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iv CONTENTS 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2 Acceleration 31 2.1 The law of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.1.1 Inertial reference frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... | University Physics I Classical Mechanics_Page_7_Chunk4201 |
CONTENTS v 3.3 Extended systems and center of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.3.1 Center of mass motion for an isolated system . . . . . . . . . . . . . . . . . . 59 3.3.2 Recoil and rocket propulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.4 In summary . . . . . . . .... | University Physics I Classical Mechanics_Page_8_Chunk4202 |
vi CONTENTS 5.1 Conservative interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.1.1 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 5.1.2 Potential energy functions and “energy landscapes” . . . . . . . . . . . . . . 94 5.2 Dissipation of energy a... | University Physics I Classical Mechanics_Page_9_Chunk4203 |
CONTENTS vii 6.3.2 Normal forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 6.3.3 Static and kinetic friction forces . . . . . . . . . . . . . . . . . . . . . . . . . 123 6.3.4 Air resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 6.4 Free-body diagr... | University Physics I Classical Mechanics_Page_10_Chunk4204 |
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 | University Physics I Classical Mechanics_Page_10_Chunk4205 |
viii CONTENTS 7.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 7.7.1 Braking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 7.7.2 Work, energy and the choice of system: dissipative case . . . . . . . . . . . . 154 7.7.3 Work, energy ... | University Physics I Classical Mechanics_Page_11_Chunk4206 |
reference: Centrifugal force and Coriolis force . . . . . . . 184 | University Physics I Classical Mechanics_Page_11_Chunk4207 |
CONTENTS ix 8.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 9 Rotational dynamics 191 9.1 Rotational kinetic energy, and moment of inertia . . . . . . . . . . . . . . . . . . . . 191 9.2 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... | University Physics I Classical Mechanics_Page_12_Chunk4208 |
x CONTENTS 10.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 10.4.1 Orbital dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 10.4.2 Orbital data from observations: Halley’s comet . . . . . . . . . . . . . . . . . 245 10.5 Advanced Topic... | University Physics I Classical Mechanics_Page_13_Chunk4209 |
. . . . . . . . . . . . . . 272 | University Physics I Classical Mechanics_Page_13_Chunk4210 |
CONTENTS xi 11.6.2 The Cavendish experiment: how to measure G with a torsion balance . . . . 273 11.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 12 Waves in one dimension 279 12.1 Traveling waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... | University Physics I Classical Mechanics_Page_14_Chunk4211 |
xii CONTENTS 13.2.1 Temperature and heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . 304 13.2.2 The gas thermometer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 13.2.3 The zero-th law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 13.3 Heat and the first l... | University Physics I Classical Mechanics_Page_15_Chunk4212 |
Preface Students: if this is too long, at the very least read the last four paragraphs. Thank you! For many years Eric Mazur’s Principles and Practice of Physics was the required textbook for University Physics I at the University of Arkansas. In writing this open-source replacement I have tried to preserve some of its... | University Physics I Classical Mechanics_Page_16_Chunk4213 |
xiv CONTENTS they have read. Then, in the lecture, they will have an opportunity to see the material presented again, as a sort of executive summary delivered by, typically, a different instructor, who will also be able to answer any questions they might have about the book’s presentation. Additionally, the instructor w... | University Physics I Classical Mechanics_Page_17_Chunk4214 |
CONTENTS xv One last word, for the students who may have read this far, concerning the use of equations and “proofs” in this book. It is essential to the nature of physics to be able to cast its results in mathematical terms, and to use math to explain and predict new results; hence, equations and mathematical derivati... | University Physics I Classical Mechanics_Page_18_Chunk4215 |
Chapter 1 Reference frames, displacement, and velocity 1.1 Introduction Classical mechanics is the branch of physics that deals with the study of the motion of anything (roughly speaking) larger than an atom or a molecule. That is a lot of territory, and the methods and concepts of classical mechanics are at the founda... | University Physics I Classical Mechanics_Page_19_Chunk4216 |
2 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY In classical mechanics, an “ideal” particle is an object with no appreciable size—a mathematical point. In one dimension (that is to say, along a straight line), its position can be specified just by giving a single number, the distance from some reference point,... | University Physics I Classical Mechanics_Page_20_Chunk4217 |
1.2. POSITION, DISPLACEMENT, VELOCITY 3 1.1.2 Aside: the atomic perspective As an aside, it should perhaps be mentioned that the building up of classical mechanics around this concept of ideal particles had nothing to do, initially, with any belief in “atoms,” or an atomic theory of matter. Indeed, for most 18th and 19... | University Physics I Classical Mechanics_Page_21_Chunk4218 |
4 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY 1.2.1 Position As stated in the previous section, we are initially interested only in describing the motion of a “particle,” which can be thought of as a mathematical point in space. (Later on we will see that, even for an extended object or system, it is often ... | University Physics I Classical Mechanics_Page_22_Chunk4219 |
1.2. POSITION, DISPLACEMENT, VELOCITY 5 be interpreted as the components of a vector that we call the point’s position vector, and denote by ⃗r (sometimes boldface letters are used for vectors, instead of an arrow on top; in that case, the position vector would be denoted by r). A vector is a mathematical object, with ... | University Physics I Classical Mechanics_Page_23_Chunk4220 |
its appropriate | University Physics I Classical Mechanics_Page_23_Chunk4221 |
6 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY units: seconds, minutes, hours, etc. x (m) t (s) Figure 1.2: A possible position vs. time graph for an object moving in one dimension. We will be often interested in plotting the position of an object as a function of time—that is to say, the graph of the functi... | University Physics I Classical Mechanics_Page_24_Chunk4222 |
1.2. POSITION, DISPLACEMENT, VELOCITY 7 difference between its initial value and its final value. The time interval itself will be written as ∆t and can be expressed as ∆t = tf −ti (1.2) where again ti and tf are the initial and final values of the time parameter (imagine, for instance, that you are reading time in second... | University Physics I Classical Mechanics_Page_25_Chunk4223 |
8 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY x (m) y (m) ri xi yi rf Δr xf yf Figure 1.3: The displacement vector for a particle that was initially at a point with position vector ⃗ri and ended up at a point with position vector ⃗rf is the difference of the position vectors. Figure 1.3 shows how this makes ... | University Physics I Classical Mechanics_Page_26_Chunk4224 |
1.2. POSITION, DISPLACEMENT, VELOCITY 9 work with significant figures this semester: the rule of thumb is, keep four significant figures in all intermediate calculations, and report three in the final result). The way we define average velocity is similar to average speed, but with one important difference: we use the displac... | University Physics I Classical Mechanics_Page_27_Chunk4225 |
10 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY average is taken over very short time intervals. Nevertheless, the fact is that for any reasonably well-behaved position function x(t), the limit in Eq. (1.7) is mathematically well-defined, and it equals what we call, in calculus, the derivative of the function... | University Physics I Classical Mechanics_Page_28_Chunk4226 |
1.2. POSITION, DISPLACEMENT, VELOCITY 11 Conversely, looking at the sample x-vs-t graphs in this chapter, you may notice that there are times when the tangent is horizontal, meaning it has zero slope, and so the instantaneous velocity at those times is zero (for instance, at the time t = 1.0 s in Figure 1.2). This make... | University Physics I Classical Mechanics_Page_29_Chunk4227 |
12 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY If the velocity changes with time, obtaining an expression for the position of the object as a function of time may be a nontrivial task. In the next chapter we will study an important special case, namely, when the velocity changes at a constant rate (constant... | University Physics I Classical Mechanics_Page_30_Chunk4228 |
1.2. POSITION, DISPLACEMENT, VELOCITY 13 However, Eq. (1.11) is just the area of the first rectangle shown under the curve in Figure 1.5 (the base of the rectangle has “length” t2 −t1, and its height is v1). Similarly for the second rectangle, and so on. So the sum ∆x1 +∆x2+. . . is both an approximation to the area und... | University Physics I Classical Mechanics_Page_31_Chunk4229 |
14 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY the function vx(t) is known, or equivalently to get ∆y if you know vy(t), and so on. Introducing the velocity vector at this point does cause a little bit of a notational difficulty. For quantities like x and ∆x, it is pretty obvious that they are the x component... | University Physics I Classical Mechanics_Page_32_Chunk4230 |
1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 15 with the river water, like a piece of driftwood that you could measure your progress relative to.) In any case, graphically, this will look as in Figure 1.6, which I have drawn for the two-dimensional case because I think it makes it easier to visualize what’s going o... | University Physics I Classical Mechanics_Page_33_Chunk4231 |
16 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY All these coordinates are also the components of the respective position vectors, shown in the figure and color-coded by reference frame (so, for instance, ⃗rAP is the position vector of P in the frame A), so the equations (1.14) can be written more compactly as... | University Physics I Classical Mechanics_Page_34_Chunk4232 |
1.3. REFERENCE FRAME CHANGES AND RELATIVE MOTION 17 Dividing Eq. (1.18) by ∆t we get the average velocities1, and then taking the limit ∆t →0 we get the instantaneous velocities. This applies in the same way to the y coordinates, and the result is the vector equation ⃗vAP = ⃗vBP + ⃗vAB (1.19) I have rearranged the term... | University Physics I Classical Mechanics_Page_35_Chunk4233 |
18 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY sometimes the information we have is given to us in a different way: for instance, we could be given the velocity of the object in frame A (⃗vAP ), and the velocity of frame B as seen in frame A (⃗vAB), and told to calculate the velocity of the object as seen in... | University Physics I Classical Mechanics_Page_36_Chunk4234 |
1.4. IN SUMMARY 19 with a velocity (relative to the truck) of 60 mph backwards, while the truck is moving forward at 60 mph. I think the result is worth watching. (Do not be distracted by their talk about momentum. We will get there, in time.) A very old, but also very good, educational video about different frames of r... | University Physics I Classical Mechanics_Page_37_Chunk4235 |
20 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY where ti is an arbitrarily chosen initial time and xi the position at that time. This can also be written in the form given by Eq. (1.9). The argument (t) on the left-hand side of (1.10) is optional, and ti is often set equal to zero, giving just x = xi + vt. T... | University Physics I Classical Mechanics_Page_38_Chunk4236 |
1.5. EXAMPLES 21 1.5 Examples 1.5.1 Motion with (piecewise) constant velocity You leave your house on your bicycle to go visit a friend. At your normal speed of 9 mph, you know it takes you 6 minutes to get there. This time, though, when you have traveled half the distance you realize you forgot a book at home that you... | University Physics I Classical Mechanics_Page_39_Chunk4237 |
22 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY (c) The whole trip consists, as detailed above, of 0.45 miles at 9 mph, and the rest, which is 1.35 miles, at 18 mph. Applying ∆t = ∆x/v to each of these intervals, we get a total time of ∆t = 0.45 miles 9 mph + 1.35 miles 18 mph = 0.125 hours = 0.125 × 60 min ... | University Physics I Classical Mechanics_Page_40_Chunk4238 |
1.5. EXAMPLES 23 (twice what it was earlier, but in the opposite direction). For the position graph, use Eq. (1.10) with xi = 724 m (this is half of the distance to your friend’s house, and the starting position for this interval), ti = 180 s and v = −8.046 m/s. • Third interval: from t = 270 s to t = 450 s. The veloci... | University Physics I Classical Mechanics_Page_41_Chunk4239 |
24 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY 1.5.2 Addition of velocities, relative motion This example was inspired by the “race on a moving sidewalk” demo at http://physics.bu.edu/~duffy/classroom.html. Please go take a look at it! Two girls, Ann and Becky (yes, A and B) decide to have a race while they... | University Physics I Classical Mechanics_Page_42_Chunk4240 |
1.5. EXAMPLES 25 Note that I have drawn one picture for each half of the race, and that all the information given in the text of the problem is there. The figure makes it clear also the notation I will be using for each of the girls’ velocities, and to see at a glance what is happening. You should next state what kind o... | University Physics I Classical Mechanics_Page_43_Chunk4241 |
26 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY Part (c): The quantity we want is written, in the notation of Section 1.3, vAB (“velocity of Becky relative to Ann”). To calculate this, we just need to know the velocities of both girls in some frame of reference (the same for both!), then subtract Ann’s veloc... | University Physics I Classical Mechanics_Page_44_Chunk4242 |
1.6. PROBLEMS 27 1.6 Problems Problem 1 x (m) t (s) The above figure is the position (in meters) versus time (in seconds) graph of an object in motion. Only the segments between t = 1 s and t = 2 s, and between t = 4 s and t = 5 s, are straight lines. The peak of the curve is at t = 3 s, x = 4 m. Answer the following qu... | University Physics I Classical Mechanics_Page_45_Chunk4243 |
28 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY x y Problem 3 Marshall Dillon is riding at 30 mph after the robber of the Dodge City bank, who has a head start of 15 minutes, but whose horse can only make 25 mph on a good day. How long does it take for Dillon to catch up with the bad guy, and how far from Do... | University Physics I Classical Mechanics_Page_46_Chunk4244 |
1.6. PROBLEMS 29 Problem 5 You are trying to pass a truck on the highway. The truck is driving at 55 mph, so you speed up to 60 mph and move over to the left lane. If the truck is 17 m long, and your car is 3 m long (a) how long does it take you to pass the truck completely? (b) How far (along the highway) have you tra... | University Physics I Classical Mechanics_Page_47_Chunk4245 |
30 CHAPTER 1. REFERENCE FRAMES, DISPLACEMENT, AND VELOCITY | University Physics I Classical Mechanics_Page_48_Chunk4246 |
Chapter 2 Acceleration 2.1 The law of inertia There is something funny about motion with constant velocity: it is indistinguishable from rest. Of course, you can usually tell whether you are moving relative to something else. But if you are enjoying a smooth airplane ride, without looking out the window, you have no id... | University Physics I Classical Mechanics_Page_49_Chunk4247 |
32 CHAPTER 2. ACCELERATION were dropping things from rest outdoors in a strong wind. But that is not what we experience on an airplane at all. The air, in fact, has no effect on the forward motion of the falling object. It does not push it in any way, because it is moving at the same velocity. This, in fact, reinforces ... | University Physics I Classical Mechanics_Page_50_Chunk4248 |
2.1. THE LAW OF INERTIA 33 This is why, historically, the law of inertia was not discovered until people started developing an appreciation for frictional forces, and the way they are constantly acting all around us to oppose the relative motion of any objects trying to slide past each other. This mention of relative m... | University Physics I Classical Mechanics_Page_51_Chunk4249 |
34 CHAPTER 2. ACCELERATION Again, nobody has pushed on them, and in fact what we can see in this case, from outside the car, is nothing but the law of inertia at work: the passengers were just keeping their initial velocity, when the car suddenly slowed down under and around them. So there is nothing wrong with the law... | University Physics I Classical Mechanics_Page_52_Chunk4250 |
2.2. ACCELERATION 35 2.2 Acceleration 2.2.1 Average and instantaneous acceleration Just as we defined average velocity in the previous chapter, using the concept of displacement (or change in position) over a time interval ∆t, we define average acceleration over the time ∆t using the change in velocity: aav = ∆v ∆t = vf ... | University Physics I Classical Mechanics_Page_53_Chunk4251 |
36 CHAPTER 2. ACCELERATION x (m) t (s) 0 5 5 10 Figure 2.1: A possible position vs. time graph for an object whose acceleration changes with time. Between t = 2.2 s and t = 2.5 s, as the object moves from x = 2 m to x = 4 m, the velocity does not appear to change very much, and the acceleration would correspondingly be... | University Physics I Classical Mechanics_Page_54_Chunk4252 |
2.2. ACCELERATION 37 called “concave upwards”), the acceleration is positive, whereas it is negative whenever the graph is convex (or “concave downwards”). It is (instantly) zero at those points where the curvature changes (which you may know as inflection points), as well as over stretches of time when the x-vs-t graph... | University Physics I Classical Mechanics_Page_55_Chunk4253 |
38 CHAPTER 2. ACCELERATION Position (meters) Acceleration (m/s2) Velocity (m/s) Time (seconds) 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 25 30 35 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 -5 -10 1 2 3 4 5 6 7 8 9 0 0 5 10 15 20 -5 -10 Time (seconds) Time (seconds) Figure 2.3: Sample position, velocity and acceleration vs. time graphs fo... | University Physics I Classical Mechanics_Page_56_Chunk4254 |
2.2. ACCELERATION 39 2.2.2 Motion with constant acceleration A particular kind of motion that is both relatively simple and very important in practice is motion with constant acceleration (see Figure 2.3 again for examples). If a is constant, it means that the velocity changes with time at a constant rate, by a fixed nu... | University Physics I Classical Mechanics_Page_57_Chunk4255 |
40 CHAPTER 2. ACCELERATION triangle of base ∆t and height vf −vi. Since vf −vi = a∆t, simple geometry immediately yields Eq. (2.7), or its equivalent (2.6). v (m/s) t (s) ti tf vi vf v(t) Figure 2.4: Graphical way to find the displacement for motion with constant acceleration. Lastly, consider what happens if we solve E... | University Physics I Classical Mechanics_Page_58_Chunk4256 |
2.2. ACCELERATION 41 2.2.3 Acceleration as a vector In two (or more) dimensions we introduce the average acceleration vector ⃗aav = ∆⃗v ∆t = 1 ∆t (⃗vf −⃗vi) (2.11) whose components are aav,x = ∆vx/∆t, etc.. The instantaneous acceleration is then the vector given by the limit of Eq. (2.11) as ∆t →0, and its components a... | University Physics I Classical Mechanics_Page_59_Chunk4257 |
42 CHAPTER 2. ACCELERATION will agree that an object’s velocity does not change (otherwise put, its acceleration is zero) when no forces act on it. Conversely, an accelerated frame will not be an inertial frame, because Eq. (2.14) will not hold. This is consistent with the examples I mentioned in Section 2.1 (the bounc... | University Physics I Classical Mechanics_Page_60_Chunk4258 |
2.3. FREE FALL 43 proportional to that object’s inertial mass, a quantity that we will introduce properly in the next chapter. For the time being, we will simply record here that this acceleration, near the surface of the earth, has a magnitude of approximately 9.8 m/s2, a quantity that is denoted by the symbol g. Thus... | University Physics I Classical Mechanics_Page_61_Chunk4259 |
44 CHAPTER 2. ACCELERATION Galileo’s main insight, on the theoretical side, was the realization that it was necessary to separate clearly the effect of gravity and the effect of the drag force. Experimentally, his big idea was to use an inclined plane to slow down the “fall” of an object, so as to make accurate measureme... | University Physics I Classical Mechanics_Page_62_Chunk4260 |
2.4. IN SUMMARY 45 4. Changes in velocity are detectable, and, by (1) above, are evidence of unbalanced forces acting on an object. 5. The rate of change of an object’s velocity is the object’s acceleration: the average acceleration over a time interval ∆t is aav = ∆v/∆t, and the instantaneous acceleration at a time t ... | University Physics I Classical Mechanics_Page_63_Chunk4261 |
46 CHAPTER 2. ACCELERATION 2.5 Examples 2.5.1 Motion with piecewise constant acceleration Construct the position vs. time, velocity vs. time, and acceleration vs. time graphs for the motion described below. For each of the intervals (a)–(d) you’ll need to figure out the position (height) and velocity of the rocket at th... | University Physics I Classical Mechanics_Page_64_Chunk4262 |
2.5. EXAMPLES 47 which the rocket reaches the top of its trajectory, and then starts coming down. The corresponding displacement is, by Eq. (2.16), ∆ytop = vi2(ttop −ti2) −1 2g(ttop −ti2)2 = 20.4 m so the maximum height it reaches is 30.4 m. At the end of the full 3-second interval, the rocket’s displacement is ∆y2 = v... | University Physics I Classical Mechanics_Page_65_Chunk4263 |
48 CHAPTER 2. ACCELERATION 2.6 Problems Problem 1 You get on your bicycle and ride it with a constant acceleration of 0.5 m/s2 for 20 s. After that, you continue riding at a constant velocity for a distance of 200 m. Finally, you slow to a stop, with a constant acceleration, over a distance of 20 m. v (m/s) t (s) 0 0 x... | University Physics I Classical Mechanics_Page_66_Chunk4264 |
2.6. PROBLEMS 49 (a) How far did you travel while you were accelerating at 0.5 m/s2, and what was your velocity at the end of that interval? (b) After that, how long did it take you to cover the next 200 m? (c) What was your acceleration while you were slowing down to a stop, and how long did it take you to come to a s... | University Physics I Classical Mechanics_Page_67_Chunk4265 |
50 CHAPTER 2. ACCELERATION | University Physics I Classical Mechanics_Page_68_Chunk4266 |
Chapter 3 Momentum and Inertia 3.1 Inertia In everyday language, we speak of something or someone “having a large inertia” to mean, essen- tially, that they are very difficult to set in motion. This usage of the word “inertia” is consistent with the “law of inertia” we introduced in the previous chapter (which states, am... | University Physics I Classical Mechanics_Page_69_Chunk4267 |
52 CHAPTER 3. MOMENTUM AND INERTIA when two identical objects (presumably, therefore, having the same inertia) collide: if the collision is head-on (so the motion, before and after, is confined to a straight line), they basically exchange velocities. For instance, a billiard ball hitting another one will stop dead and t... | University Physics I Classical Mechanics_Page_70_Chunk4268 |
3.1. INERTIA 53 reliable, repeatable measure? Will it work for any kind of collision (within reason, of course: we clearly need to stay in one dimension, and eliminate external influences such as friction), and for any initial velocity? To begin with, we have reason to expect that it does not matter whether we shoot obj... | University Physics I Classical Mechanics_Page_71_Chunk4269 |
54 CHAPTER 3. MOMENTUM AND INERTIA in these examples, namely, 1 m/s. However, unless we do the experiments we cannot really predict what will happen if we increase (or decrease) their relative velocity. In fact, we could imagine smashing the two objects at very high speed, so they might even become seriously mangled in... | University Physics I Classical Mechanics_Page_72_Chunk4270 |
3.1. INERTIA 55 3.1.2 Inertial mass: definition and properties At this point, it would seem reasonable to assume that this ratio, ∆v2/∆v1, is, in fact, telling us something about an intrinsic property of the two objects, what we have called above their “relative inertia.” It is easy, then, to see how one could assign a ... | University Physics I Classical Mechanics_Page_73_Chunk4271 |
56 CHAPTER 3. MOMENTUM AND INERTIA of the standard. Suppose that we have two objects, to which we have assigned masses m1 and m2 by arranging for each to collide with the “standard object” independently. If we now arrange for a collision between objects 1 and 2 directly, will we actually find that the ratio of their vel... | University Physics I Classical Mechanics_Page_74_Chunk4272 |
3.2. MOMENTUM 57 the system is pi = m1v1i + m2v2i, and similarly if the final velocities are v1f and v2f, the total final momentum will be pf = m1v1f + m2v2f. We then assert that the total momentum of the system is not changed by the collision. Mathemat- ically, this means pi = pf (3.4) or m1v1i + m2v2i = m1v1f + m2v2f (... | University Physics I Classical Mechanics_Page_75_Chunk4273 |
58 CHAPTER 3. MOMENTUM AND INERTIA What this means, in turn, is that each separate component (x, y and z) of the momentum will be separately conserved (so Eq. (3.7) is equivalent to three scalar equations, in three dimensions). When we get to study the vector nature of forces, we will see an interesting implication of ... | University Physics I Classical Mechanics_Page_76_Chunk4274 |
3.3. EXTENDED SYSTEMS AND CENTER OF MASS 59 of all of them; otherwise, it will tend to be closer to the more massive particle(s). The “particles” in question could be spread apart, or they could literally be the “parts” into which we choose to subdivide, for computational purposes, a single extended object. If the part... | University Physics I Classical Mechanics_Page_77_Chunk4275 |
60 CHAPTER 3. MOMENTUM AND INERTIA objects of Figure 3.1. I have assumed that object 1 starts out at x1i = −5 mm at t = 0, and object 2 starts at x2i = 0 at t = 0. Because object 2 has twice the inertia of object 1, the position of the center of mass, as given by Eq. (3.8), will always be xcm = x1/3 + 2x2/3 that is to ... | University Physics I Classical Mechanics_Page_78_Chunk4276 |
3.4. IN SUMMARY 61 3.3.2 Recoil and rocket propulsion As we have just seen, you cannot alter the motion of your center of mass without relying on some external force—which is to say, some kind of external support. This is actually something you may have experienced when you are resting on a very slippery surface and yo... | University Physics I Classical Mechanics_Page_79_Chunk4277 |
62 CHAPTER 3. MOMENTUM AND INERTIA 5. The momentum of an object of inertial mass m moving with a velocity ⃗v is defined as ⃗p = m⃗v. The total momentum of a system of objects is defined as the (vector) sum of all the individual momenta. 6. (Conservation of momentum) The momentum of an isolated system remains always con- ... | University Physics I Classical Mechanics_Page_80_Chunk4278 |
3.5. EXAMPLES 63 3.5 Examples 3.5.1 Reading a collision graph The graph shows a collision between two carts (possibly equipped with magnets so that they repel each other before they actually touch) on an air track. The inertia (mass) of cart 1 is 1 kg. Note: this is a position vs. time graph! (a) What are the initial v... | University Physics I Classical Mechanics_Page_81_Chunk4279 |
64 CHAPTER 3. MOMENTUM AND INERTIA for suitable intervals. In this way one gets v1i = −1 m s v2i = 0.5 m s (b) Similarly, one gets v1f = 1 m s v2f = −0.5 m s (c) Use this equation, or equivalent (conservation of momentum is OK) m2 m1 = −∆v1 ∆v2 m2 m1 = −1 −(−1) −0.5 −0.5 = 2 so the mass of the second cart is 2 kg. (d) ... | University Physics I Classical Mechanics_Page_82_Chunk4280 |
3.5. EXAMPLES 65 3.5.2 Collision in different reference frames, center of mass, and recoil An 80-kg hockey player (call him player 1), moving at 3 m/s to the right, collides with a 90-kg player (player 2) who was moving at 2 m/s to the left. For a brief moment, they are stuck sliding together as they grab at each other.... | University Physics I Classical Mechanics_Page_83_Chunk4281 |
66 CHAPTER 3. MOMENTUM AND INERTIA that to convert all the Earth-frame velocities to the reference frame of player 3, we just need to subtract 1.5 m/s from them. This gives us v31,i = 3 m s −1.5 m s = 1.5 m s v32,i = −2 m s −1.5 m s = −3.5 m s v31,f = v32,f = 0.353 m s −1.5 m s = −1.147 m s (3.17) The total initial mom... | University Physics I Classical Mechanics_Page_84_Chunk4282 |
3.6. PROBLEMS 67 3.6 Problems Problem 1 This figure shows the position vs. time graph for two objects before and after they collide. Assume that they form an isolated system. (a) What are the velocities of the two objects before and after the collision? (Hint: you will get a more accurate result if you choose the initia... | University Physics I Classical Mechanics_Page_85_Chunk4283 |
68 CHAPTER 3. MOMENTUM AND INERTIA (a) What was the velocity of the bullet just before impact? (b) In order to shoot a bullet at this speed, what must have been the recoil speed of the gun? Problem 4 A 2-kg object, moving at 1 m/s, collides with a 1-kg object that is initially at rest. After the collision, the two obje... | University Physics I Classical Mechanics_Page_86_Chunk4284 |
Chapter 4 Kinetic Energy 4.1 Kinetic Energy For a long time in the development of classical mechanics, physicists were aware of the existence of two different quantities that one could define for an object of inertia m and velocity v. One was the momentum, mv, and the other was something proportional to mv2. Despite thei... | University Physics I Classical Mechanics_Page_87_Chunk4285 |
70 CHAPTER 4. KINETIC ENERGY For a system of particles, we will treat kinetic energy as an additive quantity, just like we did for momentum, so the total kinetic energy of a system will just be the sum of the kinetic energies of all the particles making up the system. Note that, unlike momentum, this is a scalar (not a... | University Physics I Classical Mechanics_Page_88_Chunk4286 |
4.1. KINETIC ENERGY 71 v2i = 0, v1f = −1/3 m/s, v2f = 2/3 m/s, and so the kinetic energies are K1i = 1 2 J, K2i = 0; K1f = 1 18 J, K2f = 4 9 J Note that 1/18 + 4/9 = 9/18 = 1/2, and so Ksys,i = K1i + K2i = 1 2 J = K1f + K2f = Ksys,f In words, we find that, in this collision, the final value of the total kinetic energy is... | University Physics I Classical Mechanics_Page_89_Chunk4287 |
72 CHAPTER 4. KINETIC ENERGY from the values we had in the previous example, but note that once again the total kinetic energy after the collision equals the total kinetic energy before—namely, 1 J in this case1 . Things are, however, very different when we consider the third collision example shown in Chapter 3, namely... | University Physics I Classical Mechanics_Page_90_Chunk4288 |
4.1. KINETIC ENERGY 73 special case of inelastic collision is the one called totally inelastic, where the two objects end up stuck together, as in Figure 4.3. As we shall see later, the kinetic energy “deficit” is largest in that case. I have said above that in an elastic collision the kinetic energy is “recovered,” and... | University Physics I Classical Mechanics_Page_91_Chunk4289 |
74 CHAPTER 4. KINETIC ENERGY 4.1.2 Relative velocity and coefficient of restitution An interesting property of elastic collisions can be disclosed from a careful study of figures 4.1 and 4.2. In both cases, as you can see, the relative velocity of the two objects colliding has the same magnitude (but opposite sign) before... | University Physics I Classical Mechanics_Page_92_Chunk4290 |
4.1. KINETIC ENERGY 75 This immediately allows us to cancel out the corresponding factors in Eq (4.7), so we are left with v1i + v1f = v2i + v2f, which can be rewritten as v1f −v2f = v2i −v1i (4.8) and this is equivalent to (4.4). So, in an elastic collision the speed at which the two objects move apart is the same as ... | University Physics I Classical Mechanics_Page_93_Chunk4291 |
76 CHAPTER 4. KINETIC ENERGY Although, as I just mentioned, for most “normal” collisions the coefficient of restitution will be a positive number between 1 and 0, there can be exceptions to this. If one of the objects passes through the other (like a bullet through a target, for instance), the value of e will be negative... | University Physics I Classical Mechanics_Page_94_Chunk4292 |
4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 77 Figure 4.5 shows that the greatest loss of kinetic energy happens for the totally inelastic collision, which, as we will see in a moment, is, in fact, a general result. That being the case, the figure also shows that it may not be always be possible to bring the t... | University Physics I Classical Mechanics_Page_95_Chunk4293 |
78 CHAPTER 4. KINETIC ENERGY made to vanish entirely in an inelastic collision3: Kconv = 1 2 m1m2 m1 + m2 v2 12 = 1 2µv2 12 (4.13) The last equation implicitly defines a useful quantity that we call the reduced mass of a system of two particles, and denote by µ: µ = m1m2 m1 + m2 (4.14) Equation (4.11), with the definitio... | University Physics I Classical Mechanics_Page_96_Chunk4294 |
4.2. “CONVERTIBLE” AND “TRANSLATIONAL” KINETIC ENERGY 79 Although we have derived the decomposition (4.11) for the very restricted situation of two objects moving in one dimension, the basic result is quite general: first, everything in the derivation works if v1 and v2 are replaced by vectors ⃗v1 and ⃗v2, so the result... | University Physics I Classical Mechanics_Page_97_Chunk4295 |
80 CHAPTER 4. KINETIC ENERGY another we just add or subtract from all the velocities the relative velocity of the two frames. This operation, however, will not change any of the relative velocities of the parts of the system, since these are all differences to begin with. Mathematically, (v2 + v′) −(v1 + v′) = v2 −v1 re... | University Physics I Classical Mechanics_Page_98_Chunk4296 |
4.3. IN SUMMARY 81 7. Besides the cases considered above, one may have collisions where the objects pass through each other, giving e < 0, and “explosive collisions,” where e > 1. In these latter collisions some internal source of energy is converted into additional kinetic energy when the objects interact. The extreme... | University Physics I Classical Mechanics_Page_99_Chunk4297 |
82 CHAPTER 4. KINETIC ENERGY 4.4 Examples 4.4.1 Collision graph revisited Look again at the collision graph from example 3.5.1 from the point of view of the kinetic energy of the two carts. (a) What is the initial kinetic energy of the system? (b) How much of this is in the center of mass motion, and how much of is con... | University Physics I Classical Mechanics_Page_100_Chunk4298 |
4.4. EXAMPLES 83 On the other hand, it is also clear that Kconv is fully recovered after the collision is over, since the relative velocity just changes sign: v12,i = v2i −v1i = 0.5 m s −(−1) m s = 1.5 m s v12,f = v2f −v1f = −0.5 m s −1 m s = −1.5 m s (4.19) Therefore Kconv,f = 1 2µv2 12,f = 1 2µv2 12,i = Kconv,i (d) S... | University Physics I Classical Mechanics_Page_101_Chunk4299 |
84 CHAPTER 4. KINETIC ENERGY This is Kcm throughout, as well as Ksys right after the collision, since the collision is totally inelastic and that means that Kconv drops to zero. Also, subtracting this from (4.20) will give us the initial value of the convertible energy, without the need for a separate calculation, so K... | University Physics I Classical Mechanics_Page_102_Chunk4300 |
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