Calculus-Based Physics I by Jeffrey W. Schnick

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1 Chapter Matheatical Prelude Calculus-ased Physics I by Jeffrey W. Schnick cbphysicsia8.doc Copyright , Jeffrey W. Schnick, Creatie Coons Attribution Share-Alike License 3.0. You can copy, odify, and rerelease this work under the sae license proided you gie attribution to the author. See Table of Contents This book is dedicated to Marie, Sara, and Natalie. Matheatical Prelude... Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy Conseration of Mechanical Energy II: Springs, Rotational Kinetic Energy Conseration of Moentu Conseration of Angular Moentu One-Diensional Motion (Motion Along a Line): Definitions and Matheatics One-Diensional Motion: The Constant Acceleration Equations One-Diensional Motion: Collision Type II One-Diensional Motion Graphs Constant Acceleration Probles in Two Diensions...5 Relatie Velocity...6 Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law Freefall, a.k.a. Projectile Motion Newton s Laws #: Using Free ody Diagras Newton s Laws #: Kinds of Forces, Creating Free ody Diagras Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings The Uniersal Law of Graitation Circular Motion: Centripetal Acceleration... 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration Torque & Circular Motion...4 Vectors: The Cross Product & Torque...3 Center of Mass, Moent of Inertia Statics Work and Energy Potential Energy, Conseration of Energy, Power Ipulse and Moentu Oscillations: Introduction, Mass on a Spring Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion Waes: Characteristics, Types, Energy Wae Function, Interference, Standing Waes Strings, Air Coluns eats, The Doppler Effect Fluids: Pressure, Density, Archiedes Principle Pascal s Principle, the Continuity Equation, and ernoulli s Principle Teperature, Internal Energy, Heat, and Specific Heat Capacity Heat: Phase Changes The First Law of Therodynaics...66

2 Chapter Matheatical Prelude Matheatical Prelude Just below the title of each chapter is a tip on what I perceie to be the ost coon istake ade by students in applying aterial fro the chapter. I include these tips so that you can aoid aking the istakes. Here s the first one: The reciprocal of + is not + y. Try it in the case of soe siple nubers. y Suppose = and y=4. Then + y = + 4 = = 3 4 and the reciprocal of 4 3 is 3 4 which is clearly not 6 (which is what you obtain if you take the reciprocal of + to be +4). So what is the reciprocal of +? The 4 y reciprocal of + is. y + y This book is a physics book, not a atheatics book. One of your goals in taking a physics course is to becoe ore proficient at soling physics probles, both conceptual probles inoling little to no ath, and probles inoling soe atheatics. In a typical physics proble you are gien a description about soething that is taking place in the unierse and you are supposed to figure out and write soething ery specific about what happens as a result of what is taking place. More iportantly, you are supposed to counicate clearly, copletely, and effectiely, how, based on the description and basic principles of physics, you arried at your conclusion. To sole a typical physics proble you hae to: () for a picture based on the gien description, quite often a oing picture, in your ind, () concoct an appropriate atheatical proble based on the picture, (3) sole the atheatical proble, and (4) interpret the solution of the atheatical proble. The physics occurs in steps,, and 4. The atheatics occurs in step 3. It only represents about 5% of the solution to a typical physics proble. You ight well wonder why we start off a physics book with a chapter on atheatics. The thing is, the atheatics coered in this chapter is atheatics you are supposed to already know. The proble is that you ight be a little bit rusty with it. We don t want that rust to get in the way of your learning of the physics. So, we try to knock the rust off of the atheatics that you are supposed to already know, so that you can concentrate on the physics. As uch as we ephasize that this is a physics course rather than a atheatics course, there is no doubt that you will adance your atheatical knowledge if you take this course seriously. You will use atheatics as a tool, and as with any tool, the ore you use it the better you get at using it. Soe of the atheatics in this book is epected to be new to you. The atheatics that is epected to be new to you will be introduced in recitation on an as-needed basis. It is anticipated that you will learn and use soe calculus in this course before you eer see it in a atheatics course. (This book is addressed ost specifically to students who hae neer had a

3 Chapter Matheatical Prelude physics course before and hae neer had a calculus course before but are currently enrolled in a calculus course. If you hae already taken calculus, physics, or both, then you hae a wellearned adantage.) Two points of ephasis regarding the atheatical coponent of your solutions to physics probles that hae a atheatical coponent are in order: () You are required to present a clear and coplete analytical solution to each proble. This eans that you will be anipulating sybols (letters) rather than nubers. () For any physical quantity, you are required to use the sybol which is conentionally used by physicists, and/or a sybol chosen to add clarity to your solution. In other words, it is not okay to use the sybol to represent eery unknown. Aside fro the calculus, here are soe of the kinds of atheatical probles you hae to be able to sole: Probles Inoling Percent Change A cart is traeling along a track. As it passes through a photogate its speed is easured to be 3.40 /s. Later, at a second photogate, the speed of the cart is easured to be 3.5 /s. Find the percent change in the speed of the cart. The percent change in anything is the change diided by the original, all ties 00%. (I e ephasized the word original because the ost coon istake in these kinds of probles is diiding the change by the wrong thing.) The change in a quantity is the new alue inus the original alue. (The ost coon istake here is reersing the order. If you forget which way it goes, think of a siple proble for which you know the answer and see how you ust arrange the new and original alues to ake it coe out right. For instance, suppose you gained kg oer the suer. You know that the change in your ass is + kg. You can calculate the difference both ways we re talking trial and error with at ost two trials. You ll quickly find out that it is the new alue inus the original alue a.k.a. final inus initial that yields the correct alue for the change.) Okay, now let s sole the gien proble change % Change = 00 % (-) original Recalling that the change is the new alue inus the original alue we hae A photogate is a deice that produces a bea of light, senses whether the bea is blocked, and typically sends a signal to a coputer indicating whether the bea is blocked or not. When a cart passes through a photogate, it teporarily blocks the bea. The coputer can easure the aount of tie that the bea is blocked and use that and the known length of the cart to deterine the speed of the cart as it passes through the photogate. 3

4 Chapter Matheatical Prelude new original % Change = 00% (-) original While it s certainly okay to eorize this by accident because of failiarity with it, you should concentrate on being able to derie it using coon sense (rather than working at eorizing it). Substituting the gien alues for the case at hand we obtain % Change = s s s 00% % Change = 3. 5% Probles Inoling Right Triangles Eaple -: The length of the shorter side of a right triangle is and the length of the hypotenuse is r. Find the length of the longer side and find both of the angles, aside fro the right angle, in the triangle. Solution: Draw the triangle such that it is obious which side is the shorter side ϕ y r θ Subtract Pythagorean Theore r = + y fro both sides of the equation r = Swap sides y = r Take the square root of both sides of the equation y = r y y definition, the sine of θ is the side opposite θ diided by the hypotenuse Take the arcsine of both sides of the equation in order to get θ by itself y definition, the cosine of ϕ is the side adjacent to ϕ diided by the hypotenuse Take the arccosine of both sides of the equation in order to get ϕ by itself sin θ = r θ = sin cos ϕ = r ϕ = cos r r 4

5 Chapter Matheatical Prelude To sole a proble like the one aboe, you need to eorize the relations between the sides and the angles of a right triangle. A conenient neonic for doing so is SOHCAHTOA pronounced as a single word. Referring to the diagra aboe right: Hypotenuse θ Adjacent Opposite SOH reinds us that: Opposite sin θ = (-3) Hypotenuse CAH reinds us that: Adjacent cos θ = (-4) Hypotenuse TOA reinds us that: Opposite tan θ = (-5) Adjacent Points to reeber:. The angle θ is neer the 90 degree angle.. The words opposite and adjacent designate sides relatie to the angle. For instance, the cosine of θ is the length of the side adjacent to θ diided by the length of the hypotenuse. You also need to know about the arcsine and the arccosine functions to sole the eaple proble aboe. The arcsine function is the inerse of the sine function. The answer to the question, What is the arcsine of 0.44? is, that angle whose sine is There is an arcsine button on your calculator. It is typically labeled sin -, to be read, arcsine. To use it you probably hae to hit the inerse button or the second function button on your calculator first. The inerse function of a function undoes what the function does. Thus: sin sinθ = θ (-6) Furtherore, the sine function is the inerse function to the arcsine function and the cosine function is the inerse function to the arccosine function. For the forer, this eans that: ( sin ) = sin (-7) A neonic is soething easy to reeber that helps you reeber soething that is harder to reeber. 5

6 Chapter Matheatical Prelude Probles Inoling the Quadratic Forula First coes the quadratic equation, then coes the quadratic forula. The quadratic forula is the solution to the quadratic equation: a + b + c = 0 (-8) in which: is the ariable whose alue is sought, and a, b, and c are constants The goal is to find the alue of that akes the left side 0. That alue is gien by the quadratic forula: to be read/said: b ± b 4ac = (-9) a equals inus b, plus-or-inus the square root of b squared inus four a c, all oer two a. So, how do you know when you hae to use the quadratic forula? There is a good chance that you need it when the square of the ariable for which you are soling, appears in the equation you are soling. When that is the case, carry out the algebraic steps needed to arrange the ters as they are arranged in equation -8 aboe. If this is ipossible, then the quadratic forula is not to be used. Note that in the quadratic equation you hae a ter with the ariable to the second power, a ter with the ariable to the first power, and a ter with the ariable to the zeroth power (the constant ter). If additional powers also appear, such as the one-half power (the square root), or the third power, then the quadratic forula does not apply. If the equation includes additional ters in which the ariable whose alue is sought appears as the arguent of a special function such as the sine function or the eponential function, then the quadratic forula does not apply. Now suppose that there is a square ter and you can get the equation that you are soling in the for of equation -8 aboe but that either b or c is zero. In such a case, you can use the quadratic forula, but it is oerkill. If b in equation -8 aboe is zero then the equation reduces to a + b = 0 The easy way to sole this proble is to recognize that there is at least one in each ter, and to factor the out. This yields: ( a + b) = 0 Then you hae to realize that a product of two ultiplicands is equal to zero if either ultiplicand is equal to zero. Thus, setting either ultiplicand equal to zero and soling for yields a solution. We hae two ultiplicands inoling, so, there are two solutions to the 6

7 Chapter Matheatical Prelude equation. The second ultiplicand in the epression ( a + b) = 0 is itself, so = 0 is a solution to the equation. Setting the first ter equal to zero gies: a + b a = 0 = b = Now suppose the b in the quadratic equation a + b + c = 0, equation -8, is zero. In that case, the quadratic equation reduces to: a b a + c = 0 which can easily be soled without the quadratic forula as follows: a = c = c a = ± c a where we hae ephasized the fact that there are two square roots to eery alue by placing a plus-or-inus sign in front of the radical. Now, if upon arranging the gien equation in the for of the quadratic equation (equation -8): a + b + c = 0 you find that a, b, and c are all non-zero, then you should use the quadratic forula. Here we present an eaple of a proble whose solution inoles the quadratic forula: 7

8 Chapter Matheatical Prelude Eaple -: Quadratic Forula Eaple Proble Gien find = (-0) + At first glance, this one doesn t look like a quadratic equation, but as we begin isolating, as we always strie to do in soling for, (hey, once we hae all by itself on the left side of the equation, with no on the right side of the equation, we hae indeed soled for that s what it eans to sole for ) we quickly find that it is a quadratic equation. Wheneer we hae the unknown in the denoinator of a fraction, the first step in isolating that unknown is to ultiply both sides of the equation by the denoinator. In the case at hand, this yields ( + ) (3 + ) = 4 Multiplying through on the left we find = 4 At this point it is pretty clear that we are dealing with a quadratic equation so our goal becoes getting it into the standard for of the quadratic equation, the for of equation -8, naely: a + b + c = 0. Cobining the ters inoling on the left and rearranging we obtain = 4 Subtracting 4 fro both sides yields + 4 = 0 which is indeed in the standard quadratic equation for. Now we just hae to use inspection to identify which alues in our gien equation are the a, b, and c that appear in the standard quadratic equation (equation -8) a + b + c = 0. Although it is not written, the constant ultiplying the, in the case at hand, is just. So we hae a =, b = 4, and c =. Substituting these alues into the quadratic forula (equation -9): yields which results in b ± b 4ac = a 4 ± = 4 4()( ) () = 3, = 7 8

9 Chapter Matheatical Prelude as the solutions to the proble. As a quick check we substitute each of these alues back into the original equation, equation -0: = + and find that each substitution leads to an identity. (An identity is an equation whose alidity is triially obious, such as 6 = 6.) This chapter does not coer all the non-calculus atheatics you will encounter in this course. I e kept the chapter short so that you will hae tie to read it all. If you aster the concepts in this chapter (or re-aster the if you already astered the in high school) you will be on your way to astering all the non-calculus atheatics you need for this course. Regarding reading it all: y the tie you coplete your physics course, you are supposed to hae read this book fro coer to coer. Reading physics aterial that is new to you is supposed to be slow going. y the word reading in this contet, we really ean reading with understanding. Reading a physics tet inoles not only reading but taking the tie to ake sense of diagras, taking the tie to ake sense of atheatical deelopents, and taking the tie to ake sense of the words theseles. It inoles rereading. The ethod I use is to push y way through a chapter once, all the way through at a noel-reading pace, picking up as uch as I can on the way but not allowing yself to slow down. Then, I really read it. On the second tie through I pause and ponder, study diagras, and ponder oer phrases, looking up words in the dictionary and working through eaples with pencil and paper as I go. I try not to go on to the net paragraph until I really understand what is being said in the paragraph at hand. That first read, while of little alue all by itself, is of great benefit in answering the question, Where is the author going with this?, while I a carrying out the second read. 9

10 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy Physics professors often assign conseration of energy probles that, in ters of atheatical copleity, are ery easy, to ake sure that students can deonstrate that they know what is going on and can reason through the proble in a correct anner, without haing to spend uch tie on the atheatics. A good before-and-after-picture correctly depicting the configuration and state of otion at each of two well-chosen instants in tie is crucial in showing the appropriate understanding. A presentation of the reainder of the conceptualplus-atheatical solution of the proble starting with a stateent in equation for that the energy in the before picture is equal to the energy in the after picture, continuing through to an analytical solution and, if nuerical alues are proided, only after the analytical solution has been arried at, substituting alues with units, ealuating, and recording the result is alost as iportant as the picture. The proble is that, at this stage of the course, students often think that it is the final answer that atters rather than the counication of the reasoning that leads to the answer. Furtherore, the chosen probles are often so easy that students can arrie at the correct final answer without fully understanding or counicating the reasoning that leads to it. Students are unpleasantly surprised to find that correct final answers earn little to no credit in the absence of a good correct before-andafter picture and a well-written reainder of the solution that starts fro first principles, is consistent with the before and after picture, and leads logically, with no steps oitted, to the correct answer. Note that students who focus on correctly counicating the entire solution, on their own, on eery hoework proble they do, stand a uch better chance of successfully doing so on a test than those that just try to get the right nuerical answer on hoework probles. Mechanical Energy Energy is a transferable physical quantity that an object can be said to hae. If one transfers energy to a aterial particle that is initially at rest, the speed of that particle changes to a alue which is an indicator of how uch energy was transferred. Energy has units of joules, abbreiated J. Energy can t be easured directly but when energy is transferred to or fro an object, soe easurable characteristic (or characteristics) of that object changes (change) such that, easured alues of that characteristic or those characteristics (in cobination with one or ore characteristics such as ass that do not change by any easurable aount) can be used to deterine how uch energy was transferred. Energy is often categorized according to which easurable characteristic changes when energy is transferred. In other words, we categorize energy in accord with the way it reeals itself to us. For instance, when the easurable characteristic is teperature, we call the energy theral energy; when the easurable quantity is speed, we call the energy kinetic energy. While it can be argued that there is only one for or kind of energy, in the jargon of physics we call the energy that reeals itself one way one kind or for of energy (such as theral energy) and the energy that reeals itself another way another kind or for of energy (such as kinetic energy). In physical processes it often occurs that the 0

11 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy way in which energy is reealing itself changes. When that happens we say that energy is transfored fro one kind of energy to another. Kinetic Energy is energy of otion. An object at rest has no otion; hence, it has no kinetic energy. The kinetic energy K of a non-rotating rigid object in otion depends on the ass and speed of the object as follows : K = (-) The ass of an object is a easure of the object s inertia, the object s inherent tendency to aintain a constant elocity. The inertia of an object is what akes it hard to get that object oing. The words ass and inertia both ean the sae thing. Physicists typically use the word inertia when talking about the property in general conceptual ters, and the word ass when they are assigning a alue to it, or using it in an equation. Mass has units of kilogras, abbreiated kg. The speed has units of eters per second, abbreiated /s. Check out the units in equation -: K = On the left we hae the kinetic energy which has units of joules. On the right we hae the product of a ass and the square of a elocity. Thus the units on the right are kg and we s can deduce that a joule is a kg. s Potential Energy is energy that depends on the arrangeent of atter. Here, we consider one type of potential energy: The Graitational Potential Energy of an object near the surface of the earth is the energy (relatie to the graitational potential energy that the object has when it is at the reference leel about to e entioned) that the object has because it is "up high" aboe a reference leel such as the ground, the floor, or a table top. In characterizing the relatie graitational potential energy of an object it is iportant to specify what you are using for a reference leel. In using the concept of near-earth graitational potential energy to sole a physics proble, although you are free to choose whateer you want to as a reference leel, it is iportant to stick with one and the sae reference leel throughout the proble. The relatie graitational potential energy U g of In classical physics we deal with speeds uch saller than the speed of light c = /s. The classical physics epression K = is an approiation (a fantastic approiation at speeds uch saller than the speed of light the saller the better) to the relatiistic epression K = ( / / c ) c which is alid for all speeds. We call the potential energy discussed here the graitational potential energy of the object. Actually, it is the graitational potential energy of the object-plus-earth syste taken as a whole. It would be ore accurate to ascribe the potential energy to the graitational field of the object and the graitational field of the earth. In lifting an object, it is as if you are stretching a weird inisible spring weird in that it doesn t pull harder the ore you stretch it as an ordinary spring does and the energy is being stored in that inisible spring. For energy accounting purposes howeer, it is easier to ascribe the graitational potential energy of an object near the surface of the earth, to the object, and that is what we do in this book. This is siilar to calling the graitational force eerted on an object by the earth s graitational field the weight of the object as if it were a property of the object, rather than what it really is, an eternal influence acting on the object.

12 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy an object near the surface of the earth depends on the object's height y aboe the chosen reference leel, the object's ass, and the agnitude g of the earth s graitational field, which to a good approiation has the sae alue g = 9.80 N eerywhere near the surface of the kg earth, as follows: U g = gy (-) N The N in g = 9.80 stands for newtons, the unit of force. (Force is an ongoing push or pull.) kg Since it is an energy, the units of U g are joules, and the units on the right side of equation -, with the height y being in eters, work out to be newtons ties eters. Thus a joule ust be a newton eter, and indeed it is. Just aboe we showed that a joule is a kg s a newton eter then a newton ust be a kg. s A Special Case of the Conseration of Mechanical Energy. If a joule is also Energy is ery useful for aking predictions about physical processes because it is neer created or destroyed. To borrow epressions fro econoics, that eans we can use siple bookkeeping or accounting to ake predictions about physical processes. For instance, suppose we create, for purposes of aking such a prediction, an iaginary boundary enclosing part of the unierse. Then any change in the total aount of energy inside the boundary will correspond eactly to energy transfer through the boundary. If the total energy inside the boundary increases by E, then eactly that sae aount of energy E ust hae been transferred through the boundary into the region enclosed by the boundary fro outside that region. And if the total energy inside the boundary decreases by E, then eactly that aount of energy E ust hae been transferred through the boundary out of the region enclosed by the boundary fro inside that region. Oddly enough, in keeping book on the energy in such an enclosed part of the unierse, we rarely if eer know or care what the oerall total aount of energy is. It is sufficient to keep track of changes. What can ake the accounting difficult is that there are so any different ways in which energy can anifest itself (what we call the different fors of energy), and there is no siple energy eter that tells us how uch energy there is in our enclosed region. Still, there are processes for which the energy accounting is relatiely siple. For instance, it is relatiely siple when there is no (or negligible) transfer of energy into or out of the part of the unierse that is of interest to us, and when there are few fors of energy for which the aount of energy changes. The two kinds of energy discussed aboe (the kinetic energy of a rigid non-rotating object and graitational potential energy) are both eaples of echanical energy, to be contrasted with, for eaple, theral energy. Under certain conditions the total echanical energy of a syste of objects does not change een though the configuration of the objects does. This represents a special case of the ore general principle of the conseration energy. The conditions under which the total echanical energy of a syste doesn t change are:

13 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy () No energy is transferred to or fro the surroundings. () No energy is conerted to or fro other fors of energy (such as theral energy). Consider a couple of processes in which the total echanical energy of a syste does not reain the sae: Case # A rock is dropped fro shoulder height. It hits the ground and coes to a coplete stop. The "syste of objects" in this case is just the rock. As the rock falls, the graitational potential energy is continually decreasing. As such, the kinetic energy of the rock ust be continually increasing in order for the total energy to be staying the sae. On the collision with the ground, soe of the kinetic energy gained by the rock as it falls through space is transferred to the ground and the rest is conerted to theral energy and the energy associated with sound. Neither condition (no transfer and no transforation of energy) required for the total echanical energy of the syste to reain the sae is et; hence, it would be incorrect to write an equation setting the initial echanical energy of the rock (upon release) equal to the final echanical energy of the rock (after landing). Can the idea of an unchanging total aount of echanical energy be used in the case of a falling object? The answer is yes. The difficulties associated with the preious process occurred upon collision with the ground. You can use the idea of an unchanging total aount of echanical energy to say soething about the rock if you end your consideration of the rock before it hits the ground. For instance, gien the height fro which it is dropped, you can use the idea of an unchanging total aount of echanical energy to deterine the speed of the rock at the last instant before it strikes the ground. The "last instant before" it hits the ground corresponds to the situation in which the rock has not yet touched the ground but will touch the ground in an aount of tie that is too sall to easure and hence can be neglected. It is so close to the ground that the distance between it and the ground is too sall to easure and hence can be neglected. It is so close to the ground that the additional speed that it would pick up in continuing to fall to the ground is too sall to be easured and hence can be neglected. The total aount of echanical energy does not change during this process. It would be correct to write an equation setting the initial echanical energy of the rock (upon release) equal to the final echanical energy of the rock (at the last instant before collision). Case # A block, in contact with nothing but a sidewalk, slides across the sidewalk. The total aount of echanical energy does not reain the sae because there is friction between the block and the sidewalk. In any case inoling friction, echanical energy is conerted into theral energy; hence, the total aount of echanical energy after the sliding, is not equal to the total aount of echanical energy prior to the sliding. 3

14 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy Applying the Principle of the Conseration of Energy for the Special Case in which the Mechanical Energy of a Syste does not Change In applying the principle of conseration of echanical energy for the special case in which the echanical energy of a syste does not change, you write an equation which sets the total echanical energy of an object or syste objects at one instant in tie equal to the total echanical energy at another instant in tie. Success hangs on the appropriate choice of the two instants. The principal applies to all pairs of instants of the tie interal during which energy is neither transferred into or out of the syste nor transfored into non-echanical fors. You characterize the conditions at the first instant by eans of a "efore Picture" and the conditions at the second instant by eans of an "After Picture. In applying the principle of conseration of echanical energy for the special case in which the echanical energy of a syste does not change, you write an equation which sets the total echanical energy in the efore Picture equal to the total echanical energy in the After Picture. (In both cases, the total echanical energy in question is the aount the syste has relatie to the echanical energy it would hae if all objects were at rest at the reference leel.) To do so effectiely, it is necessary to sketch a efore Picture and a separate After Picture. After doing so, the first line in one's solution to a proble inoling an unchanging total of echanical energy always reads Energy efore = Energy After (-3) We can write this first line ore sybolically in seeral different anners: E = or E i = E f or E = E (-4) E The first two ersions use subscripts to distinguish between "before picture" and "after picture" energies and are to be read "E-sub-one equals E-sub-two" and "E-sub-i equals E-sub-f." In the latter case the sybols i and f stand for initial and final. In the final ersion, the prie sybol is added to the E to distinguish "after picture" energy fro "before picture" energy. The last equation is to be read "E equals E-prie." (The prie sybol is soeties used in atheatics to distinguish one ariable fro another and it is soeties used in atheatics to signify the deriatie with respect to. It is neer used it to signify the deriatie in this book.) The unpried/prie notation is the notation that will be used in the following eaple: 4

15 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy Eaple -: A rock is dropped fro a height of.6 eters. How fast is the rock falling just before it hits the ground? Solution: Choose the "before picture" to correspond to the instant at which the rock is released, since the conditions at this instant are specified ("dropped" indicates that the rock was released fro rest its speed is initially zero, the initial height of the rock is gien). Choose the "after picture" to correspond to the last instant before the rock akes contact with the ground since the question pertains to a condition (speed) at this instant. Rock of ass EFORE = 0 y =.6 AFTER Reference Leel =? Note that we hae oitted the subscript g (for graitational ) fro both U and U. When you are dealing with only one kind of potential energy, you don t need to use a subscript to distinguish it fro other kinds. E = 0 (since at rest) K + U = gy = = = = = K gy gy (9. 80 /s ) E s + 0 (since at ground leel) U kg Note that the unit, newton, abbreiated as N, is. Hence, the agnitude of the earth s s N near-surface graitational field g = 9.80 can also be epressed as g = 9.80 as we hae kg s done in the eaple for purposes of working out the units. 5

16 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy The solution presented in the eaple proides you with an eaple of what is required of students in soling physics probles. In cases where student work is ealuated, it is the solution which is ealuated, not just the final answer. In the following list, general requireents for solutions are discussed, with reference to the solution of the eaple proble:. Sketch (the before and after pictures in the eaple). Start each solution with a sketch or sketches appropriate to the proble at hand. Use the sketch to define sybols and, as appropriate, to assign alues to sybols. The sketch aids you in soling the proble and is iportant in counicating your solution to the reader. Note that each sketch depicts a configuration at a particular instant in tie rather than a process which etends oer a tie interal.. Write the "Concept Equation" ( E = E in the eaple). 3. Replace quantities in the "Concept Equation" with ore specific representations of the sae quantities. Repeat as appropriate. In the eaple gien, the sybol E representing total echanical energy in the before picture is replaced with "what it is, naely, the su of the kinetic energy and the potential energy K + U of the rock in the before picture. On the sae line E has been replaced with what it is, naely, the su of the kinetic energy and the potential energy K + U in the after picture. Quantities that are obiously zero hae slashes drawn through the and are oitted fro subsequent steps. This step is repeated in the net line ( gy = ) in which the graitational potential energy in the before picture, U, has been replaced with what it is, naely gy, and on the right, the kinetic energy in the after picture has been replaced with what it is, naely, The sybol that appears in this step is defined in the diagra.. 4. Sole the proble algebraically. The student is required to sole the proble by algebraically anipulating the sybols rather than substituting alues and siultaneously ealuating and anipulating the. The reasons that physics teachers require students taking college leel physics courses to sole the probles algebraically in ters of the sybols rather than working with the nubers are: (a) College physics teachers are epected to proide the student with eperience in "the net leel" in abstract reasoning beyond working with the nubers. To gain this eperience, the students ust sole the probles algebraically in ters of sybols. (b) Students are epected to be able to sole the ore general proble in which, whereas certain quantities are to be treated as if they are known, no actual alues are gien. Solutions to such probles are often used in coputer progras which enable the user to obtain results for any different alues of the "known quantities. Actual alues are 6

17 Chapter Conseration of Mechanical Energy I: Kinetic Energy & Graitational Potential Energy assigned to the known quantities only after the user of the progra proides the to the progra as input long after the algebraic proble is soled. (c) Many probles ore coplicated than the gien eaple can ore easily be soled algebraically in ters of the sybols. Eperience has shown that students accustoed to substituting nuerical alues for sybols at the earliest possible stage in a proble are unable to sole the ore coplicated probles. In the eaple, the algebraic solution begins with the line gy =. The 's appearing on both sides of the equation hae been canceled out (this is the algebraic step) in the solution proided. Note that in the eaple, had the 's not canceled out, a nuerical answer to the proble could not hae been deterined since no alue for was gien. The net two lines represent the additional steps necessary in soling algebraically for the final speed. The final line in the algebraic solution ( = gy in the eaple) always has the quantity being soled for all by itself on the left side of the equation being set equal to an epression inoling only known quantities on the right side of the equation. The algebraic solution is not coplete if unknown quantities (especially the quantity sought) appear in the epression on the right hand side. Writing the final line of the algebraic solution in the reerse order, e.g. gy =, is unconentional and hence unacceptable. If your algebraic solution naturally leads to that, you should write one ore line with the algebraic answer written in the correct order. 5) Replace sybols with nuerical alues with units, = (9. 80 ). 6 in the eaple; the s units are the units of easureent: and in the eaple). s No coputations should be carried out at this stage. Just copy down the algebraic solution but with sybols representing known quantities replaced with nuerical alues with units. Use parentheses and brackets as necessary for clarity. 6) Write the final answer with units ( = 5.6 in the eaple). s Nuerical ealuations are to be carried out directly on the calculator and/or on scratch paper. It is unacceptable to clutter the solution with arithetic and interediate nuerical answers between the preious step and this step. Units should be worked out and proided with the final answer. It is good to show soe steps in working out the units but for siple cases units (not algebraic solutions) ay be worked out in your head. In the eaple proided, it is easy to see that upon taking the square root of the product of and, one obtains hence no additional s s steps were depicted. 7

18 Chapter 3 Conseration of Mechanical Energy II: Springs, Rotational Kinetic Energy 3 Conseration of Mechanical Energy II: Springs, Rotational Kinetic Energy A coon istake inoling springs is using the length of a stretched spring when the aount of stretch is called for. Gien the length of a stretched spring, you hae to subtract off the length of that sae spring when it is neither stretched nor copressed to get the aount of stretch. Spring Potential Energy is the potential energy stored in a spring that is copressed or stretched. The spring energy depends on how stiff the spring is and how uch it is stretched or copressed. The stiffness of the spring is characterized by the force constant of the spring, k. k is also referred to as the spring constant for the spring. The stiffer the spring, the bigger its alue of k is. The sybol is typically used to characterize the aount by which a spring is copressed or stretched. It is iportant to note that is not the length of the stretched or copressed spring. Instead, it is the difference between the length of the stretched or copressed spring and the length of the spring when it is neither stretched nor copressed. The aount of energy U S stored in a spring with a force constant (spring constant) k that has either been stretched by an aount or copressed by an aount is: U s = k (3-) Rotational Kinetic Energy is the energy that a spinning object has because it is spinning. When an object is spinning, eery bit of atter aking up the object is oing in a circle (ecept for those bits on the ais of rotation). Thus, eery bit of atter aking up the object has soe kinetic energy where the is the speed of the bit of atter in question and is its ass. The thing is, in the case of an object that is just spinning, the object itself is not going anywhere, so it has no speed, and the different bits of ass aking up the object hae different speeds, so there is no one speed that we can use for the speed of the object in our old epression for kinetic energy K =. The aount of kinetic energy that an object has because it is spinning can be epressed as: K = Iw (3-) where the Greek letter oega w (please don t call it double-u) is used to represent the agnitude of the angular elocity of the object and the sybol I is used to represent the oent of inertia, a.k.a. rotational inertia, of the object. The agnitude of the angular elocity of the object is how fast the object is spinning and the oent of inertia of the object is a easure of the object s natural tendency to spin at a constant rate. The greater the oent of inertia of an object, the harder it is to change how fast that object is spinning. The agnitude of the angular elocity, the spin rate w, is easured in units of radians per second where the radian is a unit of angle. An angle is a fraction of a rotation and hence a unit of angle is a fraction of a rotation. If we diide a rotation up into 360 parts then each part is of a 360 rotation and we call each part a degree. In the case of radian easure, we diide the rotation up 8

19 Chapter 3 Conseration of Mechanical Energy II: Springs, Rotational Kinetic Energy into π parts and call each part a radian. Thus a radian is of a rotation. The fact that an π angle is a fraction of a rotation eans that an angle is really a pure nuber and the word radian abbreiated rad, is a reinder about how any parts the rotation has been diided up into, rather than a true unit. In working out the units in cases inoling radians, one can siply erase the word radian. This is not the case for actual units such as eters or joules. The oent of inertia I has units of kg. The units of the right hand side of equation 3-, rad K = Iw, thus work out to be kg. Taking adantage of the fact that a radian is not a s true unit, we can siply erase the units rad leaing us with units of kg, a cobination s that we recognize as a joule which it ust be since the quantity on the left side of the equation K = Iw (equation 3-) is an energy. Energy of Rolling An object which is rolling is both oing through space and spinning so it has both kinds of kinetic energy, the and the I w. The oeent of an object through space is called translation. To contrast it with rotational kinetic energy, the ordinary kinetic energy K = is referred to as translational kinetic energy. So, the total kinetic energy of an object that is rolling can be epressed as Rolling Translation Rotation K = K + K (3-3) K Rolling = + Iw (3-4) Now you probably recognize that an object that is rolling without slipping is spinning at a rate that depends on how fast it is going forward. That is to say that the alue of w depends on the alue of. Let s see how. When an object that is rolling without slipping copletes one rotation, it oes a distance equal to its circuference which is π ties the radius of that part of the object on which the object is rolling. Distance traeledin one rotation = πr (3-5) Now if we diide both sides of this equation by the aount of tie that it takes for the object to coplete one rotation we obtain on the left, the speed of the object and, on the right, we can interpret the π as π radians and, since π radians is one rotation the π radians diided by the tie it takes for the object to coplete one rotation is just the agnitude of the angular elocity w. Hence we arrie at = wr which is typically written: = rw (3-6) 9

20 Chapter 4 Conseration of Moentu 4 Conseration of Moentu A coon istake inoling conseration of oentu crops up in the case of totally inelastic collisions of two objects, the kind of collision in which the two colliding objects stick together and oe off as one. The istake is to use conseration of echanical energy rather than conseration of oentu. One way to recognize that soe echanical energy is conerted to other fors is to iagine a spring to be in between the two colliding objects such that the objects copress the spring. Then iagine that, just when the spring is at aiu copression, the two objects becoe latched together. The two objects oe off together as one as in the case of a typical totally inelastic collision. After the collision, there is energy stored in the copressed spring so it is clear that the total kinetic energy of the latched pair is less than the total kinetic energy of the pair prior to the collision. There is no spring in a typical inelastic collision. The echanical energy that would be stored in the spring, if there was one, results in peranent deforation and a teperature increase of the objects inoled in the collision. The oentu of an object is a easure of how hard it is to stop that object. The oentu of an object depends on both its ass and its elocity. Consider two objects of the sae ass, e.g. two baseballs. One of the is coing at you at 0 ph, and the other at 00 ph. Which one has the greater oentu? Answer: The faster baseball is, of course, harder to stop, so it has the greater oentu. Now consider two objects of different ass with the sae elocity, e.g. a Ping-Pong ball and a cannon ball, both coing at you at 5 ph. Which one has the greater oentu? The cannon ball is, of course, harder to stop, so it has the greater oentu. The oentu p of an object is equal to the product of the object s ass and elocity : p = (4-) Moentu has direction. Its direction is the sae as that of the elocity. In this chapter we will liit ourseles to otion along a line (otion in one diension). Then there are only two directions, forward and backward. An object oing forward has a positie elocity/oentu and one oing backward has a negatie elocity/oentu. In soling physics probles, the decision as to which way is forward is typically left to the proble soler. Once the proble soler decides which direction is the positie direction, she ust state what her choice is (this stateent, often ade by eans of notation in a sketch, is an iportant part of the solution), and stick with it throughout the proble. The concept of oentu is iportant in physics because the total oentu of any syste reains constant unless there is a net transfer of oentu to that syste, and if there is an ongoing oentu transfer, the rate of change of the oentu of the syste is equal to the rate at which oentu is being transferred into the syste. As in the case of energy, this eans that one can ake predictions regarding the outcoe of physical processes by eans of This classical physics epression is alid for speeds sall copared to the speed of light c = /s. The relatiistic epression for oentu is p = / c. At speeds that are ery sall copared to the speed of light, the classical physics epression p = is a fantastic approiation to the relatiistic epression. 0

21 Chapter 4 Conseration of Moentu siple accounting (bookkeeping) procedures. The case of oentu is coplicated by the fact that oentu has direction, but in this initial encounter with the conseration of oentu you will deal with cases inoling otion along a straight line. When all the otion is along one and the sae line, there are only two possible directions for the oentu and we can use algebraic signs (plus and inus) to distinguish between the two. The principle of Conseration of Moentu applies in general. At this stage in the course howeer, we will consider only the special case in which there is no net transfer of oentu to (or fro) the syste fro outside the syste. Conseration of Moentu in One Diension for the Special Case in which there is No Transfer of Moentu to or fro the Syste fro Outside the Syste In any process inoling a syste of objects which all oe along one and the sae line, as long as none of the objects are pushed or pulled along the line by anything outside the syste of objects (it s okay if they push and pull on each other), the total oentu before, during, and after the process reains the sae. The total oentu of a syste of objects is just the algebraic su of the oenta of the indiidual objects. That adjectie "algebraic" eans you hae to pay careful attention to the plus and inus signs. If you define "to the right" as your positie direction and your syste of objects consists of two objects, one oing to the right with a oentu of kg /s and the other oing to the left with oentu 5 kg /s, then the total oentu is (+ kg /s) + ( 5 kg /s) which is +7 kg /s. The plus sign in the final answer eans that the total oentu is directed to the right. Upon reading this selection you'll be epected to be able to apply conseration of oentu to two different kinds of processes. In each of these two classes of processes, the syste of objects will consist of only two objects. In one class, called collisions, the two objects bup into each other. In the other class, anti-collisions the two objects start out together, and spring apart. Soe further breakdown of the collisions class is pertinent before we get into eaples. The two etree types of collisions are the copletely inelastic collision, and the copletely elastic collision. Upon a copletely inelastic collision, the two objects stick together and oe off as one. This is the easy case since there is only one final elocity (because they are stuck together, the two objects obiously oe off at one and the sae elocity). Soe echanical energy is conerted to other fors in the case of a copletely inelastic collision. It would be a big istake to apply the principle of conseration of echanical energy to a copletely inelastic collision. Mechanical energy is not consered. The words "copletely inelastic" tell you that both objects hae the sae elocity (as each other) after the collision. In a copletely elastic collision (often referred to siply as an elastic collision), the objects bounce off each other in such a anner that no echanical energy is conerted into other fors in the collision. Since the two objects oe off independently after the collision there are two final elocities. If the asses and the initial elocities are gien, conseration of oentu yields one equation with two unknowns naely, the two final elocities. Such an equation

22 Chapter 4 Conseration of Moentu cannot be soled by itself. In such a case, one ust apply the principle of conseration of echanical energy. It does apply here. The epression "copletely elastic" tells you that conseration of echanical energy does apply. In applying conseration of oentu one first sketches a before and an after picture in which one defines sybols by labeling objects and arrows (indicating elocity), and defines which direction is chosen as the positie direction. The first line in the solution is always a stateent that the total oentu in the before picture is the sae as the total oentu in the after picture. This is typically written by eans an equation of the for: = p p (4-) The Σ in this epression is the upper case Greek letter siga and is to be read the su of. Hence the equation reads: The su of the oenta to the right in the before picture is equal to the su of the oenta to the right in the after picture. In doing the su, a leftward oentu counts as a negatie rightward oentu. The arrow subscript is being used to define the positie direction. Eaples Now let's get down to soe eaples. We'll use the eaples to clarify what is eant by collisions and anti-collisions; to introduce one ore concept, naely, relatie elocity (soeties referred to as uzzle elocity); and of course, to show the reader how to apply conseration of oentu.

23 Chapter 4 Conseration of Moentu Eaple 4- Two objects oe on a horizontal frictionless surface along the sae line in the sae direction which we shall refer to as the forward direction. The trailing object of ass.0 kg has a elocity of 5 /s forward. The leading object of ass 3. kg has a elocity of /s forward. The trailing object catches up with the leading object and the two objects eperience a copletely inelastic collision. What is the final elocity of each of the two objects? EFORE AFTER = 5 s = s =.0 kg = 3. kg Σp p + p = Σp = p + = ( +. 0 kg (5 / s) + 3. kg ( / s) =. 0kg + 3. kg =. 54 = 3 s s ) + = + The final elocity of each of the objects is 3 forward. s 3

24 Chapter 4 Conseration of Moentu 4 Eaple 4-: A cannon of ass C, resting on a frictionless surface, fires a ball of ass. The ball is fired horizontally. The uzzle elocity is M. Find the elocity of the ball and the recoil elocity of the cannon. NOTE: This is an eaple of an anti-collision proble. It also inoles the concept of relatie elocity. The uzzle elocity is the relatie elocity between the ball and the cannon. It is the elocity at which the two separate. If the elocity of the ball relatie to the ground is to the right, and the elocity of the cannon relatie to the ground is C to the left, then the elocity of the ball relatie to the cannon, also known as the uzzle elocity of the ball, is C M + =. In cases not inoling guns or cannons one typically uses the notation rel for "relatie elocity" or, relating to the eaple at hand, C for "elocity of the ball relatie to the cannon." C C 0 + = = Σp Σp M C C M C C M C C C M C M C ) ( 0 ) ( 0 + = = + = = + = Now substitute this result into equation () aboe. This yields: EFORE AFTER C Also, fro the definition of uzzle elocity: M C C M = + = Substituting this result into equation () yields: () () M C C C M C M M C C C M C M C C M C C M C ) ( + = + + = + + = + = C

25 Chapter 5 Conseration of Angular Moentu 5 Conseration of Angular Moentu Much as in the case of linear oentu, the istake that tends to be ade in the case of angular oentu is not using the principle of conseration of angular oentu when it should be used, that is, applying conseration of echanical energy in a case in which echanical energy is not consered but angular oentu is. Consider the case, for instance, in which one drops a disk (fro a negligible height) that is not spinning, onto a disk that is spinning, and after the drop, the two disks spin together as one. The together as one part tips you off that this is a copletely inelastic (rotational) collision. Soe echanical energy is conerted into theral energy (and other fors not accounted for) in the collision. It s easy to see that echanical energy is conerted into theral energy if the two disks are CD s and the botto one is initially spinning quite fast (but is not being drien). When you drop the top one onto the botto one, there will be quite a bit of slipping before the top disk gets up to speed and the two disks spin as one. During the slipping, it is friction that increases the spin rate of the top CD and slows the botto one. Friction conerts echanical energy into theral energy. Hence, the echanical energy prior to the drop is less than the echanical energy after the drop. The angular oentu of an object is a easure of how difficult it is to stop that object fro spinning. For an object rotating about a fied ais, the angular oentu depends on how fast the object is spinning, and on the object's rotational inertia (also known as oent of inertia) with respect to that ais. Rotational Inertia (a.k.a. Moent of Inertia) The rotational inertia of an object with respect to a gien rotation ais is a easure of the object's tendency to resist a change in its angular elocity about that ais. The rotational inertia depends on the ass of the object and how that ass is distributed. You hae probably noticed that it is easier to start a erry-go-round spinning when it has no children on it. When the kids clib on, the ass of what you are trying to spin is greater, and this eans the rotational inertia of the object you are trying to spin is greater. Hae you also noticed that if the kids oe in toward the center of the erry-go-round it is easier to start it spinning than it is when they all sit on the outer edge of the erry-go-round? It is. The farther, on the aerage, the ass of an object is distributed away fro the ais of rotation, the greater the object's oent of inertia with respect to that ais of rotation. The rotational inertia of an object is represented by the sybol I. During this initial coerage of angular oentu, you will not be required to calculate I fro the shape and ass of the object. You will either be gien I or epected to calculate it by applying conseration of angular oentu (discussed below). 5

26 Chapter 5 Conseration of Angular Moentu Angular Velocity The angular elocity of an object is a easure of how fast it is spinning. It is represented by the Greek letter oega, written w, (not to be confused with the letter w which, unlike oega, is pointed on the botto). The ost conenient easure of angle in discussing rotational otion is the radian. Like the degree, a radian is a fraction of a reolution. ut, while one degree is 360 of a reolution, one radian is π second or, in notational for, of a reolution. The units of angular elocity are then radians per rad. Angular elocity has direction or sense of rotation s associated with it. If one defines a rotation which is clockwise when iewed fro aboe as a positie rotation, then an object which is rotating counterclockwise as iewed fro aboe is said to hae a negatie angular elocity. In any proble inoling angular elocity, one is free to choose the positie sense of rotation, but then one ust stick with that choice throughout the proble. Angular Moentu The angular oentu L of an object is gien by: L =Iw (5-) Note that this is consistent with our original definition of angular oentu as a easure of the degree of the object's tendency to keep on spinning, once it is spinning. The greater the rotational inertia of the object, the ore difficult it is to stop the object fro spinning, and the greater the angular elocity of the object, the ore difficult it is to stop the object fro spinning. The direction of angular oentu is the sae as the direction of the corresponding angular elocity. Torque We define torque by analogy with force which is an ongoing push or pull on an object. When there is a single force acting on a particle, the oentu of that particle is changing. A torque is what you are eerting on the lid of a jar when you are trying to reoe the lid. When there is a single torque acting on a rigid object, the angular oentu of that object is changing. Conseration of Angular Moentu Angular Moentu is an iportant concept because, if there is no angular oentu transferred to or fro a syste, the total angular oentu of that syste does not change, and if there is angular oentu being transferred to a syste, the rate of change of the angular oentu of the syste is equal to the rate at which angular oentu is being transferred to the syste. As in the case of energy and oentu, this eans we can use siple accounting (bookkeeping) procedures for aking predictions on the outcoes of physical processes. In this chapter we focus on the special case in which there are no eternal torques which eans that no angular oentu is transferred to or fro the syste. 6

27 Chapter 5 Conseration of Angular Moentu Conseration of Angular Moentu for the Special Case in which no Angular Moentu is Transferred to or fro the Syste fro Outside the Syste In any physical process inoling an object or a syste of objects free to rotate about an ais, as long as there are no eternal torques eerted on the syste of objects, the total angular oentu of that syste of objects reains the sae throughout the process. Eaples The application of the conseration of angular oentu in soling physics probles for cases inoling no transfer of angular oentu to or fro the syste fro outside the syste (no eternal torque) is ery siilar to the application of the conseration of energy and to the application of the conseration of oentu. One selects two instants in tie, defines the earlier one as the before instant and the later one as the after instant, and akes corresponding sketches of the object or objects in the syste. Then one writes L = L (5-) eaning "the angular oentu in the before picture equals the angular oentu in the after picture." Net, one replaces each L with what it is in ters of the oents of inertia and angular elocities in the proble and soles the resulting algebraic equation for whateer is sought. 7

28 Chapter 5 Conseration of Angular Moentu Eaple 5- A skater is spinning at 3.0 rad/s with her ars and legs etended outward. In this position her oent of inertia with respect to the ertical ais about which she is spinning is kg. She pulls her ars and legs in close to her body changing her oent of inertia to 7. 5kg. What is her new angular elocity? EFORE rad w = 3.0 s AFTER w =? I = 45.6 kg I = 7.5 kg L = L Iw = I w I w = w I kg w = 7. 5 kg 3. 0 rad / s w = rad s 8

29 Chapter 5 Conseration of Angular Moentu Eaple 5- A horizontal disk of rotational inertia 4. 5kg with respect to its ais of syetry is spinning counterclockwise about its ais of syetry, as iewed fro aboe, at 5.5 reolutions per second on a frictionless assless bearing. A second disk, of rotational inertia. 80 kg with respect to its ais of syetry, spinning clockwise as iewed fro aboe about the sae ais (which is also its ais of syetry) at 4. reolutions per second, is dropped on top of the first disk. The two disks stick together and rotate as one about their coon ais of syetry at what new angular elocity (in units of radians per second)? w I =.80 kg I = 4.5 kg w w Soe preliinary work (epressing the gien angular elocities in units of rad/s): re π rad w = 5. 5 = s re rad s re π rad w = 4. = 89. s re rad s Now we apply the principle of conseration of angular oentu for the special case in which there is no transfer of angular oentu to or fro the syste fro outside the syste. Referring to the diagra: I w I w L = L = ( I + I ) w We define counterclockwise, as iewed fro aboe, to be the + sense of rotation. I w = w Iw I + I w = (4. 5kg ) rad /s (. 80kg 4. 5kg +. 80kg ) 89. rad / s w = 4. 9 rad s (Counterclockwise as iewed fro aboe.) 9

30 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics A istake that is often ade in linear otion probles inoling acceleration, is using the elocity at the end of a tie interal as if it was alid for the entire tie interal. The istake crops up in constant acceleration probles when folks try to use the definition of aerage elocity = in the solution. Unless you are asked specifically about t aerage elocity, you will neer need to use this equation to sole a physics proble. Aoid using this equation it will only get you into trouble. For constant acceleration probles, use the set of constant acceleration equations proided you. Here we consider the otion of a particle along a straight line. The particle can speed up and slow down and it can oe forward or backward but it does not leae the line. While the discussion is about a particle (a fictitious object which at any instant in tie is at a point in space but has no etent in space no width, height, length, or diaeter) it also applies to a rigid body that oes along a straight line path without rotating, because in such a case, eery particle of the body undergoes one and the sae otion. This eans that we can pick one particle on the body and when we hae deterined the otion of that particle, we hae deterined the otion of the entire rigid body. So how do we characterize the otion of a particle? Let s start by defining soe ariables: t How uch tie t has elapsed since soe initial tie. The initial tie is often referred to as the start of obserations and een ore often assigned the alue 0. We will refer to the aount of tie t that has elapsed since tie zero as the stopwatch reading. A tie interal t (to be read delta t ) can then be referred to as the difference between two stopwatch readings. Where the object is along the straight line. To specify the position of an object on a line, one has to define a reference position (the start line) and a forward direction. Haing defined a forward direction, the backward direction is understood to be the opposite direction. It is conentional to use the sybol to represent the position of a particle. The alues that can hae, hae units of length. The SI unit of length is the eter. (SI stands for Systee International, the international syste of units.) The sybol for the eter is. The physical quantity can be positie or negatie where it is understood that a particle which is said to be inus fie eters forward of the start line (ore concisely stated as = 5 ) is actually fie eters behind the start line. 30

31 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics a How fast and which way the particle is going the elocity of the object. ecause we are considering an object that is oing only along a line, the which way part is either forward or backward. Since there are only two choices, we can use an algebraic sign ( + or ) to characterize the direction of the elocity. y conention, a positie alue of elocity is used for an object that is oing forward, and a negatie alue is used for an object that is oing backward. Velocity has both agnitude and direction. The agnitude of a physical quantity that has direction is how big that quantity is, regardless of its direction. So the agnitude of the elocity of an object is how fast that object is going, regardless of which way it is going. Consider an object that has a elocity of 5 /s. The agnitude of the elocity of that object is 5 /s. Now consider an object that has a elocity of 5 /s. (It is going backward at 5 /s.) The agnitude of its elocity is also 5 /s. Another nae for the agnitude of the elocity is the speed. In both of the cases just considered, the speed of the object is 5 /s despite the fact that in one case the elocity was 5 /s. To understand the how fast part, just iagine that the object whose otion is under study has a built-in speedoeter. The agnitude of the elocity, a.k.a. the speed of the object, is siply the speedoeter reading. Net we hae the question of how fast and which way the elocity of the object is changing. We call this the acceleration of the object. Instruentally, the acceleration of a car is indicated by how fast and which way the tip of the speedoeter needle is oing. In a car, it is deterined by how far down the gas pedal is pressed or, in the case of car that is slowing down, how hard the drier is pressing on the brake pedal. In the case of an object that is oing along a straight line, if the object has soe acceleration, then the speed of the object is changing. Looking ahead: The elocity of an object in the one-diensional world of this chapter is, in the three-diensional world in which we lie, the coponent of the elocity of the object. For the case of an object whose elocity has only an coponent, to get the elocity of the object through three diensional space, you just hae to ultiply the coponent of the elocity by the unit ector i. In ector notation, saying that an object has a elocity of 5/si eans the object is oing with a speed of 5/s in the + direction and saying that an object has a elocity of 5/si eans the object is oing with a speed of 5/s in the direction. 3

32 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics Okay, we e got the quantities used to characterize otion. Soon, we re going to deelop soe useful relations between those ariables. While we re doing that, I want you to keep these four things in ind:. We re talking about an object oing along a line.. eing in otion eans haing your position change with tie. 3. You already hae an intuitie understanding of what instantaneous elocity is because you hae ridden in a car. You know the difference between going 65 ph and 5 ph and you know ery well that you neither hae to go 65 iles nor trael for an hour to be going 65 ph. In fact, it is entirely possible for you to hae a speed of 65 ph for just an instant (no tie interal at all) it s how fast you are going (what your speedoeter reading is) at that instant. To be sure, the speedoeter needle ay be just swinging through that reading, perhaps because you are in the process of speeding up to 75 ph fro soe speed below 65 ph, but the 65 ph speed still has eaning and still applies to that instant when the speedoeter reading is 65 ph. Take this speed concept with which you are so failiar, tack on soe directional inforation, which for otion on a line just eans, specify forward or backward; and you hae what is known as the instantaneous elocity of the object whose otion is under consideration. A lot of people say that the speed of an object is how far that object traels in a certain aount of tie. No! That s a distance. Speed is a rate. Speed is neer how far, it is how fast. So if you want to relate it to a distance you ight say soething like, Speed is what you ultiply by a certain aount of tie to deterine how far an object would go in that aount of tie if the speed stayed the sae for that entire aount of tie. For instance, for a car with a speed of 5 ph, you could say that 5 ph is what you ultiply by an hour to deterine how far that car would go in an hour if it aintained a constant speed of 5 ph for the entire hour. ut why eplain it in ters of position? It is a rate. It is how fast the position of the object is changing. If you are standing on a street corner and a car passes you going 35 ph, I bet that if I asked you to estiate the speed of the car that you would get it right within 5 ph one way or the other. ut if we were looking oer a landscape on a day with unliited isibility and I asked you to judge the distance to a ountain that was 35 iles away just by looking at it, I think the odds would be ery uch against you getting it right to within 5 iles. In a case like that, you hae a better feel for how fast than you do for how far. So why define speed in ters of distance when you can just say that the speed of an object is how fast it is going? 4. You already hae an intuitie understanding of what acceleration is. You hae been in a car when it was speeding up. You know what it feels like to speed up gradually (sall acceleration) and you know what it feels like to speed up rapidly (big, pedalto-the-etal, acceleration). 3

33 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics All right, here coes the analysis. We hae a start line (=0) and a positie direction (eaning the other way is the negatie direction). 0 Consider a oing particle that is at position when the clock reads t and at position when the clock reads t. 0 The displaceent of the particle is, by definition, the change in position = of the particle. The aerage elocity is, by definition, = (6-) t where t = t t is the change in clock reading. Now the aerage elocity is not soething that one would epect you to hae an intuitie understanding for, as you do in the case of instantaneous elocity. The aerage elocity is not soething that you can read off the speedoeter, and frankly, it s typically not as interesting as the actual (instantaneous) elocity, but it is easy to calculate and we can assign a eaning to it (albeit a hypothetical eaning). It is the constant elocity at which the particle would hae to trael if it was to undergo the sae displaceent = in the sae tie t = t t at constant elocity. The iportance of the aerage elocity in this discussion lies in the fact that it facilitates the calculation of the instantaneous elocity. Calculating the instantaneous elocity in the case of a constant elocity is easy. Looking at what we ean by aerage elocity, it is obious that if the elocity isn t changing, the instantaneous elocity is the aerage elocity. So, in the case of a constant elocity, to calculate the instantaneous elocity, all we hae to do is calculate the aerage elocity, using any displaceent with its corresponding tie interal, that we want. Suppose we hae position s. tie data on, for instance, a car traeling a straight path at 4 /s. 33

34 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics Here s soe idealized fictitious data for just such a case. Data Reading Tie [seconds] Position [eters] Nuber Reeber, the speedoeter of the car is always reading 4 /s. (It should be clear that the car was already oing as it crossed the start line at tie zero think of tie zero as the instant a stopwatch was started and the ties in the table as stopwatch readings.) The position is the distance forward of the start line. Note that for this special case of constant elocity, you get the sae aerage elocity, the known alue of constant speed, no atter what tie interal you choose. For instance, if you choose the tie interal fro.00 seconds to 0.0 seconds: = (Aerage elocity.) t = = t 3 3 t s s = 3.0 s and if you choose the tie interal 0.00 seconds to 00.0 seconds: = = = t t 4 4 t s s = 3.0 s 34

35 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics The points that need ephasizing here are that, if the elocity is constant then the calculation of the aerage speed yields the instantaneous speed (the speedoeter reading, the speed we hae an intuitie feel for), and when the elocity is constant, it doesn t atter what tie interal you use to calculate the aerage elocity; in particular, a sall tie interal works just as well as a big tie interal. So how do we calculate the instantaneous elocity of an object at soe instant when the instantaneous elocity is continually changing? Let s consider a case in which the elocity is continually increasing. Here we show soe idealized fictitious data (consistent with the way an object really oes) for just such a case. Data Reading Nuber Tie since object was at start line. [s] Position (distance ahead of start line) [] Velocity (This is what we are trying to calculate. Here are the correct answers.) [/s] What I want to do with this fictitious data is to calculate an aerage elocity during a tie interal that begins with t = s and copare the result with the actual elocity at tie t = s. The plan is to do this repeatedly, with each tie interal used being saller than the preious one. Aerage elocity fro t = s to t = 5 s: = = t t 5 5 t = s s = 34 s Note that this alue is quite a bit larger than the correct alue of the instantaneous elocity at t = s (naely 8 /s). It does fall between the instantaneous elocity of 8 /s at t = s and the instantaneous elocity of 50 /s at t = 5 seconds. That akes sense since, during the tie interal, the elocity takes on arious alues which for s < t < 5 s are all greater than 8 /s but less than 50 /s. 35

36 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics For the net two tie interals in decreasing tie interal order (calculations not shown): Aerage elocity fro t = to t = s: /s Aerage elocity fro t = to t =. s: 8.4 /s And for the last tie interal, we do show the calculation: = = t t t =. 0s s = 8.04 s Here I copy all the results so that you can see the trend: Aerage elocity fro t = to t = 5 s: 34 /s Aerage elocity fro t = to t = s: /s Aerage elocity fro t = to t =. s: 8.4 /s Aerage elocity fro t = to t =.0 s: 8.04 /s Eery answer is bigger than the instantaneous elocity at t = s (naely 8 /s). Why? ecause the distance traeled in the tie interal under consideration is greater than it would hae been if the object oed with a constant elocity of 8 /s. Why? ecause the object is speeding up, so, for ost of the tie interal the object is oing faster than 8 /s, so, the aerage alue during the tie interal ust be greater than 8 /s. ut notice that as the tie interal (that starts at t = s) gets saller and saller, the aerage elocity oer the tie interal gets closer and closer to the actual instantaneous elocity at t = s. y induction, we conclude that if we were to use een saller tie interals, as the tie interal we chose to use was ade saller and saller, the aerage elocity oer that tiny tie interal would get closer and closer to the instantaneous elocity, so that when the tie interal got to be so sall as to be irtually indistinguishable fro zero, the alue of the aerage elocity would get to be indistinguishable fro the alue of the instantaneous elocity. We write that: = li t 0 t (Note the absence of the bar oer the. This is the instantaneous elocity.) This epression for is, by definition, the deriatie of with respect to t. The deriatie of with respect to t is d written as which eans that dt 36

37 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics d = (6-) dt d Note that, as entioned, is the deriatie of with respect to t. It is not soe ariable d dt ties all diided by d ties t. It is to be read dee e by dee tee or, better yet, the deriatie of with respect to t. Conceptually what it eans is, starting at that alue of tie t at which you wish to find the elocity, let t change by a ery sall aount. Find the (also ery sall) aount by which changes as a result of the change in t and diide the tiny change in by the tiny change in t. Fortunately, gien a function that proides the position for any tie t, we don t hae to go through all of that to get, because the branch of atheatics known as differential calculus gies us a uch easier way of deterining the deriatie of a function that can be epressed in equation for. A function, in this contet, is an equation inoling two ariables, one of which is copletely alone on the left side of the equation, the other of which, is in a atheatical epression on the right. The ariable on the left is said to be a function of the ariable on the right. Since we are currently dealing with how the position of a particle depends on tie, we use and t as the ariables in the functions discussed in the reainder of this chapter. In the eaple of a function that follows, we use the sybols o, o, and a to represent constants: + at = o + o t (6-3) The sybol t represents the reading of a running stopwatch. That reading changes so t is a ariable. For each different alue of t, we hae a different alue of, so is also a ariable. Soe folks think that any sybol whose alue is not specified is a ariable. Not so. If you know that the alue of a sybol is fied, then that sybol is a constant. You don t hae to know the alue of the sybol for it to be a constant; you just hae to know that it is fied. This is the case for o, o, and a in equation 6-3 aboe. Acceleration At this point you know how to calculate the rate of change of soething. Let s apply that knowledge to acceleration. Acceleration is the rate of change of elocity. If you are speeding up, then your acceleration is how fast you are speeding up. To get an aerage alue of acceleration oer a tie interal t, we deterine how uch the elocity changes during that tie interal and diide the change in elocity by the change in stopwatch reading. Calling the elocity change, we hae a = (6-4) t To get the acceleration at a particular tie t we start the tie interal at that tie t and ake it an infinitesial tie interal. That is: 37

38 Chapter 6 One-Diensional Motion (Motion Along a Line): Definitions and Matheatics a = li t 0 t The right side is, of course, just the deriatie of with respect to t: d a = (6-5) dt 38

39 Chapter 7 One-Diensional Motion: The Constant Acceleration Equations 7 One-Diensional Motion: The Constant Acceleration Equations The constant acceleration equations presented in this chapter are only applicable to situations in which the acceleration is constant. The ost coon istake inoling the constant acceleration equations is using the when the acceleration is changing. In chapter 6 we established that, by definition, d a = dt (which we called equation 6-5) where a is the acceleration of an object oing along a straight line path, is the elocity of the object and t, which stands for tie, represents the reading of a stopwatch. This equation is called a differential equation because that is the nae that we gie to equations inoling deriaties. It s true for any function that gies a alue of a for each alue of t. An iportant special case is the case in which a is siply a constant. Here we derie soe relations between the ariables of otion for just that special case, the case in which a is constant. d Equation 6-5, a =, with a stipulated to be a constant, can be considered to be a relationship dt between and t. Soling it is equialent to finding an epression for the function that gies the d alue of for each alue of t. So our goal is to find the function whose deriatie is a dt constant. The deriatie, with respect to t, of a constant ties t is just the constant. Recalling that we want that constant to be a, let s try: = at d We ll call this our trial solution. Let s plug it into equation 6-5, a =, and see if it works. dt Equation 6-5 can be written: d a = dt and when we plug our trial solution = at into it we get: a = d dt a = a d dt a = a (at) t 39

40 Chapter 7 One-Diensional Motion: The Constant Acceleration Equations a = a That is, our trial solution = at leads to an identity. Thus, our trial solution is indeed a solution d to the equation a =. Let s see how this solution fits in with the linear otion situation under dt study. In that situation, we hae an object oing along a straight line and we hae defined a onediensional coordinate syste which can be depicted as 0 and consists of nothing ore than an origin and a positie direction for the position ariable. We iagine that soeone starts a stopwatch at a tie that we define to be tie zero, t = 0, a tie that we also refer to as the start of obserations. Rather than liit ourseles to the special case of an object that is at rest at the origin at tie zero, we assue that it could be oing with any elocity and be at any position on the line at tie zero and define the constant o to be the position of the object at tie zero and the constant o to be the elocity of the object at tie zero. d Now the solution = at to the differential equation a = yields the alue = 0 when t = 0 dt d (just plug t = 0 into = at to see this). So, while = at does sole a =, it does not eet the dt conditions at tie zero, naely that = o at tie zero. We can fi the initial condition proble easily enough by siply adding o to the original solution yielding + at (7-) This certainly akes it so that ealuates to o when t = 0. ut is it still a solution to Let s try it. If + at, then = o = o d a =? dt d d d d d a = = o o 0 = dt dt dt dt dt ( + at ) = + ( at) = + a t a. d = o + at, when substituted into a = leads to an identity so = o + at is a solution to dt d a =. What we hae done is to take adantage of the fact that the deriatie of a constant is dt zero, so if you add a constant to a function, you do not change the deriatie of that function. d The solution = o + at is not only a solution to the equation a = (with a stipulated to be a dt constant) but it is a solution to the whole proble since it also eets the initial alue condition that = o at tie zero. The solution, equation 7-: 40

41 Chapter 7 One-Diensional Motion: The Constant Acceleration Equations + at is the first of a set of four constant acceleration equations to be deeloped in this chapter. = o The other definition proided in the last section was equation 6-: d = dt which in words can be read as: The elocity of an object is the rate of change of the position of the object (since the deriatie of the position with respect to tie is the rate of change of the position). Substituting our recently-found epression for elocity yields which can be written as: + o d dt d at = dt + at (7-) = o d We seek a function that gies a alue of for eery alue of t, whose deriatie is the su dt of ters + at. Gien the fact that the deriatie of a su will yield a su of ters, naely o the su of the deriaties, let s try a function represented by the epression = +. This d works if d is o and is at. Let s focus on first. Recall that o is a constant. Further dt dt recall that the deriatie-with-respect-to-t of a constant ties t, yields that constant. So check out = o t. Sure enough, the deriatie of o t with respect to t is o, the first ter in equation 7- aboe. So far we hae = o t + (7-3) d Now let s work on. We need to be at. Knowing that when we take the deriatie of dt soething with t in it we get soething with t in it we try = constant t. The deriatie of that is constant t which is equal to a t if we choose a for the constant. If the constant is a then our trial solution for is = at. Plugging this in for in equation 7-3, = o t +, yields: = o t + at Now we are in a situation siilar to the one we were in with our first epression for (t). This epression for does sole 4

42 Chapter 7 One-Diensional Motion: The Constant Acceleration Equations d dt + at (7-4) = o but it does not gie o when you plug 0 in for t. Again, we take adantage of the fact that you can add a constant to a function without changing the deriatie of that function. This tie we add the constant o so = o + o t + at (7-5) d This eets both our criteria: It soles equation 7-4, = o + at, and it ealuates to o when dt t = 0. We hae arried at the second equation in our set of four constant acceleration equations. The two that we hae so far are, equation 7-5: and equation 7-: = o + o t + at + at = o These two are enough, but to siplify the solution of constant acceleration probles, we use algebra to coe up with two ore constant acceleration equations. Soling equation 7-, = o + at, for a yields a = o and if you substitute that into equation 7-5 you quickly t arrie at the third constant acceleration equation o + = o + t (7-6) o Soling equation 7-, = o + at, for t yields t = and if you substitute that into equation a 7-5 you quickly arrie at the final constant acceleration equation: = o + a( o ) (7-7) For your conenience, we copy down the entire set of constant acceleration equations that you are epected to use in your solutions to probles inoling constant acceleration: = o + o t + at o + = o + t + at = o Constant Acceleration Equations = o + a( o ) 4

43 Chapter 8 One-Diensional Motion: Collision Type II 8 One-Diensional Motion: Collision Type II A coon istake one often sees in incorrect solutions to collision type two probles is using a different coordinate syste for each of the two objects. It is tepting to use the position of object at tie 0 as the origin for the coordinate syste for object and the position of object at tie 0 as the origin for the coordinate syste for object. This is a istake. One should choose a single origin and use it for both particles. (One should also choose a single positie direction.) We define a Collision Type II proble to be one in which two objects are oing along one and the sae straight line and the questions are, When and where are the two objects at one and the sae position? In soe probles in this class of probles, the word collision can be taken literally, but the objects don t hae to actually crash into each other for the proble to fall into the Collision Type II category. Furtherore, the restriction that both objects trael along one and the sae line can be relaed to coer for instance, a case in which two cars are traeling in adjacent lanes of a straight flat highway. The easiest way to ake it clear what we ean here is to gie you an eaple of a Collision Type II proble. Eaple 8-: A Collision Type II Proble A car traeling along a straight flat highway is oing along at 4.0 /s when it passes a police car standing on the side of the highway s after the speeder passes it, the police car begins to accelerate at a steady 5.00 /s. The speeder continues to trael at a steady 4.0 /s. (a) How long does it take for the police car to catch up with the speeder? (b) How far does the police car hae to trael to catch up with the speeder? (c) How fast is the police car going when it catches up with the speeder? We are going to use this eaple to illustrate how, in general, one soles a Collision Type II Proble. The first step in any Collision Type II proble is to establish one and the sae coordinate syste for both objects. Since we are talking about one-diensional otion, the coordinate syste is just a single ais, so what we are really saying is that we hae to establish a start line (the zero alue for the position ariable ) and a positie direction, and we hae to use the sae start line and positie direction for both objects. A conenient start line in the case at hand is the initial position of the police car. Since both cars go in the sae direction, the obious choice for the positie direction is the direction in which both cars go. We didn t nae it that at the tie since it was the only collision proble you were faced with then, but we define the Collision Type I Proble to be the kind you soled in your study of oentu, the kind of proble (and ariations on sae) in which two objects collide, and gien the initial elocity and the ass of each object, you are supposed to find the final elocity of each object. 43

44 Chapter 8 One-Diensional Motion: Collision Type II Net, we establish one and the sae tie ariable t for both objects. More specifically, we establish what we ean by tie zero, a tie zero that applies to both objects. To choose tie zero wisely, we actually hae to think ahead to the net step in the proble, a step in which we use the constant acceleration equations to write an epression for the position of each object in ters of the tie t. We want to choose a tie zero, t = 0, such that for all positie alues of t, that is for all future ties, the acceleration of each object is indeed constant. In the case at hand, the first choice that suggests itself to e is the instant at which the speeder first passes the police car. ut if we start the stopwatch at that instant, we find that as tie passes, the acceleration of the police car is not constant; rather, the police car has an acceleration of zero for three seconds and then, fro then on, it has an acceleration of 5.00 /s. So we wouldn t be able to use a single constant acceleration equation to write down an epression for the position of the police car that would be alid for all ties t 0. Now the net instant that suggests itself to e as a candidate for tie zero is the instant at which the police car starts accelerating. This turns out to be the right choice. Fro that instant on, both cars hae constant acceleration (which is 0 in the case of the speeder and 5.00 /s in the case of the police car). Furtherore, we hae inforation on the conditions at that instant. For instance, based on our start line, we know that the position of the police car is zero, the elocity of the police car is zero, and the acceleration of the police car is 5.00 /s at that instant. These becoe our initial alues when we choose tie zero to be the instant at which the police car starts accelerating. The one thing we don t know at that instant is the position of the speeder. ut we do hae enough inforation to deterine the position of the speeder at the instant that we choose to call tie zero. Our choice of tie zero actually causes the gien proble to break up into two probles: () Find the position of the speeder at tie 0, and () Sole the Collision Type II proble. The solution of the preliinary proble, finding the position of the speeder at tie 0, is quite easy in this case because the speed of the speeder is constant. Thus the distance traeled is just the speed ties the tie. d = t s d d = 4. 0 (3. 00s) s = 3 I used the sybol t' here to distinguish this tie fro the tie t that we will use in the Collision Type II part of the proble. We can think of the proble as one that requires two stopwatches: One stopwatch, we start at the instant the speeder passes the police car. This one is used for the preliinary proble and we use the sybol t' to represent the alue of its reading. The second one is used for the Collision Type II proble. It is started at the instant the police car starts accelerating and we will use the sybol t to represent the alue of its reading. Note that d = 3 is the position of the speeder, relatie to our established start line, at t = 0. Now we are in a position to sole the Collision Type II proble. We begin by aking a sketch of the situation. The sketch is a critical part of our solution. Sketches are used to define 44

45 Chapter 8 One-Diensional Motion: Collision Type II constants and ariables. The required sketch for a Collision Type II proble is one that depicts the initial conditions. 0 = 0 0 = 0 a = 5.00 /s (constant) We hae defined the speeder s car to be car and the police car to be car. Fro the constant acceleration equation (the one that gies the position of an object as a function of tie) we hae for the speeder: 0 = t + at = + t (8-) 0 0 where we hae incorporated the fact that a is zero. For the police car: 0 0 = + t a = 0 (constant) 0 = 4.0 /s ( is constant at 4.0 /s) 0 = 3 a t = at (8-) where we hae incorporated the fact that 0 = 0 and the fact that 0 = 0. Note that both equations (8- and 8-) hae the sae tie ariable t. The epression for (equation 8-), gies the position of the speeder s car for any tie t. You tell e the tie t, and I can tell you where the speeder s car is at that tie t just by plugging it into equation 8-. Siilarly, equation 8- for gies the position of the police car for any tie t. Now there is one special tie t, let s call it t* when both cars are at the sae position. The essential part of soling a Collision Type II proble is finding that that special tie t* which we refer to as the collision tie. Okay, now here coes the big central point for the Collision Type II proble. At the special tie t*, = (8-3) This sall siple equation is the key to soling eery Collision Type II proble. Substituting our epressions for and in equations and aboe, and designating the tie as the collision tie t* we hae t* = at* This yields a single equation in a single unknown, naely, the collision tie t*. We note that t* appears to the second power. This eans that the equation is a quadratic equation so we will probably (and in this case it turns out that we do) need the quadratic forula to sole it. Thus, 45

46 Chapter 8 One-Diensional Motion: Collision Type II we need to rearrange the ters as necessary to get the equation in the for of the standard quadratic equation a + b + c = 0 (recognizing that our ariable is t* rather than ). Subtracting + * fro both sides, swapping sides, and reordering the ters yields 0 0 t a t* 0 t* 0 = 0 which is the standard for for the quadratic equation. The quadratic forula b ± b 4ac = then yields a t* = ( ) ± 0 ( ) 0 4 a a ( 0 ) which siplifies eer so slightly to t* = 0 ± 0 + a a 0 Substituting alues with units yields: t * = 4 0. ± (4. 0 ) s s s s 3 Ealuation gies two results for t*, naely t* = 9.0 s and t* =.59 s. While the negatie alue is a alid solution to the atheatical equation, it corresponds to a tie in the past and our epressions for the physical positions of the cars were written to be alid fro tie 0 on. Prior to tie 0, the police car had a different acceleration than the that we used in the s epression for the position of the police car. ecause we know that our equation is not alid for ties earlier that t = 0 we ust discard the negatie solution. We are left with t* = 9.0 s for the tie when the police car catches up with the speeder. Once you find the collision tie in a Collision Type II proble, the rest is easy. Referring back to the proble stateent, we note that the collision tie itself t* = 9.0 s is the answer to part a, How long does it take for the police car to catch up with the speeder? Part b asks, How far ust the police car trael to catch up with the speeder? At this point, to answer that, all we hae to do is to substitute the 46

47 Chapter 8 One-Diensional Motion: Collision Type II collision tie t* into equation 8-, the equation that gies the position of the police car at any tie: = at*. s = 5 00 (9 0s). = 90 Finally, in part c of the proble stateent we are asked to find the elocity of the police car when it catches up with the speeder. First we turn to the constant acceleration equations to get an epression for the elocity of the police car at as a function of tie: = 0 + a The elocity of the police car at tie zero is 0 yielding: t = a t To get the elocity of the police car at the collision tie, we just hae to ealuate this at t = t* = 9.0 s. This yields: s s = =. for the elocity of the police car when it catches up with the speeder. s 47

48 Chapter 9 One-Diensional Motion Graphs 9 One-Diensional Motion Graphs Consider an object undergoing otion along a straight-line path, where the otion is characterized by a few consecutie tie interals during each of which the acceleration is constant but typically at a different constant alue than it is for the adjacent specified tie interals. The acceleration undergoes abrupt changes in alue at the end of each specified tie interal. The abrupt change leads to a jup discontinuity in the Acceleration s. Tie Graph and a discontinuity in the slope (but not in the alue) of the Velocity s. Tie Graph (thus, there is a corner or a kink in the trace of the Velocity s. Tie graph). The thing is, the trace of the Position s. Tie graph etends soothly through those instants of tie at which the acceleration changes. Een folks that get quite proficient at generating the graphs hae a tendency to erroneously include a kink in the Position s. Tie graph at a point on the graph corresponding to an instant when the acceleration undergoes an abrupt change. Your goals here all pertain to the otion of an object that oes along a straight line path at a constant acceleration during each of seeral tie interals but with an abrupt change in the alue of the acceleration at the end of each tie interal (ecept for the last one) to the new alue of acceleration that pertains to the net tie interal. Your goals for such otion are: () Gien a description (in words) of the otion of the object; produce a graph of position s. tie, a graph of elocity s. tie, and a graph of acceleration s. tie, for that otion. () Gien a graph of elocity s. tie, and the initial position of the object; produce a description of the otion, produce a graph of position s. tie, and produce a graph of acceleration s. tie. (3) Gien a graph of acceleration s. tie, the initial position of the object, and the initial elocity of the object; produce a description of the otion, produce a graph of position s. tie, and produce a graph of elocity s. tie. The following eaple is proided to ore clearly counicate what is epected of you and what you hae to do to eet those epectations: Eaple 9- A car oes along a straight stretch of road upon which a start line has been painted. At the start of obserations, the car is already 5 ahead of the start line and is oing forward at a steady 5 /s. The car continues to oe forward at 5 /s for 5.0 seconds. Then it begins to speed up. It speeds up steadily, obtaining a speed of 35 /s after another 5.0 seconds. As soon as its speed gets up to 35 /s, the car begins to slow down. It slows steadily, coing to rest after another 0.0 seconds. Sketch the graphs of position s. tie, elocity s. tie, and acceleration s. tie pertaining to the otion of the car during the period of tie addressed in the description of the otion. Label the key alues on your graphs of elocity s. tie and acceleration s. tie. 48

49 Chapter 9 One-Diensional Motion Graphs Okay, we are asked to draw three graphs, each of which has the tie, the sae stopwatch readings plotted along the horizontal ais. The first thing I do is to ask yself whether the plotted lines/cures are going to etend both aboe and below the tie ais. This helps to deterine how long to draw the aes. Reading the description of otion in the case at hand, it is eident that: () The car goes forward of the start line but it neer goes behind the start line. So, the s. t graph will etend aboe the tie ais (positie alues of ) but not below it (negatie alues of ). () The car does take on positie alues of elocity, but it neer backs up, that is, it neer takes on negatie alues of elocity. So, the s. t graph will etend aboe the tie ais but not below it. (3) The car speeds up while it is oing forward (positie acceleration), and it slows down while it is oing forward (negatie acceleration). So, the a s. t graph will etend both aboe and below the tie ais. Net, I draw the aes, first for s. t, then directly below that set of aes, the aes for s. t, and finally, directly below that, the aes for a s. t. Then I label the aes, both with the sybol used to represent the physical quantity being plotted along the ais and, in brackets, the units for that quantity. Now I need to put soe tick arks on the tie ais. To do so, I hae to go back to the question to find the releant tie interals. I e already read the question twice and I getting tired of reading it oer and oer again. This tie I ll take soe notes: At t = 0: = 5 = 5 /s 0-5 s: = 5 /s (constant) 5-0 s: increases steadily fro 5 /s to 35 /s 0-0 s: decreases steadily fro 35 /s to 0 /s Fro y notes it is eident that the ties run fro 0 to 0 seconds and that labeling eery 5 seconds would be conenient. So I put four tick arks on the tie ais of s. t. I label the origin 0, 0 and label the tick arks on the tie ais 5, 0, 5, and 0 respectiely. Then I draw ertical dotted lines, etending y tie ais tick arks up and down the page through all the graphs. They all share the sae ties and this helps e ensure that the graphs relate properly to each other. In the following diagra we hae the aes and the graph. Ecept for the labeling of key alues I hae described y work in a series of notes. To follow y work, please read the nubered notes, in order, fro to 0. How does one reeber what goes on which ais? Here s a neonic that applies to all y s. graphs. See that in s.? Yes, it is really the first letter of the word ersus, but you should think of it as standing for ertical. The physical quantity that is closer to the in s. gets plotted along the ertical ais. For instance, in a graph of Position s. Tie, the Position is plotted along the ertical ais (a.k.a. the y-ais) leaing the Tie for the horizontal ais (a.k.a. the -ais). Incidentally, the word neonic eans eory deice, a trick, word, jingle, or iage that one can use to help reeber soething. One ore thing: You probably know this, but just in case: a.k.a. stands for also known as. 49

50 Chapter 9 One-Diensional Motion Graphs [] (8) Put point here since = 5 at t = (9) is slope of s. t, so, being constant and + eans s. t is a straight line with + slope. (0) (below) is increasing eaning slope of s. t is increasing eaning cured up () Value of is decreasing eaning slope of s. t is decreasing eaning it is cured down. Note that alues of are still + so is increasing () s. t just becoes horizontal here, like the top of a hill. Must be horizontal here since = 0 at t = 0 s. t [s] [ s ] () Put point here since = 5 /s at t = 0. 5 /s () Draw horizontal line here since is constant at 5 /s. (3) Steady increase in eans straight line fro alue at 5 s to alue at 0 s. 35 /s (4) Steady decrease in eans straight line fro alue at 0 s to alue at 0 s t [s] a [ s] 0 (6) a is slope of s. t aboe; s. t is straight line with + slope, so, a is constant alue eaning a s. t is horizontal. (5) Constant eans 0 acceleration. 4 /s (7) Aboe, s. t is a straight line with slope, so, a is constant and. When a is constant, a s. t is horizontal. 3.5 /s t [s] The key alues on the s. t graph are giens so the only ystery, about the diagra aboe, that reains is, How were the key alues on a s. t obtained? Here are the answers: 50

51 Chapter 9 One-Diensional Motion Graphs On the tie interal fro t = 5 seconds to t = 0 seconds, the elocity changes fro to 35. Thus, on that tie interal the acceleration is gien by: s 35 5 a = = f s s t t t 0s 5s = f i = 4 i s 5 s On the tie interal fro t = 0 seconds to t = 0 seconds, the elocity changes fro to 0. Thus, on that tie interal the acceleration is gien by: s 35 s i a = = f t t t f i = 0 35 s s 0s 0s = 3. 5 s 5

52 Chapter 0 Constant Acceleration Probles in Two Diensions 0 Constant Acceleration Probles in Two Diensions In soling probles inoling constant acceleration in two diensions, the ost coon istake is probably iing the and y otion. One should do an analysis of the otion and a separate analysis of the y otion. The only ariable coon to both the and the y otion is the tie. Note that if the initial elocity is in a direction that is along neither ais, one ust first break up the initial elocity into its coponents. In the last few chapters we hae considered the otion of a particle that oes along a straight line with constant acceleration. In such a case, the elocity and the acceleration are always directed along one and the sae line, the line on which the particle oes. Here we continue to restrict ourseles to cases inoling constant acceleration (constant in both agnitude and direction) but lift the restriction that the elocity and the acceleration be directed along one and the sae line. If the elocity of the particle at tie zero is not collinear with the acceleration, then the elocity will neer be collinear with the acceleration and the particle will oe along a cured path. The cured path will be confined to the plane that contains both the initial elocity ector and the acceleration ector, and in that plane, the trajectory will be a parabola. (The trajectory is just the path of the particle.) You are going to be responsible for dealing with two classes of probles inoling constant acceleration in two diensions: () Probles inoling the otion of a single particle. () Collision Type II probles in two diensions We use saple probles to illustrate the concepts that you ust understand in order to sole two-diensional constant acceleration probles. Eaple 0- A horizontal square of edge length.0 is situated on a Cartesian coordinate syste such that one corner of the square is at the origin and the corner opposite that corner is at (.0,.0 ). A particle is at the origin. The particle has an initial elocity of.0 /s directed toward the corner of the square at (.0,.0 ) and has a constant acceleration of 4.87 /s in the + direction. Where does the particle hit the perieter of the square? 5

53 Chapter 0 Constant Acceleration Probles in Two Diensions Solution and Discussion Let s start with a diagra. y (.0,.0 ) o a Now let s ake soe conceptual obserations on the otion of the particle. Recall that the square is horizontal so we are looking down on it fro aboe. It is clear that the particle hits the right side of the square because: It starts out with a elocity directed toward the far right corner. That initial elocity has an coponent and a y coponent. The y coponent neer changes because there is no acceleration in the y direction. The coponent, howeer, continually increases. The particle is going rightward faster and faster. Thus, it will take less tie to get to the right side of the square then it would without the acceleration and the particle will get to the right side of the square before it has tie to get to the far side. An iportant aside on the trajectory (path) of the particle: Consider an ordinary checker on a huge square checkerboard with squares of ordinary size (just a lot ore of the then you find on a standard checkerboard). Suppose you start with the checker on the etree left square of the end of the board nearest you (square ) and eery second, you oe the checker right one square and forward one square. This would correspond to the checker oing toward the far right corner at constant elocity. Indeed you would be oing the checker along the diagonal. Now let s throw in soe acceleration. Return the checker to square and start oing it again. This tie, each tie you oe the checker forward, you oe it rightward one ore square than you did on the preious oe. So first you oe it forward one square and rightward one square. Then you oe it forward another square but rightward two ore squares. Then forward one square and rightward three squares. And so on. With each passing second, the rightward oe gets bigger. (That s what we ean when we say the rightward elocity is continually increasing.) So what would the path of the checker look like? Let s draw a picture. 53

54 Chapter 0 Constant Acceleration Probles in Two Diensions As you can see, the checker oes on a cured path. Siilarly, the path of the particle in the proble at hand is cured. Now back to the proble at hand. The way to attack these two-diensional constant acceleration probles is to treat the otion and the y otion separately. The difficulty with that, in the case at hand, is that the initial elocity is neither along nor along y but is indeed a iture of both otion and y otion. What we hae to do is to separate it out into its and y coponents. Let s proceed with that. Note that, by inspection, the angle that the elocity ector akes with the ais is

55 Chapter 0 Constant Acceleration Probles in Two Diensions y o =.0 /s oy θ = 45 o o cosθ = o o = o cosθ s o o =.0 cos y inspection (because the angle is 45.0 ): So: oy = o = 556 oy. s = 556 o. s Now we are ready to attack the otion and the y otion separately. efore we do, let s consider our plan of attack. We hae established, by eans of conceptual reasoning, that the particle will hit the right side of the square. This eans that we already hae the answer to half of the question Where does the particle hit the perieter of the square? It hits it at =.0 and y =?. All we hae to do is to find out the alue of y. We hae established that it is the otion that deterines the tie it takes for the particle to hit the perieter of the square. It hits the perieter of the square at that instant in tie when achiees the alue of.0. So our plan of attack is to use one or ore of the -otion constant acceleration equations to deterine the tie at which the particle hits the perieter of the square and to plug that tie into the appropriate y-otion constant acceleration equation to get the alue of y at which the particle hits the side of the square. Let s go for it. 55

56 Chapter 0 Constant Acceleration Probles in Two Diensions otion We start with the equation that relates position and tie: 0 + = o + ot a t (We need to find the tie that akes =.0.) The coponent of the acceleration is the total acceleration, that is a = a. Thus, = t + o at Recognizing that we are dealing with a quadratic equation we get it in the standard for of the quadratic equation. Now we apply the quadratic forula: at + ot = 0 t = o ± o 4 a a ( ) t = o ± o + a a Substituting alues with units (and, in this step, doing no ealuation) we obtain: t =. ± s s s s. 0 Ealuating this epression yields: t = s, and t =.09 s. We are soling for a future tie so we eliinate the negatie result on the grounds that it is a tie in the past. We hae found that the particle arries at the right side of the square at tie t = s. Now the question is, What is the alue of y at that tie? 56

57 Chapter 0 Constant Acceleration Probles in Two Diensions y-otion Again we turn to the constant acceleration equation relating position to tie, this tie writing it in ters of the y ariables: 0 0 y = yo + oy t + ay t We note that y 0 is zero because the particle is at the origin at tie 0 and a y is zero because the acceleration is in the + direction eaning it has no y coponent. Rewriting this: Substituting alues with units, y = oy y =. 556 ( s) s ealuating, and rounding to three significant figures yields: t y = Thus, the particle hits the perieter of the square at (.0, ) Net, let s consider a -D Collision Type II proble. Soling a typical -D Collision Type II proble inoles finding the trajectory of one of the particles, finding when the other particle crosses that trajectory, and establishing where the first particle is when the second particle crosses that trajectory. If the first particle is at the point on its own trajectory where the second particle crosses that trajectory then there is a collision. In the case of objects rather than particles, one often has to do soe further reasoning to sole a -D Collision Type II proble. Such reasoning is illustrated in the following eaple inoling a rocket. 57

58 Chapter 0 Constant Acceleration Probles in Two Diensions Eaple 0- The positions of a particle and a thin (treat it as being as thin as a line) rocket of length 0.80 are specified by eans of Cartesian coordinates. At tie 0 the particle is at the origin and is oing on a horizontal surface at 3.0 /s at 5.0. It has a constant acceleration of.43 /s in the +y direction. At tie 0 the rocket is at rest and it etends fro (.80, 50.0 ) to (0, 50.0 ), but it has a constant acceleration in the + direction. What ust the acceleration of the rocket be in order for the particle to hit the rocket? Solution ased on the description of the otion, the rocket traels on the horizontal surface along the line y = Let s figure out where and when the particle crosses this line. Then we ll calculate the acceleration that the rocket ust hae in order for the nose of the rocket to be at that point at that tie and repeat for the tail of the rocket. Finally, we ll quote our answer as being any acceleration in between those two alues. When and where does the particle cross the line y = 50.0? We need to treat the particle s otion and the y otion separately. Let s start by breaking up the initial elocity of the particle into its and y coponents. y o = 3.0 /s oy θ = 5.0 o o cosθ = o = o cosθ 3.. s o o = 0 cos 5 0 o = 4 47 o. s oy sinθ = oy = o sinθ 3.. s o oy = 0 sin 5 0 o = 7 87 oy. s 58

59 Chapter 0 Constant Acceleration Probles in Two Diensions Now in this case, it is the y otion that deterines when the particle crosses the trajectory of the rocket because it does so when y = So let s address the y otion first. y otion of the particle 0 y = yo + oy t + ay t Note that we can t just assue that we can cross out y o but in this case the tie zero position of the particle was gien as (0, 0) eaning that y o is indeed zero for the case at hand. Now we sole for t : y = oyt + ayt a y t + oy t y = 0 t = oy ± oy 4 a ay y ( y) t = oy ± oy + a y a y y t = ± s s s. 43 s 50.0 t =.405 s, and t = 7. s. Again, we throw out the negatie solution because it represents an instant in the past and we want a future instant. Now we turn to the otion to deterine where the particle crosses the trajectory of the rocket. 59

60 Chapter 0 Constant Acceleration Probles in Two Diensions otion of the particle Again we turn to the constant acceleration equation relating position to tie, this tie writing it in ters of the ariables: 0 (because the particle starts at the origin) 0 (because the acceleration is in the y direction) = + t + a t o o = o 47 s t = 4. (. 405 s) = So the particle crosses the rocket s path at (34.80, 50.0 ) at tie t =.450 s. Let s calculate the acceleration that the rocket would hae to hae in order for the nose of the rocket to be there at that instant. The rocket has otion only. It is always on the line y = Motion of the Nose of the Rocket 0 0 n = on + ont + a nt where we use the subscript n for nose and a prie to indicate rocket. We hae crossed out on because the nose of the rocket is at (0, 50.0 ) at tie zero, and we hae crossed out on because the rocket is at rest at tie zero. n = a nt Soling for a n yields: n a n = t Now we just hae to ealuate this epression at t =.405 s, the instant when the particle crosses the trajectory of the rocket, and at n = = 34.80, the alue of at which the particle crosses the trajectory of the rocket. ( ) a n = (. 405 s) a n =.0 s It should be ephasized that the n for nose is not there to iply that the nose of the rocket has a different acceleration than the tail; rather; the whole rocket ust hae the acceleration 60

61 Chapter 0 Constant Acceleration Probles in Two Diensions a n =.0 in order for the particle to hit the rocket in the nose. Now let s find the acceleration s a that the entire rocket ust hae in order for the particle to hit the rocket in the tail. t Motion of the Tail of the Rocket 0 t = ot + ot t + a t t where we use the subscript t for tail and a prie to indicate rocket. We hae crossed out ot because the rocket is at rest at tie zero, but ot is not zero because the tail of the rocket is at (.80, 50.0 ) at tie zero. t = ot + a t t Soling for a t yields: [ t ot ] a t = t Ealuating at t =.405 s and t = = yields a = t [ ( )] (. 405 s) a t =. s as the acceleration that the rocket ust hae in order for the particle to hit the tail of the rocket. Thus: The acceleration of the rocket ust be soewhere between. 0 and., inclusie, in s s order for the rocket to be hit by the particle. 6

62 Chapter Relatie Velocity Relatie Velocity Vectors add like ectors, not like nubers. Ecept in that ery special case in which the ectors you are adding lie along one and the sae line, you can t just add the agnitudes of the ectors. Iagine that you hae a dart gun with a uzzle elocity of 45 ph. Further iagine that you are on a bus traeling along a straight highway at 55 ph and that you point the gun so that the barrel is leel and pointing directly forward, toward the front of the bus. Assuing no recoil, as it leaes the uzzle of the gun, how fast is the dart traeling relatie to the road? That s right! 00 ph. The dart is already traeling forward at 55 ph relatie to the road just because it is on a bus that is oing at 55 ph relatie to the road. Add to that the elocity of 45 ph that it acquires as a result of the firing of the gun and you get the total elocity of the dart relatie to the road. This proble is an eaple of a class of ector addition probles that coe under the heading of Relatie Velocity. It is a particularly easy ector addition proble because both elocity ectors are in the sae direction. The only challenge is the ector addition diagra, since the resultant is right on top of the other two. We displace it to one side a little bit in the diagra below so that you can see all the ectors. Defining to be the elocity of the bus relatie to the road, R to be the elocity of the dart relatie to the bus, and D to be the elocity of the dart relatie to the road; we hae DR R DR D FORWARD The ector addition proble this illustrates is = DR + R D If we define the forward direction to be the positie direction, R DR D FORWARD Positie Direction then, because the ectors we are adding are both in the sae direction, we are indeed dealing with that ery special case in which the agnitude of the resultant is just the su of the agnitudes of the ectors we are adding: The uzzle elocity of any gun is the elocity, relatie to the gun, with which the bullet,, or dart eits the barrel of the gun. The barrel eit, the opening at the front end of the gun, is called the uzzle of the gun, hence the nae, uzzle elocity. 6

63 Chapter Relatie Velocity = + DR R D DR = R + D DR = 55 ph + 45 ph DR = 00 ph = 00 ph in the direction in which the bus is traeling DR You already know all the concepts you need to know to sole relatie elocity probles (you know what elocity is and you know how to do ector addition) so the best we can do here is to proide you with soe ore worked eaples. We e just addressed the easiest kind of relatie elocity proble, the kind in which all the elocities are in one and the sae direction. The second easiest kind is the kind in which the two elocities to be added are in opposite directions. Eaple - A bus is traeling along a straight highway at a constant 55 ph. A person sitting at rest on the bus fires a dart gun that has a uzzle elocity of 45 ph straight backward, (toward the back of the bus). Find the elocity of the dart, relatie to the road, as it leaes the gun. Again defining: to be the elocity of the bus relatie to the road, R to be the elocity of the dart relatie to the bus, and D to be the elocity of the dart relatie to the road, and DR defining the forward direction to be the positie direction; we hae DR R D = + DR R D DR = R D DR = 55 ph 45 ph DR = 0 ph FORWARD Positie Direction = 0 ph in the direction in which the bus is traeling DR 63

64 Chapter Relatie Velocity It would be odd looking at that dart fro the side of the road. Relatie to you it would still be oing in the direction that the bus is traeling, tail first, at 0 ph. The net easiest kind of ector addition proble is the kind in which the ectors to be added are at right angles to each other. Let s consider a relatie elocity proble inoling that kind of ector addition proble. Eaple - A boy sitting in a car that is traeling due north at 65 ph ais a gun (a gun which uses a copressed gas to fire a sall etal or plastic ball called a ), with a uzzle elocity of 85 ph, due east, and pulls the trigger. Recoil (the backward oeent of the gun resulting fro the firing of the gun) is negligible. In what copass direction does the go? Defining to be the elocity of the car relatie to the road, CR to be the elocity of the relatie to the car, and C to be the elocity of the relatie to the road; we hae R NORTH C = 85 ph EAST CR = 65 ph θ R tanθ = θ = tan C CR C CR 85 ph θ = tan 65 ph θ = 70.6 The traels in the direction for which the copass heading is

65 Chapter Relatie Velocity Eaple -3 A boat is traeling across a rier that flows due east at 8.50 /s. The copass heading of the boat is 5.0. Relatie to the water, the boat is traeling straight forward (in the direction in which the boat is pointing) at. /s. How fast and which way is the boat oing relatie to the banks of the rier? Okay, here we hae a situation in which the boat is being carried downstrea by the oeent of the water at the sae tie that it is oing relatie to the water. Note the gien inforation eans that if the water was dead still, the boat would be going. /s at 5.0 East of North. The water, howeer, is not still. Defining to be the elocity of the water relatie to the ground, WG to be the elocity of the boat relatie to the water, and W to be the elocity of the boat relatie to the ground; we hae G NORTH G φ = 5.0 W =. /s EAST θ W G = 8.50 /s Soling this proble is just a atter of following the ector addition recipe. First we define + to be eastward and +y to be northward. Then we draw the ector addition diagra for. WG reaking it up into coponents is triial since it lies along the -ais: 65

66 Chapter Relatie Velocity y, North W G = 8.50 /s, East y inspection: WG = 8.50 /s W Gy = 0 reaking does inole a little bit of work: W y, North W sinθ = W W Wy W = W sinθ φ = 5.0 W =. /s W =. s sin(5.0 o ) W =. 899 s cosθ = Wy W, East Wy Wy Wy = W o =. cos(5. 0 ) s = 0. 8 cosθ s Now we add the coponents to get the -coponent of the resultant 66

67 Chapter Relatie Velocity G = WG + W G G = = s s s and we add the y coponents to get the y-coponent of the resultant: Gy = WGy + Wy Gy = 0 s s Gy = 0. 8 s Now we hae both coponents of the elocity of the boat relatie to the ground. We need to draw the ector coponent diagra for G to deterine the direction and agnitude of the elocity of the boat relatie to the ground. y, North G = 0 8. G y s θ = 99. G s, East We then use the Pythagorean Theore to get the agnitude of the elocity of the boat relatie to the ground, 67

68 Chapter Relatie Velocity G = G + Gy G = (. 99 /s) + (0. 8 /s) G = /s and the definition of the tangent to deterine the direction of G : tanθ = θ = tan θ = tan θ = Gy G o Gy G 0. 8 /s. 99 /s Hence, G = 5.6 /s at 43.8 North of East. 68

69 Chapter Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law Soe folks think that eery object near the surface of the earth has an acceleration of 9.8 /s downward relatie to the surface of the earth. That just isn t so. In fact, as I look around the roo in which I write this sentence, all the objects I see hae zero acceleration relatie to the surface of the earth. Only when it is in freefall, that is, only when nothing is touching or pushing or pulling on the object ecept for the graitational field of the earth, will an object eperience an acceleration of 9.8 /s downward relatie to the surface of the earth. Graitational Force near the Surface of the Earth We all lie in the inisible graitational field of the earth. Mass is always accopanied by a surrounding graitational field. Any object that has ass, including the earth, is surrounded by a graitational field. The greater the ass of the object, the stronger the field is. The earth has a huge ass; hence, it creates a strong graitational field in the region of space around it. The graitational field is a force-per-ass at each and eery point in the region around the object, always ready and able to eert a force on any particle that finds itself in the graitational field. The earth s graitational field eists eerywhere around the earth, not only eerywhere in the air, but out beyond the atosphere in outer space, and inside the earth as well. The effect of the graitational field is to eert a force on any particle, any icti, that finds itself in the field. The force on the icti depends on both a property of the icti itself, naely its ass, and on a property of the point in space at which the particle finds itself, the force-per-ass of the graitational field at that point. The force eerted on the icti by the graitational field is just the ass of the icti ties the force-per-ass alue of the graitational field at the location of the icti. Hold a rock in the pal of your hand. You can feel that soething is pulling the rock downward. It causes the rock to ake a teporary indentation in the pal of your hand and you can tell that you hae to press upward on the botto of the rock to hold it up against that downward pull. The soething is the field that we hae been talking about. It is called the graitational field of the earth. It has both agnitude and direction so we use a ector ariable, the sybol g to represent it. In general, the agnitude and the direction of a graitational field both ary fro point to point in the region of space where the graitational field eists. The graitational field of the earth, near the surface of the earth is howeer, to a ery good approiation, uch sipler than that. To a ery good approiation, the graitational field of the earth has the sae alue at all points near the surface of the earth, and it always points toward the center of the earth, a direction that we norally think of as downward. To a ery good approiation, N g = 9.80 downward (-) kg at all points near the surface of the earth. The fact that the graitational field is a force-per-ass at eery point in space eans that it ust hae units of force-per-ass. Indeed, the N (newton) 69

70 Chapter Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law N appearing in the alue is the SI unit of force (how strong the push or pull on the object kg is) and the kg (kilogra) is the SI unit of ass, so the N/kg is indeed a unit of force-per-ass. The graitational force eerted on an object by the earth s graitational field (or that of another planet when the object is near the surface of that other planet) is soeties called the weight of the object. To stress that the graitational force is a force that is being eerted on the object, rather than a property of the object itself, we will refer to it as the graitational force. The graitational force Fg eerted on an object of ass by the graitational field of the earth is gien by F = g (-) g The product of a scalar and a ector is a new ector in the sae direction as the original ector. Hence the earth s graitational force is in the sae direction as the graitational field, naely downward, toward the center of the earth. The agnitude of the product of a scalar and a ector is the product of the absolute alue of the scalar and the agnitude of the ector. [Recall that the agnitude of a ector is how big it is. A ector has both agnitude (how big) and direction (which way). So for instance, the agnitude of the force ector F = 5 N downward, is F = 5 newtons.] Hence, F g = g (-3) relates the agnitude of the graitational force to the agnitude of the graitational field. The botto line is that eery object near the surface of the earth eperiences a downward-directed graitational force whose agnitude is gien by F g = g where is the ass of the object and g is N kg When the Graitational Force is the Only Force on an Object If there is a non-zero net force on an object, that object is eperiencing acceleration in the sae direction as that net force. How uch acceleration depends on how big the net force is and on the ass of the object whose acceleration we are talking about, the object upon which the net force acts. In fact, the acceleration is directly proportional to the force. The constant of proportionality is the reciprocal of the ass of the object. = a F (-4) Equation -4 is known as Newton s nd Law. Newton s nd dp Law can also be written F = where p is the dt oentu of the object. This latter equation is alid at all possible speeds, een at speeds close to the speed of light. To derie equation -4 fro it, we use p = which is only alid for speeds sall copared to the speed of light. Hence, equation -4 is only alid for speeds sall copared to the speed of light ( /s). Also, 70

71 Chapter Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law The epression F eans the su of the forces acting on the object. It is a ector su. It is the net force acting on the object. The ass is the inertia of the object, the object s inherent resistance to a change in its elocity. (Inherent eans of itself. ) Note that the factor in equation -4: = a F eans that the bigger the ass of the object, the saller its acceleration will be, for a gien net force. Equation -4 is a concise stateent of a ultitude of eperiental results. It is referred to as Newton s nd Law. Here, we want to apply it to find the acceleration of an object in freefall near the surface of the earth. Wheneer you apply Newton s nd Law, you are required to draw a free body diagra of the object whose acceleration is under inestigation. In a free body diagra, you depict the object (in our case it is an arbitrary object, let s think of it as a rock) free fro all its surroundings, and then draw an arrow on it for each force acting on the object. Draw the arrow with the tail touching the object, and the arrow pointing in the direction of the force. Label the arrow with the sybol used to represent the agnitude of the force. Finally, draw an arrow near, but not touching, the object. Draw the arrow so that it points in the direction of the acceleration of the object and label it with a sybol chosen to represent the agnitude of the acceleration. Here we use the sybol a g for the acceleration to reind us that it is the acceleration due to the earth s graitational field g. Free ody Diagra for an Object in Freefall near the Surface of the Earth a g F g The net step in applying Newton s nd Law is to write it down. a = F (-5) the concept of force proes ipractical on the atoic scale and saller (distances less than about 0-9 ). Such sall scales are the real of quantu echanics where energy and oentu still play a ajor role. 7

72 Chapter Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law Note that equation -4: = a F is a ector equation. As such it can be considered to be three equations in one one equation for each of a total of three possible utually-orthogonal (eaning perpendicular to each other) coordinate directions in space. In the case at hand, all the ectors, (hey, there are only two, the graitational force ector and the acceleration ector) are parallel to one and the sae line, naely the ertical, so we only need one of the equations. In equation -5, a = F we use arrows as subscripts the arrow shaft alignent specifies the line along which we are suing the forces and the arrowhead specifies the direction along that line that we choose to call the positie direction. In the case at hand, referring to equation -5, we note that the shafts of the arrows are ertical, eaning that we are suing forces along the ertical and that we are dealing with an acceleration along the ertical. Also in equation -5, we note that the arrowheads are pointing downward eaning that I hae chosen to call downward the positie direction, which, by default, eans that upward is the negatie direction. (I chose to call downward positie because both of the ectors in the free body diagra are downward.) Net we replace a with what it is in the free body diagra, a g F g naely a g, and we replace F with the su of the ertical forces in the free body diagra, counting downward forces as positie contributions to the su, and upward forces as negatie contributions to the su. This is an easy substitution in the case at hand because there is only one force on the free body diagra, naely the graitational force, the downward force of agnitude F g. The result of our substitutions is: a = (-6) g F g The F g in equation -6 is the agnitude of the graitational force, that force which you already read about at the start of this chapter. It is gien in ters of the ass of the object and the agnitude of the earth s graitational field g by equation -3, F g = g. Replacing the F g in equation -6, a g = Fg, with the g to which it is equialent we hae 7

73 Chapter Graitational Force Near the Surface of the Earth, First rush with Newton s nd Law ag = g (-7) Now the that appears in the fraction is the inertia of the object. It is the aount of inherent resistance that the object has to a change in its elocity and is a easure of the total aount of aterial aking up the object. The appearing in the g part of the epression (equation -7) is the graitational ass of the object, the quantity that, in concert with the graitational field at the location of the object deterines the force on the object. It is also a easure of the total aount of aterial aking up the object. As it turns out, the inertial ass and the graitational ass of the sae object are identical (which is why we use one and the sae sybol for each) and, in equation -7, they cancel. Thus, a =g (-8) Now g is the agnitude of the earth s graitational field ector at the location of the object. g = 9.80 N/kg and a g, being an acceleration has to hae units of acceleration, naely, /s. kg Fortunately a newton is a s Thus g so the units of g, naely N/kg, do indeed work out to be. s a g = 9.80 (-9) s Now this is wild! The acceleration of an object in freefall does not depend on its ass. You saw the asses cancel. The sae thing that akes an object heay akes it sluggish. One-Diensional Free-Fall, a.k.a., One-Diensional Projectile Motion If you throw an object straight up, or siply release it fro rest, or throw it straight down; assuing that the force of air resistance is negligibly sall copared to the graitational force: the object will be in freefall fro the instant it loses contact with your hand until the last instant before it hits the ground (or whateer it does eentually hit), and the object will trael along a straight line path with a constant acceleration of downward. s Consider the case in which the object is thrown straight up. The whole tie it is in freefall, the object eperiences an acceleration of downward. While the object is on the way up, the s downward acceleration eans that the object is slowing down. At the top of its otion, when the elocity changes fro being an upward elocity to being a downward elocity, and hence, for an instant is zero, the downward acceleration eans that the elocity is changing fro zero to a non-zero downward elocity. And on the way down, the downward acceleration eans that the elocity is increasing in the downward direction. 73

74 Chapter 3 Freefall, a.k.a. Projectile Motion 3 Freefall, a.k.a. Projectile Motion The constant acceleration equations apply fro the first instant in tie after the projectile leaes the launcher to the last instant in tie before the projectile hits soething, such as the ground. Once the projectile akes contact with the ground, the ground eerts a huge force on the projectile causing a drastic change in the acceleration of the projectile oer a ery short period of tie until, in the case of a projectile that doesn t bounce, both the acceleration and the elocity becoe zero. To take this zero alue of elocity and plug it into constant acceleration equations that are deoid of post-ground-contact acceleration inforation is a big istake. In fact, at that last instant in tie during which the constant acceleration equations still apply, when the projectile is at ground leel but has not yet ade contact with the ground, (assuing that ground leel is the lowest eleation achieed by the projectile) the agnitude of the elocity of the projectile is at its biggest alue, as far fro zero as it eer gets! Consider an object in freefall with a non-zero initial elocity directed either horizontally forward; or both forward and ertically (either upward or downward). The object will oe forward, and upward or downward perhaps upward and then downward while continuing to oe forward. In all cases of freefall, the otion of the object (typically referred to as the projectile when freefall is under consideration) all takes place within a single ertical plane. We can define that plane to be the -y plane by defining the forward direction to be the direction and the upward direction to be the y direction. One of the interesting things about projectile otion is that the horizontal otion is independent of the ertical otion. Recall that in freefall, an object continually eperiences a downward acceleration of but has no horizontal acceleration. This eans that if you fire a s projectile so that it is approaching a wall at a certain speed, it will continue to get closer to the wall at that speed, independently of whether it is also oing upward and/or downward as it approaches the wall. An interesting consequence of the independence of the ertical and horizontal otion is the fact that, neglecting air resistance, if you fire a bullet horizontally fro, say, shoulder height, oer flat leel ground, and at the instant the bullet eerges fro the gun, you drop a second bullet fro the sae height, the two bullets will hit the ground at the sae tie. The forward otion of the fired bullet has no effect on its ertical otion. The ost coon istake that folks ake in soling projectile otion probles is cobining the and y otion in one standard constant-acceleration equation. Don t do that. Treat the -otion and the y-otion separately. In soling projectile otion probles, we take adantage of the independence of the horizontal () otion and the ertical (y) otion by treating the separately. The one thing that is coon to both the otion and the y otion is the tie. The key to the solution of any projectile otion probles is finding the total tie of flight. For eaple, consider the following saple proble: 74

75 Chapter 3 Freefall, a.k.a. Projectile Motion Eaple 3-: A projectile is launched with a elocity of /s at an angle of 8 aboe the horizontal oer flat leel ground fro a height of.0 aboe ground leel. How far forward does it go before hitting the ground? (Assue that air resistance is negligible.) efore getting started, we better clearly establish what we are being asked to find. We define the forward direction as the direction so what we are looking for is a alue of. More specifically, we are looking for the distance, easured along the ground, fro that point on the ground directly below the point at which the projectile leaes the launcher, to the point on the ground where the projectile hits. This distance is known as the range of the projectile. It is also known as the range of the launcher for the gien angle of launch and the downrange distance traeled by the projectile. Okay, now that we know what we re soling for, let s get started. An initial elocity of /s at 8 aboe the horizontal, eh? Uh oh! We e got a dilea. The key to soling projectile otion probles is to treat the otion and the y otion separately. ut we are gien an initial elocity o which is a i of the two of the. We hae no choice but to break up the initial elocity into its and y coponents. y = /s θ = 5 y = cosθ = cosθ = cos 5 s = s o y = sinθ = sinθ y y = sin 5 s = y s o Now we re ready to get started. We ll begin with a sketch which defines our coordinate syste, thus establishing the origin and the positie directions for and y. 75

76 Chapter 3 Freefall, a.k.a. Projectile Motion y [] 3 o o = 9.97 /s oy = 4.65 /s [] Recall that in projectile otion probles, we treat the and y otion separately. Let s start with the otion. It is the easier part because there is no acceleration. otion 0 0 = o + o t + at = t (3-) Note that for the -otion, we start with the constant acceleration equation that gies the position as a function of tie. (Iagine haing started a stopwatch at the instant the projectile lost contact with the launcher. The tie ariable t represents the stopwatch reading.) As you can see, because the acceleration in the direction is zero, the equation quickly siplifies to = o t. We are stuck here because we hae two unknowns, and t, and only one equation. It's tie to turn to the y otion. It should be eident that it is the y otion that yields the tie, the projectile starts off at a known eleation (y =.0 ) and the projectile otion ends when the projectile reaches another known eleation, naely, y = 0. o y-otion y = yo + oy t + ay t (3-) This equation tells us that the y alue at any tie t is the initial y alue plus soe other ters that depend on t. It s alid for any tie t, starting at the launch tie t = 0, while the object is in projectile otion. In particular, it is applicable to that special tie t, the last instant before the object akes contact with the ground, that instant that we are ost interested in, the tie when y = 0. What we can do, is to plug 0 in for y, and sole for that special tie t that, when plugged into equation 3-, akes y be 0. When we rewrite equation 3- with y set to 0, the sybol t takes on a new eaning. Instead of being a ariable, it becoes a special tie, the tie that akes the y in the actual equation 3- ( y = yo + oy t + ay t ) zero. 76

77 Chapter 3 Freefall, a.k.a. Projectile Motion 0 = y o + oy t + ay t (3-3) * * To ephasize that the tie in equation 3-3 is a particular instant in tie rather than the ariable tie since launch, I hae written it as t to be read t star. Eerything in equation 3-3 is a * gien ecept t so we can sole equation 3-3 for t. Recognizing that equation 3-3 is a * * quadratic equation in t we first rewrite it in the for of the standard quadratic equation * a + b + c = 0. This yields: a t + oy t + yo * * y = 0 Then we use the quadratic forula b ± b 4ac = which for the case at hand appears as: a t = * oy ± oy 4 ay yo ay which siplifies to oy ± oy t = * a y a y y o Substituting alues with units yields: t = * s ± s s s which ealuates to t = 0. 3s and t =. 7 s * * We discard the negatie answer because we know that the projectile hits the ground after the launch, not before the launch. Recall that * t is the stopwatch reading when the projectile hits the ground. Note that the whole tie it has been oing up and down, the projectile has been oing forward in accord with 77

78 Chapter 3 Freefall, a.k.a. Projectile Motion equation 3-, and ealuate: = o t. At this point, all we hae to do is plug t =. 7 s into equation 3- * = t o * = (. 7s) s = 3 This is the answer. The projectile traels 3 forward before it hits the ground. 78

79 Chapter 4 Newton s Laws #: Using Free ody Diagras 4 Newton s Laws #: Using Free ody Diagras If you throw a rock upward in the presence of another person, and you ask that other person what keeps the rock going upward, after it leaes your hand but before it reaches its greatest height, that person ay incorrectly tell you that the force of the person s hand keeps it going. This illustrates the coon isconception that force is soething that is gien to the rock by the hand and that the rock has while it is in the air. It is not. A force is all about soething that is being done to an object. We hae defined a force to be an ongoing push or a pull. It is soething that an object can be a icti to, it is neer soething that an object has. While the force is acting on the object, the otion of the object is consistent with the fact that the force is acting on the object. Once the force is no longer acting on the object, there is no such force, and the otion of the object is consistent with the fact that the force is absent. (As reealed in this chapter, the correct answer to the question about what keeps the rock going upward, is, Nothing. Continuing to go upward is what it does all by itself if it is already going upward. You don t need anything to ake it keep doing that. In fact, the only reason the rock does not continue to go upward foreer is because there is a downward force on it. When there is a downward force and only a downward force on an object, that object is eperiencing a downward acceleration. This eans that the upward-oing rock slows down, then reerses its direction of otion and oes downward eer faster.) Iagine that the stars are fied in space so that the distance between one star and another neer changes. (They are not fied. The stars are oing relatie to each other.) Now iagine that you create a Cartesian coordinate syste; a set of three utually orthogonal aes that you label, y, and z. Your Cartesian coordinate syste is a reference frae. Now as long as your reference frae is not rotating and is either fied or oing at a constant elocity relatie to the (fictitious) fied stars, then your reference frae is an inertial reference frae. Note that elocity has both agnitude and direction and when we stipulate that the elocity of your reference frae ust be constant in order for it to be an inertial reference frae, we aren t just saying that the agnitude has to be constant but that the direction has to be constant as well. The agnitude of the elocity is the speed. So, for the agnitude of the elocity to be constant, the speed ust be constant. For the direction to be constant, the reference frae ust oe along a straight line path. So an inertial reference frae is one that is either fied or oing at a constant speed along a straight line path, relatie to the (fictitious) fied stars. The concept of an inertial reference frae is iportant in the study of physics because it is in inertial reference fraes that the laws of otion known as Newton s Laws of Motion apply. Here are Newton s three laws of otion, obsered to be adhered to by any particle of atter in an inertial reference frae: I. If there is no net force acting on a particle, then the elocity of that particle is not changing. II. If there is a net force on a particle, then that particle is eperiencing an acceleration that is directly proportional to the force, with the constant of proportionality being the reciprocal of the ass of the particle. 79

80 Chapter 4 Newton s Laws #: Using Free ody Diagras III. Anytie one object is eerting a force on a second object, the second object is eerting an equal but opposite force back on the first object. Discussion of Newton s st Law Despite the nae, it was actually Galileo that cae up with the first law. He let a ball roll down a rap with another rap facing the other way in front of it so that, after it rolled down one rap, the ball would roll up the other. He noted that the ball rolled up the second rap, slowing steadily until it reached the sae eleation as the one fro which the ball was originally released fro rest. He then repeatedly reduced the angle that the second rap ade with the horizontal and released the ball fro rest fro the original position for each new inclination of the second rap. The saller the angle, the ore slowly the speed of the ball was reduced on the way up the second rap and the farther it had to trael along the surface of the second rap before arriing at its starting eleation. When he finally set the angle to zero, the ball did not appear to slow down at all on the second rap. He didn t hae an infinitely long rap, but he induced that if he did, with the second rap horizontal, the ball would keep on rolling foreer, neer slowing down because no atter how far it rolled, it would neer gain any eleation, so it would neer get up to the starting eleation. His conclusion was that if an object was oing, then if nothing interfered with its otion it would keep on oing at the sae speed in the sae direction. So what keeps it going? The answer is nothing. That is the whole point. An object doesn t need anything to keep it going. If it is already oing, going at a constant elocity is what it does as long as there is no net force acting on it. In fact, it takes a force to change the elocity of an object. It s not hard to see why it took a huge chunk of huan history for soeone to realize that if there is no net force on a oing object, it will keep oing at a constant elocity, because the thing is, where we lie, on the surface of the Earth, there is ineitably a net force on a oing object. You throw soething up and the Earth pulls downward on it the whole tie the object is in flight. It s not going to keep traeling in a straight line upward, not with the Earth pulling on it. Een if you try sliding soething across the sooth surface of a frozen pond where the downward pull of the Earth s graitational field is cancelled by the ice pressing up on the object, you find that the object slows down because of a frictional force pushing on the object in the direction opposite that of the object s elocity and indeed a force of air resistance doing the sae thing. In the presence of these ubiquitous forces, it took huankind a long tie to realize that if there were no forces, an object in otion would stay in otion along a straight line path, at constant speed, and that an object at rest would stay at rest. Discussion of Newton s nd Law Galileo induced soething else of interest fro his ball-on-the-rap eperients by focusing his attention on the first rap discussed aboe. Obseration of a ball released fro rest reealed to hi that the ball steadily sped up on the way down the rap. Try it. As long as you don t ake the rap too steep, you can see that the ball doesn t just roll down the rap at soe fied speed, 80

81 Chapter 4 Newton s Laws #: Using Free ody Diagras it accelerates the whole way down. Galileo further noted that the steeper the rap was, the faster the ball would speed up on the way down. He did trial after trial, starting with a slightly inclined plane and gradually aking it steeper and steeper. Each tie he ade it steeper, the ball would, on the way down the rap, speed up faster than it did before, until the rap got so steep that he could no longer see that it was speeding up on the way down the rap it was siply happening too fast to be obsered. ut Galileo induced that, as he continued to ake the rap steeper, the sae thing was happening. That is that the ball s speed was still increasing on its way down the rap and the greater the angle, the faster the ball would speed up. In fact, he induced that if he increased the steepness to the ultiate angle, 90, that the ball would speed up the whole way down the rap faster than it would at any saller angle but that it would still speed up on the way down. Now, when the rap is tilted at 90, the ball is actually falling as opposed to rolling down the rap, so Galileo s conclusion was that when you drop an object (for which air resistance is negligible), what happens is that the object speeds up the whole way down, until it hits the Earth. Galileo thus did quite a bit to set the stage for Sir Isaac Newton, who was born the sae year that Galileo died. It was Newton who recognized the relationship between force and otion. He is the one that realized that the link was between force and acceleration, ore specifically, that wheneer an object is eperiencing a net force, that object is eperiencing an acceleration in the sae direction as the force. Now, soe objects are ore sensitie to force than other objects we can say that eery object coes with its own sensitiity factor such that the greater the sensitiity factor, the greater the acceleration of the object for a gien force. The sensitiity factor is the reciprocal of the ass of the object, so we can write that = a F (4-) where a is the acceleration of the object, is the ass of the object, and F is the ector su of all the forces acting on the object, that is to say that F is the net force acting on the object. Discussion of Newton s Third Law In realizing that wheneer one object is in the act of eerting a force on a second object, the second object is always in the act of eerting an equal and opposite force back on the first object, Newton was recognizing an aspect of nature that, on the surface, sees quite siple and straightforward, but quickly leads to conclusions that, howeer correct they ay be, and indeed they are correct, are quite counterintuitie. Newton s 3 rd law is a stateent of the fact that any force whatsoeer is just one half of an interaction where an interaction in this sense is the utual pushing or pulling that quite often occurs when one object is in the icinity of another. 8

82 Chapter 4 Newton s Laws #: Using Free ody Diagras In soe cases, where the effect is obious, the alidity of Newton s third law is fairly eident. For instance if two people who hae the sae ass are on roller skates and are facing each other and one pushes the other, we see that both skaters go rolling backward, away fro each other. It ight at first be hard to accept the fact that the second skater is pushing back on the hands of the first skater, but we can tell that the skater that we think of as the pusher, ust also be a pushee, because we can see that she eperiences a backward acceleration. In fact, while the pushing is taking place, the force eerted on her ust be just as great as the force she eerts on the other skater because we see that her final backward speed is just as great as that of the other (saeass) skater. ut how about those cases where the effect of at least one of the forces in the interaction pair is not at all eident? Suppose for instance that you hae a broo leaning up against a slippery wall. Aside fro our knowledge of Newton s laws, how can we conince ourseles that the broo is pressing against the wall, that is, that the broo is continually eerting a force on the wall; and; how can we conince ourseles that the wall is eerting a force back on the broo? One way to conince yourself is to let your hand play the role of the wall. Moe the broo and put your hand in the place of the wall so that the broo is leaning against the pal of your hand at the sae angle that it was against the wall with the pal of your hand facing directly toward the tip of the handle. You can feel the tip of the handle pressing against the pal of your hand. In fact, you can see the indentation that the tip of the broo handle akes in your hand. You can feel the force of the broo handle on your hand and you can induce that when the wall is where your hand is, relatie to the broo, the broo handle ust be pressing on the wall with the sae force. How about this business of the wall eerting a force on (pushing on) the tip of the broo handle? Again, with your hand playing the role of the wall, quickly oe your hand out of the way. The broo, of course, falls down. efore oing your hand, you ust hae been applying a force on the broo or else the broo would hae fallen down then. You ight argue that your hand wasn t necessarily applying a force but rather that your hand was just in the way. Well I here to tell you that being in the way is all about applying a force. When the broo is leaning up against the wall, the fact that the broo does not fall oer eans that the wall is eerting a force on the broo that cancels the other forces so that they don t ake the broo fall oer. In fact, if the wall was not strong enough to eert such a force, the wall would break. Still, it would be nice to get a isceral sense of the force eerted on the broo by the wall. Let your hand play the role of the wall, but this tie, let the broo lean against your pinky, near the tip of your finger. To keep the broo in the sae orientation as it was when it was leaning against the wall, you can feel that you hae to eert a force on the tip of the broo handle. In fact, if you increase this force a little bit, the broo handle tilts ore upward, and if you decrease it, it tilts ore downward. Again, you can feel that you are pushing on the tip of the broo handle when you are causing the broo handle to reain stationary at the sae orientation it had when it was leaning against the wall, and you can induce that when the wall is where your hand is, relatie to the broo, the wall ust be pressing on the broo handle with the sae force. Note that the direction in which the wall is pushing on the broo is away fro the wall at right angles to the wall. Such a force is eerted on any object that is in contact with a solid surface. This contact force eerted by a solid surface on an object in contact with that surface is called a noral 8

83 Chapter 4 Newton s Laws #: Using Free ody Diagras force because the force is perpendicular to the surface and the word noral eans perpendicular. Using Free ody Diagras The key to the successful solution of a Newton s nd Law proble is to draw a good free body diagra of the object whose otion is under study and then to use that free body diagra to epand Newton s nd Law, that is, to replace the F with an the actual ter-by-ter su of the forces. Note that Newton s nd Law = F a is a ector equation and hence, in the ost general case (3 diensions) is actually three scalar equations in one, one for each of the three possible utually orthogonal directions in space. (A scalar is a nuber. Soething that has agnitude only, as opposed to a ector which has agnitude and direction.) In your physics course, you will typically be dealing with forces that all lie in the sae plane, and hence, you will typically get two equations fro = a F. Regarding the Free ody Diagras: The hard part is creating the fro a description of the physical process under consideration; the easy part is using the. In what little reains of this chapter, we will focus on the easy part: Gien a Free ody Diagra, use it to find an unknown force or unknown forces, and/or use it to find the acceleration of the object. 83

84 Chapter 4 Newton s Laws #: Using Free ody Diagras For eaple, gien the free body diagra Upward a F N Leftward F kf = 3 newtons F P = 3 newtons Rightward for an object of ass.00 kg, find the agnitude of the noral force F N and find the agnitude of the acceleration a. (Note that we define the sybols that we use to represent the coponents of forces and the coponent of the acceleration, in the free body diagra. We do this by drawing an arrow whose shaft represents a line along which the force lies, and whose arrowhead we define to be the positie direction for that force coponent, and then labeling the arrow with our chosen sybol. A negatie alue for a sybol thus defined, siply eans that the corresponding force or acceleration is in the direction opposite to the direction in which the arrow is pointing. Solution: Note that the acceleration and all of the forces lie along one or the other of two iaginary lines (one of which is horizontal and the other of which is ertical) that are perpendicular to each other. The acceleration along one line is independent of any forces perpendicular to that line so we can consider one line at a tie. Let s deal with the horizontal line first. We write Newton s nd Law for the horizontal line as a = F (4-) in which the shafts of the arrows indicate the line along which we are suing forces (the shafts in equation 4- are horizontal so we ust be suing forces along the horizontal) and the arrowhead indicates which direction we consider to be the positie direction (any force in the opposite direction enters the su with a inus sign). The net step is to replace a with the sybol that we hae used in the diagra to represent the rightward acceleration and the F with an actual ter-by-ter su of the forces which includes only horizontal forces and in which rightward forces enter with a + and leftward forces enter with a. This yields: a = ( F P F kf ) Substituting alues with units and ealuating gies: F g = 9.6 newtons Downward a = (3 N 3 N) = kg s 84

85 Chapter 4 Newton s Laws #: Using Free ody Diagras Now we turn our attention to the ertical direction. For your conenience, the free body diagra is replicated here: Upward a F N Leftward F kf = 3 newtons F P = 3 newtons Rightward F g = 9.6 newtons Downward Again we start with Newton s nd Law, this tie written for the ertical direction: a = F We replace a with what it is and we replace F with the ter-by-ter su of the forces with a + for downward forces and a for upward forces. Note that the only a in the free body diagra is horizontal. Whoeer cae up with that free body diagra is telling us that there is no acceleration in the ertical direction, that is, that a = 0. Thus: Soling this for F N yields ( F 0 = g FN F N = F g Substituting alues with units results in a final answer of: F N = 9.6 newtons. ) 85

86 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras There is no force of otion acting on an object. Once you hae the force or forces eerted on the object by eerything (including any force-per-ass field at the location of the object) that is touching the object, you hae all the forces. Do not add a bogus force of otion to your free body diagra. It is especially tepting to add a bogus force when there are no actual forces in the direction in which an object is going. Keep in ind, howeer, that an object does not need a force on it to keep going in the direction in which it is going; oing along at a constant elocity is what an object does when there is no net force on it. Now that you e had soe practice using free body diagras it is tie to discuss how to create the. As you draw a free body diagra, there are a couple of things you need to keep in ind: () Include only those forces acting ON the object whose free body diagra you are drawing. Any force eerted Y the object on soe other object belongs on the free body diagra of the other object. () All forces are contact forces and eery force has an agent. The agent is that which is eerting the force. In other words, the agent is the life for or thing that is doing the pushing or pulling on the object. No agent can eert a force on an object without being in contact with the object. (We are using the field point of iew, rather than the action-at-adistance point of iew for the fundaental forces of nature. Thus, for instance, it is the earth s graitational field at the location of the object, rather than the aterial earth itself, that eerts the graitational force on an object.) We are going to introduce the arious kinds of forces by eans of eaples. Here is the first eaple: Eaple 5- A rock is thrown up into the air by a person. Draw the free body diagra of the rock while it is up in the air. (Your free body diagra is applicable for any tie after the rock leaes the thrower s hand, until the last instant before the rock akes contact with whateer it is destined to hit.) Neglect any forces that ight be eerted on the rock by the air. If you see the rock flying through the air, it ay ery well look to you like there is nothing touching the rock. ut the earth s graitational field is eerywhere in the icinity of the earth. It can t be blocked. It can t be shielded. It is in the air, in the water, een in the dirt. It is in direct contact with eerything in the icinity of the earth. It eerts a force on eery object near the surface of the earth. We call that force the graitational force. You hae already studied the graitational force. We gie a brief synopsis of it here. 86

87 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras The Graitational Force Eerted on Objects Near the Surface of the Earth. ecause it has ass, the earth has a graitational field. The graitational field is a force-per-ass field. It is inisible. It is not atter. It is an infinite set of force-per-ass ectors, one at eery point in space in the icinity of the surface of the earth. Each forceper-ass ector is directed downward, toward the center of the earth and, near the surface N of the earth, has a agnitude of The sybol used to represent the earth s kg graitational field ector at any point where it eists is g. Thus, N g = 9.80 Downward. The effect of the earth s graitational field is to eert a force on kg any object that is in the earth s graitational field. The force is called the graitational force and is equal to the product of the ass of the object and the earth s graitational field ector: F = g. The agnitude of the graitational force is gien by g F g = g (5-) N where g = 9.80 is the agnitude of the earth s graitational field ector. The kg direction of the near-earth s-surface graitational force is downward, toward the center of the earth. Here is the free body diagra and the corresponding table of forces for Eaple 5-: F g a Table of Forces Sybol =? Nae Agent Victi The Earth s F g = g Graitational The Graitational Force Rock Field Note: ) The only thing touching the object while it is up in the air (neglecting the air itself) is the earth s graitational field. So there is only one force on the object, naely the graitational force. The arrow representing the force ector is drawn so that the tail of the arrow is touching the object, and the arrow etends away fro the object in the direction of the force. ) Unless otherwise stipulated and labeled on the diagra, upward is toward the top of the page and downward is toward the botto of the page. 3) The arrow representing the acceleration ust be near but not touching the object. (If it is touching the object, one ight istake it for a force.) 4) There is no elocity inforation on a free body diagra. 87

88 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras 5) There is no force of the hand acting on the object because, at the instant in question, the hand is no longer touching the object. When you draw a free body diagra, only forces that are acting on the object at the instant depicted in the diagra are included. The acceleration of the object depends only on the currently-acting forces on the object. The force of the hand is of historical interest only. 6) Regarding the table of forces: a) Make sure that for any free body diagra you draw, you are capable of aking a coplete table of forces. You are not required to proide a table of forces with eery free body diagra you draw, but you should epect to be called upon to create a table of forces ore than once. b) In the table of forces, the agent is the life for or thing that is eerting the force and the icti is the object on which the force is being eerted. Make sure that, in eery case, the icti is the object for which the free body diagra is being drawn. c) In the case at hand, there is only one force so there is only one entry in the table of forces. d) For any object near the surface of the earth, the agent of the graitational force is the earth s graitational field. It is okay to abbreiate that to Earth because the graitational field of the earth can be considered to be an inisible part of the earth, but it is NOT okay to call it graity. Graity is a subject heading corresponding to the kind of force the graitational force is, graity is not an agent. Eaple 5- A ball of ass hangs at rest, suspended by a string. Draw the free body diagra for the ball, and create the corresponding table of forces. To do this proble, you need the following inforation about strings: The Force Eerted by a Taut String on an Object to Which it is Affied (This also applies to ropes, cables, chains, and the like.) The force eerted by a string, on an object to which it is attached, is always directed away fro the object, along the length of the string. Note that the force in question is eerted by the string, not for instance, by soe person pulling on the other end of the string. The force eerted by a string on an object is referred to as a tension force and its agnitude is conentionally represented by the sybol F T. Note: There is no forula to tell you what the tension force is. If it is not gien, the only way to get it is to use Newton s nd Law. 88

89 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras Here is the free body diagra of the ball, and the corresponding table of forces for Eaple 5-: F T F g a = 0 Table of Forces Sybol=? Nae Agent Victi F Tension T Force The String The all F g =g Graitational The Earth s Force Graitational Field The all Eaple 5-3 A sled of ass is being pulled forward oer a horizontal frictionless surface by eans of a horizontal rope attached to the front of the sled. Draw the free body diagra of the sled and proide the corresponding table of forces. Aside fro the rope and the earth s graitational field, the sled is in contact with a solid surface. The surface eerts a kind of force that we need to know about in order to create the free body diagra for this eaple. The Noral Force When an object is in contact with a surface, that surface eerts a force on the object. The surface presses on the object. The force on the object is away fro the surface, and it is perpendicular to the surface. The force is called the noral force because noral eans perpendicular, and as entioned, the force is perpendicular to the surface. We use the sybol F to represent the agnitude of the noral force. N Note: There is no forula to tell you what the noral force is. If it is not gien, the only way to get it is to use Newton s nd Law. Here is the free body diagra of the sled as well as the corresponding table of forces. F N F g a F T Table of Forces Sybol=? Nae Agent Victi The F Noral The N Horizontal Force Sled Surface F T F g =g Tension Force Graitational Force The Rope The Earth s Graitational Field The Sled The Sled 89

90 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras Note: The word Free in Free ody Diagra refers to the fact that the object is drawn free of its surroundings. Do not include the surroundings (such as the horizontal surface on which the sled is sliding in the case at hand) in your Free ody Diagra. Eaple 5-4 A block of ass rests on a frictionless horizontal surface. The block is due west of a west-facing wall. The block is attached to the wall by an ideal assless uncopressed/unstretched spring whose force constant is k. The spring is perpendicular to the wall. A person pulls the block a distance directly away fro the wall and releases it fro rest. Draw the free body diagra of the block appropriate for the first instant after release. Proide the corresponding table of forces. Now, for the first tie, we hae a spring eerting a force on the object for which we are drawing the free body diagra. So, we need to know about the force eerted by a spring. The Force Eerted by a Spring The force eerted by an ideal assless spring on an object in contact with one end of the spring is directed along the length of the spring, and away fro the object if the spring is stretched, toward the object if the spring is copressed. For the spring to eert a force on the object in the stretched-spring case, the object ust be attached to the end of the spring. Not so in the copressed-spring case. The spring can push on an object whether or not the spring is attached to the object. The force depends on the aount by which the spring is stretched or copressed, and on a easure of the stiffness of the spring known as the force constant of the spring a.k.a. the spring constant and represented by the sybol k. The agnitude of the spring force is typically represented by the sybol F S. The spring force is directly proportional to the aount of stretch. The spring constant k is the constant of proportionality. Thus, F S = k (5-) Here is the free body diagra of the block, and the corresponding table of forces for Eaple 5-4: 90

91 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras UP F N F g a F S EAST EWARE: y chance, in the eaples proided in this chapter, the noral force is upward. Neer assue it to be upward. The noral force is perpendicular to, and away fro, the surface eerting it. It happens to be upward in the eaples because the object is in contact with the top of a horizontal surface. If the surface is a wall, the noral force is horizontal; if it is a ceiling, downward; if an incline, perpendicular to and away fro the incline. Neer assue the noral force to be upward. Table of Forces Sybol=? Nae Agent Victi F Noral The Horizontal The N Force Surface lock The F S = k Spring Force The Spring lock F g =g Graitational The Earth s The Force Graitational Field lock Eaple 5-5 Fro your antage point, a crate of ass is sliding rightward on a flat leel concrete floor. Nothing solid is in contact with the crate ecept for the floor. Draw the free body diagra of the crate. Proide the corresponding table of forces. Fro our eperience with objects sliding on concrete floors, we know that the crate is slowing down at the instant under consideration. It is slowing because of kinetic friction. Kinetic Friction A surface, upon which an object is sliding, eerts (in addition to the noral force) a retarding force on that object. The retarding force is in the direction opposite that of the elocity of the object. In the case of an object sliding on a dry surface of a solid body (such as a floor) we call the retarding force a kinetic frictional force. Kinetic eans otion and we include the adjectie kinetic to ake it clear that we are dealing with an object that is in otion. The kinetic frictional forula gien below is an epirical result. This eans that it is 9

92 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras deried directly fro eperiental results. It works only in the case of objects sliding on dry surfaces. It does not apply, for instance, to the case of an object sliding on a greased surface. We use the sybol F k f for the kinetic friction force. The kinetic frictional forula reads F k f = µ K F N (5-3) F N is the agnitude of the noral force. Its presence in the forula indicates that the ore strongly the surface is pressing on the object, the greater the frictional force. µ K (u-sub-k) is called the coefficient of kinetic friction. Its alue depends on the aterials of which both the object and the surface are ade as well as the soothness of the two contact surfaces. It has no units. It is just a nuber. The agnitude of the kinetic frictional force is soe fraction of the agnitude of the noral force; µ K is that fraction. Values of µ K for arious pairs of aterials can be found in handbooks. They tend to fall between 0 and. The actual alue for a gien pair of aterials depends on the soothness of the surface and is typically quoted with but a single significant digit. IMPORTANT: µ K is a coefficient (with no units) used in calculating the frictional force. It is not a force itself. Here is the free body diagra and the table of forces for the case at hand. The crate is oing rightward and slowing down it has a leftward acceleration. a F k f F N F g Table of Forces Sybol Nae Agent Victi F The Concrete The N Noral Force Floor Crate F k f = µ K F Kinetic The Concrete The N Friction Force Floor Crate The Earth s F g = g Graitational The Graitational Force Crate Field Note: Not eery object has a noral ( perpendicular-to-the-surface ) force acting on it. If the object is not in contact with a surface, then there is no noral force acting on the object. 9

93 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras Eaple 5-6 A person has pushed a brick along a tile floor toward an eastward-facing wall trapping a spring of unstretched length Lo and force constant k between the wall and the end of the brick that is facing the wall. That end of the brick is a distance d fro the wall. The person has released the brick, but the spring is unable to budge it the brick reains eactly where it was when the person released it. Draw the free body diagra for the brick and proide the corresponding table of forces. A frictional force is acting on an object at rest. Typically, an object at rest clings ore strongly to the surface with which it is in contact than the sae object does when it is sliding across the sae surface. What we hae here is a case of static friction. Static Friction Force A surface that is not frictionless can eert a static friction force on an object that is in contact with that surface. The force of static friction is parallel to the surface. It is in the direction opposite the direction of ipending otion of the stationary object. The direction of ipending otion is the direction in which the object would accelerate if there was no static friction. In general, there is no forula for calculating static friction to sole for the force of static friction, you use Newton s nd Law. The force of static friction is whateer it has to be to ake the net parallel-to-the-surface force zero. We use the sybol F sf to represent the agnitude of the static friction force. SPECIAL CASE: Iagine trying to push a refrigerator across the floor. Iagine that you push horizontally, and that you gradually increase the force with which you are pushing. Initially, the harder you push, the bigger the force of static friction. ut it can t grow foreer. There is a aiu possible static friction force agnitude for any such case. Once the agnitude of your force eceeds that, the refrigerator will start sliding. The aiu possible force of static friction is gien by: F = µ F (5-4) a possible sf S N The unitless quantity µ S is the coefficient of static friction specific to the type of surface the object is sliding on and the nature of the surface of the object. Values of µ S tend to fall between 0 and. For a particular pair of surfaces, µ S is at least as large as, and typically larger than, µ K. 93

94 Chapter 5 Newton s Laws #: Kinds of Forces, Creating Free ody Diagras a possible Clearly, this forula ( F = µ F ) is only applicable when the sf S question is about the aiu possible force of static friction. You can use this forula if the object is said to be on the erge of slipping, or if the question is about how hard one ust push to budge an object. It also coes in handy when you want to know whether or not an object will stay put. In such a case you would use Newton s nd to find out the agnitude of the force of static friction needed to keep the object fro accelerating. Then you would copare that agnitude with the aiu possible agnitude of the force of static friction. N Here is the free body diagra of the brick and the table of forces for Eaple 5-6: Up a=0 F N F S East F sf F g Table of Forces Sybol=? Nae Agent Victi F Noral The N The Tile Floor Force rick F sf Static The Friction The Tile Floor rick Force F g = g Graitational Force The Earth s Graitational Field F S = k Spring Force The Spring The rick The rick 94

95 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings When, in the case of a tilted coordinate syste, you break up the graitational force ector into its coponent ectors, ake sure the graitational force ector itself fors the hypotenuse of the right triangle in your ector coponent diagra. All too often, folks draw one of the coponents of the graitational force ector in such a anner that it is bigger than the graitational force ector it is supposed to be a coponent of. The coponent of a ector is neer bigger than the ector itself. Haing learned how to use free body diagras, and then haing learned how to create the, you are in a pretty good position to sole a huge nuber of Newton s nd Law probles. An understanding of the considerations in this chapter will enable to you to sole an een larger class of probles. Again, we use eaples to coney the desired inforation. Eaple 6- A professor is pushing on a desk with a force of agnitude F at an acute angle θ below the horizontal. The desk is on a flat, horizontal tile floor and it is not oing. For the desk, draw the free body diagra that facilitates the direct and straightforward application of Newton s nd Law of otion. Gie the table of forces. While not a required part of the solution, a sketch often akes it easier to coe up with the correct free body diagra. Just ake sure you don t cobine the sketch and the free body diagra. In this proble, a sketch helps clarify what is eant by at an acute angle θ below the horizontal. Pushing with a force that is directed at soe acute angle below the horizontal is pushing horizontally and downward at the sae tie. 95

96 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings Here is the initial free body diagra and the corresponding table of forces. θ F N a = 0 F P F sf F g Table of Forces Sybol=? Nae Agent Victi F Noral N Force The Floor Desk Static F sf Friction The Floor Desk Force F g = g Graitational The Earth s Force Graitational Field Desk F Force of P Professor Hands of Professor Desk Note that there are no two utually perpendicular lines to which all of the forces are parallel. The best choice of utually perpendicular lines would be a ertical and a horizontal line. Three of the four forces lie along one or the other of such lines. ut the force of the professor does not. We cannot use this free body diagra directly. We are dealing with a case which requires a second free body diagra. Cases Requiring a Second Free ody Diagra in Which One of More of the Forces that was in the First Free ody Diagra is Replaced With its Coponents Establish a pair of utually perpendicular lines such that ost of the ectors lie along one or the other of the two lines. After haing done so, break up each of the other ectors, the ones that lie along neither of the lines, (let s call these the rogue ectors) into coponents along the two lines. (reaking up ectors into their coponents inoles drawing a ector coponent diagra.) Draw a second free body diagra, identical to the first, ecept with rogue ectors replaced by their coponent ectors. In the new free body diagra, draw the coponent ectors in the direction in which they actually point and label the with their agnitudes (no inus signs). 96

97 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings For the case at hand, our rogue force is the force of the professor. We break it up into coponents as follows: y θ F P F Py F P F F P = P cosθ F F Py P = sinθ F P = F P cosθ F Py = F P sinθ Then we draw a second free body diagra, the sae as the first, ecept with F P replaced by its coponent ectors: F N a = 0 F P F sf F Py F g 97

98 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings Eaple 6- A wooden block of ass is sliding down a flat etal incline (a flat etal rap) that akes an acute angle θ with the horizontal. The block is slowing down. Draw the directly-usable free body diagra of the block. Proide a table of forces. We choose to start the solution to this proble with a sketch. The sketch facilitates the creation of the free body diagra but in no way replaces it. θ Since the block is sliding in the down-the-incline direction, the frictional force ust be in the upthe-incline direction. Since the block s elocity is in the down-the-incline direction and decreasing, the acceleration ust be in the up-the-incline direction. a F k f θ F g F N Table of Forces Sybol=? Nae Agent Victi F Noral The N The Rap Force lock Kinetic F kf = µ KF The N Friction The Rap lock Force F g =g Graitational Force The Earth s Graitational Field The lock No atter what we choose for a pair of coordinate aes, we cannot ake it so that all the ectors in the free body diagra are parallel to one or the other of the two coordinate aes lines. At best, the pair of lines, one line parallel to the frictional force and the other perpendicular to the rap, leaes one rogue ector, naely the graitational force ector. Such a coordinate syste is tilted on the page. 98

99 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings Cases Inoling Tilted Coordinate Systes For effectie counication purposes, students drawing diagras depicting phenoena occurring near the surface of the earth are required to use either the conention that downward is toward the botto of the page (corresponding to a side iew) or the conention that downward is into the page (corresponding to a top iew). If one wants to depict a coordinate syste for a case in which the direction downward is parallel to neither coordinate ais line, the coordinate syste ust be drawn so that it appears tilted on the page. In the case of a tilted-coordinate syste proble requiring a second free body diagra of the sae object, it is a good idea to define the coordinate syste on the first free body diagra. Use dashed lines so that the coordinate aes do not look like force ectors. Here we redraw the first free body diagra. (When you get to this stage in a proble, just add the coordinate aes to your eisting diagra.) a F kf θ y F N F g Now we break up Fg into its, y coponent ectors. This calls for a ector coponent diagra. 99

100 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings y θ θ Horizontal Line F gy θ 90 θ F g F g F g F g = sinθ F g y F g = cosθ F g = Fgsinθ F g y = Fgcosθ Net, we redraw the free body diagra with the graitational force ector replaced by its coponent ectors. a F N F kf θ F g F g y 00

101 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings Eaple 6-3 A solid brass cylinder of ass is suspended by a assless string which is attached to the top end of the cylinder. Fro there, the string etends straight upward to a assless ideal pulley. The string passes oer the pulley and etends downward, at an acute angle theta to the ertical, to the hand of a person who is pulling on the string with force F T. The pulley and the entire string segent, fro the cylinder to hand, lie in one and the sae plane. The cylinder is accelerating upward. Proide both a free body diagra and a table of forces for the cylinder. A sketch coes in handy for this one: To proceed with this one, we need soe inforation on the effect of an ideal assless pulley on a string that passes oer the pulley. Effect of an Ideal Massless Pulley The effect of an ideal assless pulley on a string that passes oer the pulley is to change the direction in which the string etends, without changing the tension in the string. y pulling on the end of the string with a force of agnitude F T, the person causes there to be a tension F T in the string. (The force applied to the string by the hand of the person, and the tension force of the string pulling on the hand of the person, are a Newton s-3 rd -law interaction pair of forces. They are equal in agnitude and opposite in direction. We choose to use one and the sae sybol F T for the agnitude of both of these forces. The directions are opposite each other.) The tension is the sae throughout the string, so, where the string is attached to the brass cylinder, the string eerts a force of agnitude F T directed away fro the cylinder along the length of the string. Here is the free body diagra and the table of forces for the cylinder: 0

102 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings F T F g a Table of Forces Sybol=? Nae Agent Victi F Tension The T The String Force Cylinder F g = g Graitational The Earth s The Force Graitational Field Cylinder Eaple 6-4 A cart of ass C is on a horizontal frictionless track. One end of an ideal assless string segent is attached to the iddle of the front end of the cart. Fro there the string etends horizontally, forward and away fro the cart, parallel to the centerline of the track, to a ertical pulley. The string passes oer the pulley and etends downward to a solid etal block of ass. The string is attached to the block. A person was holding the cart in place. The block was suspended at rest, well aboe the floor, by the string. The person has released the cart. The cart is now accelerating forward and the block is accelerating downward. Draw a free body diagra for each object. A sketch will help us to arrie at the correct answer to this proble. C Recall fro the last eaple that there is only one tension in the string. Call it F T. ased on our knowledge of the force eerted on an object by a string, iewed so that the apparatus appears as it does in the sketch, the string eerts a rightward force F T on the cart, and an upward force of agnitude F T on the block. 0

103 Chapter 6 Newton s Laws #3: Coponents, Friction, Raps, Pulleys, and Strings There is a relationship between each of seeral ariables of otion of one object attached by a taut string, which reains taut throughout the otion of the object, and the corresponding ariables of otion of the second object. The relationships are so siple that you ight consider the to be triial, but they are critical to the solution of probles inoling objects connected by a taut spring. The Relationships Aong the Variables of Motion For Two Objects, One at One End and the Other at the Other End, of an always-taut, Unstretchable String Consider the following diagra. ecause they are connected together by a string of fied length, if object goes downward 5 c, then object ust go rightward 5 c. So if object goes downward at 5 c/s then object ust go rightward at 5 c/s. In fact, no atter how fast object goes downward, object ust go rightward at the eact sae speed (as long as the string does not break, stretch, or go slack). In order for the speeds to always be the sae, the accelerations hae to be the sae as each other. So if object is picking up speed in the downward direction at, for instance, 5 c/s, then object ust be picking up speed in the rightward direction at 5 c/s. The agnitudes of the accelerations are identical. The way to deal with this is to use one and the sae sybol for the agnitude of the acceleration of each of the objects. The ideas releant to this siple eaple apply to any case inoling two objects, one on each end of an inetensible string, as long as each object oes only along a line collinear with the string segent to which the object is attached. Let s return to the eaple proble inoling the cart and the block. The two free body diagras follow: F N a F T c F T a F gc F g Note that the use of the sae sybol F T in both diagras is iportant, as is the use of the sae sybol a in both diagras. 03

104 Chapter 7 The Uniersal Law of Graitation 7 The Uniersal Law of Graitation Consider an object released fro rest an entire oon s diaeter aboe the surface of the oon. Suppose you are asked to calculate the speed with which the object hits the oon. This proble typifies the kind of proble in which students use the uniersal law of graitation to get the force eerted on the object by the graitational field of the oon, and then istakenly use one or ore of the constant acceleration equations to get the final elocity. The proble is: the acceleration is not constant. The closer the object gets to the oon, the greater the graitational force, and hence, the greater the acceleration of the object. The istake lies not in using Newton s second law to deterine the acceleration of the object at a particular point in space; the istake lies in using that one alue of acceleration, good for one object-to-oon distance, as if it were alid on the entire path of the object. The way to go on a proble like this, is to use conseration of energy. ack in chapter, where we discussed the near-surface graitational field of the earth, we talked about the fact that any object that has ass creates an inisible force-per-ass field in the region of space around it. We called it a graitational field. Here we talk about it in ore detail. Recall that when we say that an object causes a graitational field to eist, we ean that it creates an inisible force-per-ass ector at eery point in the region of space around itself. The effect of the graitational field on any particle (call it the icti) that finds itself in the region of space where the graitational field eists, is to eert a force, on the icti, equal to the force-perass ector at the icti s location, ties the ass of the icti. Now we proide a quantitatie discussion of the graitational field. (Quantitatie eans, inoling forulas, calculations, and perhaps nubers. Contrast this with qualitatie which eans descriptie/conceptual.) We start with the idealized notion of a point particle of atter. eing atter, it has ass. Since it has ass it has a graitational field in the region around it. The direction of a particle s graitational field at point P, a distance r away fro the particle, is toward the particle and the agnitude of the graitational field is gien by g = G (7-) r where: N G is the uniersal graitational constant: G = kg is the ass of the particle, and r is the distance that point P is fro the particle. In that point P can be any epty (or occupied) point in space whatsoeer, this forula gies the agnitude of the graitational field of the particle at all points in space. Equation 7- is the equation for of Newton s Uniersal Law of Graitation. Newton s Uniersal Law of Graitation is an approiation to Einstein s far ore coplicated General Theory of Relatiity. The approiation is fantastic for orbital echanics for space ehicles and ost planets, but we need to use the General Theory of Relatiity for a coplete eplanation of the orbit of Mercury. 04

105 Chapter 7 The Uniersal Law of Graitation The Total Graitational Field Vector at an Epty Point in Space Suppose that you hae two particles. Each has its own graitational field at all points in space. Let s consider a single epty point in space. Each of the two particles has its own graitational field ector at that epty point in space. We can say that each particle akes its own ector contribution to the total graitational field at the epty point in space in question. So how do you deterine the total graitational field at the epty point in space? You guessed it! Just add the indiidual contributions. And because the contribution due to each particle is a ector, the two contributions add like ectors. What the Graitational Field does to a Particle that is in the Graitational Field Now suppose that you hae the agnitude and direction of the graitational field ector g at a particular point in space. The graitational field eerts a force on any icti particle that happens to find itself at that location in space. Suppose, for instance, that a particle of ass finds itself at a point in space where the graitational field (of soe other particle or particles) is g. Then the particle of ass is subject to a force F = g (7-) G The Graitational Effect of one Particle on Another Let s put the two preceding ideas together. Particle of ass has, aong its infinite set of graitational field ectors, a graitational field ector at a location a distance r away fro itself, a point in space that happens to be occupied by another particle, particle, of ass. The graitational field of particle eerts a force on particle. The question is, how big and which way is the force? Let s start by identifying the location of particle as point P. Point P is a distance r away fro particle. Thus, the agnitude of the graitational field ector (of particle ) at point P is: g = G (7-3) r The stateent that the arious contributions to the graitational field add like ectors is known as the superposition principle for the graitational field. Einstein, in his General Theory of Relatiity, showed that the superposition principle for the graitational field is actually an approiation that is ecellent for the graitational fields you will deal with in this course but breaks down in the case of ery strong graitational fields. 05

106 Chapter 7 The Uniersal Law of Graitation Now the force eerted on particle by the graitational field of particle is gien by equation 7-, g F G =. Using F for FG to ephasize the fact that we are talking about the force of on, and writing the equation relating the agnitudes of the ectors we hae F = g Replacing the g with the epression we just found for the agnitude of the graitational field due to particle nuber we hae F = G r which, with soe inor reordering can be written as F = G (7-4) r This equation gies the force of the graitational field of particle on particle. Neglecting the iddlean (the graitational field) we can think of this as the force of particle on particle. You can go through the whole arguent again, with the roles of the particles reersed, to find that the sae epression applies to the force of particle on particle, or you can siply inoke Newton s 3 rd Law to arrie at the sae conclusion. Objects, Rather than Point Particles The ector su of all the graitational field ectors due to a spherically syetric distribution of point particles (for instance, a spherically syetric solid object), at a point outside the distribution (e.g. outside the object), is the sae as the graitational field ector due to a single particle, at the center of the distribution, whose ass is equal to the su of the asses of all the particles. Also, for purposes of calculating the force eerted by the graitational field of a point object on a spherically syetric icti, one can treat the icti as a point object at the center of the icti. Finally, regarding either object in a calculation of the graitational force eerted on a rigid object by the another object: if the separation of the objects is ery large copared to the diensions of the object, one can treat the object as a point particle located at the center of ass of the object and haing the sae ass as the object. This goes for the graitational potential energy, discussed below, as well. 06

107 Chapter 7 The Uniersal Law of Graitation How Does this Fit in with g = 9.80 N/kg? When we talked about the earth s near-surface graitational field before, we used a single alue for its agnitude, naely 9.80 N/kg and said that it always directed downward, toward the center of the earth N/kg is actually an aerage sea leel alue. g aries within about a half a percent of that alue oer the surface of the earth and the handbook alue includes a tiny centrifugal pseudo force-per-ass field contribution (affecting both agnitude and direction) steing fro the rotation of the earth. So how does the alue 9.80 N/kg for the agnitude of the graitational field near the surface of the earth relate to Newton s Uniersal Law of Graitation? Certainly the direction is consistent with our understanding of what it should be: The earth is roughly spherically syetric so for purposes of calculating the graitational field outside of the earth we can treat the earth as a point particle located at the center of the earth. The direction of the graitational field of a point particle is toward that point particle, so, anywhere outside the earth, including at any point just outside the earth (near the surface of the earth), the graitational field, according to the Uniersal Law of Graitation, ust be directed toward the center of the earth, a direction we earthlings call downward. ut how about the agnitude? Shouldn t it ary with eleation according to the Uniersal Law of Graitation? First off, how does the agnitude calculated using g = G copare with r N/kg at, for instance, sea leel. A point at sea leel is a distance r = fro the center of the earth. The ass of the earth is = kg. Substituting these alues into our epression for g (equation 7-4 which reads g = G ) we find: r g = N kg N g = 9.83 kg ( which does agree with the alue of 9.80 N/kg to within about 0.3 percent. (The difference is due in part to the centrifugal pseudo force-per-ass field associated with the earth s rotation.) We 6 can actually see, just fro the way that the alue of the radius of the earth, , is written, that increasing our eleation, een by a ile (.6 k), is not going to change the alue 6 6 of g to three significant figures. We d be increasing r fro to which, to three significant figures is still This brings up the question, How high N aboe the surface do you hae to go before g = 9.80 is no longer within a certain percentage kg of the alue obtained by eans of the Uniersal Law of Graitation? Let s inestigate that question for the case of a percent difference. In other words, how high aboe sea leel would 4 6 kg ) 07

108 Chapter 7 The Uniersal Law of Graitation we hae to go to ake g = 99% of g at sea leel, that is g = (.99)(9.80 /s ) = 9.70 /s. Letting r = r E + h where r E is the radius of the earth, and using g = 9.70 N/kg so that we can find the h that akes 9.70 N/kg we hae: Soling this for h yields GE g = ( r + h) E h = G E g r E h = N kg N kg kg h = That is to say that at any altitude aboe 40 k aboe the surface of the earth, you should use Newton s Uniersal Law of Graitation directly in order for your results to be good to within %. In suary, g = 9.80 N/kg for the near-earth-surface graitational field agnitude is an approiation to the Uniersal Law of Graitation good to within about % anywhere within about 40 k of the surface of the earth. In that region, the alue is approiately a constant because changes in eleation represent such a tiny fraction of the total earth s-center-to-surface distance as to be negligible. 08

109 Chapter 7 The Uniersal Law of Graitation The Uniersal Graitational Potential Energy So far in this course you hae becoe failiar with two kinds of potential energy, the near-earth s-surface graitational potential energy U = gy and the spring potential energy U S = k. Here we introduce another epression for graitational potential energy. This one is pertinent to situations for which the Uniersal Law of Graitation is appropriate. g U G G r = (7-5) This is the graitational potential energy of a pair of particles, one of ass and the other of ass, which are separated by a distance of r. Note that for a gien pair of particles, the graitational potential energy can take on alues fro negatie infinity up to zero. Zero is the highest possible alue and it is the alue of the graitational potential energy at infinite separation. That is to say that U G 0. The lowest conceiable alue is negatie li r infinity and it would be the alue of the graitational potential energy of the pair of particles if one could put the so close together that they were both at the sae point in space. In atheatical notation: U G. li r 0 Conseration of Mechanical Energy Probles Inoling Uniersal Graitational Potential Energy for the Special Case in which the Total Aount of Mechanical Energy does not Change You sole fied aount of echanical energy probles inoling uniersal graitational potential energy just as you soled fied aount of echanical energy probles inoling other fors of potential energy back in chapters and 3. Draw a good before and after picture, write an equation stating that the energy in the before picture is equal to the energy in the after picture, and proceed fro there. To reiew these procedures, check out the eaple proble on the net page: 09

110 Chapter 7 The Uniersal Law of Graitation Eaple 7-: How great would the uzzle elocity of a gun on the surface of the oon hae to be in order to shoot a bullet to an altitude of 0 k? Solution: We ll need the following lunar data: 6 Mass of the oon: = kg ; Radius of the oon: r = =? = 0 h Moon r = r r = r r Moon r = r + h r = r = E = E K + U G + r = K + U G = r ( is the ass of the oon and is the ass of the bullet.) G = r G = r G r G r = G r r = = G r r N kg kg = 556 s 0

111 Chapter 8 Circular Motion: Centripetal Acceleration 8 Circular Motion: Centripetal Acceleration There is a tendency to beliee that if an object is oing at constant speed then it has no acceleration. This is indeed true in the case of an object oing along a straight line path. On the other hand, a particle oing on a cured path is accelerating whether the speed is changing or not. Velocity has both agnitude and direction. In the case of a particle oing on a cured path, the direction of the elocity is continually changing, and thus the particle has acceleration. We now turn our attention to the case of an object oing in a circle. We ll start with the siplest case of circular otion, the case in which the speed of the object is a constant, a case referred to as unifor circular otion. For the oent, let s hae you be the object. Iagine that you are in a car that is traeling counterclockwise, at say 40 ph, as iewed fro aboe, around a fairly sall circular track. You are traeling in a circle. Your elocity is not constant. The agnitude of your elocity is not changing (constant speed), but the direction of your elocity is continually changing, you keep turning left! Now if you are continually turning left then you ust be continually acquiring soe leftward elocity. In fact, your acceleration has to be eactly leftward, at right angles to your elocity because, if your speed is not changing, but d your elocity is continually changing, eaning you hae soe acceleration a =, then for dt eery infinitesial change in clock reading dt, the change in elocity d that occurs during that infinitesial tie interal ust be perpendicular to the elocity itself. (If it wasn t perpendicular, then the speed would be increasing or decreasing.) So no atter where you are in the circle (around which you are traeling counterclockwise as iewed fro aboe) you hae an acceleration directed eactly leftward, perpendicular to the direction of your elocity. Now what is always directly leftward of you if you are traeling counterclockwise around a circle? Precisely! The center of the circle is always directly leftward of you. Your acceleration is thus, always, center directed. We call the center-directed acceleration associated with circular otion centripetal acceleration because the word centripetal eans center-directed. Note that if you are traeling around the circle clockwise as iewed fro aboe, you are continually turning right and your acceleration is directed rightward, straight toward the center of the circle. These considerations apply to any object an object oing in a circle has centripetal (center-directed) acceleration. We hae a couple of ways of characterizing the otion of a particle that is oing in a circle. First, we characterize it in ters of how far the particle has traeled along the circle. If we need a position ariable, we establish a start point on the circle and a positie direction. For instance, for a circle centered on the origin of an -y plane we can define the point where the circle intersects the positie ais as the start point, and define the direction in which the particle ust oe to go counterclockwise around the circle as the positie direction. The nae gien to this position ariable is s. The position s is the total distance, easured along the circle, that the ds particle has traeled. The speed of the particle is then the rate of change of s, and the dt direction of the elocity is tangent to the circle. The circle itself is defined by its radius. The second ethod of characterizing the otion of a particle is to describe it in ters of an

112 Chapter 8 Circular Motion: Centripetal Acceleration iaginary line segent etending fro the center of a circle to the particle. To use this ethod, one also needs to define a reference line segent the positie ais is the conentional choice for the case of a circle centered on the origin of an -y coordinate syste. Then, as long as you know the radius r of the circle, the angle θ that the line to the particle akes with the reference line copletely specifies the location of the particle. r s θ In geoetry, the position ariable s, defines an arc length on the circle. Recall that, by definition, the angle θ in radians is the ratio of the arc length to the radius: Soling for s we hae: θ = s r s =r θ (8-) in which we interpret the s to be the position-on-the-circle of the particle and the θ to be the angle that an iaginary line segent, fro the center of the circle to the particle, akes with a reference line segent, such as the positie -ais. Clearly, the faster the particle is oing, the faster the angle theta is changing, and indeed we can get a relation between the speed of the particle and the rate of change of θ just by taking the tie deriatie of both sides of equation 8-. Let s do that.

113 Chapter 8 Circular Motion: Centripetal Acceleration Starting with equation 8-: s = r θ we take the deriatie of both sides with respect to tie: and then rewrite the result as ds dt. dθ = r dt. s = r θ just to get the reader used to the idea that we represent the tie deriatie of a ariable, that is the rate of change of that ariable, by the writing the sybol for the ariable with a dot oer it. Then we rewrite the result as. = r θ (8-) to ephasize the fact that the rate of change of the position-on-the-circle is the speed of the particle (the agnitude of the elocity of the particle). Finally, we define the ariable w dθ. ( oega ) to be the rate of change of the angle, eaning that w is and w is θ. It should be dt clear that w is the spin rate for the iaginary line fro the center of the circle to the particle. We call that spin rate the agnitude of the angular elocity of the line segent. (The epression angular elocity, w, is ore coonly used to characterize how fast and which way a rigid.. body, rather than an iaginary line, is spinning.) Rewriting = r θ with θ replaced by w yields: = rw (8-3) 3

114 Chapter 8 Circular Motion: Centripetal Acceleration How the Centripetal Acceleration Depends on the Speed of the Particle and the Size of the Circle We are now in a position to derie an epression for that center-directed (centripetal) acceleration we were talking about at the start of this chapter. Consider a short tie interal t. (We will take the liit as t goes to zero before the end of this chapter.) During that short tie interal, the particle traels a distance s along the circle and the angle that the line, fro the center of the circle to the particle, akes with the reference line changes by an aount θ. r s θ s θ Furtherore, in that tie t, the elocity of the particle changes fro to, a change defined by = + depicted in the following ector diagra (in which the arrows representing the ectors and hae been copied fro aboe with no change in orientation or length). Note that the sall angle θ appearing in the ector addition diagra is the sae θ that appears in the diagra aboe. θ 4

115 Chapter 8 Circular Motion: Centripetal Acceleration While is a new ector, different fro, we hae stipulated that the speed of the particle is a constant, so the ector has the sae agnitude as the ector. That is, =. We redraw the ector addition diagra labeling both elocity ectors with the sae sybol. θ The agnitude of the centripetal acceleration, by definition, can be epressed as a c = li t 0 t Look at the triangle in the ector addition diagra aboe. It is an isosceles triangle. The two unlabeled angles in the triangle are equal to each other. Furtherore, in the liit as t approaches 0, θ approaches 0, and as θ approaches 0, the other two angles ust each approach 90 in order for the su of the angles to reain 80, as it ust, because the su of the interior angles for any triangle is 80. Thus in the liit as t approaches 0, the triangle is a right triangle and in that liit we can write: = tan( θ ) = tan( θ ) Substituting this into our epression for a c we hae: a c tan( θ ) = li (8-4) t 0 t Now we inoke the sall angle approiation fro the atheatics of plane geoetry, an approiation which becoes an actual equation in the liit as θ approaches zero. The Sall Angle Approiation For any angle that is ery sall copared to π radians (the saller the angle the better the approiation), the tangent of the angle is approiately equal to the angle itself, epressed in radians; and the sine of the angle is approiately equal to the angle itself, epressed in radians. In fact, where θ is in radians. tan ( θ ) θ, and sin ( θ ) θ θ 0 The sall angle approiation allows us to write θ 0 5

116 Chapter 8 Circular Motion: Centripetal Acceleration a c θ = li [where we hae replaced the tan( θ ) in equation 8-4 aboe with θ ]. t 0 t θ θ The constant can be taken outside the liit yielding ac = li. ut the li is the t 0 t t 0 t rate of change of the angle θ, which is, by definition, the angular elocity w. Thus a c =w According to equation 8-3, = rw. Soling that for w we find that w =. Substituting this r into our epression for a c yields c = a (8-5) r Please sound the dru roll! This is the result we hae been seeking. Note that by substituting r w for, we can also write our result as a c = rw (8-6) It should be pointed out that, despite the fact that we hae been focusing our attention on the case in which the particle oing around the circle is oing at constant speed, the particle has centripetal acceleration whether the speed is changing or not. If the speed of the particle is changing, the centripetal acceleration at any instant is (still) gien by 8-5 with the being the speed of the particle at that instant (and in addition to the centripetal acceleration, the particle also has soe along-the-circular-path acceleration known as tangential acceleration). The case that we hae inestigated is, howeer the rearkable case. Een if the speed of the particle is constant, the particle has soe acceleration just because the direction of its elocity is continually changing. What s ore, the centripetal acceleration is not a constant acceleration because its direction is continually changing. Visualize it. If you are driing counterclockwise (as iewed fro aboe) around a circular track, the direction in which you see the center of the circle is continually changing (and that direction is the direction of the centripetal acceleration). When you are on the easternost point of the circle the center is to the west of you. When you are at the northernost point of the circle the center is to the south of you. When you are at the westernost point of the circle, the center is to the east of you. And when you are at the southernost point of the circle, the center is to the north of you. 6

117 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration ecause so uch of the effort that we deote to dealing with angles inoles acute angles, when we go to the opposite etree, e.g. to angles of thousands of degrees, as we often do in the case of objects spinning with a constant angular acceleration, one of the ost coon istakes we huans tend to ake is siply not to recognize that when soeone asks us; starting fro tie zero, how any reolutions, or equialently how any turns or rotations an object akes; that soeone is asking for the alue of the angular displaceent θ. To be sure, we typically calculate θ in radians, so we hae to conert the result to reolutions before reporting the final answer, but the nuber of reolutions is siply the alue of θ. In the last chapter we found that a particle in unifor circular otion has centripetal acceleration gien by equations 8-5 and 8-6: a c = r a = rw c It is iportant to note that any particle undergoing circular otion has centripetal acceleration, not just those in unifor (constant speed) circular otion. If the speed of the particle (the alue of in a c = ) is changing, then the alue of the centripetal acceleration is clearly changing. r One can still calculate it at any instant at which one knows the speed of the particle. If, besides the acceleration that the particle has just because it is oing in a circle, the speed of the particle is changing, then the particle also has soe acceleration directed along (or in the eact opposite direction to) the elocity of the particle. Since the elocity is always tangent to the circle on which the particle is oing, this coponent of the acceleration is referred to as the tangential acceleration of the particle. The agnitude of the tangential acceleration of a particle in circular otion is siply the absolute alue of the rate of change of the speed of the particle d at =. The direction of the tangential acceleration is the sae as that of the elocity if the dt particle is speeding up, and in the direction opposite that of the elocity if the particle is slowing down. Recall that, starting with our equation relating the position s of the particle along the circle to the angular position θ of a particle, s =rθ, we took the deriatie with respect to tie to get the relation = rw. If we take a second deriatie with respect to tie we get d dw = r dt dt 7

118 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration dw On the left we hae the tangential acceleration a t of the particle. The on the right is the dt tie rate of change of the angular elocity of the object. The angular elocity is the spin rate, so dw a non-zero alue of eans that the iaginary line segent that etends fro the center of dt dw the circle to the particle is spinning faster or slower as tie goes by. In fact, is the rate at dt which the spin rate is changing. We call it the angular acceleration and use the sybol a (the d dw Greek letter alpha) to represent it. Thus, the relation = r can be epressed as dt dt a =ra (9-) t A Rotating Rigid ody The characterization of the otion of a rotating rigid body has a lot in coon with that of a particle traeling on a circle. In fact, eery particle aking up a rotating rigid body is undergoing circular otion. ut different particles aking up the rigid body oe on circles of different radii and hence hae speeds and accelerations that differ fro each other. For instance, each tie the object goes around once, eery particle of the object goes all the way around its circle once, but a particle far fro the ais of rotation goes all the way around circle that is bigger than the one that a particle that is close to the ais of rotation goes around. To do that, the particle far fro the ais of rotation ust be oing faster. ut in one rotation of the object, the line fro the center of the circle that any particle of the object is on, to the particle, turns through eactly one rotation. In fact, the angular otion ariables that we hae been using to characterize the otion of a line etending fro the center of a circle to a particle that is oing on that circle can be used to characterize the otion of a spinning rigid body as a whole. There is only one spin rate for the whole object, the angular elocity w, and if that spin rate is changing, there is only one rate of change of the spin rate, the angular acceleration a. To specify the angular position of a rotating rigid body, we need to establish a reference line on the rigid body, etending away fro a point on the ais of rotation in a direction perpendicular to the ais of rotation. This reference line rotates with the object. Its otion is the angular otion of the object. We also need a reference line segent that is fied in space, etending fro the sae point on the ais, and away fro the ais in a direction perpendicular to the ais. This one does not rotate with the object. Iagining the two lines to hae at one tie been collinear, the net angle through which the first line on the rigid body has turned relatie to the fied line is the angular position θ of the object. The Constant Angular Acceleration Equations While physically, there is a huge difference, atheatically, the rotational otion of a rigid body is identical to otion of a particle that only oes along a straight line. As in the case of linear otion, we hae to define a positie direction. We are free to define the positie direction whicheer way we want for a gien proble, but we hae to stick with that definition throughout the proble. Here, we establish a iewpoint soe distance away fro the rotating rigid body, 8

119 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration but on the ais of rotation, and state that, fro that iewpoint, counterclockwise is the positie sense of rotation, or alternatiely, that clockwise is the positie sense of rotation. Whicheer way we pick as positie, will be the positie sense of rotation for angular displaceent (change in angular position), angular elocity, angular acceleration, and angular position relatie to the reference line that is fied in space. Net, we establish a zero for the tie ariable; we iagine a stopwatch to hae been started at soe instant that we define to be tie zero. We call alues of angular position and angular elocity, at that instant, the initial alues of those quantities. Gien these criteria, we hae the following table of corresponding quantities. Note that a rotational otion quantity is in no way equal to its linear otion counterpart, it siply plays a role in rotational otion that is atheatically siilar to the role played by its counterpart in linear otion. Linear Motion Quantity a Corresponding Angular Motion Quantity θ w a The one ariable that the two different kinds of otion do hae in coon is the stopwatch reading t. Recall that, by definition, and dθ w = dt dw a = dt While it is certainly possible for a to be a ariable, any cases arise in which a is a constant. Such a case is a special case. The following set of constant angular acceleration equations apply in the special case of constant angular acceleration: (The deriation of these equations is atheatically equialent to the deriation of the constant linear acceleration equations. Rather than derie the again, we siply present the results.) θ = θo +w o t + a t (9-) The Constant Angular Acceleration Equations w o +w θ = θ + t (9-3) o w = w o + a t (9-4) w = w + a θ (9-5) o 9

120 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration Eaple 9- The rate at which a sprinkler head spins about a ertical ais increases steadily for the first.00 seconds of its operation such that, starting fro rest, the sprinkler copletes 5.0 reolutions clockwise (as iewed fro aboe) during that first.00 seconds of operation. A nozzle, on the sprinkler head, at a distance of.0 c fro the ais of rotation of the sprinkler head, is initially due west of the ais of rotation. Find the direction and agnitude of the acceleration of the nozzle at the instant the sprinkler head copletes its second (good to three significant figures) rotation. Solution: We re told that the sprinkler head spin rate increases steadily, eaning that we are dealing with a constant angular acceleration proble, so, we can use the constant angular acceleration equations. The fact that there is a non-zero angular acceleration eans that the nozzle will hae soe tangential acceleration a t. Also, the sprinkler head is spinning at the instant in question so the nozzle will hae soe centripetal acceleration a c. We ll hae to find both a t and a c and add the like ectors to get the total acceleration of the nozzle. Let s get started by finding the angular acceleration a. We start with the first constant angular acceleration equation (equation 9-): 0 0 θ =θ +w t + o o The initial angular elocity wo is gien as zero. We hae defined the initial angular position to be zero. This eans that, at tie t =.00 s, the angular position θ is π rad 5.0 re = 5. 0re = 94.5 rad. re Soling equation 9- aboe for a yields: a t a = θ t a = (94. 5rad) (. 00 s) Substituting this result into equation 9-: rad a = 47. s gies us a t a t =ra = (. 0 ) 47. rad/s 0

121 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration which ealuates to a t = 5.8. s Now we need to find the angular elocity of the sprinkler head at the instant it copletes.00 reolutions. The angular acceleration a that we found is constant for the first fifteen reolutions, so the alue we found is certainly good for the first two turns. We can use it in the fourth constant angular acceleration equation (equation 9-5): 0 w = w + a θ where π rad θ = re =. 00 re = 4. 00π rad re o w = a θ w = (94. 5rad/s ) 4. 00π rad w = rad / s (at that instant when the sprinkler head copletes its nd turn) Now that we hae the angular elocity, to get the centripetal acceleration we can use equation 8-6: a = rw c a =.0. c (48 67 rad /s) a c = Gien that the nozzle is initially at a point due west of the ais of rotation, at the end of.00 reolutions it will again be at that sae point. s NORTH a t a c EAST

122 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration Now we just hae to add the tangential acceleration and the centripetal acceleration ectorially to get the total acceleration. This is one of the easier kinds of ector addition probles since the ectors to be added are at right angles to each other. North a θ a c = 60.6 /s a t = 5.8 /s East Fro Pythagorean s theore we hae a = + a c at a = ( 60 6 / s ).. + (5 8 / s ) a = 6 /s Fro the definition of the tangent of an angle as the opposite oer the adjacent: tanθ = at a c θ = tan at a c θ = tan 5. 8/s /s o θ =.4 Thus a = 6 / s at.4 North of East

123 Chapter 9 Rotational Motion Variables, Tangential Acceleration, Constant Angular Acceleration When the Angular Acceleration is not Constant The angular position of a rotating body undergoing constant angular acceleration is gien, as a function of tie, by our first constant angular acceleration equation, equation 9-: θ = θ + w t + o o a t If we take the nd deriatie of this with respect to tie, we get the constant a. (Recall that the dw first deriatie yields the angular elocity w and that a =.) The epression on the right dt side of θ = θ + w t + a t contains three ters: a constant, a ter with t to the first power, and o o a ter with t to the nd power. If you are gien θ in ters of t, and it cannot be rearranged so that it appears as one of these ters or as a su of two or all three such ters; then; a is not a constant and you cannot use the constant angular acceleration equations. Indeed, if you are being asked to find the angular elocity at a particular instant in tie, then you ll want to take dθ the deriatie and ealuate the result at the gien stopwatch reading. Alternatiely, if you dt are being asked to find the angular acceleration at a particular instant in tie, then you ll want to d θ take the second deriatie and ealuate the result at the gien stopwatch reading. dt Corresponding arguents can be ade for the case of w. If you are gien w as a function of t and the epression cannot be ade to look like the constant angular acceleration equation w = w o + a t then you are not dealing with a constant angular acceleration situation and you should not use the constant angular acceleration equations. 3

124 Chapter 0 Torque & Circular Motion 0 Torque & Circular Motion The istake that crops up in the application of Newton s nd Law for Rotational Motion inoles the replaceent of the su of the torques about soe particular ais, τ o, with a su of ters that are not all torques. Oftenties, the errant su will include forces with no oent ars (a force ties a oent ar is a torque, but a force by itself is not a torque) and in other cases the errant su will include a ter consisting of a torque ties a oent ar (a torque is already a torque, ultiplying it by a oent ar yields soething that is not a torque). Folks that are in the habit of checking units will catch their istake as soon as they plug in alues with units and ealuate. We hae studied the otion of spinning objects without any discussion of torque. It is tie to address the link between torque and rotational otion. First, let s reiew the link between force and translational otion. (Translational otion has to do with the otion of a particle through space. This is the ordinary otion that you e worked with quite a bit. Until we started talking about rotational otion we called translational otion otion. Now, to distinguish it fro rotational otion, we call it translational otion.) The real answer to the question of what causes otion to persist, is nothing a oing particle with no force on it keeps on oing at constant elocity. Howeer, wheneer the elocity of the particle is changing, there is a force. The direct link between force and otion is a relation between force and acceleration. The relation is known as Newton s nd Law of Motion which we hae written as equation 4-: in which, = a F a is the acceleration of the object, how fast and which way its elocity is changing is the ass, a.k.a. inertia, of the object. can be iewed as a sluggishness factor, the bigger the ass, the saller the alue of and hence the saller the acceleration of the object for a gien net force. ( Net in this contet just eans total.) F is the ector su of the forces acting on the object, the net force. 4

125 Chapter 0 Torque & Circular Motion We find a copletely analogous situation in the case of rotational otion. The link in the case of rotational otion is between the angular acceleration of a rigid body and the torque being eerted on that rigid body. in which, a = τ (0-) I a is the angular acceleration of the rigid body, how fast and which way the angular elocity is changing I is the oent of inertia, a.k.a. the rotational inertia (but not just plain old inertia, which is ass) of the rigid body. It is the rigid body s inherent resistance to a change in how fast it (the rigid body) is spinning. ( Inherent eans of itself, part of its own being. ) I can be iewed as a sluggishness factor, the bigger the rotational inertia I, the saller the alue of I and hence the saller the angular acceleration of the object for a gien net torque. τ is the net torque acting on the object. (A torque is what you are applying to a bottle cap or jar lid when you are trying to unscrew it.) 5

126 Chapter 0 Torque & Circular Motion The Vector Nature of Torque and Angular Velocity You e surely noticed the arrows oer the letters used to represent torque, angular acceleration, and angular elocity; and as you know, the arrows ean that the quantities in question are ector quantities. That eans that they hae both agnitude and direction. Soe eplanation about the direction part is in order. Let s start with the torque. As entioned, it is a twisting action such as that which you apply to bottle cap to loosen or tighten the bottle cap. There are two ways to specify the direction associated with torque. One way is to identify the ais of rotation about which the torque is being applied, then to establish a iewpoint, a position on the ais, at a location that is in one direction or the other direction away fro the object. Then either state that the torque is clockwise, or state that it is counterclockwise, as iewed fro the specified iewpoint. Note that it is not sufficient to identify the ais and state clockwise or counterclockwise without giing the iewpoint a torque which is clockwise fro one of the two iewpoints is counterclockwise fro the other. The second ethod of specifying the direction is to gie the torque ector direction. The conention for the torque ector is that the ais of rotation is the line on which the torque ector lies, and the direction is in accord with the right hand rule for soething curly soething straight. y the the right hand rule for soething curly soething straight, if you point the thub of your cupped right hand in the direction of the torque ector, the fingers will be curled around in that direction which corresponds to the sense of rotation (counterclockwise as iewed fro the head of the torque ector looking back along the shaft) of the torque. The angular acceleration ector a and the angular elocity ector w obey the sae conention. These ectors, which point along the ais about which the rotation they represent occurs, are referred to as aial ectors. 6

127 Chapter 0 Torque & Circular Motion The Torque Due to a Force When you apply a force to a rigid body, you are typically applying a torque to that rigid body at the sae tie. Consider an object that is free to rotate about a fied ais. We depict the object as iewed fro a position on the ais, soe distance away fro the object. Fro that iewpoint, the ais looks like a dot. We gie the nae point O to the position at which the ais of rotation appears in the diagra and label it O in the diagra to ake it easier to refer to later in this discussion. There is a force F acting on the object. Ais of Rotation O F The agnitude of the torque due to a force is the agnitude of the force ties the oent ar r (read r perp ) of the force. The oent ar r is the perpendicular distance fro the ais of rotation to the line of action of the force. The line of action of the force is a line that contains the force ector. Here we redraw the gien diagra with the line of action of the force drawn in. Line of Action of the Force Ais of Rotation O F 7

128 Chapter 0 Torque & Circular Motion Net we etend a line segent fro the ais of rotation to the line of action of the force, in such a anner that it eets the line of action of the force at right angles. The length of this line segent is the oent ar r. The Moent Ar Line of Action of the Force Ais of Rotation O r F The agnitude of the torque about the specified ais of rotation is just the product of the oent ar and the force. τ = r F (0-) Applying Newton s Second Law for Rotational Motion in Cases Inoling a Fied Ais Starting on the net page, we tell you what steps are required (and what diagra is required) in the solution of a fied-ais Newton s nd Law for Rotational Motion proble by eans of an eaple. 8

129 Chapter 0 Torque & Circular Motion Eaple 0-: A flat etal rectangular plate lies on a flat horizontal frictionless surface with (at the instant in question) one corner at the origin of an -y coordinate syste and the opposite corner at point P which is at (93, 45 ). The plate is pin connected to the horizontal surface at (0.0 c, 0.0 c). A counterclockwise (as iewed fro aboe) torque, with respect to the pin, of 5.0 N, is being applied to the plate and a force of.0 N in the y direction is applied to the corner of the plate at point P. The oent of inertia of the plate, with respect to the pin, is.8 kg. Find the angular acceleration of the plate at the instant for which the specified conditions preail. We start by drawing a pseudo free body diagra of the object as iewed fro aboe (downward is into the page ): y F =.0 N I =. 8 kg τ = 5.0 N a O 0.93 We refer to the diagra as a pseudo free body diagra rather than a free body diagra because: a. We oit forces that are parallel to the ais of rotation (because they do not affect the rotation of the object about the ais of rotation). In the case at hand, we hae oitted the force eerted on the plate by the graitational field of the earth (which would be into the page in the diagra) as well as the noral force eerted by the frictionless surface on the plate ( out of the page ). b. We ignore forces eerted on the plate by the pin. (Such forces hae no oent ar and hence do not affect the rotation of the plate about the ais of rotation. Note, a pin can, howeer, eert a frictional torque assue it to be zero unless otherwise specified.) Pin connected to the horizontal surface eans that there is a short ertical ale fied to the horizontal surface and passing through a sall round hole in the plate so that the plate is free to spin about the ale. Assue that there is no frictional torque eerted on a pin-connected object unless otherwise specified in the case at hand. 9

130 Chapter 0 Torque & Circular Motion Net we annotate the pseudo free body diagra to facilitate the calculation of the torque due to the force F : y F =.0 N I =. 8 kg τ = 5.0 N a O r = Line of Action of the Force Now we go ahead and apply Newton s nd Law for Rotational Motion, equation 0- : a = τ I As in the case of Newton s nd Law (for translational otion) this equation is three scalar equations in one, one equation for each of three utually perpendicular aes about which rotation, under the ost general circustances, could occur. In the case at hand, the object is constrained to allow rotation about a single ais. In our solution, we need to indicate that we are suing torques about that ais, and we need to indicate which of the two possible rotational senses we are defining to be positie. We do that by eans of the subscript o to be read counterclockwise about point O. Newton s nd Law for rotational otion about the ertical ais (perpendicular to the page and represented by the dot labeled O in the diagra) reads: a o = τ (0-3) I o 30

131 Chapter 0 Torque & Circular Motion Now, when we replace the epression τ with the actual ter-by-ter su of the torques, I o we note that τ is indeed counterclockwise as iewed fro aboe (and hence positie) but that the force F, where it is applied, would tend to cause a clockwise rotation of the plate, eaning that the torque associated with force F is clockwise and hence, ust enter the su with a inus sign. Substituting alues with units yields: a =. 8 kg a = ( τ T r I F ) [ 5. 0 N 0. 93( 0N) ]. Ealuating and rounding the answer to three significant figures gies us the final answer: Regarding the units we hae: rad a = 8.55 (counterclockwise as iewed fro aboe) s kg N = kg = kg s s = rad s kg where we hae taken adantage of the fact that a newton is a s is not a true unit but rather a arker that can be inserted as needed. and the fact that the radian 3

132 Chapter Vectors: The Cross Product & Torque Vectors: The Cross Product & Torque Do not use your left hand when applying either the right-hand rule for the cross product of two ectors (discussed in this chapter) or the right-hand rule for soething curly soething straight discussed in the preceding chapter. There is a relational operator for ectors that allows us to bypass the calculation of the oent ar. The relational operator is called the cross product. It is represented by the sybol read cross. The torque τ can be epressed as the cross product of the position ector r for the point of application of the force, and the force ector F itself: τ =r F (-) efore we begin our atheatical discussion of what we ean by the cross product, a few words about the ector r are in order. It is iportant for you to be able to distinguish between the position ector r for the force, and the oent ar, so we present the below in one and the sae diagra. We use the sae eaple that we hae used before: Ais of Rotation O Position of the Point of Application of the Force F in which we are looking directly along the ais of rotation (so it looks like a dot) and the force lies in a plane perpendicular to that ais of rotation. We use the diagraatic conention that, the point at which the force is applied to the rigid body is the point at which one end of the arrow in the diagra touches the rigid body. Now we add the line of action of the force and the oent ar r to the diagra, as well as the position ector r of the point of application of the force. You are uch ore failiar with relational operators then you ight realize. The + sign is a relational operator for scalars (nubers). The operation is addition. Applying it to the nubers and 3 yields +3=5. You are also failiar with the relational operators,, and for subtraction, ultiplication, and diision (of scalars) respectiely. 3

133 Chapter Vectors: The Cross Product & Torque The Moent Ar Line of Action of the Force r O r Position Vector for the Point of Application of the Force F The oent ar can actually be defined in ters of the position ector for the point of application of the force. Consider a tilted -y coordinate syste, haing an origin on the ais of rotation, with one ais parallel to the line of action of the force and one ais perpendicular to the line of action of the force. We label the ais for perpendicular and the y ais for parallel. O r F 33

134 and Chapter Vectors: The Cross Product & Torque Now we break up the position ector r into its coponent ectors along the aes. r O r r F Fro the diagra it is clear that the oent ar r is just the agnitude of the coponent ector, in the perpendicular-to-the-force direction, of the position ector of the point of application of the force. 34

135 Chapter Vectors: The Cross Product & Torque Now let s discuss the cross product in general ters. Consider two ectors, A and that are neither parallel nor anti-parallel to each other. Two such ectors define a plane. Let that plane be the plane of the page and define θ to be the saller of the two angles between the two ectors when the ectors are drawn tail to tail. θ The agnitude of the cross product ector A A is gien by = A sinθ The direction of the cross product ector A is gien by the right-hand rule for the cross product of two ectors 3. To apply this right-hand rule, etend the fingers of your right hand so that they are pointing directly away fro your right elbow. Etend your thub so that it is at right angles to your fingers. A (-) Two ectors that are anti-parallel are in eact opposite directions to each other. The angle between the is 80 degrees. Anti-parallel ectors lie along parallel lines or along one and the sae line, but point in opposite directions. 3 You need to learn two right-hand rules for this course: the right-hand rule for soething curly soething straight, and this one, the right-hand rule for the cross product of two ectors. 35

136 Chapter Vectors: The Cross Product & Torque Keeping your fingers aligned with your forear, point your fingers in the direction of the first ector (the one that appears before the in the atheatical epression for the cross product; e.g. the A in A ). A Now rotate your hand, as necessary, about an iaginary ais etending along your forear and along your iddle finger, until your hand is oriented such that, if you were to close your fingers, they would point in the direction of the second ector. This thub is pointing straight out of the page, right at you! Your thub is now pointing in the direction of the cross product ector. C = A. The cross product ector C is always perpendicular to both of the ectors that are in the cross product (the A and the in the case at hand). Hence, if you draw the so that both of the ectors that are in the cross product are in the plane of the page, the cross product ector will always be perpendicular to the page, either straight into the page, or straight out of the page. In the case at hand, it is straight out of the page. A 36

137 Chapter Vectors: The Cross Product & Torque When we use the cross product to calculate the torque due to a force F whose point of application has a position ector r, relatie to the point about which we are calculating the torque, we get an aial torque ector τ. To deterine the sense of rotation that such a torque ector would correspond to, about the ais defined by the torque ector itself, we use The Right Hand Rule For Soething Curly Soething Straight. Note that we are calculating the torque with respect to a point rather than an ais the ais about which the torque acts, coes out in the answer. Calculating the Cross Product of Vectors that are Gien in i, j, k Notation Unit ectors allow for a straightforward calculation of the cross product of two ectors under een the ost general circustances, e.g. circustances in which each of the ectors is pointing in an arbitrary direction in a three-diensional space. To take adantage of the ethod, we need to know the cross product of the Cartesian coordinate ais unit ectors i, j, and k with each other. First off, we should note that any ector crossed into itself gies zero. This is eident fro equation -: A = A sinθ, because if A and are in the sae direction, then θ = 0, and since sin 0 = 0, we hae A = 0. Regarding the unit ectors, this eans that: i i = 0 j j = 0 k k = 0 Net we note that the agnitude of the cross product of two ectors that are perpendicular to each other is just the ordinary product of the agnitudes of the ectors. This is also eident fro equation -: A = A sinθ, because if A is perpendicular to then θ = 90 and sin 90 = so A = A Now if A and are unit ectors, then their agnitudes are both, so, the product of their agnitudes is also. Furtherore, the unit ectors i, j, and k are all perpendicular to each other so the agnitude of the cross product of any one of the with any other one of the is the product of the two agnitudes, that is,. 37

138 Chapter Vectors: The Cross Product & Torque Now how about the direction? Let s use the right hand rule to get the direction of i j : z y k j i Figure With the fingers of the right hand pointing directly away fro the right elbow, and in the sae direction as i, (the first ector in i j ) to ake it so that if one were to close the fingers, they would point in the sae direction as j, the pal ust be facing in the +y direction. That being the case, the etended thub ust be pointing in the +z direction. Putting the agnitude (the agnitude of each unit ector is ) and direction (+z) inforation together we see 4 that i j = k. Siilarly: j k = i, k i = j, j i = k, k j = i, and i k = j. One way of reebering this is to write i, j, k twice in succession: i, j, k, i, j, k. Then, crossing any one of the first three ectors into the ector iediately to its right yields the net ector to the right. ut crossing any one of the last three ectors into the ector 4 You ay hae picked up on a bit of circular reasoning here. Note that in Figure, if we had chosen to hae the z ais point in the opposite direction (keeping and y as shown) then i j would be pointing in the z direction. In fact, haing chosen the + and +y directions, we define the +z direction as that direction that akes i j = k. Doing so fors what is referred to as a right-handed coordinate syste which is, by conention, the kind of coordinate syste that we use in science and atheatics. If i j = k then you are dealing with a left-handed coordinate syste, soething to be aoided. 38

139 Chapter Vectors: The Cross Product & Torque 39 iediately to its left yields the negatie of the net ector to the left (left-to-right +, but right-to-left ). Now we re ready to look at the general case. Any ector A can be epressed in ters of unit ectors: k j i z y A A A + + = A Doing the sae for a ector then allows us to write the cross product as: ) ( ) ( k j i k j i z y z y A A A = A Using the distributie rule for ultiplication we can write this as: ) ( ) ( ) ( k j i k k j i j k j i i z y z z y y z y A A A = A k k j k i k k j j j i j k i j i i i z z y z z z y y y y z y A A A A A A A A A = A Using, in each ter, the coutatie rule and the associatie rule for ultiplication we can write this as: ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( k k j k i k k j j j i j k i j i i i = z z y z z z y y y y z y A A A A A A A A A A Now we ealuate the cross product that appears in each ter: (0) ) ( ) ( ) ( (0) ) ( ) ( ) ( (0) z z y z z z y y y y z y A A A A A A A A A = i j i k j k A Eliinating the zero ters and grouping the ters with i together, the ters with j together, and the ters with k together yields:

140 Chapter Vectors: The Cross Product & Torque 40 ) ( ) ( ) ( ) ( ) ( ) ( k k j j i i = y y z z y z z y A A A A A A A Factoring out the unit ectors yields: ) ( ) ( ) ( k j i y y z z y z z y A A A A A A + + = A which can be written on one line as: k j i ) ( ) ( ) ( y y z z y z z y A A A A A A + + = A (-3) This is our end result. We can arrie at this result uch ore quickly if we borrow a tool fro that branch of atheatics known as linear algebra (the atheatics of atrices). We for the 3 3 atri z y z y A A A k j i by writing i, j, k as the first row, then the coponents of the first ector that appears in the cross product as the second row, and finally the coponents of the second ector that appears in the cross product as the last row. It turns out that the cross product is equal to the deterinant of that atri. We use absolute alue signs on the entire atri to signify the deterinant of the atri. So we hae: z y z y A A A k j i = A (-4) To take the deterinant of a 3 3 atri you work your way across the top row. For each eleent in that row you take the product of the eleents along the diagonal that etends down and to the right, inus the product of the eleents down and to the left; and you add the three results (one result for each eleent in the top row) together. If there are no eleents down and to the appropriate side, you oe oer to the other side of the atri (see below) to coplete the diagonal.

141 Chapter Vectors: The Cross Product & Torque For the first eleent of the first row, the i, take the product down and to the right, ( this yields i A y z ) inus the product down and to the left i A j A y y k A z z i A j A y y k A z z ( the product down-and-to-the-left is i A z y ). For the first eleent in the first row, we thus hae: i A i A which can be written as: ( y z z y A A )i. Repeating the process for the second and third eleents in the first row (the j and the k) we get ( z z y A A ) j and A A ) k respectiely. Adding the three results, ( y y to for the deterinant of the atri results in: A = as we found before, the hard way. ( Ay z Azy z z y y z ) i + ( A A ) j + ( A A ) k z y (-3) 4

142 Chapter Center of Mass, Moent of Inertia Center of Mass, Moent of Inertia A istake that crops up in the calculation of oents of inertia, inoles the Parallel Ais Theore. The istake is to interchange the oent of inertia of the ais through the center of ass, with the one parallel to that, when applying the Parallel Ais Theore. Recognizing that the subscript CM in the parallel ais theore stands for center of ass will help one aoid this istake. Also, a check on the answer, to ake sure that the alue of the oent of inertia with respect to the ais through the center of ass is saller than the other oent of inertia, will catch the istake. Center of Mass Consider two particles, haing one and the sae ass, each of which is at a different position on the ais of a Cartesian coordinate syste. y # # Coon sense tells you that the aerage position of the aterial aking up the two particles is idway between the two particles. Coon sense is right. We gie the nae center of ass to the aerage position of the aterial aking up a distribution, and the center of ass of a pair of sae-ass particles is indeed idway between the two particles. How about if one of the particles is ore assie than the other? One would epect the center of ass to be closer to the ore assie particle, and again, one would be right. To deterine the position of the center of ass of the distribution of atter in such a case, we copute a weighted su of the positions of the particles in the distribution, where the weighting factor for a gien particle is that fraction, of the total ass, that the particle s own ass is. Thus, for two particles on the ais, one of ass, at, and the other of ass, at, y (, 0 ) (, 0 ) 4

143 Chapter Center of Mass, Moent of Inertia the position of the center of ass is gien by = + (-) + + Note that each weighting factor is a proper fraction and that the su of the weighting factors is always. Also note that if, for instance, is greater than, then the position of particle will count ore in the su, thus ensuring that the center of ass is found to be closer to the ore assie particle (as we know it ust be). Further note that if =, each weighting factor is, as is eident when we substitute for both and in equation -: = = + = + The center of ass is found to be idway between the two particles, right where coon sense tells us it has to be. The Center of Mass of a Thin Rod Quite often, when the finding of the position of the center of ass of a distribution of particles is called for, the distribution of particles is the set of particles aking up a rigid body. The easiest rigid body for which to calculate the center of ass is the thin rod because it etends in only one diension. (Here, we discuss an ideal thin rod. A physical thin rod ust hae soe nonzero diaeter. The ideal thin rod, howeer, is a good approiation to the physical thin rod as long as the diaeter of the rod is sall copared to its length.) In the siplest case, the calculation of the position of the center of ass is triial. The siplest case inoles a unifor thin rod. A unifor thin rod is one for which the linear ass density µ, the ass-per-length of the rod, has one and the sae alue at all points on the rod. The center of ass of a unifor rod is at the center of the rod. So, for instance, the center of ass of a unifor rod that etends along the ais fro = 0 to = L is at ( L/, 0 ). The linear ass density µ, typically called linear density when the contet is clear, is a easure of how closely packed the eleentary particles aking up the rod are. Where the linear density is high, the particles are close together. 43

144 Chapter Center of Mass, Moent of Inertia To picture what is eant by a non-unifor rod, a rod whose linear density is a function of position, iagine a thin rod ade of an alloy consisting of lead and aluinu. Further iagine that the percentage of lead in the rod aries soothly fro 0% at one end of the rod to 00% at the other. The linear density of such a rod would be a function of the position along the length of the rod. A one-illieter segent of the rod at one position would hae a different ass than that of a one-illieter segent of the rod at a different position. People with soe eposure to calculus hae an easier tie understanding what linear density is than calculus-depried indiiduals do because linear density is just the ratio of the aount of ass in a rod segent to the length of the segent, in the liit as the length of the segent goes to zero. Consider a rod that etends fro 0 to L along the ais. Now suppose that s () is the ass of that segent of the rod etending fro 0 to where 0 but < L. Then, the linear d density of the rod at any point along the rod, is just s ealuated at the alue of in d question. Now that you hae a good idea of what we ean by linear ass density, we are going to illustrate how one deterines the position of the center of ass of a non-unifor thin rod by eans of an eaple. Eaple - Find the position of the center of ass of a thin rod that etends fro 0 to.890 along the ais of a Cartesian coordinate syste and has a linear density gien kg by µ ( ) = In order to be able to deterine the position of the center of ass of a rod with a gien length and a gien linear density as a function of position, you first need to be able to find the ass of such a rod. To do that, one ight be tepted to use a ethod that works only for the special case of a unifor rod, naely, to try using = µ L with L being the length of the rod. The proble with this is, that µ aries along the entire length of the rod. What alue would one use for µ? One ight be tepted to ealuate the gien µ at = L and use that, but that would be acting as if the linear density were constant at µ = µ (L). It is not. In fact, in the case at hand, µ (L) is the aiu linear density of the rod, it only has that alue at one point on the rod. What we can do is to say that the infinitesial aount of ass d in a segent d of the rod is µ d. Here we are saying that at soe position on the rod, the aount of ass in the infinitesial length d of the rod is the alue of µ at that alue, ties the infinitesial length d. Here we don t hae to worry about the fact that µ changes with position since the segent d is infinitesially long, eaning, essentially, that it has zero length, so the whole segent is essentially at one position and hence the alue of µ at that is good for the whole segent d. d = µ () d (-) 44

145 Chapter Center of Mass, Moent of Inertia y z d (L, 0) d = µ d Now this is true for any alue of, but it just coers an infinitesial segent of the rod at. To get the ass of the whole rod, we need to add up all such contributions to the ass. Of course, since each d corresponds to an infinitesial length of the rod, we will hae an infinite nuber of ters in the su of all the d s. An infinite su of infinitesial ters, is an integral. L d = 0 µ ( ) d (-3) where the alues of hae to run fro 0 to L to coer the length of the rod, hence the liits on the right. Now the atheaticians hae proided us with a rich set of algoriths for ealuating integrals, and indeed we will hae to reach into that toolbo to ealuate the integral on the right, but to ealuate the integral on the left, we cannot, should not, and will not turn to such an algorith. Instead, we use coon sense and our conceptual understanding of what the integral on the left eans. In the contet of the proble at hand, d eans the su of all the infinitesial bits of ass aking up the rod. Now, if you add up all the infinitesial bits of ass aking up the rod, you get the ass of the rod. So d is just the ass of the rod, which we will call. Equation -3 then becoes L = µ ( ) d (-4) 0 kg Replacing µ () with the gien epression for the linear density µ = which I choose kg to write as µ = b with b being defined by b we obtain 3 Factoring out the constant yields L = b 0 d 3 45

146 Chapter Center of Mass, Moent of Inertia = b 0 L d When integrating the ariable of integration raised to a power all we hae to do is increase the power by one and diide by the new power. This gies 3 = b 3 L 0 Ealuating this at the lower and upper liits yields = b 3 3 L = b 3 3 L bl = kg The alue of L is gien as and we defined b to be the constant in the gien 3 kg epression for µ, µ = , so kg ( ) 3 = = 0.57 kg That s a alue that will coe in handy when we calculate the position of the center of ass. Now, when we calculated the center of ass of a set of discrete particles (where a discrete particle is one that is by itself, as opposed, for instance, to being part of a rigid body) we just carried out a weighted su in which each ter was the position of a particle ties its weighting factor and the weighting factor was that fraction, of the total ass, represented by the ass of the particle. We carry out a siilar procedure for a continuous distribution of ass such as that which akes up the rod in question. Let s start by writing one single ter of the su. We ll consider an infinitesial length d of the rod at a position along the length of the rod. The position, as just stated, is, and the weighting factor is that fraction of the total ass of the rod 46

147 Chapter Center of Mass, Moent of Inertia that the ass d of the infinitesial length d represents. That eans the weighting factor is d, so, a ter in our weighted su of positions looks like: d Now, d can be epressed as µ d so our epression for the ter in the weighted su can be written as µ d That s one ter in the weighted su of positions, the su that yields the position of the center of ass. The thing is, because the alue of is unspecified, that one ter is good for any infinitesial segent of the bar. Eery ter in the su looks just like that one. So we hae an epression for eery ter in the su. Of course, because the epression is for an infinitesial length d of the rod, there will be an infinite nuber of ters in the su. So, again we hae an infinite su of infinitesial ters. That is, again we hae an integral. Our epression for the position of the center of ass is: L µ d = kg Substituting the gien epression µ ( ) = for µ, which we again write as µ = b kg with b being defined by b 0.650, yields = L 0 b d Rearranging and factoring the constants out gies Net we carry out the integration. = b L 0 3 d = b 4 4 L 0 47

148 Chapter Center of Mass, Moent of Inertia = b 4 4 L bl = 4 Now we substitute alues with units; the ass of the rod that we found earlier, the constant b that we defined to siplify the appearance of the linear density function, and the gien length L of the rod: kg (0 890 ) 3. = 4(0. 57 kg) = This is our final answer for the position of the center of ass. Note that it is closer to the denser end of the rod, as we would epect. The reader ay also be interested to note that had we substituted the epression 3 bl = that we deried for the ass, rather than the alue we 3 obtained when we ealuated that epression, our epression for would hae siplified to which ealuates to = 0.668, the sae result as the one aboe. 3 4 L Moent of Inertia a.k.a. Rotational Inertia You already know that the oent of inertia of a rigid object, with respect to a specified ais of rotation, depends on the ass of that object, and how that ass is distributed relatie to the ais of rotation. In fact, you know that if the ass is packed in close to the ais of rotation, the object will hae a saller oent of inertia than it would if the sae ass was ore spread out relatie to the ais of rotation. Let s quantify these ideas. (Quantify, in this contet, eans to put into equation for.) We start by constructing, in our inds, an idealized object for which the ass is all concentrated at a single location which is not on the ais of rotation: Iagine a assless disk rotating with angular elocity w about an ais through the center of the disk and perpendicular to its faces. Let there be a particle of ass ebedded in the disk at a distance r fro the ais of rotation. Here s what it looks like fro a iewpoint on the ais of rotation, soe distance away fro the disk: 48

149 Chapter Center of Mass, Moent of Inertia w O r Particle of ass Massless Disk where the ais of rotation is arked with an O. ecause the disk is assless, we call the oent of inertia of the construction, the oent of inertia of a particle, with respect to rotation about an ais fro which the particle is a distance r. Knowing that the elocity of the particle can be epressed as = r w you can show yourself how I ust be defined in order for the kinetic energy epression K = Iw for the object, iewed as a spinning rigid body, to be the sae as the kinetic energy epression K = for the particle oing through space in a circle. Either point of iew is alid so both iewpoints ust yield the sae kinetic energy. Please go ahead and derie what I ust be and then coe back and read the deriation below. Here is the deriation: Gien that K =, we replace with r w. This gies K = which can be written as K = ( r ) w For this to be equialent to we ust hae K = Iw I = r ( rw) (-5) This is our result for the oent of inertia of a particle of ass, with respect to an ais of rotation fro which the particle is a distance r. 49

150 Chapter Center of Mass, Moent of Inertia Now suppose we hae two particles ebedded in our assless disk, one of ass at a distance r fro the ais of rotation and another of ass at a distance r fro the ais of rotation. w O r r Massless Disk The oent of inertia of the first one by itself would be I = r and the oent of inertia of the second particle by itself would be I = r The total oent of inertia of the two particles ebedded in the assless disk is siply the su of the two indiidual oents of inertial. I = I + I r r I = + This concept can be etended to include any nuber of particles. For each additional particle, one siply includes another i r i ter in the su where i is the ass of the additional particle and r i is the distance that the additional particle is fro the ais of rotation. In the case of a rigid object, we subdiide the object up into an infinite set of infinitesial ass eleents d. Each ass eleent contributes an aount of oent of inertia di =r d (-6) to the oent of inertia of the object, where r is the distance that the particular ass eleent is fro the ais of rotation. 50

151 Chapter Center of Mass, Moent of Inertia Eaple - Find the oent of inertia of the rod in Eaple - with respect to rotation about the z ais. kg In Eaple -, the linear density of the rod was gien as µ = To reduce the nuber of ties we hae to write the alue in that epression, we will write it as µ = b with b kg being defined as b The total oent of inertia of the rod is the infinite su of the infinitesial contributions fro each and eery ass eleent d aking up the rod. 3 di =r d (-6) y z d (L, 0) d = µ d In the diagra, we hae indicated an infinitesial eleent d of the rod at an arbitrary position on the rod. The z ais, the ais of rotation, looks like a dot in the diagra and the distance r in di =r d, the distance that the bit of ass under consideration is fro the ais of rotation, is siply the abscissa of the position of the ass eleent. Hence, equation -6 for the case at hand can be written as di = d 5

152 Chapter Center of Mass, Moent of Inertia which we copy here di = d y definition of the linear ass density µ, the infinitesial ass d can be epressed as d = µ d. Substituting this into our epression for d I yields d I = µ d Now µ was gien as b (with b actually being the sybol that I chose to use to represent the kg gien constant ). Substituting b in for µ in our epression for d I yields 3 d I = ( b ) d di = b This epression for the contribution of an eleent d of the rod to the total oent of inertia of the rod is good for eery eleent d of the rod. The infinite su of all such infinitesial contributions is thus the integral di = Again, as with our last integration, on the left, we hae not bothered with liits of integration the infinite su of all the infinitesial contributions to the oent of inertia is siply the total oent of inertia. I = On the right we use the liits of integration 0 to L to include eery eleent of the rod which etends fro = 0 to = L, with L gien as Factoring out the constant b gies us Now we carry out the integration: L 0 I = b L 0 b L 0 4 b 4 4 d 4 d d d I = b 5 L 5 0 I = b 5 5 L

153 Chapter Center of Mass, Moent of Inertia I = b Substituting the gien alues of b and L yields: 5 L 5 I = kg ( ) 5 5 I = kg The Parallel Ais Theore We state, without proof, the parallel ais theore: in which: I =ICM + d (-7) I is the oent of inertia of an object with respect to an ais fro which the center of ass of the object is a distance d. I CM is the oent of inertia of the object with respect to an ais that is parallel to the first ais and passes through the center of ass. is the ass of the object. d is the distance between the two aes. The parallel ais theore relates the oent of inertia I CM of an object, with respect to an ais through the center of ass of the object, to the oent of inertia I of the sae object, with respect to an ais that is parallel to the ais through the center of ass and is at a distance d fro the ais through the center of ass. A conceptual stateent ade by the parallel ais theore is one that you probably could hae arried at by eans of coon sense, naely that the oent of inertia of an object with respect to an ais through the center of ass is saller than the oent of inertia about any ais parallel to that one. As you know, the closer the ass is packed to the ais of rotation, the saller the oent of inertia; and; for a gien object, per definition of the center of ass, the ass is packed ost closely to the ais of rotation when the ais of rotation passes through the center of ass. 53

154 Chapter Center of Mass, Moent of Inertia Eaple -3 Find the oent of inertia of the rod fro eaples - and -, with respect to an ais that is perpendicular to the rod and passes through the center of ass of the rod. Recall that the rod in question etends along the ais fro = 0 to = L with L = and kg that the rod has a linear density gien by µ = bl with b The ais in question can be chosen to be one that is parallel to the z ais, the ais about which, in soling eaple -, we found the oent of inertia to be I = kg. In soling eaple - we found the ass of the rod to be = 0.57 kg and the center of ass of the rod to be at a distance d = away fro the z ais. Here we present the solution to the proble: y z d The center of ass of the rod I I =I I CM CM + d =I d C M = kg kg ( ) I CM = kg 54

155 Chapter 3 Statics 3 Statics It bears repeating: Make sure that any force that enters the torque equilibriu equation is ultiplied by a oent ar, and that any pure torque (such as τ o in the solution of eaple 3- on page 5) that enters the torque equilibriu equation is NOT ultiplied by a oent ar. For any rigid body, at any instant in tie, Newton s nd Law for translational otion and Newton s nd Law for Rotational otion = a F a = τ I both apply. In this chapter we focus on rigid bodies that are in equilibriu. This topic, the study of objects in equilibriu, is referred to as statics. eing in equilibriu eans that the acceleration and the angular acceleration of the rigid body in question are both zero. When a = 0, Newton s nd Law for translational otion boils down to and when a = 0, Newton s nd Law for Rotational otion becoes F = 0 (3-) τ = 0 (3-) These two ector equations are called the equilibriu equations. They are also known as the equilibriu conditions. In that each of the ectors has three coponents, the two ector equations actually represent a set of si scalar equations: F = 0 F y = 0 F z = 0 τ < = 0 τ y< = 0 τ z< = 0 55

156 Chapter 3 Statics In any cases, all the forces lie in one and the sae plane, and if there are any torques aside fro the torques resulting fro the forces, those torques are about an ais perpendicular to that plane. If we define the plane in which the forces lie to be the -y plane, then for such cases, the set of si scalar equations reduces to a set of 3 scalar equations (in that the other 3 are triial 0=0 identities): F = 0 (3-3) F y = 0 (3-4) τ z< = 0 (3-5) Statics probles represent a subset of Newton s nd Law probles. You already know how to sole Newton s nd Law probles so there is not uch new for you to learn here, but a couple of details regarding the way in which objects are supported will be useful to you. Many statics probles inole beas and coluns. eas and coluns are referred to collectiely as ebers. The analysis of the equilibriu of a eber typically entails soe approiations which inole the neglect of soe short distances. As long as these distances are sall copared to the length of the bea, the approiations are ery good. One of these approiations is that, unless otherwise specified, we neglect the diensions of the cross section of the eber (for instance, the width and height of a bea). We do not neglect the length of the eber. Pin-Connected Mebers A pin is a short ale. A eber which is pin-connected at one end, is free to rotate about the pin. The pin is perpendicular to the direction in which the eber etends. In practice, in the case of a eber that is pin-connected at one end, the pin is not really right at the end of the eber, but unless the distance fro the pin to the end (the end that is ery near the pin) of the eber is specified, we neglect that distance. Also, the echanis by which a bea is pinconnected to, for instance, a wall, causes the end of the bea to be a short distance fro the wall. Unless otherwise specified, we are supposed to neglect this distance as well. A pin eerts a force on the eber. The force lies in the plane that contains the eber and is perpendicular to the pin. eyond that, the direction of the force, initially, is unknown. On a free body diagra of the eber, one can include the pin force as an unknown force at an unknown angle, or one can include the unknown and y coponents of the pin force. 56

157 Chapter 3 Statics Eaple 3- One end of a bea of ass 6.9 kg and of length.00 is pin-connected to a wall. The other end of the bea rests on a frictionless floor at a point that is.80 away fro the wall. The bea is in a plane that is perpendicular to both the wall and the floor. The pin is perpendicular to that plane. Find the force eerted by the pin on the bea, and find the noral force eerted on the bea by the floor. Solution First let s draw a sketch: L =.00 =.80 Now we draw a free body diagra of the eber: F Py O F P F g F N 57

158 Chapter 3 Statics We are going to need to apply the torque equilibriu condition to the bea so I a going to add oent ars to the diagra. My plan is to su the torques about point O so I will depict oent ars with respect to an ais through point O. F r = =.80 N F Py r Fg = / O F P F g F N Now let s apply the equilibriu conditions: F = 0 F P = There are three unknown force alues depicted in the free body diagra and we hae already found one of the! Let s apply another equilibriu condition: 0 F = 0 FP y Fg + FN = 0 F g + F 0 (3-6) P y N = There are two unknowns in this equation. We can t sole it but it ay proe useful later on. Let s apply the torque equilibriu condition. τ O< = 0 F + F g g + F N = N =

159 Chapter 3 Statics Here, I copy that last line for you before proceeding: g + F g F N = N = 0 F N 6. 9 kg (9. 80 newtons/kg) = F = 3 9 N. newtons g We can use this result ( F N = ) in equation 3-6 (the one that reads FP y g + FN = 0 ) to obtain a alue for F Py : F g P y + FN = 0 F Py = g F N F Py F Py = g g g F Py = 6. 9 kg (9. 80 newtons/kg) = F = 3 9 P y. newtons Recalling that we found F P to be zero, we can write, for our final answer: FP = 3.9 newtons, straight upward and = 3.9 newtons, straight upward F N 59

160 Chapter 3 Statics Fi-Connected Mebers A fi-connected eber is one that is rigidly attached to a structure (such as a wall) that is eternal to the object whose equilibriu is under study. An eaple would be a etal rod, one end of which is welded to a etal wall. A fied connection can apply a force in any direction, and it can apply a torque in any direction. When all the other forces lie in a plane, the force applied by the fied connection will be in that plane. When all the other torques are along or parallel to a particular line, then the torque eerted by the fied connection will be along or parallel to that sae line. Eaple 3- A horizontal bar of length L and ass is fi connected to a wall. Find the force and the torque eerted on the bar by the wall. Solution First a sketch: 60

161 Chapter 3 Statics then a free body diagra: F oy O F o τ o L/ F g = g followed by the application of the equilibriu conditions to the free body diagra: F = 0 F = 0 o That was quick. Let s see what setting the su of the ertical forces equal to zero yields: F = 0 F oy F = 0 g F = oy F g F oy = g Now for the torque equilibriu condition: τ O< = 0 L τ F o g = L τ g o = 0 0 τ o = gl The wall eerts an upward force of agnitude g, and a counterclockwise (as iewed fro that position for which the free end of the bar is to the right) torque of agnitude g L on the bar. 6

162 Chapter 4 Work and Energy 4 Work and Energy You hae done quite a bit of proble soling using energy concepts. ack in chapter we defined energy as a transferable physical quantity that an object can be said to hae and we said that if one transfers energy to a aterial particle that is initially at rest, that particle acquires a speed which is an indicator of how uch energy was transferred. We said that an object can hae energy because it is oing (kinetic energy), or due to its position relatie to soe other object (potential energy). We said that energy has units of joules. You hae dealt with translational kinetic energy K =, rotational kinetic energy K = Iw, spring potential energy U = k, near-earth s-surface graitational potential energy U = gy, and the uniersal graitational potential energy U = G r corresponding to the Uniersal Law of Graitation. The principle of the conseration of energy is, in the opinion of this author, the central ost iportant concept in physics. Indeed, at least one dictionary defines physics as the study of energy. It is iportant because it is consered and the principle of conseration of energy allows us to use siple accounting procedures to ake predictions about the outcoes of physical processes that hae yet to occur and to understand processes that hae already occurred. According to the principle of conseration of energy, any change in the total aount of energy of a syste can be accounted for in ters of energy transferred fro the iediate surroundings to the syste or to the iediate surroundings fro the syste. Physicists recognize two categories of energy transfer processes. One is called work and the other is called heat flow. In this chapter we focus our attention on work. Conceptually, positie work is what you are doing on an object when you push or pull on it in the sae direction in which the object is oing. You do negatie work on an object when you push or pull on it in the direction opposite the direction in which the object is going. The neonic for reebering the definition of work that helps you reeber how to calculate it is Work is Force ties Distance. The neonic does not tell the whole story. It is good for the case of a constant force acting on an object that oes on a straight line path when the force is in the sae eact direction as the direction of otion. A ore general, but still not copletely general, how-to-calculate-it definition of work applies to the case of a constant force acting on an object that oes along a straight line path (when the force is not necessarily directed along the path). In such a case, the work W done on the object, when it traels a certain distance along the path, is: the along-the-path coponent of the force F ties the length of the path segent r. W = F r (4-) Een this case still needs soe additional clarification: If the force coponent ector along the path is in the sae direction as the object s displaceent ector, then F is positie, so the work is positie; but if the force coponent ector along the path is in the opposite direction to that of As entioned before, the potential energy is actually the energy of the syste of the objects and their fields as a whole, but it is coon to assign it to part of the syste for bookkeeping purposes as I do in this book. 6

163 Chapter 4 Work and Energy the object s displaceent ector, then F is negatie, so the work is negatie. Thus, if you are pushing or pulling on an object in a direction that would tend to ake it speed up, you are doing positie work on the object. ut if you are pushing or pulling on an object in a direction that would tend to slow it down, you are doing negatie work on the object. In the ost general case in which the coponent of the force along the path is continually changing because the force is continually changing (such as in the case of an object on the end of a spring) or because the path is not straight, our how-to-calculate-it definition of the work becoes: For each infinitesial path segent aking up the path in question, we take the product of the along-the-path force coponent and the infinitesial length of the path segent. The work is the su of all such products. Such a su would hae an infinite nuber of ters. We refer to such a su as an integral. The Relation etween Work and Motion Let s go back to the siplest case, the case in which a force F is the only force acting on a particle of ass which oes a distance r (while the force is acting on it) in a straight line in the eact sae direction as the force. The plan here is to inestigate the connection between the work on the particle and the otion of the particle. We ll start with Newton s nd Law. Free ody Diagra F a a = F Soling for F, we arrie at: a = F F = a On the left, we hae the agnitude of the force. If we ultiply that by the distance r, we get the work done by the force on the particle as it oes the distance r along the path, in the sae direction as the force. If we ultiply the left side of the equation by r then we hae to ultiply the right by the sae thing to aintain the equality. F r = a r 63

164 Chapter 4 Work and Energy On the left we hae the work W, so: W = a r On the right we hae two quantities used to characterize the otion of a particle so we hae certainly et our goal of relating work to otion, but we can untangle things on the right a bit if we recognize that, since we hae a constant force, we ust hae a constant acceleration. This eans the constant acceleration equations apply, in particular, the one that (in ters of r rather than ) reads: = + a r Soling this for a r gies o a r = o Substituting this into our epression for W aboe (the one that reads W = a r ) we obtain which can be written as W = W = o o Of course we recognize the o as the kinetic energy of the particle before the work is done on the particle and the as the kinetic energy of the particle after the work is done on it. To be consistent with the notation we used in our early discussion of the conseration of echanical energy we change to the notation in which the prie sybol ( ) signifies after and no super- or subscript at all (rather than the subscript o ) represents before. Using this notation and the definition of kinetic energy, our epression for W becoes: W = K K Since the after kinetic energy inus the before kinetic energy is just the change in kinetic energy K, we can write the epression for W as: W = K (4-) This is indeed a siple relation between work and otion. The cause, work on a particle, on the left, is eactly equal to the effect, a change in the kinetic energy of the particle. This result is so iportant that we gie it a nae, it is the Work-Energy Relation. It also goes by the nae: The Work-Energy Principle. It works for etended rigid bodies as well. In the case of a rigid body that rotates, it is the displaceent of the point of application of the force, along the path of said point of application, that is used (as the r ) in calculating the work done on the object. 64

165 Chapter 4 Work and Energy In the epression W = K, the work is the net work (the total work) done by all the forces acting on the particle or rigid body. The net work can be calculated by finding the work done by each force and adding the results, or by finding the net force and using it in the definition of the work. Calculating the Work as the Force-Along-the-Path Ties the Length of the Path Consider a block on a flat frictionless incline that akes an angle θ with the ertical. The block traels fro a point A near the top of the incline to a point, a distance d in the down-theincline direction fro A. Find the work done, by the graitational force, on the block. A θ d F g = g We e drawn a sketch of the situation (not a free body diagra). We note that the force for which we are supposed to calculate the work is not along the path. So, we define a coordinate syste with one ais in the down-the-incline direction and the other perpendicular to that ais A θ F g = g and break the graitational force ector up into its coponents with respect to that coordinate syste. 65

166 Chapter 4 Work and Energy F g = g θ F g F g F F g = F cosθ = g cosθ g = F sinθ = g sinθ g g Now we redraw the sketch with the graitational force replaced by its coponents: A θ d F g F g F g, being perpendicular to the path does no work on the block as the block oes fro A to. The work done by the graitational force is gien by W = F d W = F g d W = g(cosθ ) d W = g d cosθ While this ethod for calculating the work done by a force is perfectly alid, there is an easier way. It inoles another product operator for ectors (besides the cross product), called the dot product. To use it, we need to recognize that the length of the path, cobined with the direction of otion, is none other than the displaceent ector (for the point of application of the force). Then we just need to find the dot product of the force ector and the displaceent ector. 66

167 Chapter 4 Work and Energy The Dot Product of Two Vectors The dot product of the ectors A and is written A and is epressed as: A = A cosθ (4-3) where θ, just as in the case of the cross product, is the angle between the two ectors after they hae been placed tail to tail. θ A The dot product can be interpreted as either A (the coponent of A along, ties, the agnitude of ) or A (the coponent of along A, ties, the agnitude of A ), both of which ealuate to one and the sae alue. This akes the dot product perfect for calculating the work. Since F r = F r and F r is W, we hae W = F r (4-4) y eans of the dot product, we can sole the eaple in the last section uch ore quickly than we did before. A θ d Find the work done on the block by the graitational force when the object oes fro point A to Point. F g = g 67

168 Chapter 4 Work and Energy We define the displaceent ector d to hae a agnitude equal to the distance fro point A to point with a direction the sae as the direction of otion (the down-the-rap direction). Using our definition of work as the dot product of the force and the displaceent, equation 4-4: W r = F with the graitational force ector Fg being the force, and d being the displaceent, the work can be written as: W F d. = g Using the definition of the dot product we find that: W = F dcosθ. g Replacing the agnitude of the graitational force with g we arrie at our final answer: W = gdcosθ. This is the sae answer that we got prior to our discussion of the dot product. In cases in which the force and the displaceent ectors are gien in i, j, k notation, finding the work is straightforward. The Dot Product in Unit Vector Notation The siple dot product relations aong the unit ectors akes it easy to ealuate the dot product of two ectors epressed in unit ector notation. Fro what aounts to our definition of the dot product, equation 4-3: A = A cosθ we note that a ector dotted into itself is siply the square of the agnitude of the ector. This is true because the angle between a ector and itself is 0 and cos 0 is. o A A = AA cos0 = A Since the unit ectors all hae agnitude, any unit ector dotted into itself yields () which is just. i i =, j j =, and k k = Now the angle between any two different Cartesian coordinate ais unit ectors is 90 and the cos 90 is 0. Thus, the dot product of any Cartesian coordinate ais unit ector into any other Cartesian coordinate ais unit ector is zero. 68

169 Chapter 4 Work and Energy 69 So, if k j i z y A A A + + = A and k j i z y + + = then A is just ) ( ) ( k j i k j i z y z y A A A = A ) ( ) ( ) ( k j i k k j i j k j i i z y z z y y z y A A A = A k k j k i k k j j j i j k i j i i i z z y z z z y y y y z y A A A A A A A A A = A k k j k i k k j j j i j k i j i i i = z z y z z z y y y y z y A A A A A A A A A A z z y y A A A + + = A The end result is that the dot product of two ectors is siply the su of: the product of the two ectors coponents, the product of their y coponents, and the product of their z coponents. Energy Transfer Work s. Center of Mass Pseudo-Work I introduced the topic of work by stating that it represents one category of energy transfer to a syste. As such, work is energy transfer work. There is a quantity that is calculated in uch the sae way as work, with one subtle difference. I going to call the quantity center of ass pseudo-work and I going to use a couple of specific processes inoling a frictionless horizontal surface, a spring, and a block (and in one case, another block) to distinguish energy transfer work fro center of ass pseudo-work. Suppose we attach the spring to the wall so that the spring sticks out horizontally and then push the block toward the wall in such a anner as to copress the spring. Then we release the block fro rest and start our obserations at the first instant subsequent to release. Let our syste be the block. The spring pushes the block away fro the wall. The spring transfers energy to the block while the spring is in contact with the block. The work done can be calculated as the integral of ector force dot ector infinitesial displaceent which I'll loosely state as the integral of force ties distance. The distance in this case is the displaceent (the infinite set of infinitesial displaceents) of the point of application of the force. This kind of work is energy transfer work. It is the aount of energy transferred to the block by the spring.

170 Chapter 4 Work and Energy Now let's disconnect the spring fro the wall and attach the spring to the block so that the spring sticks out horizontally fro the block and again push the block up against the wall, copressing the spring, and release the block fro rest. Let our syste be the block plus spring. The block goes sliding off as before, this tie with the spring attached. The wall does no energy-transfer work on the syste because the part of the wall that is eerting the force on the syste is not oing there is no displaceent. Howeer, we find soething useful if we calculate the integral of the ector force (eerted by the wall) dot ector infinitesial displaceent of the center of ass of the syste loosely stated, force ties distance oed by center of ass. I' calling that "soething useful" the center of ass pseudo-work eperienced by the syste. It's useful because our Newton s Law deriation shows that quantity to be equal to the change in the center of ass kinetic energy of the syste. In this case the syste gained soe center of ass kinetic energy een though no energy was transferred to it. How did that happen? Energy that was already part of the syste, energy stored in the spring, was conerted to center of ass kinetic energy. So what is the subtle difference? In both cases we are, loosely speaking, calculating force ties distance. ut in the case of energy transfer work, the distance is the distance oed by that eleent of the agent of the force that is in contact with the icti at eactly that point where the force is being applied, whereas, in the case of center of ass work, the distance in force ties distance is the distance oed by the center of ass. For a particle, there is no difference. For a truly rigid body undergoing purely translational otion (no rotation) there is no difference. ut beware, a truly rigid body is an idealized object in which no bit of the body can oe relatie to any other bit of the body. Een for such a body, if there is rotation, there will be a difference between the energy transfer work and the center of ass pseudo-work done on the object. Consider for instance a block at rest on a horizontal frictionless surface. You apply an off-center horizontal force to the block for a short distance by pressing on the block with your finger. The work you do is the integrated force ties distance oer which you oe the tip of your finger. It will be greater than the integrated force ties the distance oer which the center of ass oes. Soe of the work you do goes into increasing the center of ass kinetic energy of the rigid body and soe of it goes into increasing the rotational kinetic energy of the rigid body. In this case the energy transfer work is greater than the center of ass pseudo-work. Concluding Rearks At this point you hae two ways of calculating the work done on an object. If you are gien inforation about the force and the path you will use the force ties distance definition of work. ut if you are gien inforation on the effect of the work (the change in kinetic energy) then you will deterine the alue of the change in kinetic energy and substitute that into the work energy relation, equation 4-: W = K to deterine the work (or center of ass pseudo-work as applicable). There is yet another ethod for calculating the work. Like the first ethod, it is good for cases in which you hae inforation on the force and the path. It only works for certain kinds of forces, but when it does work, to use it, the only thing you need to know about the path is the positions of the endpoints. This third ethod for calculating the work inoles the potential energy, the ain topic of our net chapter. 70

171 Chapter 5 Potential Energy, Conseration of Energy, Power 5 Potential Energy, Conseration of Energy, Power The work done on a particle by a force acting on it as that particle oes fro point A to point under the influence of that force, for soe forces, does not depend on the path followed by the particle. For such a force there is an easy way to calculate the work done on the particle as it oes fro point A to point. One siply has to assign a alue of potential energy (of the particle ) to point A (call that alue U A ) and a alue of potential energy to point (call that alue U ). One chooses the alues such that the work done by the force in question is just the negatie of the difference between the two alues. W = ( U ) U A W = U (5-) U = U U A is the change in the potential energy eperienced by the particle as it oes fro point A to point. The inus sign in equation 5- ensures that an increase in potential energy corresponds to negatie work done by the corresponding force. For instance for the case of nearearth s-surface graitational potential energy, the associated force is the graitational force, a.k.a. the graitational force. If we lift an object upward near the surface of the earth, the graitational force does negatie work on the object since the (downward) force is in the opposite direction to the (upward) displaceent. At the sae, tie, we are increasing the capacity of the particle to do work so we are increasing the potential energy. Thus, we need the sign in W = U to ensure that the change in potential energy ethod of calculating the work gies the sae algebraic sign for the alue of the work that the force-along-the path ties the length of the path gies. Note that in order for this ethod of calculating the work to be useful in any case that ight arise, one ust assign a alue of potential energy to eery point in space where the force can act on a particle so that the ethod can be used to calculate the work done on a particle as the particle oes fro any point A to any point. In general, this eans we need a alue for each of an infinite set of points in space. This assignent of a alue of potential energy to each of an infinite set of points in space ight see daunting until you realize that it can be done by eans of a siple algebraic epression. For instance, we hae already written the assignent for a particle of ass for the case of the uniersal graitational force due to a particle of ass. It was equation 7-5: U G = r The potential energy is actually the potential energy of the syste consisting of the particle, whateer the particle is interacting with, and the releant field. For instance, if we are talking about a particle in the graitational field of the earth, the potential energy under discussion is the potential energy of the earth plus particle and graitational field of the earth plus particle. For accounting purposes, it is conenient to ascribe the potential energy to the particle and that is what I do in this book. 7

172 Chapter 5 Potential Energy, Conseration of Energy, Power N in which G is the uniersal graitational constant G = and r is the distance that kg particle is fro particle. Note that considering particle to be at the origin of a coordinate syste, this equation assigns a alue of potential energy to eery point in the unierse! The alue, for any point, siply depends on the distance that the point is fro the origin. Suppose we want to find the work done by the graitational force due to particle, on particle as particle oes fro point A, a distance r A fro particle to point, a distance r fro particle. The graitational force eerted on it (particle ) by the graitational field of particle does an aount of work, on particle, gien by (starting with equation 5-): W = U W = ( U ) U A W G r G r = A W = G r r A The Relation etween a Conseratie Force and the Corresponding Potential While this business of calculating the work done on a particle as the negatie of the change in its potential energy does ake it a lot easier to calculate the work, we do hae to be careful to define the potential such that this ethod is equialent to calculating the work as the force-along-the-path ties the length of the path. Rather than jup into the proble of finding the potential energy at all points in a threediensional region of space for a kind of force known to eist at all points in that threediensional region of space, let s look into the sipler proble of finding the potential along a line. We define a coordinate syste consisting of a single ais, let s call it the -ais, with an origin and a positie direction. We put a particle on the line, a particle that can oe along the line. We assue that we hae a force that acts on the particle whereer the particle is on the line and that the force is directed along the line. While we will also address the case of a force which has the sae alue at different points along the line, we assue that, in general, the force aries with position. Reeber this fact so that you can find the flaw discussed below. ecause we want to define a potential for it, it is iportant that the work done on the particle by the force being eerted on the particle, as the particle oes fro point A to point does not depend on how the particle gets fro point A to point. Our goal is to define a potential energy function for the force such that we get the sae alue for the work done on the particle by the force whether we use the force-along-the-path ethod to calculate it or the negatie of the change of 7

173 Chapter 5 Potential Energy, Conseration of Energy, Power potential energy ethod. Suppose the particle undergoes a displaceent along the line under the influence of the force. See if you can see the flaw in the following, before I point it out: We write W = F for the work done by the force, calculated using the force-along-the-path ties the length of the path idea, and then W = U for the work done by the force calculated using the negatie of the change in potential energy concept. Setting the two epressions equal to each U other, we hae, F = U which we can write as F = for the relation between the potential energy and the -coponent of the force. Do you see where we went wrong? While the ethod will work for the special case in which the force is a constant, we were supposed to coe up with a relation that was good for the general case in which the force aries with position. That eans that for each alue of in the range of alues etending fro the initial alue, let s call it A, to the alue at the end of the displaceent A +, there is a different alue of force. So the epression W = F is inappropriate. Gien a nuerical proble, there is no one alue to plug in for F, because F aries along the. To fi things, we can shrink to infinitesial size, so sall that, A and A + are, for all practical purposes, one and the sae point. That is to say, we take the liit as 0. Then our relation becoes F U li 0 = which is the sae thing as F U = li 0 The liit of U du that appears on the right is none other than the deriatie, so: d F du = ( (5-) d To ephasize the fact that force is a ector, we write it in unit ector notation as: du F = i ( (5-3) d Let s ake this ore concrete by using it to deterine the potential energy due to a force with which you are failiar the force due to a spring. 73

174 Chapter 5 Potential Energy, Conseration of Energy, Power Consider a block on frictionless horizontal surface. The block is attached to one end of a spring. The other end of the spring is attached to a wall. The spring etends horizontally away fro the wall, at right angles to the wall. Define an -ais with the origin at the equilibriu position of that end of the spring which is attached to the block. Consider the away-fro-the-wall direction to be the positie direction. Eperientally, we find that the force eerted by the spring on the block is gien by: F = k i ( (5-4) where k is the force constant of the spring. (Note: A positie, corresponding to the block haing been pulled away fro the wall, thus stretching the spring, results in a force in the negatie direction. A negatie, copressed spring, results in a force in the + direction, consistent with du coon sense.) y coparison with equation 5-3 (the one that reads F = i ) we note that d the potential energy function has to be defined so that du = k d This is such a siple case that we can pretty uch guess what U has to be. U has to be defined such that when we take the deriatie of it we get a constant (the k) ties to the power of. Now when you take the deriatie of to a power, you reduce the power by one. For that to result in a power of, the original power ust be. Also, the deriatie of a constant ties soething yields that sae constant ties the deriatie, so, there ust be a factor of k in the potential energy function. Let s try U = k and see where that gets us. The deriatie of k is k. Ecept for that factor of out front, that is eactly what we want. Let s aend our guess by ultiplying it by a factor of, to eentually cancel out the that coes down when we take the deriatie. With du U = k we get = k which is eactly what we needed. Thus d U = k (5-5) is indeed the potential energy for the force due to a spring. You used this epression back in chapter. Now you know where it coes fro. We hae considered two other conseratie forces. For each, let s find the potential energy du function U that eets the criterion that we hae written as, F = i. d 74

175 Chapter 5 Potential Energy, Conseration of Energy, Power First, let s consider the near-earth s-surface graitational force eerted on an object of ass, by the earth. We choose our single ais to be directed ertically upward with the origin at an arbitrary but clearly specified and fied eleation for the entire proble that one ight sole using the concepts under consideration here. y conention, we call such an ais the y ais rather than the ais. Now we know that the graitational force is gien siply (again, this is an eperiental result) by F = gj where the g is the known agnitude of the graitational force and the j is the downward direction. Equation 5-3, written for the case at hand is: du F = dy For the last two equations to be consistent with each other, we need U to be defined such that j du = g dy For the deriatie of U with respect to y to be the constant g, U ust be gien by U = gy (5-6) and indeed this is the equation for the earth s near-surface graitational potential energy. Please erify that when you take the deriatie of it with respect to y, you do indeed get the agnitude of the graitational force, g. Now let s turn our attention to the Uniersal Law of Graitation. Particle nuber of ass creates a graitational field in the region of space around it. Let s define the position of particle nuber to be the origin of a three-diensional Cartesian coordinate syste. Now let s assue that particle nuber is at soe position in space, a distance r away fro particle. Let s define the direction that particle is in, relatie to particle, as the + direction. Then, the coordinates of particle are (r, 0, 0). r is then the coponent of the position ector for particle, a quantity that we shall now call. That is, is defined such that = r. In ters of the coordinate syste thus defined, the force eerted by the graitational field of particle, on particle, is gien by: G F = i 75

176 Chapter 5 Potential Energy, Conseration of Energy, Power which I rewrite here: Copare this with equation 5-3: G F = du F = d i i Cobining the two equations, we note that our epression for the potential energy U in ters of ust satisfy the equation du G = d It s easier to deduce what U ust be if we write this as du d = G For the deriatie of U with respect to to be a constant (G ) ties a power ( ) of, U itself ust be that sae constant (G ) ties to the net higher power ( ), diided by the alue of the latter power. G U = which can be written G = U Recalling that the in the denoinator is siply the distance fro particle to particle which we hae also defined to be r, we can write this in the for in which it is ore coonly written: G = (5-7) r U This is indeed the epression for the graitational potential that we gae you (without any justification for it) back in Chapter 7, the chapter on the Uniersal Law of Graitation. 76

177 Chapter 5 Potential Energy, Conseration of Energy, Power Conseration of Energy Reisited Recall the work-energy relation, equation 4- fro last chapter, W = K, the stateent that work causes a change in kinetic energy. Now consider a case in which all the work is done by conseratie forces, so, the work can be epressed as the negatie of the change in potential energy. U = K Further suppose that we are dealing with a situation in which a particle oes fro point A to point under the influence of the force or forces corresponding to the potential energy U. Then, the preceding epression can be written as: ( U = K K UA ) A U + U A = K K A K + U = K + U A A Switching oer to notation in which we use pried ariables to characterize the particle when it is at point and unpried ariables at A we hae: K + U = K + U Interpreting E = K + U as the energy of the syste at the before instant, and E = K + U as the energy of the syste at the after instant, we see that we hae deried the conseration of echanical energy stateent for the special case of no net energy transfer to or fro the surroundings and no conersion of energy within the syste fro echanical energy to other fors or ice ersa. In equation for, the stateent is E = E (5-8) an equation to which you were introduced in chapter. Note that you would be well adised to reiew chapter now, because for the current chapter, you are again responsible for soling any of the chapter--type probles (reebering to include, and correctly use, before and after diagras) and answer any of the chapter--type questions. 77

178 Chapter 5 Potential Energy, Conseration of Energy, Power Power In this last section on energy we address a new topic. As a separate and iportant concept, it would desere its own chapter ecept for the fact that it is such a siple, straightforward concept. Power is the rate of energy transfer, energy conersion, and in soe cases, the rate at which transfer and conersion of energy are occurring siultaneously. When you do work on an object, you are transferring energy to that object. Suppose for instance that you are pushing a block across a horizontal frictionless surface. You are doing work on the object. The kinetic energy of the object is increasing. The rate at which the kinetic energy is increasing is referred to as power. The rate of change of any quantity (how fast that quantity is changing) can be calculated as the deriatie of that quantity with respect to tie. In the case at hand, the power P can be epressed as dk P = (5-9) dt the tie deriatie of the kinetic energy. Since K = we hae P = d dt P = P d dt d = dt d P = dt P = a P = F P = F (5-0) where a is the acceleration coponent parallel to the elocity ector. The perpendicular coponent changes the direction of the elocity but not the agnitude. esides the rate at which the kinetic energy is changing, the power is the rate at which work is being done on the object. In an infinitesial tie interal dt, you do an infinitesial aount of work dw = F d on the object. Diiding both sides by dt, we hae 78

179 which again is Chapter 5 Potential Energy, Conseration of Energy, Power dw dt d = F dt F P = as it ust be since, in accord with the work-energy relation, the rate at which you do work on the object has to be the rate at which the kinetic energy of the object increases. If you do work at a steady rate for a finite tie interal, the power is constant and can siply be calculated as the aount of work done during the tie interal diided by the tie interal itself. For instance, when you clib stairs, you conert cheical energy stored in your body to graitational potential energy. The rate at which you do this is power. If you clib at a steady rate for a total increase of graitational potential energy of U oer a tie interal t then the constant alue of your power during that tie interal is U P = (5-) t If you know that the power is constant, you know the alue of the power P, and you are asked to find the total aount of work done, the total aount of energy transferred, and/or the total aount of energy conerted during a particular tie interal t, you just hae to ultiply the power P by the tie interal t. Energy = P t (5-) One could include at least a dozen forulas on your forula sheet for power, but they are all so siple that, if you understand what power is, you can coe up with the specific forula you need for the case on which you are working. We include but one forula on the forula sheet, de P = (5-3) dt which should reind you what power is. Since power is the rate of change of energy, the SI units of power ust be s J. This cobination unit is gien a nae, the watt, abbreiated W. W J s 79

180 Chapter 6 Ipulse and Moentu 6 Ipulse and Moentu First, a Few More Words on Work and Energy, for Coparison Purposes Iagine a gigantic air hockey table with a whole bunch of pucks of arious asses, none of which eperiences any friction with the horizontal surface of the table. Assue air resistance to be negligible. Now suppose that you coe up and gie each puck a shoe, where the kind of shoe that you gie the first one is special in that the whole tie you are pushing on that puck, the force has one and the sae alue; and the shoe that you gie each of the other pucks is siilar in the following respect: To each puck you apply the sae force that you applied to the first puck, oer the sae eact distance. Since you gie each of the pucks a siilar shoe, you ight epect the otion of the pucks (after the shoe) to hae soething in coon and indeed we find that, while the pucks (each of which, after the shoe, oes at its own constant elocity) hae speeds that differ fro one another (because they hae different asses), they all hae the sae alue of the product and indeed if you put a ½ in front of that product and call it kinetic energy K, the coon alue of is identical to the product of the agnitude of the force used during the shoe, and the distance oer which the force is applied. This latter product is what we hae defined to be the work W and we recognize that we are dealing with a special case of the work energy principle W = K, a case in which, for each of the pucks, the initial kinetic energy is zero. We can odify our eperient to obtain ore general results, e.g. a saller constant force oer a greater distance results in the sae kinetic energy as long as the product of the agnitude of the force and the distance oer which it is applied is the sae as it was for the other pucks, but it is interesting to consider how different it would see to us, in the original eperient, as we oe fro a high-ass puck to a low ass puck. Iagine doing that. You push on the high-ass puck with a certain force, for a certain distance. Now you oe on to a low-ass puck. As you push on it fro behind, with the sae force that you used on the high-ass puck, you notice that the low-ass puck speeds up uch ore rapidly. You probably find it uch ore difficult to aintain a steady force because it is siply ore difficult to keep up with the low ass puck. And of course, it coers the specified distance in a uch shorter aount of tie. So, although you push it for the sae distance, you ust push the low-ass puck for a shorter aount of tie in order to ake it so that both pucks hae one and the sae kinetic energy. Pondering on it you recognize that if you were to push the low-ass puck for the sae aount of tie as you did the high-ass puck (with the sae force), that the low-ass puck would hae a greater kinetic energy after the shoe, because you would hae to push on it oer a greater distance, eaning you would hae done ore work on it. Still, you iagine that if you were to push on each of the pucks for the sae aount of tie (rather than distance), that their respectie otions would hae to hae soething in coon, because again, there is soething siilar about their respectie shoes. Now we Moe on to Ipulse and Moentu You decide to do the eperient you hae been thinking about. You place each of the pucks at rest on the frictionless surface. You apply one and the sae constant force to each of the pucks for one and the sae aount of tie. Once again, you find this ore difficult with the lower ass pucks. While you are pushing on it, a low-ass puck speeds up faster than a high-ass puck does. As a result you hae to keep pushing on a low-ass puck oer a greater distance and 80

181 Chapter 6 Ipulse and Moentu it is going faster when you let it go. Haing gien all the pucks a siilar shoe, you epect there to be soething about the otion of each of the pucks that is the sae as the corresponding characteristic of the otion of all the other pucks. We hae already established that the saller the ass of the puck, the greater the speed, and the greater the kinetic energy of the puck. Eperientally, we find that all the pucks hae one and the sae alue of the product, where is the post-shoe puck elocity. Further, we find that the alue of is equal to the product of the constant force F and the tie interal t for which it was applied. That is, F t = The product of the force and the tie interal for which it is applied is such an iportant quantity that we gie it a nae, ipulse, and a sybol J. J = F t (6-) Also, as you probably recall fro chapter 4, by definition, the product of the ass of an object, and its elocity, is the oentu p of the object. Thus, the results of the eperient described aboe can be epressed as J = p The eperient dealt with a special case, the case in which each object was initially at rest. If we do a siilar eperient in which, rather than being initially at rest, each object has soe known initial elocity, we find, eperientally, that the ipulse is actually equal to the change in oentu. J = p (6-) Of course if we start with zero oentu, then the change in oentu is the final oentu. Equation 6-, J = p, is referred to as the Ipulse-Moentu Relation. It is a cause and effect relationship. You apply soe ipulse (force ties tie) to an object, and the effect is a change in the oentu of the object. The result, which we hae presented as an eperiental result, can be deried fro Newton s second law of otion. Here we do so for the case in which the force acting on the object is constant during the tie interal under consideration. Note that the force which appears in the definition of ipulse is the net eternal force acting on the object. Consider the case of a particle, of ass, which has but one, constant force (which could actually be the ector su of all the forces) acting on it. 8

182 Chapter 6 Ipulse and Moentu As always, in applying Newton s second law of otion, we start by drawing a free body diagra: F a In order to keep track of the ector nature of the quantities inoled we apply Newton s nd Law in ector for (equation 4-): = a F In the case at hand the su of the forces is just the one force F, so: Soling for F, we arrie at: a = F F = a ultiplying both sides by t we obtain F t = a t Gien that the force is constant, the resulting acceleration is constant. In the case of a constant acceleration, the acceleration can be written as the ratio of the change in that occurs during the tie interal t, to the tie interal t itself. a = t Substituting this into the preceding epression yields: F t = t F t = The change in elocity can be epressed as the final elocity (the elocity at the end of the tie interal during which the force acts) inus the initial elocity (the elocity at the start of the tie interal): =. Substituting this into F t = yields which can be written as t F t = ( ) F t = 8

183 Chapter 6 Ipulse and Moentu Recognizing that is the final oentu and that is the initial oentu we realize that we hae F t = p p On the left, we hae what is defined to be the ipulse, and on the right we hae the change in oentu (equation 6-): J = p This copletes our deriation of the ipulse oentu relation fro Newton s nd Law. Conseration of Moentu Reisited Regarding the conseration of oentu, we first note that, for a particle, if the net eternal force on the particle is zero, then the ipulse, defined by J = F t, deliered to that particle during any tie interal t, is 0. If the ipulse is zero then fro J = p, the change in oentu ust be 0. This eans that the oentu p is a constant, and since p =, if the oentu is constant, the elocity ust be constant. This result siply confirs that, in the absence of a force, our ipulse oentu relation is consistent with Newton s st Law of Motion, the one that states that if there is no force on a particle, then the elocity of that particle does not change. Now consider the case of two particles in which no eternal forces are eerted on either of the particles. (For a syste of two particles, an internal force would be a force eerted by one particle on the other. An eternal force is a force eerted by soething outside the syste on soething inside the syste.) The total oentu of the pair of particles is the ector su of the oentu of one of the particles and the oentu of the other particle. Suppose that the particles are indeed eerting forces on each other during a tie interal t. To keep things siple we will assue that the force that either eerts on the other is constant during the tie interal. Let s identify the two particles as particle # and particle # and designate the force eerted by on as F. ecause this force is eerted on particle #, it will affect the otion of particle # and we can write the ipulse oentu relation as F p t = (6-3) Now particle # can t eert a force on particle without particle # eerting an equal and opposite force back on particle. That is, the force F eerted by particle # on particle # is the negatie of F. F = F Of course F ( eff of on ) affects the otion of particle only, and the ipulse-oentu relation for particle reads 83

184 Replacing F with F we obtain Chapter 6 Ipulse and Moentu F t = p F t = (6-4) p Now add equation 6-3 ( F t = p ) and equation 6-4 together. The result is: F t F t = p p + p + 0 = p On the right is the total change in oentu for the pair of particles ptotal = p + p so what we hae found is that which can be written as 0 = p TOTAL p = 0 TOTAL (6-5) Recapping: If the net eternal force acting on a pair of particles is zero, the total oentu of the pair of particles does not change. Add a third particle to the i and any oentu change that it ight eperience because of forces eerted on it by the original two particles would be canceled by the oentu changes eperienced by the other two particles as a result of the interaction forces eerted on the by the third particle. We can etend this to any nuber of particles, and since objects are ade of particles, the concept applies to objects. That is, if, during soe tie interal, the net eternal force eerted on a syste of objects is zero, then the oentu of that syste of objects will not change. As you should recall fro Chapter 4, the concept is referred to as conseration of oentu for the special case in which there is no net transfer of oentu to the syste fro the surroundings, and you apply it in the case of soe physical process such as a collision, by picking a before instant and an after instant, drawing a sketch of the situation at each instant, and writing the fact that, the oentu in the before picture has to be equal to the oentu in the after picture, in equation for: p = p. When you read this chapter, you should again consider yourself responsible for soling any of the probles, and answering any of the questions, that you were responsible for back in Chapter 4. 84

185 Chapter 7 Oscillations: Introduction, Mass on a Spring 7 Oscillations: Introduction, Mass on a Spring If a siple haronic oscillation proble does not inole the tie, you should probably be using conseration of energy to sole it. A coon tactical error in probles inoling oscillations is to anipulate the equations giing the position and elocity as a function of tie, = a cos(πf t) and = a sin(πf t) rather than applying the principle of conseration of energy. This turns an easy fie-inute proble into a difficult fifteen-inute proble. When soething goes back and forth we say it ibrates or oscillates. In any cases oscillations inole an object whose position as a function of tie is well characterized by the sine or cosine function of the product of a constant and elapsed tie. Such otion is referred to as sinusoidal oscillation. It is also referred to as siple haronic otion. Math Aside: The Cosine Function y now, you hae had a great deal of eperience with the cosine function of an angle as the ratio of the adjacent to the hypotenuse of a right triangle. This definition coers angles fro 0 radians to π radians (0 to 90 ). In applying the cosine function to siple haronic otion, we use the etended definition which coers all angles. The etended definition of the cosine of the angle θ is that the cosine of an angle is the coponent of a unit ector, the tail of which is on the origin of an -y coordinate syste; a unit ector that originally pointed in the + direction but has since been rotated counterclockwise fro our iewpoint, through the angle θ, about the origin. Here we show that the etended definition is consistent with the adjacent oer hypotenuse definition, for angles between 0 radians and π radians. For such angles, we hae: y u θ u u y in which, u, being the agnitude of a unit ector, is of course equal to, the pure nuber with no units. Now, according to the ordinary definition of the cosine of θ as the adjacent oer the hypotenuse: 85

186 Chapter 7 Oscillations: Introduction, Mass on a Spring Soling this for u we see that Recalling that u =, this eans that cos θ = u u u = ucosθ u = cosθ Recalling that our etended definition of cos θ is, that it is the coponent of the unit ector û when û akes an angle θ with the -ais, this last equation is just saying that, for the case at hand (θ between 0 and π radians) our etended definition of definition. cos θ is equialent to our ordinary π 3π At angles between and radians (90 and 70 ) we see that u takes on negatie alues (when the coponent ector is pointing in the negatie direction, the coponent alue is, by definition, negatie). According to our etended definition, cosθ takes on negatie alues at such angles as well. y u y u θ u 86

187 Chapter 7 Oscillations: Introduction, Mass on a Spring With our etended definition, alid for any angle θ, a graph of the cos θ s. θ appears as: cos θ 0 π π 3 π 4 π θ [radians] - Soe Calculus Relations Inoling the Cosine The deriatie of the cosine of θ, with respect to θ : d cosθ = sinθ dθ The deriatie of the sine of θ, with respect to θ : d sin θ = cosθ dθ Soe Jargon Inoling The Sine And Cosine Functions When you epress, define, or ealuate the function of soething, that soething is called the arguent of the function. For instance, suppose the function is the square root function and the epression in question is 3. The epression is the square root of 3, so, in that epression, 3 is the arguent of the square root function. Now when you take the cosine of soething, that soething is called the arguent of the cosine, but in the case of the sine and cosine functions, we gie it another nae as well, naely, the phase. So, when you write cosθ, the ariable θ is the arguent of the cosine function, but it is also referred to as the phase of the cosine function. 87

188 Chapter 7 Oscillations: Introduction, Mass on a Spring In order for an epression inoling the cosine function to be at all eaningful, the phase of the cosine ust hae units of angle (for instance, radians or degrees). A lock Attached to the End of a Spring Consider a block of ass on a frictionless horizontal surface. The block is attached, by eans of an ideal assless horizontal spring haing force constant k, to a wall. A person has pulled the block out, directly away fro the wall, and released it fro rest. The block oscillates back and forth (toward and away fro the wall), on the end of the spring. We would like to find equations that gie the block s position, elocity, and acceleration as functions of tie. We start by applying Newton s nd Law to the block. efore drawing the free body diagra we draw a sketch to help identify our one-diensional coordinate syste. We will call the horizontal position of the point at which the spring is attached, the position of the block. The origin of our coordinate syste will be the position at which the spring is neither stretched nor copressed. When the position is positie, the spring is stretched and eerts a force, on the block, in the direction. When the position of is negatie, the spring is copressed and eerts a force, on the block, in the + direction. k = 0 Equilibriu Position + 88

189 Chapter 7 Oscillations: Introduction, Mass on a Spring Now we draw the free body diagra of the block: F S = k F N a F g = g + and apply Newton s nd Law: a = F a = ( k) k a = This equation, relating the acceleration of the block to its position, can be considered to be an equation relating the position of the block to tie if we substitute for a using: and so d a = dt d = dt d d a = dt dt which is usually written d a = (7-) dt and read d squared by dt squared or the second deriatie of with respect to t. Substituting this epression for a into aboe) yields k a = (the result we deried fro Newton s nd Law 89

190 Chapter 7 Oscillations: Introduction, Mass on a Spring d = dt k (7-) We know in adance that the position of the block depends on tie. That is to say, is a function of tie. This equation, equation 7-, tells us that if you take the second deriatie of with respect to tie you get itself, ties a negatie constant ( k/). We can find the an epression for in ters of t that soles 7- by the ethod of guess and check. Grossly, we re looking for a function whose second deriatie is essentially the negatie of itself. Two functions eet this criterion, the sine and the cosine. Either will work. We arbitrarily choose to use the cosine function. We include soe constants in our trial solution (our guess) to be deterined during the check part of our procedure. Here s our trial solution: π rad = a cos t T Here s how we hae arried at this trial solution: Haing established that, depends on the cosine of a ultiple of the tie ariable, we let the units be our guide. We need the tie t to be part of the arguent of the cosine, but we can t take the cosine of soething unless that π rad soething has units of angle. The constant, with the constant T haing units of tie T (we ll use seconds), akes it so that the arguent of the cosine has units of radians. It is, π rad howeer, ore than just the units that otiates us to choose the ratio as the constant. T To ake the arguent of the cosine hae units of radians, all we need is a constant with units of π rad radians per second. So why write it as? Here s the eplanation: The block goes back T and forth. That is, it repeats its otion oer and oer again as tie goes by. Starting with the block at its aiu distance fro the wall, the block oes toward the wall, reaches its closest point of approach to the wall and then coes back out to its aiu distance fro the wall. At that point, it s right back where it started fro. We define the constant alue of tie T to be the aount of tie that it takes for one iteration of the otion. Now consider the cosine function. We chose it because its second deriatie is the negatie of itself, but it is looking better and better as a function that gies the position of the block as a function of tie because it too repeats itself as its phase (the arguent of the cosine) continually increases. Suppose the phase starts out as 0 at tie 0. The cosine of 0 radians is, the biggest the cosine eer gets. We can ake this correspond to the block being at its aiu distance fro the wall. As the phase increases, the cosine gets saller, then goes negatie, eentually reaching the alue when the phase is π radians. This could correspond to the block being closest to the wall. Then, as the phase continues to increase, the cosine increases until, when the phase is π, the cosine is back up to corresponding to the block being right back where it started fro. Fro here, as the phase of the cosine continues to increase fro π to 4π, the 90

191 Chapter 7 Oscillations: Introduction, Mass on a Spring cosine again takes on all the alues that it took on fro 0 to π. The sae thing happens again as the phase increases fro 4π to 6π, fro 8π to 0π, etc. Getting back to that constant our trial solution for (t): π rad T that we guessed should be in the phase of the cosine in π rad = a cos t T With T being defined as the tie it takes for the block to go back and forth once, look what happens to the phase of the cosine as the stopwatch reading continually increases. Starting fro π rad 0, as t increases fro 0 to T, the phase of the cosine, t, increases fro 0 to π radians. T So, just as the block, fro tie 0 to tie T, goes though one cycle of its otion, the cosine, fro tie 0 to tie T, goes through one cycle of its pattern. As the stopwatch reading increases fro T to T, the phase of the cosine increases fro π rad to 4π rad. The block undergoes the second cycle of its otion and the cosine function used to deterine the position of the block goes through the second cycle of its pattern. The idea holds true for any tie t as the stopwatch reading continues to increase, the cosine function keeps repeating its cycle in eact synchronization with the block, as it ust if its alue is to accurately represent the position of the π rad block as a function of tie. Again, it is no coincidence. We chose the constant in the T phase of the cosine so that things would work out this way. A few words on jargon are in order before we oe on. The tie T that it takes for the block to coplete one full cycle of its otion is referred to as the period of the oscillations of the block. π rad Now how about that other constant, the a in our educated guess = a cos t? T Again, the units were our guide. When you take the cosine of an angle, you get a pure nuber, a alue with no units. So, we need the a there to gie our function units of distance (we ll use eters). We can further relate a to the otion of the block. The biggest the cosine of the phase can eer get is, thus, the biggest a ties the cosine of the phase can eer get is a. π rad So, in the epression = a cos t, with being the position of the block at any tie t, T a ust be the aiu position of the block, the position of the block, relatie to its equilibriu position, when it is as far fro the wall as it eer gets. π rad Okay, we e gien a lot of reasons why = acos t should well describe the otion T of the block, but unless it is consistent with Newton s nd Law, that is, unless it satisfies equation 7-: 9

192 Chapter 7 Oscillations: Introduction, Mass on a Spring d dt = k which we deried fro Newton s nd Law, it is no good. So, let s plug it into equation 7- and d see if it works. First, let s take the second deriatie of our trial solution with respect to t dt (so we can plug it and itself directly into equation 7-): Gien π rad = acos t, T the first deriatie is d = dt a d π rad = dt T π rad π rad sin t T T a π rad sin t T The second deriatie is then Now we are ready to substitute this and itself, d k equation = dt substitution yields: d π rad π rad π rad = cos t a dt T T T d π rad π rad = a cos t dt T T π rad = a cos t, into the differential T (equation 7-) steing fro Newton s nd Law of Motion. The π rad T a π rad cos t = T k a π rad cos t T 9

193 Chapter 7 Oscillations: Introduction, Mass on a Spring which we copy here for your conenience. π rad T a π rad cos t = T k a π rad cos t T π rad The two sides are the sae, by inspection, ecept that where appears on the left, we T k π rad hae on the right. Thus, substituting our guess, = a cos t, into the differential T d k equation that we are trying to sole, = (equation 7-) leads to an identity if and only dt π rad k if =. This eans that the period T is deterined by the characteristics of the T spring and the block, ore specifically by the force constant (the stiffness factor ) k of the spring, and the ass (the inertia) of the block. Let s sole for T in ters of these quantities. π rad Fro = T k we find: π rad = T k T = π rad k T = π (7-3) k where we hae taken adantage of the fact that a radian is, by definition, / by siply deleting the rad fro our result. The presence of the in the nuerator eans that the greater the ass, the longer the period. That akes sense: we would epect the block to be ore sluggish when it has ore ass. On the other hand, the presence of the k in the denoinator eans that the stiffer the spring, the shorter the period. This akes sense too in that we would epect a stiff spring to result in quicker oscillations. Note the absence of a in the result for the period T. Many folks would epect that the bigger the oscillations, the longer it would take the block to coplete each oscillation, but the absence of a in our result for T shows that it just isn t so. The period T does not depend on the size of the oscillations. So, our end result is that a block of ass, on a frictionless horizontal surface, a block that is attached to a wall by an ideal assless horizontal spring, and released, at tie t = 0, fro rest, fro a position = a, a distance a fro its equilibriu position; will oscillate about the 93

194 Chapter 7 Oscillations: Introduction, Mass on a Spring equilibriu position with a period of tie will be gien by T = π. Furtherore, the block s position as a function k π rad = a cos t T (7-4) Fro this epression for (t) we can derie an epression for the elocity (t) as follows: = = d dt d dt = a a π rad cos t T π rad π rad sin t T T π rad π rad = a sin t (7-5) T T And fro this epression for (t) we can get the acceleration a(t) as follows: d a = dt d π rad π rad a = t dt a sin T T π rad π rad π rad a = a cos t T T T π rad π rad a = a cos t T T (7-6) k Note that this latter result is consistent with the relation a = between a and that we deried fro Newton s nd Law near the start of this chapter. Recognizing that the π rad a cos t T thing as π rad is and that the T is k, it is clear that equation 7-6 is the sae k a = (7-7) 94

195 Chapter 7 Oscillations: Introduction, Mass on a Spring Frequency The period T has been defined to be the tie that it takes for one coplete oscillation. In SI units we can think of it as the nuber of seconds per oscillation. The reciprocal of T is thus the nuber of oscillations per second. This is the rate at which oscillations occur. We gie it a nae, frequency, and a sybol, f. f = (7-8) T oscillations The units work out to be which we can think of as as the oscillation, uch like s s the radian is a arker rather than a true unit. A special nae has been assigned to the SI unit of oscillation frequency, is defined to be hertz, abbreiated Hz. You can think of Hz as s oscillation either or siply s. s In ters of frequency, rather than period, we can use f = to epress all our preious results in T ters of f rather than t. f = π k = a cos ( π radf t) = πf a sin π a = ( πf ) a cos ( radf t) ( π radf t) 95

196 Chapter 7 Oscillations: Introduction, Mass on a Spring y inspection of the epressions for the elocity and acceleration aboe we see that the greatest possible alue for the elocity is π f and the greatest possible alue for the acceleration is ( π f ) a. Defining a a = (πf ) (7-9) a and a a ( πf ) a = (7-0) and, oitting the units rad fro the phase (thus burdening the user with reebering that the units of the phase are radians while aking the epressions a bit ore concise) we hae: ( πf t) = cos (7-) a ( πf t) = sin (7-) a ( πf t) a = a cos (7-3) a The Siple Haronic Equation = a, we refer to the otion as siple haronic otion. In the case of a block on a spring we found that When the otion of an object is sinusoidal as in cos ( πf t) a = constant (7-4) k where the constant was and was shown to be equal to ( π f ). Written as d = ( πf ) (7-5) dt the equation is a copletely general equation, not specific to a block on a spring. Indeed, any tie you find that, for any syste, the second deriatie of the position ariable, with respect to tie, is equal to a negatie constant ties the position ariable itself, you are dealing with a case of siple haronic otion, and you can equate the absolute alue of the constant to ( π f ). 96

197 Chapter 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion Starting with the pendulu bob at its highest position on one side, the period of oscillations is the tie it takes for the bob to swing all the way to its highest position on the other side and back again. Don t forget that part about and back again. y definition, a siple pendulu consists of a particle of ass suspended by a assless unstretchable string of length L in a region of space in which there is a unifor constant graitational field, e.g. near the surface of the earth. The suspended particle is called the pendulu bob. Here we discuss the otion of the bob. While the results to be reealed here are ost precise for the case of a point particle, they are good as long as the length of the pendulu (fro the fied top end of the string to the center of ass of the bob) is large copared to a characteristic diension (such as the diaeter if the bob is a sphere or the edge length if it is a cube) of the bob. (Using a pendulu bob whose diaeter is 0% of the length of the pendulu (as opposed to a point particle) introduces a 0.05% error. You hae to ake the diaeter of the bob 45% of the pendulu length to get the error up to %.) If you pull the pendulu bob to one side and release it, you find that it swings back and forth. It oscillates. At this point, you don t know whether or not the bob undergoes siple haronic otion, but you certainly know that it oscillates. To find out if it undergoes siple haronic otion, all we hae to do is to deterine whether its acceleration is a negatie constant ties its position. ecause the bob oes on an arc rather than a line, it is easier to analyze the otion using angular ariables. θ L The bob oes on the lower part of a ertical circle that is centered at the fied upper end of the string. We ll position ourseles such that we are iewing the circle, face on, and adopt a coordinate syste, based on our point of iew, which has the reference direction straight 97

198 Chapter 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion downward, and for which positie angles are easured counterclockwise fro the reference direction. Referring to the diagra aboe, we now draw a pseudo free-body diagra (the kind we use when dealing with torque) for the string-plus-bob syste. O r = L sinθ + a θ L F g = g We consider the counterclockwise direction to be the positie direction for all the rotational otion ariables. Applying Newton s nd Law for Rotational Motion, yields: a o< = τ I o< a = gl sinθ I Net we ipleent the sall angle approiation. Doing so eans our result is approiate and that the saller the aiu angle achieed during the oscillations, the better the approiation. According to the sall angle approiation, with it understood that θ ust be in radians, sin θ θ. Substituting this into our epression for a, we obtain: a = glθ I Here coes the part where we treat the bob as a point particle. The oent of inertia of a point particle, with respect to an ais that is a distance L away, is gien by I = L. Substituting this into our epression for a we arrie at: 98

199 Chapter 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion gl a = θ L Soething profound occurs in our siplification of this equation. The asses cancel out. The ass that deterines the driing force behind the otion of the pendulu (the graitational force F g = g) in the nuerator, is eactly canceled by the inertial ass of the bob in the denoinator. The otion of the bob does not depend on the ass of the bob! Siplifying the epression for a yields: g a = θ L d θ Recalling that a, we hae: dt d θ g = θ (8-) dt L Hey, this is the siple haronic otion equation, which, in generic for, appears as d = constant (equation 7-4) in which the constant can be equated to ( π f ) where f dt is the frequency of oscillations. The position ariable in our equation ay not be, but we still hae the second deriatie of the position ariable being equal to the negatie of a constant ties the position ariable itself. That being the case, nuber : we do hae siple haronic otion, g and nuber : the constant ust be equal to ( π f ). L g = ( πf ) L Soling this for f, we find that the frequency of oscillations of a siple pendulu is gien by g f = (8-) π L Again we call your attention to the fact that the frequency does not depend on the ass of the bob! T = as in the case of the block on a spring. This relation between T and f is a definition that f applies to any oscillatory otion (een if the otion is not siple haronic otion). 99

200 Chapter 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion All the other forulas for the siple pendulu can be transcribed fro the results for the block on a spring by writing Thus, θ in place of, w in place of, and a in place of a. θ = θ cos(πf t) (8-3) a w = w sin(πf t) (8-4) a a = a cos(πf t) (8-5) a w = ( θ (8-6) a πf ) a πf ) a a = ( θ (8-7) a Energy Considerations in Siple Haronic Motion Let s return our attention to the block on a spring. A person pulls the block out away fro the wall a distance a fro the equilibriu position, and releases the block fro rest. At that instant, before the block picks up any speed at all, (but when the person is no longer affecting the otion of the block) the block has a certain aount of energy E. And since we are dealing with an ideal syste (no friction, no air resistance) the syste has that sae aount of energy fro then on. In general, while the block is oscillating, the energy E = K + U is partly kinetic energy K = and partly spring potential energy U = k. The aount of each aries, but the total reains the sae. At tie 0, the K in E = K + U is zero since the elocity of the block is zero. So, at tie 0: E = U E = k a An endpoint in the otion of the block is a particularly easy position at which to calculate the total energy since all of it is potential energy. As the spring contracts, pulling the block toward the wall, the speed of the block increases so, the kinetic energy increases while the potential energy U = k decreases because the spring becoes less and less stretched. On its way toward the equilibriu position, the syste has both kinetic and potential energy 00

201 Chapter 8 Oscillations: The Siple Pendulu, Energy in Siple Haronic Motion E = K + U with the kinetic energy K increasing and the potential energy U decreasing. Eentually the block reaches the equilibriu position. For an instant, the spring is neither stretched nor copressed and hence it has no potential energy stored in it. All the energy (the sae total that we started with) is in the for of kinetic energy, K =. 0 E = K + U E = K The block keeps on oing. It oershoots the equilibriu position and starts copressing the spring. As it copresses the spring, it slows down. Kinetic energy is being conerted into spring potential energy. As the block continues to oe toward the wall, the eer-the-sae alue of total energy represents a cobination of kinetic energy and potential energy with the kinetic energy decreasing and the potential energy increasing. Eentually, at its closest point of approach to the wall, its aiu displaceent in the direction fro its equilibriu position, at its turning point, the block, just for an instant has a elocity of zero. At that instant, the kinetic energy is zero and the potential energy is at its aiu alue: 0 E = K + U E = U Then the block starts oing out away fro the wall. Its kinetic energy increases as its potential energy decreases until it again arries at the equilibriu position. At that point, by definition, the spring is neither stretched nor copressed so the potential energy is zero. All the energy is in the for of kinetic energy. ecause of its inertia, the block continues past the equilibriu position, stretching the spring and slowing down as the kinetic energy decreases while, at the sae rate, the potential energy increases. Eentually, the block is at its starting point, again just for an instant, at rest, with no kinetic energy. The total energy is the sae total as it has been throughout the oscillatory otion. At that instant, the total energy is all in the for of potential energy. The conersion of energy, back and forth between the kinetic energy of the block and the potential energy stored in the spring, repeats itself oer and oer again as long as the block continues to oscillate (with and this is indeed an idealization no loss of echanical energy). A siilar description, in ters of energy, can be gien for the otion of an ideal (no air resistance, copletely unstretchable string) siple pendulu. The potential energy, in the case of the siple pendulu, is in the for of graitational potential energy U = gy rather than spring potential energy. The one alue of total energy that the pendulu has throughout its oscillations is all potential energy at the endpoints of the oscillations, all kinetic energy at the idpoint, and a i of potential and kinetic energy at locations in between. 0

202 Chapter 9 Waes: Characteristics, Types, Energy 9 Waes: Characteristics, Types, Energy Consider a long taut horizontal string of great length. Suppose one end is in the hand of a person and the other is fied to an iobile object. Now suppose that the person oes her hand up and down. The person causes her hand, and her end of the string, to oscillate up and down. To discuss what happens, we, in our ind, consider the string to consist of a large nuber of ery short string segents. It is iportant to keep in ind that the force of tension of a string segent eerted on any object, including another segent of the string, is directed away fro the object along the string segent that is eerting the force. (The following discussion and diagras are intentionally oersiplified. The discussion does correctly gie the gross idea of how oscillations at one end of a taut string can cause a pattern to oe along the length of the string despite the fact that the indiidual bits of string are essentially doing nothing ore than oing up and down. The person is holding one end of the first segent. She first oes her hand upward. This tilts the first segent so that the force of tension that it is eerting on the second segent has an upward coponent. This, in turn, tilts the second segent so that its force of tension on the third segent now has an upward coponent. The process continues with the 3 rd segent, the 4 th segent, etc. 0

203 Chapter 9 Waes: Characteristics, Types, Energy After reaching the top of the oscillation, the person starts oing her hand downward. She oes the left end of the first segent downward, but by this tie, the first four segents hae an upward elocity. Due to their inertia, they continue to oe upward. The downward pull of the first segent on the left end of the second segent causes it to slow down, coe to rest, and eentually start oing downward. Inertia plays a huge role in wae propagation. To propagate eans to go or to trael. Waes propagate through a ediu. 03

204 Chapter 9 Waes: Characteristics, Types, Energy 04

205 Chapter 9 Waes: Characteristics, Types, Energy Crest Trough Each ery short segent of the string undergoes oscillatory otion like that of the hand, but for any gien section, the otion is delayed relatie to the otion of the neighboring segent that is closer to the hand. The net effect of all these string segents oscillating up and down, each with the sae frequency but slightly out of synchronization with its nearest neighbor, is to create a disturbance in the string. Without the disturbance, the string would just reain on the original horizontal line. The disturbance oes along the length of the string, away fro the hand. The disturbance is called a wae. An obserer, looking at the string fro the side sees crests and troughs of the disturbance, oing along the length of the string, away fro the hand. Despite appearances, no aterial is oing along the length of the string, just a disturbance. The illusion that actual aterial is oing along the string can be eplained by the tiing with which the indiidual segents oe up and down, each about its own equilibriu position, the position it was in before the person started aking waes. Wae Characteristics In our pictorial odel aboe, we depicted a hand that was oscillating but not undergoing siple haronic otion. If the oscillations that are causing the wae do confor to siple haronic otion, then each string segent aking up the string will eperience siple haronic otion (up and down). When indiidual segents aking up the string are each undergoing siple haronic otion, the wae pattern is said to ary sinusoidally in both tie and space. We can tell that it aries sinusoidally in space because a graph of the displaceent y, the distance that a gien point on the string is aboe its equilibriu position, ersus, how far fro the end of the string the point on the string is; for all points on the string; is sinusoidal. 05

206 Chapter 9 Waes: Characteristics, Types, Energy y [eters] A IT OF STRING S DISPLACEMENT AOVE OR ELOW THAT IT S EQUILIRIUM POSITION VS. THE POSITION OF THE IT OF STRING ALONG THE LENGTH OF THE STRING 0 [eters] We say that the wae aries sinusoidally with tie because, for any point along the length of the string, a graph of the displaceent of that point fro its equilibriu position s. tie is sinusoidal: y [eters] A IT OF STRING S DISPLACEMENT AOVE OR ELOW THAT IT S EQUILIRIUM POSITION VS. TIME 0 t [seconds] There are a nuber of ways of characterizing the wae-on-a-string syste. You could probably coe up with a rather coplete list yourself: the rate at which the oscillations are occurring, how long it takes for a gien tiny segent of the string to coplete one oscillation, how big the oscillations are, the sallest length of the unique pattern which repeats itself in space, and the speed at which the wae pattern traels along the length of the string. Physicists hae, of course, gien naes to the arious quantities, in accordance with that iportant lowest leel of scientific actiity naing and categorizing the arious characteristics of that aspect of the natural world which is under study. Here are the naes: 06

207 Chapter 9 Waes: Characteristics, Types, Energy Aplitude Any particle of a string with waes traeling through it undergoes oscillations. Such a particle goes away fro its equilibriu position until it reaches its aiu displaceent fro its equilibriu position. Then it heads back toward its equilibriu position and then passes right through the equilibriu position on its way to its aiu displaceent fro equilibriu on the other side of its equilibriu position. Then it heads back toward the equilibriu position and passes through it again. This otion repeats itself continually as long as the waes are traeling through the location of the particle of the string in question. The aiu displaceent of any particle along the length of the string, fro that point s equilibriu position, is called the aplitude y a of the wae. The aplitude can be annotated on both of the two kinds of graphs gien aboe (Displaceent s. Position, and Displaceent s. Tie). Here we annotate it on the Displaceent s. Position graph: y [eters] DISPLACEMENT VS. POSITION Aplitude y a 0 Peak-to-Peak Aplitude [eters] The peak-to-peak aplitude, a quantity that is often easier to easure than the aplitude itself, has also been annotated on the graph. It should be obious that the peak-to-peak aplitude is twice the aplitude. 07

208 Chapter 9 Waes: Characteristics, Types, Energy Period The aount of tie that it takes any one particle along the length of the string to coplete one oscillation is called the period T. Note that the period is copletely deterined by the source of the waes. The tie it takes for the source of the waes to coplete one oscillation is equal to the tie it takes for any particle of the string to coplete one oscillation. That tie is the period of the wae. The period, being an aount of tie, can only be annotated on the Displaceent s. Tie graph (not on the Displaceent s. Position Along the String graph). y [eters] DISPLACEMENT VS. TIME Period T 0 t [seconds] Frequency The frequency f is the nuber-of-oscillations-per-second that any particle along the length of the string undergoes. It is the oscillation rate. Since it is the nuber-of-oscillations-per-second and the period is the nuber-of-seconds-per-oscillation, the frequency f is siply the reciprocal of the period T: f =. T Aplitude, period, and frequency are quantities that you learned about in your study of oscillations. Here, they characterize the oscillations of a point on a string. Despite the fact that the string as a whole is undergoing wae otion, the fact that the point itself, any point along the length of the string, is siply oscillating, eans that the definitions of aplitude, period, and frequency are the sae as the definitions gien in the chapter on oscillations. Thus, our discussion of aplitude, period, and frequency represents a reiew. Now, howeer, it is tie to oe on to soething new, a quantity that does not apply to siple haronic otion but does apply to waes. 08

209 Chapter 9 Waes: Characteristics, Types, Energy Waelength The distance oer which the wae pattern repeats itself once, is called the waelength λ of the wae. ecause the waelength is a distance easured along the length of the string, it can be annotated on the Displaceent s. Position Along the String graph (but not on the Displaceent s. Tie graph): y [eters] DISPLACEMENT VS. POSITON Waelength λ 0 [eters] Wae Velocity The wae elocity is the speed and direction with which the wae pattern is traeling. (It is NOT the speed with which the particles aking up the string are traeling in their up and down otion.) The direction part is straightforward, the wae propagates along the length of the string, away fro the cause (soething oscillating) of the wae. The wae speed (the constant speed with which the wae propagates) can be epressed in ters of other quantities that we hae just discussed. 09

210 Chapter 9 Waes: Characteristics, Types, Energy To get at the wae speed, what we need to do is to correlate the up-and-down otion of a point on the string, with the otion of the wae pattern oing along the string. Consider the following Displaceent s. Position graph for a wae traeling to the right. In the diagra, I hae shaded in one cycle of the wae, arked off a distance of one waelength, and drawn a dot at a point on the string whose otion we shall keep track of. y [eters] Waelength λ 0 [eters] Now, let s allow soe tie to elapse, just enough tie for the wae to oe oer one quarter of a waelength. y [eters] Waelength λ 0 [eters] In that tie we note that the point on the string arked by the dot has oed fro its equilibriu position to its aiu displaceent fro equilibriu position. As the wae has oed oer one quarter of a waelength, the point on the string has copleted one quarter of an oscillation. 0

211 Chapter 9 Waes: Characteristics, Types, Energy Let s allow the sae aount of tie to elapse again, the tie it takes for the wae to oe oer one quarter of a waelength. y [eters] Waelength λ 0 [eters] At this point, the wae has oed a total of a half a waelength oer to the right, and the point on the string arked by the dot has oed fro its equilibriu position up to its position of aiu positie displaceent and back to its equilibriu position; that is to say, it has copleted half of an oscillation. Let s let the sae aount of tie elapse again, enough tie for the wae pattern to oe oer another quarter of a waelength. y [eters] Waelength λ 0 The wae has oed oer a total distance of three quarters of a waelength and the point on the string that is arked with a dot has oed on to its aiu negatie (downward) displaceent fro equilibriu eaning that it has copleted three quarters of an oscillation.

212 Chapter 9 Waes: Characteristics, Types, Energy Now we let the sae aount of tie elapse once ore, the tie it takes for the wae to oe oer one quarter of a waelength. y [eters] Waelength λ 0 At this point, the wae has oed oer a distance equal to one waelength and the point on the string arked by a dot has copleted one oscillation. It is that point of the string whose otion we hae been keeping track of that gies us a handle on the tie. The aount of tie that it takes for the point on the string to coplete one oscillation is, by definition, the period of the wae. Now we know that the wae oes a distance of one waelength λ in a tie interal equal to one period T. For soething oing with a constant speed (zero acceleration), the speed is siply the distance traeled during a specified tie interal diided by the duration of that tie interal. So, we hae, for the wae speed : One typically sees the forula for the wae speed epressed as λ = (9-) T = λf (9-) where the relation f = between frequency and period has been used to eliinate the period. T Equation 9- ( = λf ) suggests that the wae speed depends on the frequency and the waelength. This is not at all the case. Indeed, as far as the waelength is concerned, it is the other way round the oscillator that is causing the waes deterines the frequency, and the corresponding waelength depends on the wae speed. The wae speed is predeterined by the characteristics of the string how taut it is, and how uch ass is packed into each illieter of it. Looking back on our discussion of how oscillations at one end of a taut string result in waes propagating through it, you can probably deduce that the greater the tension in the string, the faster the wae will oe along the string. When the hand oes the end of the first segent up, the force eerted on the second segent of the string by the first segent will be greater, the greater the tension in the string. Hence the second segent will eperience a greater

213 Chapter 9 Waes: Characteristics, Types, Energy acceleration. This goes for all the segents down the line. The greater the acceleration, the faster the segents pick up speed and the faster the disturbance, the wae, propagates along the string that is, the greater the wae speed. We said that the wae speed also depends on the aount of ass packed into each illieter of the string. This refers to the linear density µ, the ass-per-length, of the string. The greater the ass-per-length, the greater the ass of each segent of the string and the less rapidly the elocity will be changing for a gien force. Here we proide, with no proof, the forula for the speed of a wae in a string as a function of the string characteristics, tension F T and linear ass density µ : F = (9-3) µ T Note that this epression agrees with the conclusions that the greater the tension, the greater the wae speed; but the greater the linear ass density, the saller the wae speed. Kinds of Waes We e been talking about a wae in a string. Lots of other edia, besides strings, support waes as well. You hae probably heard, and probably heard of, sound waes in air. Sound waes trael through other gases and they trael through liquids and solids as well. Perhaps you hae heard of seisic waes as well, the waes that trael through the earth when earthquakes occur. All of these waes fall into one of two categories. Which category is deterined by the orientation of the lines along which the oscillations of the particles aking up the ediu occur relatie to the direction of propagation of the waes. If the particles oscillate along lines that are perpendicular to the direction in which the wae traels, the wae is said to be a transerse wae (because transerse eans perpendicular to). If the particles oscillate along a line that is along the direction in which the wae traels, the wae is said to be longitudinal. The wae in a horizontal string that we discussed at such length is an eaple of a transerse wae. Calling the direction in which the string etends away fro the oscillator the forward direction, we discussed the case in which the particles aking up the string were oscillating up and down. The wae traels along the string and the up and down directions are indeed perpendicular to the forward direction aking it clear that we were dealing with transerse waes. It should be noted that the oscillations could hae been fro side to side or at any angle relatie to the ertical as long as they were perpendicular to the string. Also, there is no stipulation that the string be horizontal in order for transerse waes to propagate in it. The string was said to be horizontal in our introduction to waes in order to siplify the discussion. 3

214 Chapter 9 Waes: Characteristics, Types, Energy Wae Power Consider a ery long stationary string etending fro left to right. Consider a ery short segent of the string, call it segent A, at an arbitrary distance along the string fro the left end. Now suppose that soeone is holding the left end of the long string in her hand and that she starts oscillating the left end of the string up and down. As you know, a wae will trael through the string fro left to right. efore it gets to segent A, segent A is at rest. After the front end of the wae gets to segent A, segent A will be oscillating up and down like a ass on a spring. Segent A has ass and it has elocity throughout ost of its otion, so clearly it has energy. It had none before the wae got there, so waes ust hae energy. y oscillating the end of the string the person has gien energy to the string and that energy traels along the string in the for of a wae. Here we gie an epression for the rate at which energy propagates along the string in ters of the string and wae properties. That rate is the energy-per-tie passing through any point (through which the wae is traeling) in the string. It is the power of the wae. The analysis that yields the epression for the rate at which the energy of waes in a string passes a point on the string, that is, the power of the wae, is straightforward but too lengthy to include here. The result is: π f ya P = µ F T As regards waes in a string with a gien tension and linear ass density, we note that P =f (9-4) ya This relation applies to all kinds of waes. The constant of proportionality depends on the kind of wae that you are dealing with, but the proportionality itself applies to all kinds of waes. I was reinded of this relation by a physical therapist who was using ultrasound waes to deposit energy into y back uscles. She entioned that a doubling of the frequency of the ultrasonic waes would proide deeper penetration of the sound waes, but that it would also result in a quadrupling of the rate at which energy would be deposited in the tissue. Hence, to deposit the sae total aount of energy that she deposited at a gien frequency on one occasion; a treatent at double the frequency would either last one fourth as long (at the sae aplitude) or would be carried out for the sae aount of tie at half the aplitude. 4

215 Chapter 9 Waes: Characteristics, Types, Energy Intensity Consider a tiny buzzer, suspended in air by a string. Sound waes, caused by the buzzer, trael outward in all away-fro-the-buzzer directions: These circles represent spherical wae fronts. In the diagra aboe, the black dot represents the buzzer, and the circles represent waefronts collections of points in space at which the air olecules are at their aiu displaceent away fro the buzzer. The waefronts are actually spherical shells. In a 3D odel of the, they would be well represented by soap bubbles, one inside the other, all sharing the sae center. Note that subsequent to the instant depicted, the air olecules at the location of the waefront will start oing back toward the buzzer, in their toward and away fro the buzzer oscillations, whereas the waefront itself will oe steadily outward fro the buzzer as the net layer of air olecules achiees its aiu displaceent position and then the net, etc. Now consider a fied iaginary spherical shell centered on the buzzer. The power of the wae is the rate at which energy passes through that shell. As entioned, the power obeys the relation P =f ya Note that the power does not depend on the size of the spherical shell; all the energy deliered to the air by the buzzer ust pass through any spherical shell centered on the buzzer. ut the surface of a larger spherical shell is farther fro the buzzer and our eperience tells us that the further you are fro the buzzer the less loud it sounds suggesting that the power deliered to our ear is saller. So how can the power for a large spherical shell (with its surface far fro the source) be the sae as it is for a sall spherical shell? We can say that as the energy oes away fro the source, it spreads out. So by the tie it reaches the larger spherical shell, the power passing through, say, any square illieter of the larger spherical shell is relatiely sall, but the larger spherical shell has enough ore square illieters for the total power through it to be the sae. 5

216 Chapter 9 Waes: Characteristics, Types, Energy Now iagine soebody near the source with their eardru facing the source. These circles represent spherical wae fronts. The aount of energy deliered to the ear is the power-per-area passing through the iaginary source-centered spherical shell whose surface the ear is on, ties the area of the eardru. Since the spherical shell is sall, eaning it has relatiely little surface area, and all the power fro the source ust pass through that spherical shell, the power-per-area at the location of the ear ust be relatiely large. Multiply that by the fied area of the eardru and the power deliered to the eardru is relatiely large. If the person is farther fro the source, These circles represent spherical wae fronts. the total power fro the source is distributed oer the surface of a larger spherical shell so the power-per-area is saller. Multiply that by the fied area of the eardru to get the power deliered to the ear. It is clear that the power deliered to the ear will be saller. The farther the ear is fro the source; the saller is the fraction of the total power of the source, receied by the eardru. 6

217 Chapter 9 Waes: Characteristics, Types, Energy How loud the buzzer sounds to the person is deterined by the power deliered to the ear, which, we hae noted, depends on the power-per-area at the location of the ear. The power-per-area of sound waes is gien a nae. It is called the intensity of the sound. For waes in general (rather than just sound waes) we siply call it the intensity of the wae. While the concept of intensity applies to waes fro any kind of source, it is particularly easy to calculate in the case of the sall buzzer deliering energy uniforly in all directions. For any point in space, we create an iaginary spherical shell, through that point, centered on the buzzer. Then the intensity I at the point (and at any other point on the spherical shell) is siply the power of the source diided by the area of the spherical shell: P I = (9-5) 4π r where the r is the radius of the iaginary spherical shell, but ore iportantly, it is the distance of the point at which we wish to know the intensity, fro the source. Since P =f y, we hae a f ya I = (9-6) r At a fied distance fro a source of a fied frequency, recognizing that y a is the aplitude of the waes we hae I = (Aplitude) (9-7) 7

218 Chapter 30 Wae Function, Interference, Standing Waes 30 Wae Function, Interference, Standing Waes In that two of our fie senses (sight and sound) depend on our ability to sense and interpret waes, and in that waes are ubiquitous, waes are of iense iportance to huan beings. Waes in physical edia confor to a wae equation that can be deried fro Newton s Second Law of otion. The wae equation reads: y = y t where: y is the displaceent of a point on the ediu fro its equilibriu position, is the position along the length of the ediu, t is tie, and is the wae elocity. (30-) Take a good look at this iportant equation. ecause it inoles deriaties, the wae equation is a differential equation. The wae equation says that the second deriatie of the displaceent with respect to position (treating the tie t as a constant) is directly proportional to the second deriatie of the displaceent with respect to tie (treating the position as a constant). When you see an equation for which this is the case, you should recognize it as the wae equation. In general, when the analysis of a continuous ediu, e.g. the application of Newton s second law to the eleents aking up that ediu, leads to an equation of the for y = constant t y, the constant will be an algebraic cobination of physical quantities representing properties of the ediu. That cobination can be related to the wae elocity by constant = For instance, application of Newton s Second Law to the case of a string results in a wae equation in which the constant of proportionality depends on the linear ass density µ and the string tension F T : y µ y = F t T Recognizing that the constant of proportionality square of the wae elocity, we hae µ has to be equal to the reciprocal of the F T 8

219 Chapter 30 Wae Function, Interference, Standing Waes µ = F T F T = (30-) µ relating the wae elocity to the properties of the string. The solution of the wae equation y y = can be epressed as t y = y cos( π a φ ) λ π ± t + (30-3) T where: y is the displaceent of a point in the ediu fro its equilibriu position, y a is the aplitude of the wae, is the position along the length of the ediu, λ is the waelength, t is tie, T is the period, and φ is a constant haing units of radians. φ is called the phase constant. A in the location of the ± is used in the case of a wae traeling in the + direction and a + for one traeling in the direction. Equation 30-3, the solution to the wae equation y y =, is known as the wae function. Substitute the wae function into the wae t equation and erify that you arrie at λ =, T λ a necessary condition for the wae function to actually sole the wae equation. = is the T stateent that the wae speed is equal to the ratio of the waelength to the period, a relation that we deried in a conceptual fashion in the last chapter. At position = 0 in the ediu, at tie t = 0, the wae function, equation 30-3, ealuates to ) y = y a cos( φ. The phase of the cosine boils down to the phase constant φ whose alue thus deterines the alue of y at = 0, t = 0. (Note that the phase of the cosine is the arguent of the cosine that which you are taking the cosine of.) The alue of the phase constant φ is of no releance to our 9

220 Chapter 30 Wae Function, Interference, Standing Waes present discussion so we arbitrarily set φ = 0. Also, on your forula sheet, we write the wae function only for the case of a wae traeling in the + direction, that is, we replace the ± with a. The wae function becoes y = y cos( a t) λ π T π (30-4) This is the way it appears on your forula sheet. You are supposed to know that this corresponds to a wae traeling in the + direction and that the epression for a wae traeling in the direction can be arried at by replacing the with a +. Interference Consider a case in which two waefors arrie at the sae point in a ediu at the sae tie. We ll use idealized waefors in a string to ake our points here. In the case of a string, the only way two waefors can arrie at the sae point in the ediu at the sae tie is for the waefors to be traeling in opposite directions: b h A h b The two waefors depicted in the diagra aboe are scheduled to arrie at point A at the sae tie. At that tie, based on the waefor on the left alone, point A would hae a displaceent +h, and based on the waefor on the right alone, point A would hae the displaceent h. So, what is the actual displaceent of point A when both waefors are at point A at the sae tie? To answer that, you siply add the would-be single-waefor displaceents together algebraically (taking the sign into account). One does this point for point oer the entire length of the string for any gien instant in tie. In the following series of diagras we show the point-for-point addition of displaceents for seeral instants in tie. 0

221 Chapter 30 Wae Function, Interference, Standing Waes = = = = = =

222 Chapter 30 Wae Function, Interference, Standing Waes The phenoenon in which waes traeling in different directions siultaneously arrie at one and the sae point in the wae ediu is referred to as interference. When the waefors add together to yield a bigger waefor, + = the interference is referred to as constructie interference. When the two waefors tend to cancel each other out, + = the interference is referred to as destructie interference.

223 Chapter 30 Wae Function, Interference, Standing Waes Reflection of a Wae fro the End of a Mediu Upon reflection fro the fied end of a string, the displaceent of the points on a traeling waefor is reersed. Wae Pulse Approaching Fied End Wae Pulse Receding Fro Fied End After Reflection The fied end, by definition, neer undergoes any displaceent. Now we consider a free end. A fied end is a natural feature of a taut string. A free end, on the other hand, an idealization, is at best an approiation in the case of a taut string. We approiate a free end in a physical string by eans of a drastic and abrupt change in linear density. Consider a rope, one of which is attached to the wae source, and the other end of which is attached to one end of a piece of thin, but strong, fishing line. Assue that the fishing line etends through soe large distance to a fied point so that the whole syste of rope plus fishing line is taut. A wae traeling along the rope, upon encountering the end of the rope attached to the thin fishing line, behaes approiately as if it has encountered a free end of a taut rope. Rope Fishing Line In the case of sound waes in a pipe, a free end can be approiated by an open end of the pipe. Enough said about how one ight set up a physical free end of a wae ediu, what happens when a wae pulse encounters a free end? The answer is, as in the case of the fied end, the waefor is reflected, but this tie, there is no reersal of displaceents. 3

224 Chapter 30 Wae Function, Interference, Standing Waes Wae Pulse Approaching Free End Wae Pulse Receding Fro Free End After Reflection Standing Waes Consider a piece of string, fied at both ends, into which waes hae been introduced. The configuration is rich with interference. A wae traeling toward one end of the string reflects off the fied end and interferes with the waes that were trailing it. Then it reflects off the other end and interferes with the again. This is true for eery wae and it repeats itself continuously. For any gien length, linear density, and tension of the string, there are certain frequencies for which, at, at least one point on the string, the interference is always constructie. When this is the case, the oscillations at that point (or those points) on the string are aial and the string is said to hae standing waes in it. Again, standing waes result fro the interference of the reflected waes with the transitted waes and with each other. A point on the string at which the interference is always constructie is called an antinode. Any string in which standing waes eist also has at least one point at which the interference is always destructie. Such a point on the string does not oe fro its equilibriu position. Such a point on the string is called a node. It ight see that it would be a daunting task to deterine the frequencies that result in standing waes. Suppose you want to inestigate whether a point on a string could be an antinode. Consider an instant in tie when a wae crest is at that position. You need to find the conditions that would ake it so that in the tie it takes for the crest to trael to one fied end of the string, reflect back as a trough and arrie back at the location in question; a trough, e.g. one that was trailing the original crest, propagates just the right distance so that it arries at the location in question at the sae tie. As illustrated in the net chapter, the analysis that yields the frequencies of standing waes is easier than these tiing considerations would suggest. 4

225 Chapter 3 Strings, Air Coluns 3 Strings, Air Coluns e careful not to jup to any conclusions about the waelength of a standing wae. Folks will do a nice job drawing a graph of Displaceent s. Position Along the Mediu and then interpret it incorrectly. For instance, look at the diagra on this page. Folks see that a half waelength fits in the string segent and quickly write the waelength as λ = L. ut this equation says that a whole waelength fits in half the length of the string. This is not at all the case. Rather than recognizing that the fraction is releant and quickly using that fraction in an epression for the waelength, one needs to be ore systeatic. First write what you see, in the for of an equation, and then sole that equation for the waelength. For instance, in the diagra below we see that one half a waelength λ fits in the length L of the string. Writing this in equation for yields λ = L. Soling this for λ yields λ = L. One can deterine the waelengths of standing waes in a straightforward anner and obtain the frequencies fro = λf where the wae speed is deterined by the tension and linear ass density of the string. The ethod depends on the boundary conditions the conditions at the ends of the wae ediu. (The wae ediu is the substance [string, air, water, etc.] through which the wae is traeling. The wae ediu is what is waing. ) Consider the case of waes in a string. A fied end forces there to be a node at that end because the end of the string cannot oe. (A node is a point on the string at which the interference is always destructie, resulting in no oscillations. An antinode is a point at which the interference is always constructie, resulting in aial oscillations.) A free end forces there to be an antinode at that end because at a free end the wae reflects back on itself without phase reersal (a crest reflects as a crest and a trough reflects as a trough) so at a free end you hae one and the sae part of the wae traeling in both directions along the string. The waelength condition for standing waes is that the wae ust fit in the string segent in a anner consistent with the boundary conditions. For a string of length L fied at both ends, we can eet the boundary conditions if half a waelength is equal to the length of the string. L Such a wae fits the string in the sense that wheneer a zero-displaceent part of the wae is aligned with one fied end of the string another zero-displaceent part of the wae is aligned with the other fied end of the string. 5

226 Chapter 3 Strings, Air Coluns Since half a waelength fits in the string segent we hae: λ = L λ = L Gien the wae speed, the frequency can be soled for as follows: = λf f = λ f = It should be noted that despite the fact that the wae is called a standing wae and the fact that it is typically depicted at an instant in tie when an antinode on the string is at its aiu displaceent fro its equilibriu position, all parts of the string (ecept the nodes) do oscillate about their equilibriu position. L Note that, while the interference at the antinode, the point in the iddle of the string in the case at hand, is always as constructie as possible, that does not ean that the string at that point is always at aiu displaceent. At ties, at that location, there is indeed a crest interfering with a crest, but at other ties, there is a zero displaceent part of the wae interfering with a zero-displaceent part of the wae, at ties a trough interfering with a trough, and at ties, an interediate-displaceent part of the wae interfering with the sae interediate-displaceent part of the wae traeling in the opposite direction. All of this corresponds to the antinode oscillating about its equilibriu position. 6

227 Chapter 3 Strings, Air Coluns The λ = L wae is not the only wae that will fit in the string. It is, howeer, the longest waelength standing wae possible and hence is referred to as the fundaental. There is an entire sequence of standing waes. They are naed: the fundaental, the first oertone, the second oertone, the third oertone, etc, in order of decreasing waelength, and hence, increasing frequency. Fundaental L λ = L so λ = L st Oertone λ = L nd Oertone 3 λ = L so λ = 3 L Each successie waefor can be obtained fro the preceding one by including one ore node. A wae in the series is said to be a haronic if its frequency can be epressed as an integer ties the fundaental frequency. The alue of the integer deterines which haronic ( st, nd, 3 rd, etc.) the wae is. The frequency of the fundaental wae is, of course, ties itself. The nuber is an integer so the fundaental is a haronic. It is the st haronic. 7

228 Chapter 3 Strings, Air Coluns Starting with the waelengths in the series of diagras aboe, we hae, for the frequencies, using = λf which can be rearranged to read The Fundaental f = λ λ FUND = L f = FUND λ FUND f FUND = L The st Oertone λ st O.T. = L f = st O.T. λ st O.T. f st O.T. = L The nd Oertone λ nd O.T. = f = nd O.T. 3 L λ f nd O.T. = L f nd O.T. = 3 nd O.T. 3 L 8

229 Chapter 3 Strings, Air Coluns Epressing the frequencies in ters of the fundaental frequency f FUND = we hae L f = L L FUND = = f f st O.T. = = = f L L 3 f = L L nd O.T. = = 3 3 FUND FUND f FUND Note that the fundaental is (as always) the st haronic; the st oertone is the nd haronic; and the nd oertone is the 3 rd haronic. While it is true for the case of a string that is fied at both ends (the syste we hae been analyzing), it is not always true that the set of all oertones plus fundaental includes all the haronics. For instance, consider the following eaple: Eaple 3- Solution An organ pipe of length L is closed at one end and open at the other. Gien that the speed of sound in air is s, find the frequencies of the fundaental and the first three oertones. L Fundaental λ = L 4 so λ = 4L L st Oertone 3 λ = 4 L so 4 λ = 3 L 9

230 Chapter 3 Strings, Air Coluns L nd Oertone 5 λ = 4 L so 4 λ = 5 L L 3 rd Oertone 7 λ = 4 L so 4 λ = 7 In the preceding sequence of diagras, a graph of displaceent s. position along the pipe, for an instant in tie when the air olecules at an antinode are at their aiu displaceent fro equilibriu, is a ore abstract representation then the corresponding graph for a string. The sound wae in air is a longitudinal wae, so, as the sound waes trael back and forth along the length of the pipe, the air olecules oscillate back and forth (rather than up and down as in the case of the string) about their equilibriu positions. Thus, how high up on the graph a point on the graph is, corresponds to how far to the right (using the iewpoint fro which the pipe is depicted in the diagras) of its equilibriu position the thin layer of air olecules, at the corresponding position in the pipe, is. It is conentional to draw the waefor right inside the outline of the pipe. The boundary conditions are that a closed end is a node and an open end is an antinode. Starting with the waelengths in the series of diagras aboe, we hae, for the frequencies, using s = λf which can be rearranged to read L The Fundaental f = s λ λ FUND = 4L s f FUND = λ s f FUND = 4L FUND 30

231 Chapter 3 Strings, Air Coluns The st Oertone λ st O.T. = 4 3 L s f st O.T. = λ s f st O.T. = L s f st O.T. = 4 L st O.T. The nd Oertone λ nd O.T. = 4 5 L s f nd O.T. = λ s f nd O.T. = L 4 5 5s f nd O.T. = 4 L nd O.T. The 3 rd Oertone λ 3rd O.T. = 4 7 L s f 3rd O.T. = λ s f 3rd O.T. = L 4 7 7s f 3rd O.T. = 4 L 3rd O.T. 3

232 Chapter 3 Strings, Air Coluns s Epressing the frequencies in ters of the fundaental frequency f FUND = we hae 4L f = 4L 4L s s FUND = = 3 f = 4 L 4L f s s st O.T. = = f = 4 L 4L s s nd O.T. = = f = 4 L 4L s s 3rd O.T. = = 7 7 Note that the frequencies of the standing waes are odd integer ultiples of the fundaental frequency. That is to say that only odd haronics, the st, 3 rd, 5 th, etc. occur in the case of a pipe closed at one end and open at the other. FUND f f f FUND FUND FUND Regarding, Waes, in a Mediu that is in Contact with a nd Mediu Consider a iolin string oscillating at its fundaental frequency, in air. For conenience of discussion, assue the iolin to be oriented so that the oscillations are up and down. Each tie the string goes up it pushes air olecules up. This results in sound waes in air. The iolin with the standing wae in it can be considered to be the soething oscillating that is the cause of the waes in air. Recall that the frequency of the waes is identical to the frequency of the source. Thus, the frequency of the sound waes in air will be identical to the frequency of the waes in the string. In general, the speed of the waes in air is different fro the speed of waes in the string. Fro = λf, this eans that the waelengths will be different as well. 3

233 Chapter 3 eats, The Doppler Effect 3 eats, The Doppler Effect Soe people get ied up about the Doppler Effect. They think it s about position rather than about elocity. (It is really about elocity.) If a single frequency sound source is coing at you at constant speed, the pitch (frequency) you hear is higher than the frequency of the source. How uch higher depends on how fast the source is coing at you. Folks ake the istake of thinking that the pitch gets higher as the source approaches the receier. No. That would be the case if the frequency depended on how close the source was to the receier. It doesn t. The frequency stays the sae. The Doppler Effect is about elocity, not position. The whole tie the source is oing straight at you, it will sound like it has one single unchanging pitch that is higher than the frequency of the source. Now duck! Once the object passes your position and it is heading away fro you it will hae one single unchanging pitch that is lower than the frequency of the source. eats Consider two sound sources, in the icinity of each other, each producing sound waes at its own single frequency. Any point in the air-filled region of space around the sources will receie sound waes fro both the sources. The aplitude of the sound at any position in space will be the aplitude of the su of the displaceents of the two waes at that point. This aplitude will ary because the interference will alternate between constructie interference and destructie interference. Suppose the two frequencies do not differ by uch. Consider the displaceents at a particular point in space. Let s start at an instant when two sound wae crests are arriing at that point, one fro each source. At that instant the waes are interfering constructiely, resulting in a large total aplitude. If your ear were at that location, you would find the sound relatiely loud. Let s ark the passage of tie by eans of the shorter period, the period of the higher-frequency waes. One period after the instant just discussed, the net crest (call it the second crest) fro the higher-frequency source is at the point in question, but the peak of the net crest fro the lower-frequency source is not there yet. Rather than a crest interfering with a crest, we hae a crest interfering with an interediate-displaceent part of the wae. The interference is still constructie but not to the degree that it was. When the third crest fro the higher-frequency source arries, the corresponding crest fro the lower-frequency source is een farther behind. Eentually, a crest fro the higher-frequency source is arriing at the point in question at the sae tie as a trough fro the lower-frequency source. At that instant in tie, the interference is as destructie as it gets. If your ear were at the point in question, you would find the sound to be inaudible or of ery low olue. Then the trough fro the lower-frequency source starts falling behind until, eentually a crest fro the higher-frequency source is interfering with the crest preceding the corresponding crest fro the lower-frequency source and the interference is again as constructie as possible. To a person whose ear is at a location at which waes fro both sources eist, the sound gets loud, soft, loud, soft, etc. The frequency with which the loudness pattern repeats itself is called the beat frequency. Eperientally, we can deterine the beat frequency by tiing how long it takes for the sound to get loud N ties and then diiding that tie by N (where N is an arbitrary 33

234 Chapter 3 eats, The Doppler Effect integer chosen by the eperienter the bigger the N the ore precise the result). This gies the beat period. Taking the reciprocal of the beat period yields the beat frequency. The beat frequency is to be contrasted with the ordinary frequency of the waes. In sound, we hear the beat frequency as the rate at which the loudness of the sound aries whereas we hear the ordinary frequency of the waes as the pitch of the sound. Deriation of the eat Frequency Forula Consider sound fro two different sources ipinging on one point, call it point P, in airoccupied space. Assue that one source has a shorter period T SHORT and hence a higher frequency f HIGH than the other (which has period and frequency T LONG and f LOW respectiely). The plan here is to epress the beat frequency in ters of the frequencies of the sources we get there by relating the periods to each other. As in our conceptual discussion, let s start at an instant when a crest fro each source is at point P. When, after an aount of tie T SHORT passes, the net crest fro the shorter-period source arries, the corresponding crest fro the longer-period source won t arrie for an aount of tie T = T LONG T. In fact, with the arrial of each successie short-period crest, the corresponding long-period crest is another T behind. Eentually, after soe nuber n of short periods, the long-period crest will arrie a full long period T LONG after the corresponding short-period crest arries. SHORT n T = (3-) T LONG This eans that as the short-period crest arries, the long-period crest that precedes the corresponding long-period crest is arriing. This results in constructie interference (loud sound). The tie it takes, starting when the interference is aially constructie, for the interference to again becoe aially constructie is the beat period T EAT = nt SHORT (3-) Let s use equation 3- to eliinate the n in this epression. Soling equation 3- for n we find that T n = LONG T Substituting this into equation 3- yields T EAT T = T LONG T SHORT T is just TLONG TSHORT so T EAT = T LONG TLONG T SHORT T SHORT 34

235 Chapter 3 eats, The Doppler Effect T EAT T = T LONG LONG T T SHORT SHORT Diiding top and botto by the product T T T yields LONG SHORT = EAT T SHORT T LONG Taking the reciprocal of both sides results in T EAT = T SHORT T LONG Now we use the frequency-period relation f = to replace each reciprocal period with its T corresponding frequency. This yields: f EAT = f f (3-3) HIGH LOW for the beat frequency in ters of the frequencies of the two sources. The Doppler Effect Consider a single-frequency sound source and a receier. The source is soething oscillating. It produces sound waes. They trael through air, at speed, the speed of sound in air, to the receier and cause soe part of the receier to oscillate. (For instance, if the receier is your ear, the sound waes cause your eardru to oscillate.) If the receier and the source are at rest relatie to the air, then the receied frequency is the sae as the source frequency. S R Source Receier 35

236 Chapter 3 eats, The Doppler Effect ut if the source is oing toward or away fro the receier, and/or the receier is oing toward or away fro the source, the receied frequency will be different fro the source frequency. Suppose for instance, the receier is oing toward the source with speed R. S R R Source Receier The receier eets wae crests ore frequently than it would if it were still. Starting at an instant when a waefront is at the receier, the receier and the net waefront are coing together at the rate + R (where is the speed of sound in air). The distance between the waefronts is just the waelength λ which is related to the source frequency f by = λf eaning that λ =. Fro the fact that, in the case of constant elocity, distance is just speed f ties tie, we hae: λ = + ) T ( R λ T = (3-4) + R for the period of the receied oscillations. Using T = and λ = equation 3-4 can be f f written as: / f = f + R = f + R f + R f = f (Receier Approaching Source) This equation states that the receied frequency f is a factor ties the source frequency. The epression + R is the speed at which the sound wae in air and the receier are approaching each other. If the receier is oing away fro the source at speed R, the speed at which the 36

237 Chapter 3 eats, The Doppler Effect sound waes are catching up with the receier is R and our epression for the receied frequency becoes R f = f (Receier Receding fro Source) Now consider the case in which the source is oing toward the receier. S S R The source produces a crest which oes toward the receier at the speed of sound. ut the source oes along behind that crest so the net crest it produces is closer to the first crest than it would be if the source was at rest. This is true of all the subsequent crests as well. They are all closer together than they would be if the source was at rest. This eans that the waelength of the sound waes traeling in the direction of the source is reduced relatie to what the waelength would be if the source was at rest. The distance d that the source traels toward the receier in the tie fro the eission of one crest to the eission of the net crest, that is in period T of the source oscillations, is d = T S where is the speed of the source. The waelength is what the waelength would be (λ) if the S source was at rest, inus the distance d = T that the source traels in one period S λ = λ d λ = λ S T (3-5) Now we ll use = λf soled for waelength λ = to eliinate the waelengths and f = f T soled for the period T = to eliinate the period. With these substitutions, equation 3-5 f becoes 37

238 Chapter 3 eats, The Doppler Effect = S f f f = ( f f S f = f S ) f = f (Source Approaching Receier) If the source is oing away fro the receier, the sign in front of the speed of the source is reersed eaning that S f = f (Source Receding fro Receier) + The four epressions for the receied frequency as a function of the source frequency are cobined on your forula sheet where they are written as: S f ± R = f (3-6) In soling a Doppler Effect proble, rather than copying this epression directly fro your forula sheet, you need to be able to pick out the actual forula that you need. For instance, if the receier is not oing relatie to the air you should oit the ± R. If the source is not oing relatie to the air, you need to oit the S. To get the forula just right you need to recognize that when either the source is oing toward the receier or the receier is oing toward the source, the Doppler-shifted receied frequency is higher (and you need to recognize that when either is oing away fro the other, the Doppler-shifted receied frequency is lower). You also need enough atheatical say to know which sign to choose to ake the receied frequency f coe out right. S 38

239 Chapter 33 Fluids: Pressure, Density, Archiedes Principle 33 Fluids: Pressure, Density, Archiedes Principle One istake you see in solutions to suberged-object static fluid probles, is the inclusion, in the free body diagra for the proble, in addition to the buoyant force, of a pressure-ties-area force typically epressed as F P = PA. This is double counting. Folks that include such a force, in addition to the buoyant force, don t realize that the buoyant force is the net su of all the pressure-ties-area forces eerted, on the suberged object by the fluid in which it is suberged. Gases and liquids are fluids. Unlike solids, they flow. A fluid is a liquid or a gas. Pressure A fluid eerts pressure on the surface of any substance with which the fluid is in contact. Pressure is force-per-area. In the case of a fluid in contact with a flat surface oer which the pressure of the fluid is constant, the agnitude of the force on that surface is the pressure ties the area of the surface. Pressure has units of N/. Neer say that pressure is the aount of force eerted on a certain aount of area. Pressure is not an aount of force. Een in the special case in which the pressure oer the certain aount of area is constant, the pressure is not the aount of force. In such a case, the pressure is what you hae to ultiply the area by to deterine the aount of force. The fact that the pressure in a fluid is 5 N/ in no way iplies that there is a force of 5 N acting on a square eter of surface (any ore than the fact that the speedoeter in your car reads 35 ph iplies that you are traeling 35 iles or that you hae been traeling for an hour). In fact, if you say that the pressure at a particular point underwater in a swiing pool is 5,000 N/ (fifteen thousand newtons per square eter), you are not specifying any area whatsoeer. What you are saying is that any infinitesial surface eleent that ay be eposed to the fluid at that point will eperience an infinitesial force of agnitude df that is equal to 5,000 N/ ties the area d A of the surface. When we specify a pressure, we re talking about a would-be effect on a would-be surface eleent. We talk about an infinitesial area eleent because it is entirely possible that the pressure aries with position. If the pressure at one point in a liquid is 5,000 N/ it could ery well be 6,000 N/ at a point that s less than a illieter away in one direction and 4,000 N/ at a point that s less than a illieter away in another direction. Let s talk about direction. Pressure itself has no direction. ut the force that a fluid eerts on a surface eleent, because of the pressure of the fluid, does hae direction. The force is perpendicular to, and toward, the surface. Isn t that interesting? The direction of the force resulting fro soe pressure (let s call that the pressure-ties-area force) on a surface eleent is deterined by the icti (the surface eleent) rather than the agent (the fluid). 39

240 Chapter 33 Fluids: Pressure, Density, Archiedes Principle Pressure Dependence on Depth For a fluid near the surface of the earth, the pressure in the fluid increases with depth. You ay hae noticed this, if you hae eer gone deep under water, because you can feel the effect of the pressure on your ear drus. efore we inestigate this phenoenon in depth, I need to point out that in the case of a gas, this pressure dependence on depth is, for any practical purposes, negligible. In discussing a container of a gas for instance, we typically state a single alue for the pressure of the gas in the container, neglecting the fact that the pressure is greater at the botto of the container. We neglect this fact because the difference in the pressure at the botto and the pressure at the top is so ery sall copared to the pressure itself at the top. We do this when the pressure difference is too sall to be releant, but it should be noted that een a ery sall pressure difference can be significant. For instance, a heliu-filled balloon, released fro rest near the surface of the earth would fall to the ground if it weren t for the fact that the air pressure in the icinity of the lower part of the balloon is greater (albeit only slightly greater) than the air pressure in the icinity of the upper part of the balloon. Let s do a thought eperient. (Einstein was fond of thought eperients. They are also called Gedanken eperients. Gedanken is the Geran word for thought.) Iagine that we construct a pressure gauge as follows: We cap one end of a piece of thin pipe and put a spring copletely inside the pipe with one end in contact with the end cap. Now we put a disk whose diaeter is equal to the inside diaeter of the pipe, in the pipe and bring it into contact with the other end of the spring. We grease the inside walls of the pipe so that the disk can slide freely along the length of the pipe, but we ake the fit eact so that no fluid can get past the disk. Now we drill a hole in the end cap, reoe all the air fro the region of the pipe between the disk and the end cap, and seal up the hole. The position of the disk in the pipe, relatie to its position when the spring is neither stretched nor copressed, is directly proportional to the pressure on the outer surface, the side facing away fro the spring, of the disk. We calibrate (ark a scale on) the pressure gauge that we hae just anufactured, and use it to inestigate the pressure in the water of a swiing pool. First we note that, as soon as we reoed the air, the gauge started to indicate a significant pressure (around N/ ), naely the air pressure in the atosphere. Now we oe the gauge around and watch the gauge reading. Whereer we put the gauge (we define the location of the gauge to be the position of the center point on the outer surface of the disk) on the surface of the water, we get one and the sae reading, (the air pressure reading). Net we erify that the pressure reading does indeed increase as we lower the gauge deeper and deeper into the water. Then we find, the point I wrote this paragraph to ake, that if we oe the gauge around horizontally at one particular depth, the pressure reading does not change. That s the eperiental result I want to use in the following deelopent, the eperiental fact that the pressure has one and the sae alue at all points that are at one and the sae depth in a fluid. Here we derie a forula that gies the pressure in an incopressible static fluid as a function of the depth in the fluid. Let s get back into the swiing pool. Now iagine a closed surface enclosing a olue, a region in space, that is full of water. I going to call the water in such a olue, a olue of water, and I going to gie it another nae as well. If it were ice, I would call it a chunk of ice, but since it is liquid water, I ll call it a slug of water. We re going 40

241 Chapter 33 Fluids: Pressure, Density, Archiedes Principle to derie the pressure s. depth relation by inestigating the equilibriu of an object which is a slug of water. Consider a cylindrical slug of water whose top is part of the surface of the swiing pool and whose botto is at soe arbitrary depth h below the surface. I going to draw the slug here, isolated fro its surroundings. The slug itself is, of course, surrounded by the rest of the water in the pool. P o W = g P In the diagra, we use arrows to coney the fact that there is pressure-ties-area force on eery eleent of the surface of the slug. Now the downward pressure-ties-area force on the top of the slug is easy to epress in ters of the pressure because the pressure on eery infinitesial area eleent aking up the top of the slug has one and the sae alue. In ters of the deterination of the pressure-ties-area, this is the easy case. The agnitude of the force, F o, is just the pressure P o ties the area A of the top of the cylinder. F o = A siilar arguent can be ade for the botto of the cylinder. All points on the botto of the cylinder are at the sae depth in the water so all points are at one and the sae pressure P. The botto of the cylinder has the sae area A as the top so the agnitude of the upward force F on the botto of the cylinder is gien by P A o F = PA As to the sides, if we diide the sidewalls of the cylinder up into an infinite set of equal-sized infinitesial area eleents, for eery sidewall area eleent, there is a corresponding area eleent on the opposite side of the cylinder. The pressure is the sae on both eleents because they are at the sae depth. The two forces then hae the sae agnitude, but because the eleents face in opposite directions, the forces hae opposite directions. Two opposite but equal 4

242 Chapter 33 Fluids: Pressure, Density, Archiedes Principle forces add up to zero. In such a anner, all the forces on the sidewall area eleents cancel each other out. Now we are in a position to draw a free body diagra of the cylindrical slug of water. P o A W=g PA Applying the equilibriu condition yields F = 0 PA g Po A = 0 (33-) At this point in our deriation of the relation between pressure and depth, the depth does not eplicitly appear in the equation. The ass of the slug of water, howeer, does depend on the length of the slug which is indeed the depth h. First we note that = rv (33-) where r is the density, the ass-per-olue, of the water aking up the slug and V is the olue of the slug. The olue of a cylinder is its height ties its face area so we can write = rha Substituting this epression for the ass of the slug into equation 33- yields PA r hag Po A = 0 P r hg Po = 0 P = P o + rgh (33-3) 4

243 Chapter 33 Fluids: Pressure, Density, Archiedes Principle While we hae been writing specifically about water, the only thing in the analysis that depends on the identity of the incopressible fluid is the density r. Hence, as long as we use the density of the fluid in question, equation 33-3 ( P = P o + rgh ) applies to any incopressible fluid. It says that the pressure at any depth h is the pressure at the surface plus rgh. A few words on the units of pressure are in order. We hae stated that the units of pressure are N/. This cobination of units is gien a nae. It is called the pascal, abbreiated Pa. Pa = Pressures are often quoted in ters of the non-si unit of pressure, the atosphere, abbreiated at and defined such that, on the aerage, the pressure of the earth s atosphere at sea leel is at. In ters of the pascal, N 5 at = Pa The big istake that folks ake in applying equation 33-3 ( P = P o + rgh ) is to ignore the units. They ll use at for P o and without conerting that to pascals, they ll add the product rgh to it. Of course, if one uses SI units for r, g, and h, the product rgh coes out in N/ which is a pascal which is definitely not an atosphere (but rather, about a hundred-thousandth of an atosphere). Of course one can t add a alue in pascals to a alue in atospheres. The way to go is to conert the alue of P o that was gien to you in units of atospheres, to pascals, and then add the product rgh (in SI units) to your result so that your final answer coes out in pascals. Gauge Pressure Reeber the gauge we constructed for our thought eperient? That part about eacuating the inside of the pipe presents quite the anufacturing challenge. The gauge would becoe inaccurate as air leaked in by the disk. As regards function, the description is fairly realistic in ters of actual pressure gauges in use, ecept for the puping of the air out the pipe. To ake it ore like an actual gauge that one ight purchase, we would hae to leae the interior open to the atosphere. In use then, the gauge reads zero when the pressure on the sensor end is atosphere, and in general, indicates the aount by which the pressure being easured eceeds atospheric pressure. This quantity, the aount by which a pressure eceeds atospheric pressure, is called gauge pressure (since it is the alue registered by a typical pressure gauge.) When it needs to be contrasted with gauge pressure, the actual pressure that we 43

244 Chapter 33 Fluids: Pressure, Density, Archiedes Principle hae been discussing up to this point is called absolute pressure. The absolute pressure and the gauge pressure are related by: P = P G + P O (33-4) where: P is the absolute pressure, P G is the gauge pressure, and P O is atospheric pressure. When you hear a alue of pressure (other than the so-called baroetric pressure of the earth s atosphere) in your eeryday life, it is typically a gauge pressure (een though one does not use the adjectie gauge in discussing it.) For instance, if you hear that the recoended tire pressure for your tires is 3 psi (pounds per square inch) what is being quoted is a gauge pressure. Folks that work on entilation systes often speak of negatie air pressure. Again, they are actually talking about gauge pressure, and a negatie alue of gauge pressure in a entilation line just eans that the absolute pressure is less than atospheric pressure. Archiedes Principle The net pressure-ties-area force on an object suberged in a fluid, the ector su of the forces on all the infinite nuber of infinitesial surface area eleents aking up the surface of an object, is upward because of the fact that pressure increases with depth. The upward pressureties-area force on the botto of an object is greater than the downward pressure-ties-area force on the top of the object. The result is a net upward force on any object that is either partly or totally suberged in a fluid. The force is called the buoyant force on the object. The agent of the buoyant force is the fluid. If you take an object in your hand, suberge the object in still water, and release the object fro rest, one of three things will happen: The object will eperience an upward acceleration and bob to the surface, the object will reain at rest, or the object will eperience a downward acceleration and sink. We hae ephasized that the buoyant force is always upward. So why on earth would the object eer sink? The reason is, of course, that after you release the object, the buoyant force is not the only force acting on the object. The graitational force still acts on the object when the object is suberged. Recall that the earth s graitational field pereates eerything. For an object that is touching nothing of substance but the fluid it is in, the free body diagra (without the acceleration ector being included) is always the sae (ecept for the relatie lengths of the arrows): F g and the whole question as to whether the object (released fro rest in the fluid) sinks, stays put, or bobs to the surface, is deterined by how the agnitude of the buoyant force copares with 44

245 Chapter 33 Fluids: Pressure, Density, Archiedes Principle that of the graitational force. If the buoyant force is greater, the net force is upward and the object bobs toward the surface. If the buoyant force and the graitational force are equal in agnitude, the object stays put. And if the graitational force is greater, the object sinks. So how does one deterine how big the buoyant force on an object is? First, the triial case: If the only forces on the object are the buoyant force and the graitational force, and the object reains at rest, then the buoyant force ust be equal in agnitude to the graitational force. This is the case for an object such as a boat or a log which is floating on the surface of the fluid it is in. ut suppose the object is not freely floating at rest. Consider an object that is suberged in a fluid. We hae no inforation on the acceleration of the object, but we cannot assue it to be zero. Assue that a person has, while aintaining a fir grasp on the object, suberged the object in fluid, and then, released it fro rest. We don t know which way it is going fro there, but we can not assue that it is going to stay put. To derie our epression for the buoyant force, we do a little thought eperient. Iagine replacing the object with a slug of fluid (the sae kind of fluid as that in which the object is suberged), where the slug of fluid has the eact sae size and shape as the object. Fro our eperience with still water we know that the slug of fluid would indeed stay put, eaning that it is in equilibriu. F gsf a = 0 Table of Forces Sybol=? Nae Agent Victi uoyant The Surrounding The Slug Force Fluid of Fluid F g SF = SF g Graitational Force on the Slug of Fluid The Earth s Graitational Field The Slug of Fluid 45

246 Chapter 33 Fluids: Pressure, Density, Archiedes Principle Applying the equilibriu equation F = 0 to the slug of fluid yields: F = 0 a = 0 F gsf = 0 F gsf = F gsf The last equation states that the buoyant force on the slug of fluid is equal to the graitational force on the slug of fluid. Now get this; this is the cru of the deriation: ecause the slug of fluid has the eact sae size and shape as the original object, it presents the eact sae surface to the surrounding fluid, and hence, the surrounding fluid eerts the sae buoyant force on the slug of fluid as it does on the original object. Since the buoyant force on the slug of fluid is equal in agnitude to the graitational force acting on the slug of fluid, the buoyant force on the original object is equal in agnitude to the graitational force acting on the slug of fluid. This is Archiedes principle. = the buoyant force, which is equal in agnitude to the graitational force that would be acting on that aount of fluid that would fit in the space occupied by the suberged part of the object. F g o = o g (The graitational force) Archiedes Principle states that: The buoyant force on an object that is either partly or totally suberged in a fluid is upward, and is equal in agnitude to the graitational force that would be acting on that aount of fluid that would be where the object is if the object wasn t there. For an object that is totally suberged, the olue of that aount of fluid that would be where the object is if the object wasn t there is equal to the olue of the object itself. ut for an object that is only partly suberged, the olue of that aount of fluid that would be where the object is if the object wasn t there is equal to the (typically unknown) olue of the suberged part of the object. Howeer, if the object is freely floating at rest, the equilibriu equation (instead of Archiedes Principle) can be used to quickly establish that the buoyant force (of a freely floating object such as a boat) is equal in agnitude to the graitational force acting on the object itself. 46

247 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle There are a couple of istakes that tend to crop up with soe regularity in the application of the ernoulli equation P + r + rgh = constant. Firstly, folks tend to forget to create a diagra in order to identify point and point in the diagra so that they can write the ernoulli equation in its useful for: P + r + rgh = P + r + rgh. Secondly, when both the elocities in ernoulli s equation are unknown, they forget that there is another equation that relates the elocities, naely, the continuity equation in the for A = A which states that the flow rate at position is equal to the flow rate at position. Pascal s Principle Eperientally, we find that if you increase the pressure by soe gien aount at one location in a fluid, the pressure increases by that sae aount eerywhere in the fluid. This eperiental result is known as Pascal s Principle. We take adantage of Pascal s principle eery tie we step on the brakes of our cars and trucks. The brake syste is a hydraulic syste. The fluid is oil that is called hydraulic fluid. When you depress the brake pedal you increase the pressure eerywhere in the fluid in the hydraulic line. At the wheels, the increased pressure acting on pistons attached to the brake pads pushes the against disks or drus connected to the wheels. Eaple 34- A siple hydraulic lift consists of two pistons, one larger than the other, in cylinders connected by a pipe. The cylinders and pipe are filled with water. In use, a person pushes down upon the saller piston and the water pushes upward on the larger piston. The diaeter of the saller piston is.0 centieters. The diaeter of the larger piston is.0 centieters. On top of the larger piston is a etal support and on top of that is a car. The cobined ass of the support-plus-car is 998 kg. Find the force that the person ust eert on the saller piston to raise the car at a constant elocity. Neglect the asses of the pistons. 47

248 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Solution We start our solution with a sketch. D S =.0 c D L =.0 c Now, let s find the force R N eerted on the larger piston by the car support. y Newton s 3 rd Law, it is the sae as the noral force F N eerted by the larger piston on the car support. We ll draw and analyze the free body diagra of the car-plus-support to get that. F = 0 F N F g = g F N g = 0 Table of Forces Sybol Nae Agent Victi F g =g Graitational The Earth s The Force on Graitational Support- Support-Plus-Car Field Plus-Car F N Noral Force The Large Piston The Support- Plus-Car F = g N F = 998 kg 9 80 N. newtons kg F = 9780 N newtons 48

249 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Now we analyze the equilibriu of the larger piston to deterine what the pressure in the fluid ust be in order for the fluid to eert enough force on the piston (with the car-plus-support on it) to keep it oing at constant elocity. R N = F N = 9780 a = 0 F PL = PA L F = 0 Table of Forces Sybol Nae Agent Victi R N = F Interaction N Support (That part, Partner to Large = 9780 of the hydraulic lift, Noral Force Piston newtons that the car is on.) (see aboe) F PL Pressure-Related Force on Large Piston The Water Large Piston F PL RN = 0 PA L RN = 0 RN P = (34-) A L A L is the area of the face of the larger piston. We can use the gien larger piston diaeter D L = 0.0 to deterine the area of the face of the larger piston as follows: A L DL = πr where r = is the radius of the larger piston. L L DL L A = π A A L L 0. 0 = π = Substituting this and the alue R N = F N = 9780 newtons into equation 34- aboe yields 9780 newtons P = N P = 8333 (We intentionally keep 3 too any significant figures in this interediate result.) 49

250 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Now we just hae to analyze the equilibriu of the saller piston to deterine the force that the person ust eert on the saller piston. F PERSON a = 0 F PS = PA S F = 0 F PS FPERSON = PA 0 S FPERSON = F = PA PERSON S 0 The area A S of the face of the saller piston is just pi ties the square of the radius of the saller piston where the radius is saller piston. So: D S, half the diaeter of the DS PERSON F = Pπ F PERSON = N π F PERSON = 07 N 50

251 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Fluid in Motion the Continuity Principle The Continuity Principle is a fancy nae for soething that coon sense will tell you has to be the case. It is siply a stateent of the fact that for any section of a single pipe, filled with an incopressible fluid (an idealization approached by liquids), through which the fluid with which the pipe is filled is flowing, the aount of fluid that goes in one end in any specified aount of tie is equal to the aount that coes out the other end in the sae aount of tie. If we quantify the aount of fluid in ters of the ass, this is a stateent of the conseration of ass. Haing stipulated that the segent is filled with fluid, the incoing fluid has no roo to epand in the segent. Haing stipulated that the fluid is incopressible, the olecules aking up the fluid cannot be packed closer together; that is, the density of the fluid cannot change. With these stipulations, the total ass of the fluid in the segent of pipe cannot change, so, any tie a certain ass of the fluid flows in one end of the segent, the sae ass of the fluid ust flow out the other end. Position Position.. Segent of Pipe Filled with the Fluid that is Flowing Through the Pipe This can only be the case if the ass flow rate, the nuber of kilogras-per-second passing a gien position in the pipe, is the sae at both ends of the pipe segent... = (34-) An interesting consequence of the continuity principle is the fact that, in order for the ass flow rate (the nuber of kilogras per second passing a gien position in the pipe) to be the sae in a fat part of the pipe as it is in a skinny part of the pipe, the elocity of the fluid (i.e. the elocity of the olecules of the fluid) ust be greater in the skinny part of the pipe. Let s see why this is the case. 5

252 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Here, we again depict a pipe in which an incopressible fluid is flowing. Position Position V V Segent of Pipe etween Positions and Keeping in ind that the entire pipe is filled with the fluid, the shaded region on the left represents the fluid that will flow past position in tie t and the shaded region on the right represents the fluid that will flow past position in the sae tie t. In both cases, in order for the entire slug of fluid to cross the releant position line, the slug ust trael a distance equal to its length. Now the slug labeled has to be longer than the slug labeled since the pipe is skinnier at position and by the continuity equation = (the aount of fluid that flows into the segent of the pipe between position and position is equal to the aount of fluid that flows out of it). So, if the slug at position is longer and it has to trael past the position line in the sae aount of tie as it takes for the slug at position to trael past its position line, the fluid elocity at position ust be greater. The fluid elocity is greater at a skinnier position in the pipe. Let s get a quantitatie relation between the elocity at position and the elocity at position. Starting with = we use the definition of density to replace each ass with the density of the fluid ties the releant olue: r V = r V Diiding both sides by the density tells us soething you already know: V = V As an aside, we note that if you diide both sides by t and take the liit as t goes to.. zero, we hae V = V which is an epression of the continuity principle in ters of olue flow rate. The olue flow rate is typically referred to siply as the flow rate. While we use the SI units 3 s rate epressed in units of gallons per inute. for flow rate, the reader ay be ore failiar with flow 5

253 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Now back to our goal of finding a atheatical relation between the elocities of the fluid at the two positions in the pipe. Here we copy the diagra of the pipe and add, to the copy, a depiction of the face of slug of area A and the face of slug of area A. Position Position V V A A We left off with the fact that V = V. Each olue can be replaced with the area of the face of the corresponding slug ties the length of that slug. So, A = A Recall that is not only the length of slug, it is also how far slug ust trael in order for the entire slug of fluid to get past the position line. The sae is true for slug and position. Diiding both sides by the one tie interal t yields: Taking the liit as t goes to zero results in: A = A t t A = A (34-3) This is the relation, between the elocities, that we hae been looking for. It applies to any pair of positions in a pipe copletely filled with an incopressible fluid. It can be written as A = constant (34-4) which eans that the product of the cross-sectional area of the pipe and the elocity of the fluid at that cross section is the sae for eery position along the fluid-filled pipe. To take adantage of this fact, one typically writes, in equation for, that the product A at one location is equal to the sae product at another location. In other words, one writes equation

254 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Note that the epression A, the product of the cross-sectional area of the pipe, at a particular position, and the elocity of the fluid at that sae position, haing been deried by diiding an epression for the olue of fluid V that would flow past a gien position of the pipe in tie t, by t, and taking the liit as t goes to zero, is none other than the flow rate (the olue flow rate) discussed in the aside aboe. Flow Rate = A Further note that if we ultiply the flow rate by the density of the fluid, we get the ass flow rate.. = r A (34-5) Fluid in Motion ernoulli s Principle The deriation of ernoulli s Equation represents an elegant application of the Work-Energy Theore. Here we discuss the conditions under which ernoulli s Equation applies and then siply state and discuss the result. ernoulli s Equation applies to a fluid flowing through a full pipe. The degree to which ernoulli s Equation is accurate depends on the degree to which the following conditions are et: ) The fluid ust be eperiencing steady state flow. This eans that the flow rate at all positions in the pipe is not changing with tie. ) The fluid ust be eperiencing strealine flow. Pick any point in the fluid. The infinitesial fluid eleent at that point, at an instant in tie, traeled along a certain path to arrie at that point in the fluid. In the case of strealine flow, eery infinitesial eleent of fluid that eer finds itself at that sae point traeled the sae path. (Strealine flow is the opposite of turbulent flow.) 3) The fluid ust be non-iscous. This eans that the fluid has no tendency to stick to either the sides of the pipe or to itself. (Molasses has high iscosity. Alcohol has low iscosity.) 54

255 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle Consider a pipe full of a fluid that is flowing through the pipe. In the ost general case, the cross-sectional area of the pipe is not the sae at all positions along the pipe and different parts of the pipe are at different eleations relatie to an arbitrary, but fied, reference leel. Position P Position r P h h Pick any two positions along the pipe, e.g. positions and in the diagra aboe. (You already know that, in accord with the continuity principle, A = A.) Consider the following unnaed su of ters: P + r + rgh where, at the position under consideration: P is the pressure of the fluid, r (the Greek letter rho) is the density of the fluid, is the agnitude of the elocity of the fluid, g = 9.80 N is the near-surface agnitude of the earth s graitational field, and kg h is the eleation, relatie to a fied reference leel, of the position in the pipe. The ernoulli Principle states that this unnaed su of ters has the sae alue at each and eery position along the pipe. ernoulli s equation is typically written: P + r + rgh = constant (34-6) but to use it, you hae to pick two positions along the pipe and write an equation stating that the 55

256 Chapter 34 Pascal s Principle, the Continuity Equation, and ernoulli s Principle alue of the unnaed su of ters is the sae at one of the positions as it is at the other. + r + rgh = P + r + rgh P (34-7) A particularly interesting characteristic of fluids is incorporated in this equation. Suppose positions and are at one and the sae eleation as depicted in the following diagra: Position Position P P Then h = h in equation 34-7 and equation 34-7 becoes: P + r = P + r Check it out. If > then P ust be less than P in order for the equality to hold. This equation is saying that, where the elocity of the fluid is high, the pressure is low. 56

257 Chapter 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity As you know, teperature is a easure of how hot soething is. Rub two sticks together and you will notice that the teperature of each increases. You did work on the sticks and their teperature increased. Doing work is transferring energy. So you transferred energy to the sticks and their teperature increased. This eans that an increase in the teperature of a syste is an indication of an increase in the internal energy (a.k.a. theral energy) of the syste. (In this contet the word syste is therodynaics jargon for the generalization of the word object. Indeed an object, say an iron ball, could be a syste. A syste is just the subject of our inestigations or considerations. A syste can be as siple as a saple of one kind of gas or a chunk of one kind of etal, or it can be ore coplicated as in the case of a can plus soe water in the can plus a theroeter in the water plus a lid on the can. For the case at hand, the syste is the two sticks.) The internal energy of a syste is energy associated with the otion of olecules, atos, and the particles aking up atos relatie to the center of ass of the syste, and the potential energy corresponding to the positions and elocities of the aforeentioned subicroscopic constituents of the syste relatie to each other. As usual with energy accounting, the absolute zero of energy in the case of internal energy doesn t atter only changes in internal energy hae any releance. As such, you or the publisher of a table of internal energy alues (for a gien substance, publishers actually list the internal energy per ass or the internal energy per ole of the substance under specified conditions rather that the internal energy of a saple of such a substance), are free to choose the zero of internal energy for a gien syste. In aking any predictions regarding a physical process inoling that syste, as long as you stick with the sae zero of internal energy throughout your analysis, the easurable results of your prediction or eplanation will not depend on your choice of the zero of internal energy. Another way of increasing the teperature of a pair of sticks is to bring the into contact with soething hotter than the sticks are. When you do that, the teperature of the sticks autoatically increases you don t hae to do any work on the. Again, the increase in the teperature of either stick indicates an increase in the internal energy of that stick. Where did that energy coe fro? It ust hae coe fro the hotter object. You ay also notice that the hotter object s teperature decreased when you brought it into contact with the sticks. The decrease in teperature of the hotter object is an indication that the aount of internal energy in the hotter object decreased. You brought the hotter object in contact with the sticks and energy was autoatically transferred fro the hotter object to the sticks. The energy transfer in this case is referred to as the flow of heat. Heat is energy that is autoatically transferred fro a hotter object to a cooler object when you bring the two objects in contact with each other. Heat is not soething that a syste has but rather energy that is transferred or is being transferred. Once it gets to the syste to which it is transferred we call it internal energy. The idea is to distinguish between what is being done to a syste, Work is done on the syste and/or heat is caused to flow into it, with how the syste changes as a result of what was done to it, The internal energy of the syste increases. The fact that an increase in the teperature of an object is an indication of energy transferred to that object ight suggest that anytie you transfer energy to an object its teperature increases. ut this is not the case. Try putting a hot spoon in a glass of ice water. (Here we consider a case 57

258 Chapter 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity for which there is enough ice so that not all of the ice elts.) The spoon gets as cold as the ice water and soe of the ice elts, but the teperature of the ice water reains the sae (0 C). The cooling of the spoon indicates that energy was transferred fro it, and since the spoon was in contact with the ice water the energy ust hae been transferred to the ice water. Indeed the ice does undergo an obserable change; soe of it elts. The presence of ore liquid water and less ice is an indication that there is ore energy in the ice water. Again there has been a transfer of energy fro the spoon to the ice water. This transfer is an autoatic flow of heat that takes place when the two systes are brought into contact with one another. Eidently, heat flow does not always result in a teperature increase. Eperient shows that when a higher teperature object is in contact with a lower teperature object, heat is flowing fro the higher teperature object to the lower teperature object. The flow of heat persists until the two objects are at one and the sae teperature. We define the aerage translational kinetic energy of a olecule of a syste as the su of the translational kinetic energies of all the olecules aking up the syste diided by the total nuber of olecules. When two siple ideal gas systes, each inoling a ultitude of single ato olecules interacting ia elastic collisions, are brought together, we find that heat flows fro the syste in which the aerage translational kinetic energy per olecule is greater to the syste in which the aerage translational kinetic energy per olecule is lesser. This eans the forer syste is at a higher teperature. That is to say that the higher the translational kinetic energy, on the aerage, of the particles aking up the syste, the higher the teperature. This is true for any systes. Solids consist of atos that are bound to neighboring atos such that olecules tend to be held in their position, relatie to the bulk of the solid, by electrostatic forces. A pair of olecules that are bound to each other has a lower aount of internal potential energy relatie to the sae pair of olecules when they are not bound together because we hae to add energy to the bound pair at rest to yield the free pair at rest. In the case of ice water, the transfer of energy into the ice water results in the breaking of bonds between water olecules, which we see as the elting of the ice. As such, the transfer of energy into the ice water results in an increase in the internal potential energy of the syste. The two different kinds of internal energy that we hae discussed are internal potential energy and internal kinetic energy. When there is a net transfer of energy into a syste, and the acroscopic echanical energy of the syste doesn t change (e.g. for the case of an object near the surface of the earth, the speed of object as a whole does not increase, and the eleation of the object does not increase), the internal energy (the internal kinetic energy, the internal potential energy, or both) of the syste increases. In soe, but not all, cases, the increase in the internal energy is accopanied by an increase in the teperature of the syste. If the teperature doesn t increase, then we are probably dealing with a case in which it is the internal potential energy of the syste that increases. 58

259 Chapter 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity Heat Capacity and Specific Heat Capacity Let s focus our attention on cases in which heat flow into a saple of atter is accopanied by an increase in the teperature of the saple. For any substances, oer certain teperature ranges, the teperature change is (at least approiately) proportional to the aount of heat that flows into the substance. T = Q Traditionally, the constant of proportionality is written as C so that T = C Q where the upper case C is the heat capacity. This equation is ore coonly written as Q = C T (35-) which states that the aount of heat that ust flow into a syste to change the teperature of that syste by T is the heat capacity C ties the desired teperature change T. Thus the heat capacity C is the heat-per-teperature-change. It s reciprocal is a easure of a syste s teperature sensitiity to heat flow. Let s focus our attention on the siplest kind of syste, a saple of one kind of atter, such as a certain aount of water. The aount of heat that is required to change the teperature of the saple by a certain aount is directly proportional to the ass of the single substance; e.g., if you double the ass of the saple it will take twice as uch heat to raise its teperature by, for instance, C. Matheatically, we can write this fact as C = It is traditional to use a lower case c for the constant of proportionality. Then C = c where the constant of proportionality c is the heat-capacity-per-ass of the substance in question. The heat-capacity-per-ass c is referred to as the ass specific heat capacity or siply the specific heat capacity of the substance in question. (In this contet, the adjectie specific eans per aount. ecause the aount can be specified in ore than one way we hae the epression ass specific eaning per aount of ass and the epression olar specific eaning per nuber of oles. Here, since we are only dealing with ass specific heat, we can oit the word ass without generating confusion.) The specific heat capacity c has a different alue for each different kind of substance in the unierse. (Okay, there ight be soe coincidental duplication but you get the idea.) In ters of the ass specific heat capacity, 59

260 Chapter 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity equation 35- ( Q = C T ), for the case of a syste consisting only of a saple of a single substance, can be written as Q = c T (35-) The specific heat capacity c is a property of the kind of atter of which a substance consists. As such, the alues of specific heat for arious substances can be tabulated. Specific Heat Capacity* Substance J o kg C Ice (solid water) 090 Liquid Water 486 Water Vapor (gas) 000 Solid Copper 387 Solid Aluinu 90 Solid Iron 448 *The specific heat capacity of a substance aries with teperature and pressure. The alues gien correspond to atospheric pressure. Use of these representatie constant alues for cases inoling atospheric pressure and teperature ranges between 00 C and +600 C, as applicable for the phase of the aterial, can be epected to yield reasonable results but if precision is required, or inforation on how reasonable your results are is needed, you should consult a therodynaics tetbook and therodynaics tables and carry out a ore sophisticated analysis. Note how any ore Joules of energy are needed to raise the teperature of kg of liquid water C than are required to raise the teperature of kg of a etal C. Teperature Despite the fact that you are quite failiar with it, soe ore discussion of teperature is in order. Wheneer you easure soething, you are really just coparing that soething with an arbitrarily-established standard. For instance, when you easure the length of a table with a eter stick, you are coparing the length of the table with the odern day equialent of what was historically established as one ten-thousandth of the distance fro the earth s north pole to the equator. In the case of teperature, a standard, now called the degree Celsius was established as follows: At atosphere of pressure, the teperature at which water freezes was defined to be 0 C and the teperature at which water boils was defined to be 00 C. Then a substance with a teperature-dependent easurable characteristic, such as the length of a colun of liquid ercury, was used to interpolate and etrapolate the teperature range. (Mark the position of the end of the colun of ercury on the tube containing that ercury when it is at the teperature of freezing water and again when it is at the teperature of boiling water. Diide the interal between the two arks into a hundred parts. Use the sae length of each of those parts to etend the scale in both directions and call it a teperature scale.) Note the arbitrary anner in which the zero of the Celsius scale has been established. The choice of zero is irreleant for our purposes since equations 35- ( Q = C T ) and 35-60

261 Chapter 35 Teperature, Internal Energy, Heat, and Specific Heat Capacity ( Q = c T ) relate teperature change, rather than teperature itself, to the aount of heat flow. An absolute teperature scale has been established for the SI syste of units. The zero of teperature on this scale is set at the greatest possible teperature such that it is theoretically ipossible for the teperature of any syste in equilibriu to be as low as the zero of the Kelin scale. The unit of teperature on the Kelin scale is the kelin, abbreiated K. Note the absence of the degree sybol in the unit. The Kelin scale is siilar to the Celsius scale in that a change in teperature of, say, K, is equialent to a change in teperature of C. (Note regarding units notation: The units C are used for a teperature on the Celsius scale, but the units C are used for a teperature change on the Celsius scale.) On the Kelin scale, at a pressure of one atosphere, water freezes at 73.5 K. So, a teperature in kelin is related to a teperature in C by Teperature in K = ( Teperature in C ) K K O C 6

262 Chapter 36 Heat: Phase Changes 36 Heat: Phase Changes There is a tendency to beliee that any tie heat is flowing into ice, the ice is elting. NOT SO. When heat is flowing into ice, the ice will be elting only if the ice is already at the elting teperature. When heat is flowing into the ice that is below the elting teperature, the teperature of the ice is increasing. As entioned in the preceding chapter, there are ties when you bring a hot object into contact with a cooler saple, that heat flows fro the hot object to the cooler saple, but the teperature of the cooler saple does not increase, een though no heat flows out of the cooler saple (e.g. into an een colder object). This occurs when the cooler saple undergoes a phase change. For instance, if the cooler saple happens to be H O ice or H O ice plus liquid water, at 0 C and atospheric pressure, when heat is flowing into the saple, the ice is elting with no increase in teperature. This will continue until all the ice is elted (assuing enough heat flows into the saple to elt all the ice). Then, after the last bit of ice elts at 0 C, if heat continues to flow into the saple, the teperature of the saple will be increasing. Lets reiew the question about how it can be that heat flows into the cooler saple without causing the cooler saple to war up. Energy flows fro the hotter object to the cooler saple, but the internal kinetic energy of the cooler saple does not increase. Again, how can that be? What happens is that the energy flow into the cooler saple is accopanied by an increase in the internal potential energy of the saple. It is associated with the breaking of electrostatic bonds between olecules where the negatie part of one olecule is bonded to the positie part of another. The separating of the olecules corresponds to an increase in the potential energy of the syste. This is siilar to a book resting on a table. It is graitationally bound to the earth. If you lift the book and put it on a shelf that is higher than the tabletop, you hae added soe energy to the earth/book syste, but you hae increased the potential energy with no net increase in the kinetic energy. In the case of elting ice, heat flow into the saple anifests itself as an increase in the potential energy of the olecules without an increase in the kinetic energy of the olecules (which would be accopanied by a teperature increase). The aount of heat that ust flow into a single-substance solid saple that is already at its elting teperature in order to elt the whole saple depends on a property of the substance of which the saple consists, and on the ass of the substance. The releant substance property is called the latent heat of elting. The latent heat of elting is the heat-per-ass needed to elt the substance at the elting teperature. Note that, despite the nae, the latent heat is not an aount of heat but rather a ratio of heat to ass. The sybol used to represent latent heat in general is L, and we use the subscript for elting. In ters of the latent heat of elting, the aount of heat, Q, that ust flow into a saple of a single-substance solid that is at the elting teperature, in order to elt the entire saple is gien by: In this discussion, we are treating the saple as if it had one well-defined teperature. This is an approiation. When the saple is in contact with a hotter object so that heat is flowing fro the hotter object to the saple, the part of the saple in direct contact with and in the near icinity of the hotter object will be at a higher teperature than other parts of the saple. The hotter the object, the greater the ariation in the teperature of the local bit of the saple with distance fro the object. We neglect this teperature ariation so our discussion is only appropriate when the teperature ariation is sall. 6

263 Chapter 36 Heat: Phase Changes Q = L Note the absence of a T in the epression Q = L. There is no T in the epression because there is no teperature change in the process. The whole phase change takes place at one teperature. So far, we hae talked about the case of a solid saple, at the elting teperature, which is in contact with a hotter object. Heat flows into the saple, elting it. Now consider a saple of the sae substance in liquid for at the sae teperature but in contact with a colder object. In this case, heat will flow fro the saple to the colder object. This heat loss fro the saple does not result in a decrease in the teperature. Rather, it results in a phase change of the substance of which the saple consists, fro liquid to solid. This phase change is called freezing. It also goes by the nae of solidification. The teperature at which freezing takes place is called the freezing teperature, but it is iportant to reeber that the freezing teperature has the sae alue as the elting teperature. The heat-per-ass that ust flow out of the substance to freeze it (assuing the substance to be at the freezing teperature already) is called the latent heat of fusion, or L f. The latent heat of fusion for a gien substance has the sae alue as the latent heat of elting for that substance: L f = L The aount of heat that ust flow out of a saple of ass in order to conert the entire saple fro liquid to solid is gien by: Again, there is no teperature change. Q = L f The other two phase changes we need to consider are aporization and condensation. Vaporization is also known as boiling. It is the phase change in which liquid turns into gas. It too (as in the case of freezing and elting), occurs at a single teperature, but for a gien substance, the boiling teperature is higher than the freezing teperature. The heat-per-ass that ust flow into a liquid to conert it to gas is called the latent heat of aporization L. The heat that ust flow into ass of a liquid that is already at its boiling teperature (a.k.a. its aporization teperature) to conert it entirely into gas is gien by: Q = L Condensation is the phase change in which gas turns into liquid. In order for condensation to occur, the gas ust be at the condensation teperature, the sae teperature as the boiling teperature (a.k.a. the aporization teperature). Furtherore, heat ust flow out of the gas, as it does when the gas is in contact with a colder object. Condensation takes place at a fied teperature known as the condensation teperature. (The elting teperature, the freezing teperature, the boiling teperature, and the condensation teperature are also referred to as the elting point, the freezing point, the boiling point, and the condensation point, respectiely.) The heat-per-ass that ust be etracted fro a particular kind of gas that is already at the 63

264 Chapter 36 Heat: Phase Changes condensation teperature, to conert that gas to liquid at the sae teperature, is called the latent heat of condensation L c. For a gien substance, the latent heat of condensation has the sae alue as the latent heat of aporization. For a saple of ass of a gas at its condensation teperature, the aount of heat that ust flow out of the saple to conert the entire saple to liquid is gien by: Q = L c It is iportant to note that the actual alues of the freezing teperature, the boiling teperature, the latent heat of elting, and the latent heat of aporization are different for different substances. For water we hae: Phase Change Teperature Latent Heat Melting MJ Freezing 0 C kg oiling or Vaporization MJ Condensation 00 C. 6 kg Eaple 36- How uch heat does it take to conert 444 gras of H O ice at 9.0 C to stea (H O gas) at 8.0 C? Discussion of Solution Rather than sole this one for you, we siply eplain how to sole it. To conert the ice at 9.0 C to stea at 8.0 C, we first hae to cause enough heat to flow into the ice to war it up to the elting teperature, 0 C. This step is a specific heat capacity proble. We use Q = c ice T where T is [0 C ( 9.0 C)] = 9.0 C. Now that we hae the ice at the elting teperature, we hae to add enough heat to elt it. This step is a latent heat proble. Q = L After Q + Q flows into the H O, we hae liquid water at 0 C. Net, we hae to find 64

265 Chapter 36 Heat: Phase Changes how uch heat ust flow into the liquid water to war it up to the boiling point, 00 C. Q 3 = c liquid water T where T = (00 C 0 C) = 00 C. After Q + Q + Q 3 flows into the H O, we hae liquid water at 00 C. Net, we hae to find how uch heat ust flow into the liquid water at 00 C to conert it to stea at 00 C. Q 4 = L After Q + Q + Q 3 + Q 4 flows into the H 0, we hae water apor (gas) at 00 C. Now, all we need to do is to find out how uch heat ust flow into the water gas at 00 C to war it up to 8 C. Q 5 = c stea T where T = 8 C 00 C = 8 C. So the aount of heat that ust flow into the saple of solid ice at 9.0 C in order for saple to becoe stea at 8 C (the answer to the question) is: Q total = Q + Q + Q 3 + Q 4 + Q 5 65

266 Chapter 37 The First Law of Therodynaics 37 The First Law of Therodynaics We use the sybol U to represent internal energy. That is the sae sybol that we used to represent the echanical potential energy of an object. Do not confuse the two different quantities with each other. In probles, questions, and discussion, the contet will tell you whether the U represents internal energy or it represents echanical potential energy. We end this physics tetbook as we began the physics part of it (Chapter was a atheatics reiew), with a discussion of conseration of energy. ack in Chapter, the focus was on the conseration of echanical energy; here we focus our attention on theral energy. In the case of a deforable syste, it is possible to do soe net work on the syste without causing its echanical kinetic energy + Iw to change (where is the ass of the syste, is the speed of the center of ass of the syste, I is the oent of inertia of the syste, and w is the agnitude of the angular elocity of the syste). Eaples of such work would be: the bending of a coat hanger, the stretching of a rubber band, the squeezing of a lup of clay, the copression of a gas, and the stirring of a fluid. When you do work on soething, you transfer energy to that soething. For instance, consider a case in which you push on a cart that is initially at rest. Within your body, you conert cheical potential energy into echanical energy, which, by pushing the cart, you gie to the cart. After you hae been pushing on it for a while, the cart is oing, eaning that it has soe kinetic energy. So, in the end, the cart has soe kinetic energy that was originally cheical potential energy stored in you. Energy has been transferred fro you to the cart. In the case of the cart, what happens to the energy that you transfer to the cart is clear. ut how about the case of a deforable syste whose center of ass stays put? When you do work on such a syste, you transfer energy to that syste. So what happens to the energy? Eperientally, we find that the energy becoes part of the internal energy of the syste. The internal energy of the syste increases by an aount that is equal to the work done on the syste. This increase in the internal energy can be an increase in the internal potential energy, an increase in the internal kinetic energy, or both. An increase in the internal kinetic energy would anifest itself as an increase in teperature. Doing work on a syste represents the second way, which we hae considered, of causing an increase in the internal energy of the syste. The other way was for heat to flow into the syste. The fact that doing work on a syste and/or haing heat flow into that syste will increase the internal energy of that syste, is represented, in equation for, by: U = Q +W IN 66

267 Chapter 37 The First Law of Therodynaics which we copy here for your conenience: U = Q + W IN (37-) In this equation, U is the change in the internal energy of the syste, Q is the aount of heat that flows into the syste, and W IN is the aount of work that is done on the syste. This equation is referred to as the First Law of Therodynaics. Cheists typically write it without the subscript IN on the sybol W representing the work done on the syste. (The subscript IN is there to reind us that the W IN represents a transfer of energy into the syste. In the cheistry conention, it is understood that W represents the work done on the syste no subscript is necessary.) Historically, physicists and engineers hae studied and deeloped therodynaics with the goal of building a better heat engine, a deice, such as a stea engine, designed to produce work fro heat. That is, a deice for which heat goes in and work coes out. It is probably for this reason that physicists and engineers alost always write the first law as: U = Q W (37-) where the sybol W represents the aount of work done by the syste on the eternal world. (This is just the opposite of the cheistry conention.) ecause this is a physics course, this ( U = Q W ) is the for in which the first law appears on your forula sheet. I suggest aking the first law as eplicit as possible by writing it as U = Q IN W OUT or, better yet: U = Q IN + W IN (37-3) In this for, the equation is saying that you can increase the internal energy of a syste by causing heat to flow into that syste and/or by doing work on that syste. Note that any one of the quantities in the equation can be negatie. A negatie alue of Q IN eans that heat actually flows out of the syste. A negatie alue of W IN eans that work is actually done by the syste on the surroundings. Finally, a negatie alue of U eans that the internal energy of the syste decreases. Again, the real tip here is to use subscripts and coon sense. Write the First Law of Therodynaics in a anner consistent with the facts that heat or work into a syste will increase the internal energy of the syste, and heat or work out of the syste will decrease the internal energy of the syste. 67