plasma hot and keeps the fusion reaction going. The combination of high temperature and high density requirements necessarily, means that the pressure must also be very high, to hold all of the reactants together., As in the case of fission reactions, early work on fusion was directed toward nuclear weapons. Scientists solved the problem of achieving the, ultrahigh temperature and pressure conditions necessary for fusion by using a fission (“atomic”) bomb as a trigger, to both heat up and, compress the fusion fuel. Some trigger! Using atomic bombs to power a nearby fusion reactor is not a very popular proposal., There are two less-explosive schemes that are currently being pursued to keep the superhot plasma together for fusion to occur. Magnetic, confinement uses electromagnetic fields to hold the charged particles together. Inertial confinement fusion uses a solid pellet of deuterium and, tritium that is crushed by the light pressure of perfectly timed, short-duration, high-powered laser beams from all directions. The term comes, from the fact that the particles’ own inertia keeps them in place during the laser pulse., , 710, , Copyright 2007 Kinetic Books Co. Chapter 38
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38.15 - Radioactivity and radiation, , Radioactive decay: A nucleus spontaneously, emits particles or high-energy photons and, either changes identity or becomes less excited., Transmutation: Changing of one element to, another after Į or ȕ radiation is emitted., Fission is not the only way that a nucleus with an unstable combination of protons and, neutrons can change into a more stable configuration. An unstable nucleus may instead, spontaneously emit a charged particle or a high-energy photon in order to reach a more, stable state. It changes via a nuclear reaction. Isotopes that change (decay) like this are, said to be radioactive. It is possible for the isotope to become a different chemical, element after the decay, a process known as transmutation., , Unstable nuclei, Too massive or wrong ratio of protons to, neutrons, , The outgoing radiation can be classified as alpha rays, beta rays, or gamma rays. (They, are represented by the Greek letters Į, ȕ, and Ȗ.) Different decay processes result in, different forms of radiation. Alpha and beta rays consist of matter particles, while, gamma rays are photons (light particles)., In an alpha decay, the initial radioactive isotope decays into a different element by, emitting an Į particle. An Į particle is made up of two protons and two neutrons. It is a, helium-4 nucleus., The initial isotope is known as the parent. Because it loses two protons, its atomic, number is reduced by two (the mass number is reduced by four). Since the nucleus now, has a different number of protons, it becomes another element í it has transmuted. The, newly-formed nucleus is known as the daughter., Alpha decay occurs most commonly in heavy nuclei whose ratio of protons to neutrons,, Z/N, is too large, making them unstable. An Į particle is a very stable particle, and the, daughter nucleus that is left behind is more stable (tightly bound) than the parent. To, put it another way, the net result of the radioactive decay is a reduction of the ratio Z/N., , Radioactive decay with Į or ȕ ray, Nucleus decays by emitting charged, particle, ·Radioactive element has transmuted, (changed to another element), , Beta decay is characterized by the emission of an electron or antielectron. There are, two types of beta decay, negative and positive. Negative ȕ emission is represented in, Concept 4. This occurs when a neutron inside the nucleus decays into a proton, an, electron (the beta ray), and an almost zero-mass, uncharged particle known as an, antineutrino. The antineutrino’s interaction with matter is so weak that it is very hard to, detect, and so it is customarily left out of nuclear equations., The emitted electron did not exist in the nucleus beforehand, and is not one of the, orbital electrons in the parent nucleus. When a neutron inside the nucleus turns into a, proton, electron, and neutrino (and then emits the electron as a negative beta ray), its, number of protons increases by one. The mass number stays the same. The released, electron usually zips away, leaving behind a daughter atom with a net positive charge., In positive ȕ emission, the nucleus emits an antielectron (also called a positron)., Antielectrons are essentially the same as electrons, except they have a change of, positivee. The decay process is shown in Concept 5. A proton inside the atom decays, into a neutron, an antielectron (the beta ray), and a neutrino. Like an antineutrino, a, neutrino has no charge and almost zero mass. Its interaction with matter is also weak, and it too is customarily left out of nuclear equations., In emitting either type of ȕ ray, the initial isotope changes atomic number, so ȕ decay, results in transmutation., In gamma decay, the radioactive isotope emits a high-energy photon, also known as a Ȗ, ray., , Alpha radiation, Nucleus emits alpha particle (2 protons,, 2 neutrons), Number of protons in nucleus decreases, by 2, ·Transmutation, Mass number decreases by 4, , Since the atomic number stays the same, the atom is the same chemical element after the decay. In this case it is the nuclear energy that, changes in the decay. A nucleus can have different energy states. When a nucleus changes from an excited, high energy state to a lower one,, a photon is emitted., This is similar to the case when an electron falls from one energy level to another. A notable difference is that nuclear energy levels are much, more widely spaced, on the order of millions of electron volts as opposed to say five or ten. This means the emitted photon, called a gamma, ray, is much more energetic than the photon emitted by an excited atom., A gamma decay is represented in Concept 6. An excited nucleus is denoted by an asterisk “*” after the usual symbol. How does the nucleus, get into an excited state? This usually happens after another kind of decay. In many cases of Į and ȕ decay, the product nucleus is in an, excited state, after which it emits a Ȗ ray and transitions to a lower state or to the ground state. Because the photon is electrically neutral,, transmutation does not occur during gamma radiation., , Copyright 2007 Kinetic Books Co. Chapter 38, , 711
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Negative beta radiation, Nucleus emits negative beta particle, (electron), Number of protons in nucleus increases, by 1, ·Transmutation, Nuclear mass is unchanged, , Positive beta radiation, Nucleus emits positive beta particle, (antielectron), Number of protons in nucleus decreases, by 1, ·Transmutation, Nuclear mass is unchanged, , Radioactive decay with gamma, ray, Nucleus emits gamma ray (high-energy, photon), ·Becomes less excited, ·No transmutation, ·Nuclear mass is unchanged, , 712, , Copyright 2007 Kinetic Books Co. Chapter 38
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38.16 - Sample problem: radioactive decay, A thorium-232 nucleus (Z = 90, A =, 232) is radioactive and decays by first, emitting an alpha particle, then two, negative beta particles. What are the, atomic number and the mass number, of the daughter nucleus after each of, these three decay steps?, , Variables, atomic number, number of, protons, , Z, , mass number, total number of, protons and neutrons, , A, , What is the strategy?, 1., , The Į and ȕ particles have known charge and known mass number. Subtract these from the atomic number and the mass number of, the parent nucleus to determine the daughter nucleus’s values of Z and A., , Physics principles and equations, An Į particle consists of 2 protons and 2 neutrons. Its mass number is four., A negative ȕ particle consists of one electron. When the parent nucleus emits it, the number of protons increases by one and its mass number, is unchanged., Step-by-step solution, We will use the notation Z0 and A0 to denote the initial values of the proton number and the mass number, Z1 and A1 to denote their values, after the first decay step, Z2 and A2 to denote their values after the second decay step, and Z3 and A3 to denote their values after the third and, final decay step., , Step, , Reason, , 1., , Z0 = 90, , 2., , Z1 = Z0 – 2 Į emission decreases nuclear charge, , 3., , A1 = A0 – 4 Į emission decreases nuclear mass, , initial values, , After the first decay process, the nucleus has 88 protons and 228 nucleons in total. This is radon-228, which then emits a negative ȕ particle., , Step, , Reason, , 4., , Z1 = 88, , 5., , Z2 = Z1 + 1 ȕ emission increases nuclear charge, , 6., , A2 = A1, , initial values, , ȕ emission does not affect nuclear mass, , After the second decay process, there is a nucleus with 89 protons and 228 nucleons in total. This is actinium-228, which then decays by, emitting another negative ȕ particle., , Step, , Reason, , 7., , Z2 = 89, , 8., , Z3 = Z2 + 1 ȕ emission increases nuclear charge, , 9., , A3 = A2, , initial values, , ȕ emission does not affect nuclear mass, , After the third decay process, the nucleus has 90 protons, and a total of 228 nucleons. The end result after the three decays is again thorium, because the number of its protons is again 90. After the three decays, the element is thorium-228. This lighter thorium isotope then continues, to decay in a series of Į and ȕ decays until it becomes the stable isotope lead-208., , Copyright 2007 Kinetic Books Co. Chapter 38, , 713
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38.17 - Radioactive decay and half-lives, , Half-life: The time it takes for half of a group of, radioactive atoms to decay., We have discussed the methods by which radioactive isotopes decay, by emitting Į, ȕ,, or Ȗ rays. What can be said about the rate at which they decay? If you have a sample of, radioactive atoms, how many of them will decay in, say, the next minute?, The answer depends on the radioactive isotope. For any isotope, the fraction of atoms, that will decay in a minute can be determined. A quantity of particular interest is the, half-life, which is the average time it takes for one-half of the radioactive material, present to decay., For example, let’s consider 16.00 mg of a parent nuclide, 131I, which is a radioactive, isotope of iodine. Its half-life is 8.04 days. You would find that after 8.04 days, one-half, of the parent nuclei have decayed, and 8.00 mg of 131I remains. After another 8.04, days, one-half of the remaining iodine will have decayed, and 4.00 mg remains. After a, third 8.04 days, only 2.00 mg would remain, and after 8.04 more days, 1.00 mg would, remain, and so on. Half of the remaining iodine decays every 8.04 days., , Half-life, Average time for one-half of a group of, radioactive atoms to decay, , If it is possible to know how many atoms in a sample are going to decay within a certain, time interval, is it possible to know when a particular atom will decay? Numerous, experiments have shown that the answer to that question for a particular atom is no., This is similar to the situation of flipping one thousand coins and making a prediction of, “50% heads.” The prediction will be quite accurate, though you cannot reliably predict, the outcome of any particular coin. The process of radioactive decay provided evidence, of the statistical nature of quantum mechanics, which governs processes on a, subatomic scale., , During each half-life, one-half of, remaining radioactive atoms decay, , Radioactive decay is, probabilistic, Cannot be predicted for an individual, atom, Can state the probability of any atom, decaying within a certain time, ·or what fraction of the atoms will decay, within a certain time, , 38.18 - Interactive problem: radioactive dating, Carbon-14 is a radioactive isotope of carbon that has six protons and eight neutrons in its nucleus. It is commonly used to establish a date for, organic specimens. In the first simulation, you will observe the decay of carbon-14 (C-14), and determine the half-life of that radioactive, isotope., You are equipped with a digital timer and a gauge that reports the number of parent atoms that are present. Before the simulation starts, there, are 32 billion parent atoms. Each of the spheres on the screen represents a billion atoms. A sphere changes color when a billion atoms have, decayed., When you press GO, the timer starts and the carbon begins to decay to nitrogen by emitting ȕ rays, and the number of daughter nitrogen, atoms begins to grow. The daughter atoms are stable., In the simulation, time is sped up and passes in thousands of years. The number of carbon atoms is shown both graphically with a, “thermometer”-type gauge as well as numerically. When half the initial number of parent atoms has become daughter atoms, press PAUSE, and note the elapsed time. (That is, at this moment, 16 billion of the carbon atoms remain present, and the rest have decayed.), , 714, , Copyright 2007 Kinetic Books Co. Chapter 38
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After noting the time, restart the process by pressing GO to resume the simulation and press PAUSE when half of the remaining parent atoms, have become daughter atoms í in other words, when about eight billion carbon atoms remain. How does this second time interval compare to, the first interval of time, when the number of parent atoms changed from 32 billion to 16 billion?, Repeat this process once more (or as many times as you like), each time pausing, when the number of parent atoms has fallen in half. Do you see a pattern? You are, measuring the half-life of the material, which is the average time needed for half of, the parent atoms in a radioactive sample to decay. Enter your measurement for the, half-life of carbon-14 by selecting the appropriate amount of years. Press CHECK to, see if you are correct., In the second simulation, you will use your newly acquired skills at measuring halflives to investigate a crime scene. You are an environmental investigator and a, criminal is once again dumping pure samples of a radioactive lead isotope,, , into a vacant lot. Holy ecological disaster!, You have been unable to catch the perpetrator in the act, but a security camera, filmed three suspicious-looking characters in the vacant lot at different times. If you, can determine when the radioactive waste was dumped in the lot, you will know, which of these three suspects is guilty., The factory that creates the waste is cooperating with you. They tell you that the, isotope was pure lead-209 samples that initially contained 192 billion atoms. When, you find the waste, it is about midnight. At midnight, 24 billion lead atoms remain,, which means 168 billion of the lead atoms have decayed into bismuth atoms. In, other words, one-eighth of the original lead-209 is left., Your mission has three parts. First, determine how many half-lives have elapsed, since the pure lead-209 sample was dumped. If you are having trouble with this, piece of your detective work, return to the first simulation and calculate how many, half-lives it takes for seven-eighths of the carbon atoms to decay., Second, measure the half-life of lead-209 using a technique similar to what you, used in the first simulation. You have the same tools as you had before., Third, you can use the evidence from the security camera. The camera filmed Anna, in the lot 6.51 hours before you obtained the sample. Sara was loitering in the area, about 9.76 hours before this time, and a third suspect, Katherine, was filmed there, 13.0 hours before the sample was found., To put this all together: You use the value you determined for the half life of lead,, and multiply that by the number of half-lives that have passed since the lead was dumped. That tells you how long the lead was there, so you, can nail the suspect. To confirm your conclusion, drag the handcuffs in the simulation to the dastardly dumper. The simulation (and perhaps, the suspect’s reaction) will let you know if you are correct., , 38.19 - Particle physics and GUTs, With this section, we come to the end of our discussion of the atom and the nucleus. It seems appropriate to take a peek ahead to the current, state of nuclear physics, and to discuss what you and others may be learning in the decades ahead., Particle physics, also known as high-energy physics, is one of the largest subfields of current physics. Essentially, it tries to answer the, question “What is everything made of?” Ordinary matter is composed of the particles you have encountered so far í protons, neutrons, and, electrons í and in the early 1900s, that seemed to be all that was needed to answer the big question. However, starting in the 1930s, particle, physicists began discovering new, exotic particles that were created in an energy-to-mass conversion during collisions between the known, particles. Several hundred other particles have since been discovered, most of which are unstable. Some of the particles are stable if they are, left alone, but are composed of antimatter, which annihilates ordinary matter upon contact., Early experimenters relied on cosmic rays (high-energy particles that permeate our galaxy) to initiate reactions. (The exact source of cosmic, rays is still an open question. Stars emit them during intermittent flare-ups, but supernova explosions, when stars die, are thought to be, responsible for much of the cosmic ray output.), Later, more controlled experiments were carried out in particle accelerators, also known as atom-smashers, where particles are made to collide, with higher and higher energies. Physicists now recognize that many of the heavier particles are made up from smaller building blocks called, quarks. They understand that the smorgasbord of particles is due to the fact that when particles collide, newly-created quarks can combine with, those already present to form systems of bound quarks., The ultimate goal of physicists is a theory of everything (how is that for ambition?). Historically speaking, great breakthroughs in physics have, often resulted in a simplification of our view of the universe. For example, Newton’s universal law of gravity showed that the laws governing, celestial orbits were the same as those governing the motion of objects falling under earth’s gravity. Maxwell and his generation showed how, electricity and magnetism, long thought to be unrelated, were really aspects of the same force. Physicists optimistically believe that the, universe has an underlying simplicity, and that the number of fundamental forces can be reduced further., We have discussed gravity, the electromagnetic force, and the strong force. (There is also the weak force which we have not discussed.) Albert, Einstein spent a lot of his working life trying to interpret these forces as different aspects of a single “superforce”. Historically, electricity and, magnetism were united in the 1800s, and in the latter part of the 20th century, the weak force and the electromagnetic force were also joined, theoretically. The goal of further reduction with the ultimate prize of unification still continues today., , Copyright 2007 Kinetic Books Co. Chapter 38, , 715
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As far as high-energy physicists are concerned, the goal is a Grand Unification Theory, or GUT. The current dream is to unify gravity with the, strong, weak, and electromagnetic forces. You may have heard of attempts such as superstring theory, which interpret particles, such as, electrons, as being modes of oscillation of unimaginably small “strings”. Acceptance of superstring theory demands the idea that the universe, may not consist of four dimensions (three spatial dimensions plus time), but instead ten or more dimensions. Bizarre as these ideas may seem,, they are no less bizarre than relativity, nuclear theory, and quantum physics would have seemed to a scientist of the 19th century., , 38.20 - Gotchas, All carbon atoms are the same. Not true: While carbon atoms all have six protons in their nucleus, different carbon atoms may have different, numbers of neutrons. Nuclei with six protons but different numbers of neutrons are isotopes of carbon., The nucleus is so small that an atom is about 99.99% empty space. No, but you are on the right track if you think this. Since the atomic, diameter is about 10,000 times larger than the nucleus, the atom is more like 99.9999999999% empty space., Protons and neutrons have different charge but approximately the same mass. Yes. A proton has a charge of 1.60×10í19 C and a neutron has, no net electrical charge. The neutron is about 0.1% more massive than the proton, and they are each about 1800 times as massive as the, electron., The energy that is required to disassemble a nucleus into its constituent parts is the same amount of energy that is released when the same, nucleus is assembled from separated nucleons. Yes; the amount of energy is the same and is called the binding energy. It requires energy to, break apart a stable nucleus, and energy is released when the nucleus is assembled., If half of a radioactive isotope decays during one half-life, then after two half-lives, it will all be gone. No, a half-life is the average time it takes, for one-half of the radioactive material that is present to decay. After two half-lives, there will be one-quarter of the original amount left. After, three half-lives, there will be one-eighth, and so on., , 38.21 - Summary, Elements are substances that cannot be divided or changed into other substances, using ordinary chemical methods. An atom is the smallest piece of an element that, still has its chemical and physical properties. An atom consists of electrons orbiting, a very small, very dense nucleus. The nucleus contains both protons and neutrons,, which collectively are called nucleons., The number of protons in an atom is the atomic number, Z. The number of neutrons, is the neutron number, N. The total is known as the atomic mass number, A. Atomic, masses are measured in terms of atomic mass units. The mass of a carbon-12, atom is defined to be exactly 12 u., Isotopes are forms of an element with the same atomic number (which makes them, the same element) but different numbers of neutrons (and hence different atomic, mass numbers)., A fundamental, very short-range interaction called the strong force holds the, nucleons together, counteracting the electrostatic repulsion between protons., The nucleons are nearly incompressible and are tightly packed in the nucleus. The, nuclear radius grows roughly as the cube root of the mass number., , Z = atomic (proton) number, N = number of neutrons, Atomic mass number, , A=Z+N, Atomic mass unit, , u = 1.66054×10–27 kg, Nuclear radius, , R = (1.2×10–15 m)A1/3, , Mass-energy equivalence, Nuclei are stable only for certain combinations of protons and neutrons. On a plot of, neutron number versus proton number, stable nuclei are represented as a band of, E = mc2, stability that passes through the center of the diagram. For small nuclei, the, numbers of protons and neutrons are roughly equal, but for large nuclei, the, neutrons outnumber the protons by about 50%. Heavy nuclei need an excess of neutrons to dilute the proton concentration., Binding energy is the energy that must be added to disassemble, or unbind, a nucleus into the protons and neutrons that make it up. The same, amount of energy is released if the nucleus is assembled from nucleons that are initially separated., The sum of the masses of the separate nucleons is always greater than the mass of the nucleus when it is whole. The mass difference is, related to the binding energy by Einstein’s equation for mass-energy equivalence., When the binding energy of one nucleus is greater than that of another, this means the particles are more tightly bound. For comparing how, tightly bound two nuclei are, the binding energy per nucleon is the important figure., The shape of the binding energy per nucleon vs. mass number curve is important. The curve is highest in the middle. This means that light, nuclei can undergo the fusion process and release energy as this will increase their binding energy per nucleon. Similarly, heavy nuclei can, undergo the fission process and release energy as this also will increase their binding energy per nucleon., In the process of radioactive decay, an unstable nucleus spontaneously emits particles or high-energy photons. If the parent nucleus emits a, charged particle such as an Į particle, negative ȕ, or positive ȕ, then the number of protons in the nucleus changes. Transmutation is the, changing of one element to another. If the parent nucleus emits gamma rays, which are uncharged photons, then transmutation does not take, place, because the number of protons in the nucleus has not changed., The half-life of a radioactive isotope is the average time it takes for the decay of one-half of the atoms that are present in a sample., , 716, , Copyright 2007 Kinetic Books Co. Chapter 38
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Chapter 38 Problems, , Conceptual Problems, C.1, , In Rutherford's scattering experiment, positive alpha particles are directed at a thin gold foil. What was the key piece of, evidence that led him to conclude that the positive charge and mass of the atom were concentrated in the nucleus?, i., ii., iii., iv., , C.2, , What is the approximate ratio between an atomic diameter and a nuclear diameter?, i., ii., iii., iv., , C.3, , All the alpha particles bounced straight back from the gold foil., None of the alpha particles were deflected by the gold foil., A small fraction were deflected from the gold foil., The alpha particles were absorbed by the gold foil., , 1:1, 10:1, 100:1, 10000:1, , Consider the following atoms: 12C, 14C, 14N, 14O, 16O. The atomic numbers of carbon, nitrogen, and oxygen are 6, 7, and 8,, respectively., (a) Which atoms are isotopes of each other? Check all that apply., (b) Which atoms have the same number of neutrons? Check all that apply., (c) Which atoms have the same number of protons? Check all that apply., (a), , 12C, , and 14C, , 14C, , and 14N and 14O, , 14O, , and 16O, , none of them, (b), , 12C, , and 14C, , 12C, , and 14O, , 14C, , and 16O, , 12C, , and 14N, , none of them, (c), , 12C, , and 14C, , 14C, , and 14N and 14O, , 14O, , and 16O, , none of them, C.4, , The strong force acts between which of the following types of particles? Check all that apply., proton and neutron, proton and proton, proton and electron, neutron and neutron, neutron and electron, , C.5, , What force or forces are responsible for keeping electrons in their orbits around the nucleus?, i., ii., iii., iv., , C.6, , Strong force, Electric (Coulomb) attraction, Strong force and Coulomb attraction, None of these, , Consider the following atoms: 12C, 14C, 16O. Rank these in order, from smallest to largest nuclear radii., 12C 14C 16O, 12C 16O 14C, 14C 12C 16O, 14C 16O 12C, 16O 12C 14C, 16O 14C 12C, , C.7, , Can two iron atoms fuse to form a nucleus with greater binding energy per nucleon? Explain your answer., Yes, , C.8, , No, , Explain in your own words why a chain reaction is possible during uranium fission., , Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems, , 717
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C.9, , In the sun, 4 hydrogen nuclei undergo a multi-step process and eventually form a helium-4 nucleus (plus some positrons,, neutrinos and gamma rays, all of negligible mass). What is this an example of?, i., ii., iii., iv., , Nuclear fission, Nuclear fusion, Alpha decay, Chain reaction, , C.10 A radioactive sample emits both alpha particles and negative beta particles. If a uniform external magnetic field is present in, the region, what will be observed regarding the paths of the emitted particles?, i., ii., iii., iv., , Neither path bends., The paths of both particles bend in the same direction., Only one particle's path bends., The particles' paths bend in opposite directions., , Section Problems, Section 3 - Components of the nucleus, 3.1, , A neutral atom of potassium has 19 protons in its nucleus. How many electrons does it possess?, electrons, , 3.2, , For the neutral chlorine atom, , , determine the (a) mass number, (b) number of protons, (c) number of neutrons (d), , number of electrons., (a), , 3.3, , is the mass number, , (b), , protons, , (c), , neutrons, , (d), , electrons, , For a doubly-ionized calcium ion, which is a, , atom with two electrons removed (leaving it with a net charge of +2e),, , determine the (a) mass number, (b) number of protons, (c) number of neutrons (d) number of electrons., (a), , 3.4, , is the mass number, , (b), , protons, , (c), , neutrons, , (d), , electrons, , What is the mass of an electron, measured in atomic mass units?, u, , Section 6 - Nuclear properties, 6.1, , In a Rutherford-like experiment that analyzes electrons that are scattered from a sample, a scientist measures the radius of a, nucleus in the sample to be 4.8×10í15 m. What of these is likely to be the mass number of the nucleus?, i., ii., iii., iv., v., , 6.2, , 1, 2, 4, 16, 64, , Find the mass number of a nucleus whose radius is approximately one-half that of uranium-240., , Section 7 - Sample problem: nuclear density, 7.1, , Estimate the radius of a sphere of nuclear matter whose mass is equal to that of the Earth, 5.97×1024 kg. Take 2.3×1017, kg/m3 as the density of nuclear matter., m, , 718, , Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems
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Section 8 - Stable nuclei, 8.1, , Lead is an element that has 82 protons in its nucleus. Which of the following mass numbers is most likely to correspond to a, stable isotope of lead?, i., ii., iii., iv., v., , 41, 123, 164, 206, 246, , Section 9 - Nuclear binding energy, 9.1, , What is the binding energy for a nucleus of helium-4, also known as an alpha particle? A neutron has a mass of 1.00866 u, a, proton has a mass of 1.00728 u, and an alpha particle has a mass of 4.00153 u, where u = 1.66054×10í27 kg. Express the, energy in joules to three significant figures., J, , Section 13 - Fission, 13.1 Consider the fission reaction that occurs when a neutron is captured by a uranium-235 atom and fissions into xenon-140,, strontium-94, and 2 neutrons. Calculate the energy released in this reaction. The masses are: neutron = 1.0087 u, uranium235 = 235.0439 u, xenon-140 = 139.9216 u, strontium-94 = 93.9154 u., J, , Section 14 - Fusion, 14.1 In the sun, 4 hydrogen nuclei undergo a multi-step process and eventually form a helium-4 nucleus (plus some positrons,, neutrinos and gamma rays, all of negligible mass). For each helium atom produced, about 26 MeV of energy is released. (a), Which has more mass, the 4 hydrogen atoms or the helium-4 atom? (b) What is the magnitude of the mass difference?, (a), , 4 hydrogen atoms, , (b), , 1 helium-4 atom, , kg, , 14.2 The sun radiates electromagnetic energy at the rate of 3.91×1026 W. (a) What is the change in the sun's mass during one, second? Express the answer as a positive mass. (b) The current mass of the sun is 1.99×1030 kg. What fraction of the sun's, mass is lost in a century? Express the answer as a decimal number and assume years of 365.25 days., (a), , kg, , (b), , Section 15 - Radioactivity and radiation, 15.1 An excited nucleus emits a gamma ray of energy 7.5 MeV. (a) Does the nuclear mass increase, or does it decrease? (b), What is the magnitude of the change?, (a), , Increase, , (b), , Decrease, , kg, , 15.2 A carbon-14 atom decays into nitrogen-14 (Z = 7). What particle(s) did it emit?, i., ii., iii., iv., , Positive beta, Negative beta, Gamma, Two alpha, , particle(s), , 15.3 A uranium-238 nucleus decays into a uranium-234 nucleus. What particle(s) did it emit?, i., ii., iii., iv., , Four positive beta, One alpha, Two alpha and two negative beta, One alpha and two negative beta, , particle(s), , Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems, , 719
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15.4 Lead (Z = 82, A = 214) decays in a series of steps. The nucleus emits 2 negative betas, 1 alpha, 1 negative beta, 1 alpha,, and 1 negative beta. What is the final daughter product?, i., ii., iii., iv., v., , Lead-206, Lead-205, Lead-222, Lead-210, None of these, , 15.5 Uranium (Z = 92, A = 234) decays in a series of steps. The nucleus emits 4 alphas, 1 negative beta, 2 alphas, 3 negative, betas, and finally, 1 alpha. What is the final daughter product?, i., ii., iii., iv., v., , Lead-206, Lead-208, Lead-210, Lead-212, Uranium-206, , 15.6 Uranium-238 (Z = 92) decays by alpha emission to thorium-234 (Z = 90), by emitting a helium-4 nucleus. Your friend claims, to have witnessed a new type of decay where the uranium-238 emitted a helium-3 nucleus and became a thorium-235 atom., Prove that this is impossible for an isolated U-238 atom. Here is some helpful data: U-238 has an atomic mass of 238.051 u,, the mass of a helium-3 nucleus is 3.014 u, and the mass of a Th-235 atom is 235.048 u., , Section 17 - Radioactive decay and half-lives, 17.1 Carbon-14 has a half-life of 5730 years. After four half-lives have elapsed, what percentage of an initially pure sample would, remain unchanged?, percent, , Section 18 - Interactive problem: radioactive dating, 18.1 Using the simulation in the first interactive problem in this section, determine the half-life of Carbon-14., i., ii., iii., iv., , 16000, 11460, 5730, 2865, , years, , 18.2 (a) Use the information for the second interactive problem in this section to determine the number of half-lives that elapsed, between the time the pure lead-209 was dumped and midnight. (b) Use the second simulation to measure the half-life of lead209. (c) Using the information given about when each suspect was spotted, determine who is the guilty party., (a), (b), (c), , 720, , half-lives have elapsed, hours, i. Anna, ii. Sara, iii. Katherine, , Copyright 2000-2007 Kinetic Books Co. Chapter 38 Problems
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Chapter 1 Answers, Answers to selected problems, C.5, , (a) Increases, (b) Decreases, , 2.1, , 100000 cm, , 2.3, , 9.966 teradollars, , 2.5, , 100000 tablets, , 3.1, , 3.75eí11, , 3.3, , 5.68e8 s, , 3.5, , 1.00e8, , 5.1, , Dime, , 8.1, , 25.0 mi/h, , 8.3, , 1.28e6 cubits, , 8.5, , Yes, , 8.7, , (a) 5.79e10 m, (b) 5 910 000 000 km, , 8.9, , 2.72 m, , 8.11 4.9 kg, 11.1 6.2 ft, 11.3 8 cm, 11.5 3.79 km, 12.1 430 m, 12.3 9.87 ft, 13.1 1.88 radians, , Copyright 2000-2007 Kinetic Books Co. Chapter 1 Answers, , 721
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Chapter 2 Answers, Answers to selected problems, C.1 No, C.5 (a) í15 m, (b) 15 m, C.7 Yes, Answers vary, , 7.1 (a) Right, (b) Zero, (c) Start the ball moving slowly to the left and then, increase its speed., 8.1 6.67 s, , C.9 Instantaneous velocity, , 9.1 1.90e+5 m/s2, , C.12 Zero, , 9.3 (a) 0 m/s2, (b) 9.6 m/s2, (c) 7.7 m/s2, , C.13 No, Answers vary, 0.1 Negative, 1.1 1.6 meters, 2.1 1 step(s), 2.3 (a) 5.3 km, (b) 5300 m, 3.1 0.343 km, 3.3 2.4e+3 s, 4.1 (a) 0.063 m/s, (b) 240 m/s, , 9.5 (a) 0 cm/s2, (b) í3.0 cm/s2, (c) 3.0 cm/s2, 9.7 31v/5, 10.1 (a) 1.0 s to 2.0 s, (b) 0 to 1.0 s, (c) 4.0 to 5.0 s, 13.1 The white rabbit, The black rabbit, 15.1 2.8 m/s, , 4.3 (a) 138 km, (b) 110 km/h, , 15.3 (a) 15.0 m, (b) 3.04 s, , 4.5 (a) 10 ft/s, (b) 11 km/h, , 15.5 a = 2(vt + d)/t2, , 4.7 21.0 s, 5.1 (a) 1440 m, (b) 7200 s, 5.3 (a) 3.0 m/s, (b) í3.0 m/s, (c) 0 m/s, 5.5 5.40 m/s, , 722, , 18.1 3.19 m/s, 18.3 76.7 m/s, 18.5 0.682 m/s2, 18.7 1.24 s, 18.9 í4.2 m/s, 18.11 (a) 1.22 s, (b) 9.10 m, , Copyright 2000-2007 Kinetic Books Co. Chapter 2 Answers
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Chapter 3 Answers, Answers to selected problems, C.3, , 6.0 knots Southwest, , C.5, , íD, , 0.1, , (a) (í4.0 km, 3.0 km), (b) (3.0 km, 2.0 km), (c) (í3.0 km, í2.0 km), , 1.1, , 81.2, , 1.3, , 1.41e3 kg/m3, , 3.1, , (7.00 knots, 121°), , 3.3, , ( 3.00, 90.0°), , 4.1, , (a) (3, í25) blocks, (b) No, , 4.3, , (a) A = (3, 2), (b) B = (í1, í3), (c) C = (í3, 3), , 5.3, , (a) Submit answer on paper., (b) D, (c) C, (d) F, (e) B and E, , 6.1, , (a) (18, 8), (b) (3, 6), (c) (10, 10, 10), (d) (a+d,b+e,c+f), , 6.3, , (a) a = 5, (b) b = í4, , 8.1, , (a) (18, í6, 48), (b) (9, í12, 15), (c) a2, íab, íac), (d) (í2a+18, í10í6b, í2c+12), , 9.1, , (a) (8, 230°), (b) (21, 200°), (c) (32, 80°), , Copyright 2000-2007 Kinetic Books Co. Chapter 3 Answers, , 723
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Chapter 4 Answers, Answers to selected problems, C.3, , On the Moon, Answers vary, , C.5, , (a) The speed of the river, (b) North, (c) No, , 0.1, , (a) No, (b) The speed increases, , 1.1, , (38.8,29.2) m/s, , 1.3, , (108, 40.3) m/s, , 2.1, , (a) (1.0, 4.0) m/s2, (b) (0.0, 5.0) m/s, , 2.3, , ( 180, -3.75) m/s, , 2.5, , (0.73, 1.3) m, , 3.1, , (a) 0.64 s, (b) 33 m/s, , 3.3, , 1.20 s, , 3.5, , 66.5 m, , 3.7, , (a) 10.5 s, (b) 284 m, (c) í29.3 m/s, (d) 39.8 m/s, , 3.9, , b, , 3.11 1.07 s, 3.13 1 to 9, 5.1, , At the professor's glove, Answers vary, , 7.1, , -6.06 m/s, , 7.3, , 11.0 m/s, , 7.5, , ( 2.36, 14.5) m/s, , 9.1, , At the professor's glove, Answers vary, , 10.1 24.4°, 10.3 ( 9.29, 20.6) m/s, 10.4 35.3 m/s, 10.6 (a) 1100 m/s, (b) 49.3°, 10.8 4.07 s, 10.10 (a) 1.93 m, (b) 2.29 s, 13.1 3.1 m/s, 15.1 8.0 m/s, , 724, , Copyright 2000-2007 Kinetic Books Co. Chapter 4 Answers
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Chapter 5 Answers, Answers to selected problems, C.1 Yes, Answers vary, , 11.1 (a) 1.72 N, (b) 3.34 N, , C.5 Greater between blocks 1 and 2, Answers vary, , 11.3 (a) 172 N, (b) 53.0°, , C.13 Yes, Answers vary, , 12.1 117 N, , 0.1 It can be doing any of these, 2.1 0 N (direction does not matter), 4.1 10 dumbells, 4.3 (a) 209 N, (b) 21.3 kg, , 15.1 (a) 49 N, Down, (b) 49 N, Up, (c) 78 N, Right, (d) 23 N, Left, 17.1 2.03e3 N, 17.3 260 N, , 5.1, , 515 kg, , 18.1 0 m/s2, , 5.3, , (a) 5.0 s, (b) 7.0 s, (c) 6.0 s, , 18.3 0.13, 18.5 0.0950, , 5.5, , 1.2eí3 N, , 5.7, , 22500 kg, , 5.9, , 8.10eí7 N/m2, , 5.11 (a) 0.0788 N, (b) 0.259 N, m/s2, , 22.1 (a) 59 N, Straight down, (b) 51 N, Up and to the left, (c) 23 N, Down the plane, (d) 78 N, Up the plane, 23.1 -0.301 m, 23.3 (a)1.25 m/s2, (b) Upward, , 5.13 (a) 13.4, (b) 603 N, (c) Yes, , 23.5 (a) 4.12e+3 lbs/in, (b) 7.21e+5 N/m, , 5.15 5.1 m/s2, , 24.1 1.60, , 9.1 (a) 6.54e4 N, (b) 7.57e4 N, , 25.1 (a) 12 N, Directly to the right, (b) Any magnitude greater than 10 N, Directly to the right, (c) 10 N, Directly to the right, , 10.1 í0.233 m/s2, , Copyright 2000-2007 Kinetic Books Co. Chapter 5 Answers, , 725
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Chapter 6 Answers, Answers to selected problems, 0.1, , (a) Greater, (b) To the same height as the other hill, , 1.1, , 201 J, , 1.3, , 181 J, , 4.1, , 49 times higher, , 4.3, , 1410 J, , 5.1, , (a) 8.0eí17 J, (b) 1.3e+7 m/s, , 5.3, , 1.63e+3 N, , 5.5, , (a) 8460 J, (b) 26.0 m/s, , 7.1, , 133 N, , 9.1, , 2.1e+3 W, , 9.3, , 3.41 m/s, , 9.5, , 2.24e+5 J, , 10.1 $184, 11.1 -8.58 J, 11.3 2.04e+10 J, 11.5 801 W, 14.1 42.5 kg, 14.3 1.44 m/s2, 16.1 63.8 m, 19.1 8.0 m, , 726, , Copyright 2000-2007 Kinetic Books Co. Chapter 6 Answers
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Chapter 7 Answers, Answers to selected problems, C.1, , No, Answers vary, , C.3, , (a) They have the same momentum, (b) A's kinetic energy is greater, , C.5, , Truck impulse equal, , C.7, , Center of mass, Answers vary, , 0.1, , (a) Yes, (b) No, (c) Yes, , 1.1, , 1.29 kg·m/s, , 1.3, , 31 m/s, , 1.5, , 90 kg·m/s, , 2.1, , 2.8e2 kg·m/s, , 3.1, , -1.73e+3 N, , 3.3, , 25800 N, , 3.6, , (a) 1.13e3 kg·m/s, (b) 4.50e4 N, , 3.8, , (a) 4.0 N, (b) 12 kg·m/s, (c) 80 m/s, , 3.10 2.79e+3 N, 3.12 (a) í4.09eí23 kg·m/s, (b) 0.205 N, 5.1, , -1.26e+3 m/s, , 5.3, , í3.25 m/s, , 5.5, , (a) í0.43 m/s, (b) 0.77 m/s, , 8.1, , 32 N, , 8.2, , 118 kg, , 9.1, , (a) 2.3 kg, (b) 3.1 m/s, , 9.3, , (a) 8.1 kg, (b) í1.9 m/s, , 12.1 6.0 m/s, 13.1 (a) í0.40 m/s, (b) í3.6 m/s, (c) í1.8 m/s, 13.3 0 m/s, 13.5 2.0 kg·m/s, 13.7 (a) 3.34 m/s, (b) No, 14.1 4.7e6 m, 14.3 (a) 6.0 m, (b) 6.3 m, 16.1 (a) Impossible, (b) Elastic, (c) Inelastic, , Copyright 2000-2007 Kinetic Books Co. Chapter 7 Answers, , 727
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Chapter 8 Answers, Answers to selected problems, 0.1, , (a) Yes, (b) Yes, , 2.1, , 1.31e4 m/s, , 2.3, , 2.28e11 m, , 3.1, , 1.15 m/s2, , 3.3, , (a) 0.0337 m/s2, (b) 0 m/s2, , 3.5, , 21 m/s2, , 4.1, , 9.0 m/s, , 5.1, , 4.50 "G's", , 5.3, , 1.24eí4 N, , 5.5, , 6.3 N, , 6.1, , 103 s, , 7.1, , (a) 8.0 m/s, (b) 9.5 m/s, , 728, , Copyright 2000-2007 Kinetic Books Co. Chapter 8 Answers
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Chapter 9 Answers, Answers to selected problems, C.1, , Yes, Answers vary, , 0.1, , (a) Decrease, (b) Decrease, (c) Stay the same, (d) Increase, , 1.1, , 48 m, , 2.1, , 14 rad, , 2.3, , 7.17eí4 rad, , 2.5, , (a) 3.54 s, (b) 0.315 rad, (c) 0.500 rad/s, , 3.1, , 2.0e+3 m, , 3.3, , 1.99eí7 rad/s, , 3.5, , (a) 0.10 rad/s, (b) 23 rad/s, , 4.1, , -28.0 rad/s2, , 4.3, , í0.12 rad/s2, , 7.1, , (a) 7.08eí3 rad/s, (b) 2.01eí4 m/s2, (c) 2.05eí5 g, , 7.3, , 17 m/s, , 7.5, , 0.619 rad/s2, , 8.1, , 0.083 m/s2, , Copyright 2000-2007 Kinetic Books Co. Chapter 9 Answers, , 729
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Chapter 10 Answers, Answers to selected problems, C.1, , (a) Yes, (b) Yes, , C.3, , False, Answers vary, , 0.1, , (a) Yes, (b) Farthest from the axis of rotation, , 1.1, , 376 N, , 1.3, , 0.21 N·m, , 2.1, , 1.60 rad/s2, , 2.3, , 53 rad/s, , 3.1, , 19 kg·m2, , 3.3, , 1.0e2 kg·m2, , 4.1, , 1.3 s, , 4.3, , 2.0 rad/s2, , 4.5, , 5.7eí5 kg·m2, , 4.7, , 0.21, , 6.1, , 2.85e5 N, , 8.1, , 1.93e+5 kg·m2/s, , 9.1, , 49.8 kg·m2/s, , 10.1 143 rad/s, 10.3 119 rad/s, 11.1 50 rad/s, 11.3 2.15 rad/s, , 730, , Copyright 2000-2007 Kinetic Books Co. Chapter 10 Answers
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Chapter 11 Answers, Answers to selected problems, C.1, , b, d, , C.5, , Steel,, Copper,, Titanium,, Aluminum, , 1.1, , 1.47 m, , 1.2, , 0.23 m, , 1.4, , 1.4e3 N, , 1.6, , 0.302 m, , 1.9, , 1.20 m, , 3.1, , 9.2eí2 m, , 3.3, , (a) (1.0,0.50), (b) (1.0,1.0), (c) (0.63,0.38), (d) (0.69,0), , 3.5, , 0.827 m, , 3.7, , 0.32 m, , 4.1, , (a) 950 N, (b) Negative y, (c) 1.32 m, , 6.1, , (a) 1.6e5 N/m2, (b) 2.0e6 N/m2, , 7.1, , 0.0257 m, , 7.3, , 2.72eí5 m, , 7.5, , 4.0e9 N/m2, , 7.7, , 82e9 N/m2, , 8.1, , 1.5 m3, , 8.3, , í5.6eí5, , Copyright 2000-2007 Kinetic Books Co. Chapter 11 Answers, , 731
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Chapter 12 Answers, , Chapter Assumptions, Unless stated otherwise, use the following values., , rEarth = 6.38×106 m, MEarth = 5.97×1024 kg, , Answers to selected problems, 0.1, , (a) Decrease, (b) Increase, , 1.1, , 9.04e4 N, , 1.3, , 4.11e11 m, , 1.5, , (2.0eí9,í5.2eí10) N, , 2.1, , 9.76 m/s2, , 2.3, , 24.8 m/s2, , 8.1, , 7900 m/s, , 9.1, , 7.69e3 m/s, , 9.3, , 1.02e3 m/s, , 11.1 3160 m/s, 12.1 (a) It increases, (b) It decreases, 14.1 22.7 AU, 14.3 (a) 1.00 AU, (b) 0.050, 14.5 22.9 AU, 16.1 (a) 3.74e8 s, (b) 11.9 years, 16.3 2.40e+9 s, 16.5 400 hours, 16.6 1.05e4 years, 17.1 (a) í6.32e11 J, (b) 3.16e11 J, (c) í3.16e11 J, 17.3 1.47e+13 J, 17.5 í1.14e10 J, 18.1 1.04e4 m/s, 18.3 (a) 3.55e4 m/s, (b) 3.76e4 m/s, 18.5 (a) 1.47e4 m, (b) 2.95e12 m, , 732, , Copyright 2000-2007 Kinetic Books Co. Chapter 12 Answers
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Chapter 13 Answers, , Chapter Assumptions, Unless stated otherwise, use the following values:, Atmospheric pressure at the Earth's surface: Patm, , = 1.013×105 Pa, , Density of pure water = 1000 kg/m3, Density of seawater = 1030 kg/m3, “Standard temperature and pressure” means 0°C and the atmospheric pressure stated above., , Answers to selected problems, C.3 No. The bottle will not explode., Answers vary, C.5 No, Answers vary, C.9 The new waterline mark is lower on the hull., , 5.1 (a) 1.49e5 Pa, (b) 2.50e5 Pa, (c) Ben Chapman, 6.1 (a) 1.01e3 kg/m3, (b) 1.03e5 Pa, , C.11 The ball bearing, Answers vary, , 6.3 1.10e+5 Pa, , C.14 The lake's water level rises very slightly, , 7.3 (a) 8.18eí2 m3, (b) 1.05e3 kg/m3, (c) 87.0 , , people/km2, , 2.1 (a) 31.3, (b) 42.7 people/km2, (c) 1.86 people/km2, (d) 6.10e3 people/km2, 2.3 (a) 12 kg, (b) 26 lb, , 7.1 6.76e+3 m3, , 8.1 (a) 79.3 N, (b) 12.7 N, 9.1 43.8 , 10.1 The 3.20 kg crown, , 3.1 (a) 6.85e6 Pa, (b) 1.57e5 Pa, , 11.1 63.8 N, , 3.3 (a) 7.64e4 Pa, (b) 1.95e3 Pa, , 13.1 36.6 m/s, , 3.5 52.5 N, , 16.1 (a) 29.0 u, (b) 4.81eí26 kg, (c) 8.90e4 Pa at 1000 m, 5.31e4 Pa at 5000 m, 2.78e4 Pa at 10,000 m, , 4.1 (a) 1.01e7 Pa, (b) 5.05e7 Pa, (c) 1.01e8 Pa, 4.3 (a) 10.3 m, (b) 0.760 m, (c) 760 mm, (d) 29.9 in, , 14.1 8.36e+4 Pa, , 4.5 9.34e5 Pa, , Copyright 2000-2007 Kinetic Books Co. Chapter 13 Answers, , 733
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Chapter 14 Answers, , Chapter Assumptions, The general form of the equation of motion for an object in SHM is x(t), , = A cos (Ȧt + ij)., , Answers to selected problems, C.1, , No, Answers vary, , C.2, , Remove mass, , C.3, , kg/s, , 0.1, , (a) Stay the same, (b) Sinusoidal function, , 3.1, , (a) 2.78eí4 Hz, (b) 2.31eí5 Hz, , 4.1, , 0.105 rad/s, , 5.1, , (a) 5 m, (b) 3 m, (c) 4 m, , 5.3, , x(t) = 3.5 cos ((ʌ/2)t + 2.0), , 6.1, , (a) 0.40 m, (b) 2.0 s, , 7.1, , 103 m/s, , 8.1, , 0.15 m, , 9.1, , 0.018 kg·m2, , 10.1 3.15 m, 10.3 90.0 minutes, 11.1 1.20 m, , 734, , Copyright 2000-2007 Kinetic Books Co. Chapter 14 Answers
Page 737 :
Chapter 15 Answers, Answers to selected problems, C.1, , Longitudinal, Answers vary, , C.3, , No, Answers vary, , C.5, , AM station, Answers vary, , C.7, , Wave on tauter string arrives first., , 0.1, , (a) Yes, (b) No, (c) No, , 4.1, , 2.5 m, , 5.1, , 6.0 m, , 6.1, , 0.00174 s, , 7.1, , 1.42e9 Hz, , 7.3, , 11 meters, , 7.5, , 0.497 m, , 8.1, , 347 m/s, , 8.3, , 1470 m/s, , 9.1, , 0.080 kg/m, , 10.1 í3.4eí2 m, 10.3 (a) Right to left, (b) 3.34eí3 m, (c) 4.00 m, (d) 15.0 Hz, , Copyright 2000-2007 Kinetic Books Co. Chapter 15 Answers, , 735
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Chapter 16 Answers, , Chapter Assumptions, Unless stated otherwise, use 343 m/s for the speed of sound., , Answers to selected problems, 0.1, , (a) Higher, (b) Stays the same, , 3.1, , Higher, , 4.1, , 12 m, , 4.3, , 1.72eí2 W/m2, , 5.1, , 3.2eí5 W/m2, , 5.3, , 50 dB, , 5.5, , (a) 5.0eí17 W, (b) 1.6eí10 W, (c) 5.0eí3 W, , 5.7, , 20, , 7.1, , 787 Hz, , 7.3, , 44.6 m/s, , 7.5, , 33.1 m, , 9.1, , 3.01, , 736, , Copyright 2000-2007 Kinetic Books Co. Chapter 16 Answers
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Chapter 17 Answers, Answers to selected problems, 0.1, , (a) Trough, (b) 2.00 m, (c) 1.00 m, , 4.1, , 97.1 Hz, , 4.3, , 2.29e+3 Hz, , 4.5, , 415 N, , 5.1, , (a) 1.50 m, (b) 2, , 7.1, , 1860 Hz, , 7.2, , Destructive, , 8.1, , 5.36 Hz, , 8.3, , 4.4 Hz, , 8.5, , 1.28 Hz, , Copyright 2000-2007 Kinetic Books Co. Chapter 17 Answers, , 737
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Chapter 18 Answers, Answers to selected problems, C.1, , (a) Answers vary, (b) Answers vary, (c) 99 °F, , C.3, , It does not change, , 2.1, , (a) 0 °F, (b) 96 °F, , 3.1, , (a) í173.15 °C, (b) í279.67 °F, (c) 22 °C, (d) í459.67 °F, , 3.3, , 119 °C, , 8.1, , 0.00196 m, , 8.3, , 2.062eí5 1/C°, , 8.5, , 158.3°C, , 9.1, , 4.48e8 N/m2, , 11.1 0.103 cm3, 11.3 79.3 gallons, 11.5 7.54eí3 m, 11.7 í0.830 %, 12.1 68.9 °C, 12.3 2.57e7 J, 12.5 7.6eí3 K, 12.7 (a) 2.7 kg, (b) 52°C, 12.9 1.68e6 J, 13.1 538 J/kg·K, 15.1 1.37e+4 J, 15.3 8.74e4 J, 18.1 (a) 2.7e2 W, (b) 47 W, 18.3 (a) 8.59 in, (b) 23.7 in, 18.5 2.4 W, 21.1 2.70e5 J, , 738, , Copyright 2000-2007 Kinetic Books Co. Chapter 18 Answers
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Chapter 19 Answers, , Chapter Assumptions, Use the following values for constants:, , NA = 6.02×1023, R = 8.31 J/mol·K, k = 1.38×10–23 J/K., In problems which require you to know the atomic weights of atoms or molecules, use the following:, 12.0 u for a carbon atom (C), 4.00 u for a helium atom (He), 1.00 u for a hydrogen atom (H), 14.0 for a nitrogen atom (N), 20.2 u for a neon atom (Ne), 16.0 u for an oxygen atom (O), 44.0 u for a carbon dioxide molecule (CO2), 18.0 u for a water molecule (H2O), , Answers to selected problems, 3.1, , 1.73e5 Pa, , 3.3, , 0.150 m3, , 4.1, , 1.0e+22 molecules, , 4.3, , (a) 44.0 u, (b) 4.40eí2 kg/mol, (c) 1.68e23 molecules, , 5.1, , (a) 7.2 moles, (b) 4.3e24 molecules, , 5.3, , 1.24e5 Pa, , 5.5, , í54.5 mol, , 5.7, , 741 K, , 7.1, , 2.00eí22 m3, , 9.1, , 1.1eí20 J, , 9.3, , 531 K, , Copyright 2000-2007 Kinetic Books Co. Chapter 19 Answers, , 739
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Chapter 20 Answers, Answers to selected problems, C.1, , Adiabatic, , 0.1, , (a) The heat is equal to the change in internal energy., (b) The work done is the negative of the change in internal energy., (c) The net heat transferred is equal to the sum of the change in internal energy and the work done., , 1.1, , (a) í3.4e5 J, (b) í5.1e5 J, (c) 1.7e5 J, , 1.3, , í1700 J, , 3.1, , (a) 9.50e3 J, (b) Work is done by the system, , 3.3, , (a) 5.48e5 J, (b) 7.15e5 J, , 740, , Copyright 2000-2007 Kinetic Books Co. Chapter 20 Answers
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Chapter 21 Answers, Answers to selected problems, C.1, , Stays the same, , C.3, , The room temperature goes up, Answers vary, , 1.1, , 177 J, , 1.3, , 930 J, , 4.1, , 0.222 J/K, , 5.1, , 1.45 J/K, , 5.3, , 674 J, , 7.1, , 458°C, , 8.1, , (a) 755 K, (b) 1.21e+5 J, , 8.3, , 8.88e+3 J, , 8.5, , (a) 28.7 %, (b) 53.2 %, , 9.1, , (a) 47.6 %, , 9.3, , (a) 1.22, (b) 1.33, , 10.1 (a) 16, (b) 88 J, 10.3 (a) 1.4e5 J, (b) 97 cents, 12.1 0.0281 L, , Copyright 2000-2007 Kinetic Books Co. Chapter 21 Answers, , 741
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Chapter 22 Answers, Answers to selected problems, C.1, , Richard, Answers vary, , C.3, , (a) 0 C, (b) 0 C, , C.5, , (a) Rubber is a(n) insulator., (b) Iron is a(n) conductor., (c) Copper is a(n) conductor., (d) Wood is a(n) insulator., , C.7, , the "10 cm" end, , C.9, , The force does not change., , C.11 (a) Yes, (b) Answers vary, 0.1, , (a) Away from each other, (b) Towards each other, (c) Decreases, , 1.1, , +3.20eí19 C, , 1.3, , 1.60eí20 kg, , 2.1, , (a) 0 C, (b) í1.60eí19 C, , 2.3, , 3.86e+13 electrons, , 2.5, , 1.5e20 electrons, , 3.1, , (a) íe, (b) ʌ, , 4.1, , 0C, , 6.1, , í3.6eí6 C, , 7.1, , (a) Answers vary, (b) +4.00 μC, (c) 40000 dollars, , 8.1, , 2.0eí5 C, , 9.1, , (a) Attractive, (b) 8.22eí8 N, , 9.3, , 9.32eí7 C, , 11.1 0.072 N, 11.3 3.02e3 N, 11.5 (a) NW corner íQ, SW corner +Q, SE corner íQ, (b) NE corner íQ, NW corner íQ, SW corner +Q, SE corner +Q, (c) NE corner +Q, NW corner íQ, SW corner íQ, SE corner +Q, 11.7 í167 C, 13.1 34 protons, , 742, , Copyright 2000-2007 Kinetic Books Co. Chapter 22 Answers
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Chapter 23 Answers, Answers to selected problems, C.1, , 2, , C.3, , (a) The combined field is zero, (b) To the right, (c) The combined field is zero, , C.5, , Proton, Answers vary, , 0.1, , Greatly increases, , 1.1, , (a) 1.75eí3 N, (b) í2.89 N, (c) í184000 N, (d) 1.87eí12 N, , 1.3, , 1.36eí21 N, , 1.5, , 600 N, , 2.1, , 2.61 N/C, , 2.3, , (a) 5.14e11 N/C, (b) Away from the proton, , 5.1, , (a) Closer together, (b) The direction of the force is tangent to the field lines, , 6.1, , (a) 2.59e11 N/C, to the right, (b) 1.12e11 N/C, to the left, (c) 9.55e10 N/C, to the right, , 6.3, , 16.2 C, , 7.1, , 18 lines going into í3q, , 7.3, , (a) q2, (b) q1, q3, (c) |q2| > |q1| > |q3|, (d) 3 times stronger, , 8.1, , 89.9 s, , 8.3, , (a) í8.69eí17 N, (b) í9.54e13 m/s2, (c) 1.00e6 m/s, (d) 3.33eí3, , 8.5, , 4.64e3 m/s, , 8.7, , (a) 4.79e10 m/s2, (b) 7.31eí7 s, (c) 1.28eí2 m, (d) 1.02eí18 J, , 12.1 9.76eí9 C·m, 12.3 6.59 m, , Copyright 2000-2007 Kinetic Books Co. Chapter 23 Answers, , 743
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Chapter 24 Answers, , Chapter Assumptions, Unless stated otherwise, the reference configuration for zero electric potential and zero electric potential energy is one in which there, is infinite separation between charges., , Answers to selected problems, C.1 (a) It doubles, (b) Yes, (c) No, , 3.1 2.6 J, , C.4 (a) No, (b) Answers vary, (c) No, Answers vary, , 3.5 3.0eí12 m, , 0.1 (a) Always positive, (b) Closer to, 0.3 (a) It increases, (b) A vertical line, 1.1 (a) 5.4 μJ, (b) í5.4 μJ, (c) Closer together, 1.3 (a) í52.3 J, (b) 52.3 J, (c) Farther apart, 2.1 (a) 5.09eí3 J, (b) í5.09eí3 J, (c) 5.09eí3 J, , 3.3 (a) 2e2 N, (b) 2eí13 J, , 4.1 (a) 1.44eí9 V, (b) í1.44eí9 V, (c) Near the proton, (d) Magnitudes are equal, 4.3 2.41eí9 C, 6.1 32 V, 7.1 (a) 6.34eí20 J, (b) í6.34eí20 J, (c) 0.396 V, 7.2 (a) 1.28eí20 J, (b) 8.83e25 ions, 7.4 (a) 1.04e4 J, (b) $ 2.6eí4, 7.5 í2.66eí2 C, 10.1 2.53e+8 V, , 2.3 (a) 2.75eí2 J, (b) í2.75eí2 J, (c) 2.75eí2 J, , 10.3 (a) Sphere, (b) The surface closer to the point charge, (c) 0.32, , 2.5 (a) 6.71eí4 J, (b) í6.71eí4 J, (c) 6.71eí4 J, (d) Yes, , 11.1 í9.6eí14 J, 11.3 -40.9 V, 11.5 710 V, 11.7 (a) Planes parallel to the yz plane, (b) x = 2.24 m, (c) x = 1.87 m, , 744, , Copyright 2000-2007 Kinetic Books Co. Chapter 24 Answers
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Chapter 25 Answers, , Chapter Assumptions, For any problems in this chapter that involve a household electrical outlet, assume that the voltage supplied is fixed at 120 V. While, this is not true for actual outlets, whose voltage varies and periodically exceeds 120 V, the simplification will not affect the accuracy, of your answer., , Answers to selected problems, C.1, , (a) 1, (b) 3, , C.3, , The current stays the same, Answers vary, , C.7, , Noníohmic, , C.9, , No, Answers vary, , C.11 No, Answers vary, C.13 Electric energy, 1.1, , (a) 6.00eí2 C, (b) 3.75e17 electrons, (c) 0 protons, , 1.3, , 12.3 C, , 1.5, , 7.4e-7 A, , 1.7, , 4.2eí7 s, , 1.9, , (a) 3.89eí2 C/m2, (b) Down, , 2.1, , (a) 1.1e4 s, (b) Answers vary, , 3.1, , 16.4 ȍ, , 3.3, , 4.0eí3 A, , 3.5, , 1.29 V, , 4.1, , 100 ȍ, , 5.1, , 2.4eí7 ȍ, , 5.3, , Aluminum, , 5.5, , (a) 10.0 ȍ, (b) 16.0 ȍ, (c) 2.00 ȍ, , 6.1, , 360°C, , 6.3, , (a) 0.00501°Cí1, (b) Tungsten, , 6.5, , 16.3°C, , 6.7, , 51.8 ȍ, , 7.1, , 3.1 V, , 7.3, , 20 light bulbs, , 7.5, , 14 ȍ, , 7.7, , (a) 2.4 ȍ, (b) 1.0e2 W, , 8.1, , $7.17e20, , 11.1 2.07e10 W, , Copyright 2000-2007 Kinetic Books Co. Chapter 25 Answers, , 745
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Chapter 26 Answers, Answers to selected problems, C.1, , (a) It increases, (b) It stays the same, , C.2, , (a) Increased, (b) Same amount of charge, , 0.1, , Charge is proportional to potential difference, , 1.1, , 3.2eí10 C, , 1.3, , 7.1eí6 F, , 1.5, , 7.2eí5 C, , 4.1, , 4.00eí9 F, , 4.3, , 4.45eí1 J, , 4.5, , 9.96eí7 F, , 4.7, , 1.54eí2 J, , 6.1, , 76 J, , 8.1, , 7.2 nF, , 8.3, , 1.60e8 V/m, , 8.4, , 66.0 nF, , 11.1 1.6eí7 C, , 746, , Copyright 2000-2007 Kinetic Books Co. Chapter 26 Answers
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Chapter 27 Answers, Answers to selected problems, C.4, , The same, Answers vary, , C.6, , Circuit with two light bulbs, Answers vary, , C.8, , Parallel, Answers vary, , C.12 (a) R1 is in series with the combination of R2 and R3, (b) R2 is in parallel with R3, C.14 It will decrease to zero, Answers vary, 0.1, , (a) They are equal, (b) It is the same everywhere, , 2.1, , 1.17e6 J, , 3.1, , 3.0 V, , 3.3, , (a) 6.6 W, (b) 3.6e4 J, , 3.5, , 2.0 ȍ, , 3.7, , 1.8e4 s, , 5.1, , 0.55 A, , 6.1, , 1.0 ȍ, , 6.3, , 4200 ȍ, , 6.5, , 3.23 ȍ, , 6.7, , 2.7 A, , 6.9, , 12.0 V, , 7.1, , 25.0 ȍ, , 10.1 1.67 ȍ, 10.3 0.175 ȍ, 10.5 77 ȍ, 10.7 8.75 ȍ, 10.9 34.8 V, 10.11 4, 10.13 (a) 500 ȍ, (b) 125 ȍ, 11.1 60.0 ȍ, 12.1 (a) 3.75 ȍ, (b) 5.00 ȍ, , Copyright 2000-2007 Kinetic Books Co. Chapter 27 Answers, , 747
Page 750 :
Chapter 28 Answers, , Chapter Assumptions, Elementary charge, e, Mass of electron, me, Mass of proton, mp, , = 1.60×10–19 C, , = 9.11×10–31 kg, , = 1.67×10–27 kg, , Unless stated otherwise, use 5.00×10–5 T for the strength of the Earth's magnetic field at its surface., , Answers to selected problems, C.1 Down, Answers vary, , 13.1 2.75eí3 m, , C.2 (a) The field is perpendicular to the plane of the wires., (b) Reinforce, (c) Cancel, Answers vary, , 13.5 3.67e+2 s, , C.5 (a) No, (b) Electron, C.6 No, Answers vary, C.9 No, Answers vary, C.12 (a) Toward you, (b) Left, (c) Down, (d) Down, C.14 (a) Negative, (b) Positive, (c) Answers vary, C.16 Away from you, Up, 0.1 (a) Curve, (b) No, (c) Yes, 7.1 (a) 6.08eí16 N, (b) Positive y direction, , 13.3 56.3 m/s, , 13.7 2.28eí12 T, 15.1 (a) 1.29eí2 T, (b) Decrease, 15.3 (a) 4.65eí3 m, (b) 5.65eí3 m, (c) 2.00eí3 m, 15.5 0.0286 m, 16.1 (a) 8.29 m/s, (b) 1.07 m, (c) 0.872 m/s, (d) 0.708 m, 17.1 (a) A circle, (b) A straight line, 18.1 14.1 N, 18.3 2.27 A, 18.5 35.2°, 20.1 (a), (d), 22.1 2.00eí5 T, 22.3 1.14eí3 T, 22.5 896 A, , 7.3 (a) 805 m/s, (b) East, , 22.7 (a) 1.64eí26 N, (b) 55.4 m, , 7.5 1.78 C, , 22.9 4.58eí23 N, in the same direction as the current, , 7.7 17.3°, , 22.11 (a) Opposite directions, (b) 75.0 A, , 9.1 (a) 2, (b) 180q, (c) 90.0°, 10.1 0.417 T, 10.3 0.631 T, , 22.13 8.80eí3 m, 22.14 6.67 m, 22.15 Direction: +z, Strength: 3.66eí7 T, , 10.5 5.76eí3 T, 11.1 2.90 T, 12.1 3.00e+4 m/s, 12.3 2.21eí22 J, 12.5 (a) 1.0e4 m/s, (b) 1.8eí2 m/s, , 748, , Copyright 2000-2007 Kinetic Books Co. Chapter 28 Answers
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Chapter 29 Answers, , Chapter Assumptions, Unless stated otherwise, use 5.00×10–5, Elementary charge, e, , T for the strength of the Earth's magnetic field at its surface., , = 1.60×10–19 C, , Answers to selected problems, C.1, , No, Answers vary, , C.4, , Counterclockwise, Answers vary, , C.7, , (a) Counterclockwise, (b) Clockwise, (c) Answers vary, , C.9, , (a) The same direction, Answers vary, (b) Answers vary, , 0.1, , (a) No, (b) Yes, (c) Yes, , 2.1, , (a) Yes, (b) The induced emf increases when the wire moves faster., (c) The induced emf increases when the field strength increases., (d) Answers vary, , 5.1, , 0.500 m, , 5.3, , 0.808 V, , 6.1, , 4.25 Wb, , 7.1, , 12.9 T· m2, , 7.3, , 1.13eí6 A, , 8.1, , 8.00 ms, , 11.1 (a) The emf is a maximum when the velocity vector is Perpendicular to the magnetic field., (b) The emf is zero when the velocity vector is Parallel to the magnetic field., (c) Yes, (d) Yes, 12.1 0.021 s, 13.1 (a) 0.2 A, (b) 2.0 V, (c) 1.0e6 Hz, 15.1 1.50e+3 turns, 15.3 26.2, 15.5 (a) 7.75e3 for 100,000 V, 1.74e4 for 225,000 V, 2.67e4 for 345,000 V, (b) 900 A for 100,000 V, 400 A for 225,000 V, 261 A for 345,000 V, (c) 9.07e7 W for 100,000 V, 1.79e7 W for 225,000 V, 7.63e6 W for 345,000 V, (d) Answers vary, 17.1 20 V, , Copyright 2000-2007 Kinetic Books Co. Chapter 29 Answers, , 749
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Chapter 30 Answers, , Chapter Assumptions, Unless stated otherwise, electromagnetic waves are assumed to be propagating as plane waves through a vacuum., When converting light-years to meters, assume that a year has 365 days., The speed of light in a vacuum is 3.00×108 m/s., The Sun radiates energy at the rate of 3.91×1026 W., , Answers to selected problems, C.1, , AM, Answers vary, , C.4, , (a) Northísouth, (b) Vertical, , C.6, , (a) Vertical, (b) Vertical, , 1.1, , (a) 1.28 lightíseconds, (b) 8.34 lightíminutes, (c) 2.6e20 m, , 1.3, , (a) 7.5e5 Hz, (b) AM, , 1.5, , (a) 12.2 cm, (b) 6.10 cm, (c) Yes, , 1.7, , (a) (0, 0, 0), (b) (255, 255, 255), (c) 16777216 colors, , 1.9, , (a) 19%, (b) 57%, (c) 16%, , 2.1, , Away from you, , 3.1, , 1.39eí6 T, , 750, , Copyright 2000-2007 Kinetic Books Co. Chapter 30 Answers
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Chapter 31 Answers, Answers to selected problems, C.2, , Virtual, Larger, Upright, , C.4, , Real, Smaller, Inverted, , C.6, , (a) Larger, (b) Inverts, Becomes real, , 0.1, , (a) To the left of C, In between C and F, (b) To the left of C, (c) In between C and F, To the right of F, , 3.1, , 7.50 m, , 5.1, , (a) 1.19 m, (b) 53.5°, (c) 180°, , 5.3, , 22, , 5.5, , 1.56 m, , 9.1, , -1.15 cm, , 9.3, , 10.5 cm, , 14.1 (a) No, (b) No, (c) No, (d) No, 15.1 (a) +15.2 cm, (b) í0.879, (c) Inverted, 15.3 16.5 cm, 15.5 (a) 12.5 cm, (b) 8.33 cm, 15.7 63.3 cm, 15.9 (a) Concave, (b) 6.20 cm, 15.11 í34.0 cm, 17.1 6.16 cm, , Copyright 2000-2007 Kinetic Books Co. Chapter 31 Answers, , 751
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Chapter 32 Answers, , Chapter Assumptions, Unless stated otherwise, the incident medium is air, with an index of refraction of 1.00., , Answers to selected problems, 0.1, , (a) Air, (b) Toward, , 2.1, , Diamond, , 2.3, , 1.86e8 m/s, , 3.1, , 12.0°, , 3.3, , 1.27, , 3.5, , 1.63, , 3.7, , 1.71, , 7.1, , (a) 65.8°, (b) 34.8°, , 8.1, , 24.4°, , 8.3, , 1.54, , 9.1, , 0.11°, , 752, , Copyright 2000-2007 Kinetic Books Co. Chapter 32 Answers
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Chapter 33 Answers, Answers to selected problems, C.1, , Real; larger; inverted, , C.3, , Virtual; smaller; upright, , C.5, , Inverted, Answers vary, , 0.1, , (a) Between the focal point and the lens., (b) On the opposite side of the focal point as the lens., (c) Between the focal point and the lens., , 4.1, , (a) No, (b) No, (c) No, , 7.1, , (a) Positive, (b) Positive, (c) 4.11 cm, (d) Converging, , 7.3, , 0.135 m, , 7.5, , (a) Negative, (b) Real, (c) 14.0 cm, (d) 4.67 cm, , 7.7, , 50.0 cm, , 7.9, , (a) Diverging, (b) í3.94 cm, , 7.11 0.0233 m, 7.13 1.18 m, 8.1, , (a) 10.0 cm, (b) 6.00 cm, , 10.1 (a) Diverging, (b) í40.0 cm, 13.1 0.0066 radians, 13.3 1.40e5 m, , Copyright 2000-2007 Kinetic Books Co. Chapter 33 Answers, , 753
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Chapter 34 Answers, Answers to selected problems, C.1, , Ȝ/2, , C.2, , No, Answers vary, , 3.1, , 606eí9 m, , 3.3, , 68 fringes, , 754, , Copyright 2000-2007 Kinetic Books Co. Chapter 34 Answers
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Chapter 35 Answers, , Chapter Assumptions, Assume the speed of light in a vacuum is 3.00×108 m/s for all observers., , Answers to selected problems, C.1, , No, , C.2, , No, , C.4, , V/2, , 3.1, , (a) 9.0 s, (b) 8.5 s, , 5.1, , (a) Yes, (b) Yes, (c) Yes, (d) No, , 6.1, , (a) 3.00eí7 s, (b) 3.25eí7 s, , 6.3, , 1.55e+8 m/s, , 6.5, , 2.89e+8 m/s, , 7.1, , (a) 555 m, (b) 745 m, (c) 1.85eí6 s, (d) 2.48eí6 s, (e) 1.34, (f) 2.48eí6 s, , 8.1, , (a) 6.0eí8 s, (b) 1.0eí7 s, (c) Shorter, , 9.1, , (a) 0.333c, (b) 65.0 m, , 10.2 (a) 1.20 s, (b) 1.27 s, (c) 1.69 s, 10.4 0.111c, 12.1 (a) 0.875c, (b) 1700 MeV/c, 12.3 1.32 MeV, 12.5 (a) 0.163 km, (b) 1.46e+13 years, 12.7 5.9e4, , Copyright 2000-2007 Kinetic Books Co. Chapter 35 Answers, , 755
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Chapter 36 Answers, , Chapter Assumptions, When converting between electron-volts and joules, assume that, 1 eV = 1.60×10–19 J, , Answers to selected problems, C.3, , (a) There are more electrons leaving the surface., (b) The electrons' maximum kinetic energy is still equal to E0., , C.4, , There is movement of holes in the valence band., , C.7, , (a) Positive, (a) Negative, , C.9, , Yes, Answers vary, , 2.1, , 486 nm, , 3.1, , 4.2e14 Hz up to 7.5e14 Hz, , 3.3, , 1000000, , 5.1, , 4.74 eV, , 5.3, , 12 nm, , 8.1, , shorter wavelength: 122 nm, longer wavelength: 658 nm, , 11.1 1.11eí6 m, 14.1 2.4 eV, , 756, , Copyright 2000-2007 Kinetic Books Co. Chapter 36 Answers
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Chapter 37 Answers, , Chapter Assumptions, 1 eV = 1.60×10–19 J, h = 6.63×10–34 J·s, ƫ = 1.05×10–34 J·s, mec2 = 0.511 MeV, , Answers to selected problems, C.1, , Electron A, Answers vary, , C.3, , Becomes smaller, Answers vary, , 1.1, , 1.19eí27 kg·m/s, , 2.1, , 2.01eí38 m, , 2.3, , 1.67Eí27 kg, , 2.5, , 120 V, , 3.1, , (a) 1.41eí10 m, (b) 2.80e3 m/s, , 6.1, , (a) 5.0eí6 m, (b) 1.1Eí29 kg·m/s, , 6.3, , (a) 3.3eí20 m, (b) No, , Copyright 2000-2007 Kinetic Books Co. Chapter 37 Answers, , 757
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Chapter 38 Answers, Answers to selected problems, C.1, , A small fraction were deflected from the gold foil., , C.3, , (a) 12C and 14C, and 16O, 12, (b) C and 14O, 14C and 16O, (c) 12C and 14C, 14O and 16O, 14O, , C.5, , Electric (Coulomb) attraction, , C.7, , No, Answers vary, , C.9, , Nuclear fusion, , 3.1, , 19 electrons, , 3.3, , (a) 40 is the mass number, (b) 20 protons, (c) 20 neutrons, (d) 18 electrons, , 6.1, , 64, , 7.1, , 1.8e2 m, , 8.1, , 206, , 9.1, , 4.54eí12 J, , 13.1 2.96eí11 J, 14.1 (a) 4 hydrogen atoms, (b) 4.62eí29 kg, 15.1 (a) Decrease, (b) 1.3eí29 kg, 15.3 One alpha and two negative beta particle(s), 15.5 Leadí206, 17.1 6.25 percent, 18.1 5730 years, , 758, , Copyright 2000-2007 Kinetic Books Co. Chapter 38 Answers
