Calculus-Based Physics I by Jeffrey W. Schnick

Transcription

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