Physics. 2. Physics-Problems, exercises, etc. I. Gottlieb, Michael A.
II. 莱顿，拉尔夫。III. 桑兹，马修·L.（马修·林兹）。
IV. Leighton, Robert B. V. Vogt, Rochus E. VI. Feynman, Richard Phillips.
Feynman lectures on physics. VII. Title.
QC23.F47 1989 补编
530 '.076-dc22
2005013077
ISBN 0-8053-9063-4
2006 年 Michelle Feynman，Carl Feynman，Michael Gottlieb 和 Ralph Leighton 版权所有。保留所有权利。在美利坚合众国制造。本出版物受版权保护，任何禁止的复制、存储在检索系统中或以任何形式或任何方式传输，包括电子、机械、复印、录音等，均需事先获得出版商的许可。要获得使用本作品材料的许可，请向 Pearson Education, Inc.，Permissions Department，1900 E. Lake Ave.，Glenview，IL 60025 提交书面请求。有关许可的信息，请致电(847) 486-2635。
一个特殊的例子：几年前，我在一个派对上遇到了迈克尔·戈特利布，当时主人正在电脑屏幕上展示现场图瓦人喉音歌手的谐波共振-这种活动让生活在旧金山附近如此有趣。戈特利布学过数学，对物理学非常感兴趣，所以我建议他阅读《费曼物理学讲义》-大约一年后，他花了六个月的时间仔细阅读这些讲义。正如戈特利布在自己的介绍中描述的那样，最终导致了你现在正在阅读的这本书，以及《费曼物理学讲义》的新的“最终版”。 iv - 前言
这个建议了重建缺失的讲座的想法，如果它们被证明有趣，就将它们提供给加州理工学院和 Addison-Wesley，以便包括在更完整和错误更正的《讲座》版本中。但首先我必须找到缺失的讲座，而我仍然在哥斯达黎加！通过一点演绎逻辑和调查，拉尔夫能够找到讲座笔记，这些笔记以前被藏在他父亲的办公室和加州理工学院档案馆之间的某个地方。拉尔夫还获得了缺失讲座的录音带，而在我回到加利福尼亚后在档案馆中研究勘误时，我幸运地在一盒杂项底片中发现了黑板照片（长期以来被认为丢失）。费曼的继承人慷慨地允许我们使用这些材料，因此，在费曼-莱顿-桑兹三人组中现在唯一幸存的成员马特·桑兹提供了一些有用的批评后，拉尔夫和我重建了 B 评论作为样本，并将其与《讲座》的勘误一起呈现给加州理工学院和 Addison-Wesley。
In order for me to actually draw the picture of any vector, I have to make a decision as to the scale. When we talked about forces, we said that so-andso many newtons were going to be represented by 1 inch (or 1 meter, or whatever). And here, we have to say that so-and-so many meters per second is going to be represented by 1 inch. Someone else could draw the picture with position vectors the same lengths as ours, but with the velocity vector one-third as long as ours-he's just using a different scale for his velocity vector. There's no unique way to draw the length of a vector because the choice of scale is arbitrary.
Those are some of the deviations and difficulties of differentiating vectors.
Of course, you can differentiate the derivative of a vector, then differentiate that, and so on. I called the derivative of A "velocity," but that's only because is the position; if is something else, its derivative is something other than velocity. For example, if is the momentum, the time derivative of momentum equals the force, so the derivative of would be the force. And if were the velocity, the time derivative of the velocity is the acceleration, and so on. What I've been telling you is generally true of differentiating vectors, but here I've given only the example of positions and velocities.
1-8 Line integrals
Finally, there's only one more thing that I have to talk about for vectors, and that is a horrible, complicated thing, called a "line integral":
We'll take as an example that you have a certain vector field , which you want to integrate along a curve from point to point . Now, in order for this line integral to mean something, there must be some way of defining the value of at every point on between and . If is defined as the force applied to an object at point , but you can't tell me how the force changes as you move along , at least between and , then "the integral
FIGURE 1-13 A constant force F defined on the straight-line path .
of along from to " makes no sense. (I said "at least," because could be defined anywhere else too, but at least you must define it on the part of the curve that you are integrating along.)
In a moment I'll define the line integral of an arbitrary vector field along an arbitrary curve, but first let's consider the case where is constant, and is a straight-line path from to -a displacement vector, which I'll call . (See Fig. 1-13.) Then, since is constant, we can take it outside the integral (just like ordinary integration), and the integral of srom to is just , so the answer is . That's the line integral for a constant force and a straight-line path-the easy case:
(Remember that is the component of the force in the direction of the displacement times the magnitude of the displacement; in other words, it's simply the distance along the line times the component of force in that direction. There are a lot of other ways to look at it, too: it's the component of the displacement in the direction of the force, times the magnitude of the force; it's the magnitude of the force times the magnitude of the displacement, times the cosine of the angle between them. These are all equivalent.)
More generally, the line integral is defined as follows. First, we break up the integral by dividing between and into equal segments: , . Then the integral along is the integral along plus the integral along plus the integral along , and so on. We choose large so that we can approximate each by a little displacement vector, , over which has an approximately constant value, . (See Fig. 1-14.) Then, by the "constant force straight-line path" rule, segment contributes approximately to the integral. So, if you add together for equals 1 to , that's an excellent approximation to the integral. The integral is exactly equal to this sum only if we take the limit as goes
FIGURE 1-14 A variable force defined on the curve .
to infinity: you take the segments as fine as you can; you take them a little finer than that, and you get the correct integral:
(This integral, of course, depends upon the curve-generally-though sometimes it doesn't in the physics.)
Well, then, that's all there is to the mathematics that you have to know to do the physics-for now, at least-and these things, most particularly the calculus and the early parts of the vector theory, should become second nature. Some things-like the line integral—may not be second nature now, but they will be, eventually, as you use them more; they aren't so vital yet, and that's harder. The things you "gotta get into your head good," right now, are the calculus, and the little things about taking the components of vectors in various directions.
1-9 A simple example
I'll give one example-just a very simple one-to show how to take components of vectors. Suppose we have a machine of some kind, as illustrated in Figure 1-15: it's got two rods connected by a pivot (like an elbow joint) with a big weight on it. The end of one rod is connected to the floor by a stationary pivot, and the end of the other rod has a rolling pivot that rolls along the floor in a slot-it's part of a machine, see, and it's going choochoog, choo-choog-the roller's going back and forth, the weight's going up and down, and so on.
choo-
choog
FIGURE 1-15 A simple machine.
Let's say the weight is , the rods are 0.5 meters long, and at a certain moment when the machine is standing still, the distance from the weight to the floor just happens to come out, luckily, to 0.4 meters-so that we have a 3-4-5 triangle, to make the arithmetic easier. (See Fig. 1-16.) (The arithmetic shouldn't make any difference; the real difficulty is to get the ideas right.)
The problem is to figure out what horizontal push you have to make on the roller in order to hold that weight up. Now, I'm going to make an assumption that we will need in order to do the problem. We make the assumption that when a rod has pivots at both ends, then the net force is always directed along the rod. (It turns out to be true; you may feel it's selfevident.) It would not necessarily be true if there were a pivot only at one end of the rod, because then I could push the rod sideways. But if there's a
FIGURE 1-16 What force, P, is required to hold up the weight?
pivot at both ends, I can only push along the rod. So let's suppose that we know that-that the forces must lie in the directions of the rods.
We also know something else from the physics: that the forces are equal and opposite at the ends of the rods. For example, whatever force is exerted by the rod on the roller must also be exerted by that rod, in the opposite direction, on the weight. So, that's the problem: with these ideas about the properties of rods, we try to figure out what's the horizontal force on the roller.
I think the way I'd like to try to do it is this: the horizontal force exerted on the roller by the rod is a certain component of the net force on it. (Of course, there's also a vertical component due to the "confining slot," which is unknown and uninteresting; it's part of the net force on the roller, which is exactly opposite the net force on the weight.) Therefore I can get the components of the force exerted on the roller by the rod-in particular, the horizontal component I want-if I can get the components of the force exerted by the rod on the weight. If I call the horizontal force on the weight , then the horizontal force on the roller is , and the force needed to hold the weight up is equal and opposite to that, so .
The vertical force on the weight from the rod, , is very easy: it's simply equal to the weight of the thing, which is , times , the gravitational constant. (Something else you have to know from physics-g is 9.8 , in the mks system.) is 2 times , or 19.6 newtons, so the vertical force on the roller is -19.6 newtons. Now, how can I get the horizontal force? Answer: I get it by knowing that the net force must lie along the rod. If is 19.6 , and the net force lies along the rod, then how much must be? (See Fig. 1-17.)
Well, we have the projections of the triangles, which have been designed very nicely, so that the ratio of the horizontal to the vertical sides is 3 to 4 ; that's the same ratio as is to , (I don't care about the net force, , here; I only need the force in the horizontal direction) and I already know what the vertical force is. So, the magnitude of the horizontal force-unknown-is to 19.6 as 0.3 is to 0.4 . Therefore I multiply by 19.6 and I get:
We conclude that the horizontal force on the roller needed to hold the weight up, is 14.7 newtons. That's the answer to this problem.
FIGURE 1-17 The force on the weight and the force on the roller from one rod.
Or is it?
You see, you can't do physics just by plugging in the formulas: you'll never get anywhere without having something else besides knowing the rules, the formulas for projections, and all that stuff; you have to have a certain feeling for the real situation! I'll make some more remarks about that in a minute, but here, in this particular problem, the difficulty is the following: the net force on the weight is not only from one rod, there's also a force exerted on it by the other rod, in some direction, and I left that out when I made the analysis-so it's all wrong!
I also have to worry about the force that the rod with the stationary pivot exerts on the weight. Now it's getting complicated: how can I figure out what that force is? Well, what is the net force of everything on the weight? Just the gravity-it just balances the gravity; there is no force horizontally on the weight. So the clue by which I can find out how much "juice" there is along the rod with the stationary pivot, is to notice that it must exert just enough horizontally to balance the horizontal force that the other rod is exerting.
Therefore, if I were to draw the force that the rod with the stationary pivot exerts, its horizontal component would be exactly opposite the horizontal component that the rod with the roller exerts, and the vertical components would be equal because of the identical 3-4-5 triangles the rods make: both rods are pushing up the same amount because their horizontal
components must balance-if the rods were different lengths, you'd have a little more work to do, but it's the same idea.
So, let's start out with the weight again: the forces from the rods on the weight are the first things to get straightened out. So, let's look at the forces from the rods on the weight. The reason I keep repeating this to myself is because otherwise I get the signs all mixed up: The force from the weight on the rods is the opposite of the force from the rods on the weight. I always have to start over after I get all balled up like this; I have to think it out again, and make up my mind as to what I want to talk about. So I say, "Look at the forces from the rods on the weight: there's a force , which is in the direction of one rod. Then there's a force , in the direction of the other rod. Those are the only two forces, and they are in the directions of the rods."
Now, the net of these two forces-ahhhh! I'm beginning to see the light! The net of these two forces has no horizontal component, and a vertical component of 19.6 newtons. Ah! Let me draw the picture again, since I did it wrong before. (See Fig. 1-18.)
FIGURE 1-18 The force on the weight and the forces on the roller and pivot, from both rods.
The horizontal forces balance, therefore the vertical components add, and the 19.6 newtons is not just the vertical component of the force from one rod, but the total from both; since each rod contributes half, the vertical component from the rod with the roller is only 9.8 newtons.
Now when we take the horizontal projection of this force, multiplying it by as we did before, we get the horizontal component of force from the rod with the roller on the weight, and that takes care of that:
1-10 Triangulation
I have a few moments left, so I'd like to make a little speech about the relation of the mathematics to the physics-which, in fact, was well illustrated by this little example. It will not do to memorize the formulas, and to say to yourself, "I know all the formulas; all I gotta do is figure out how to put 'em in the problem!"
Now, you may succeed with this for a while, and the more you work on memorizing the formulas, the longer you'll go on with this method-but it doesn't work in the end.
You might say, "I'm not gonna believe him, because I've always been successful: that's the way I've always done it; I'm always gonna do it that way."
You are not always going to do it that way: you're going to flunknot this year, not next year, but eventually, when you get your job, or something-you're going to lose along the line somewhere, because physics is an enormously extended thing: there are millions of formulas! It's impossible to remember all the formulas-it's impossible!
And the great thing that you're ignoring, the powerful machine that you're not using, is this: suppose Figure 1-19 is a map of all the physics formulas, all the relations in physics. (It should have more then two dimensions, but let's suppose it's like that.)
Now, suppose that something happened to your mind, that somehow all the material in some region was erased, and there was a little spot of missing goo in there. The relations of nature are so nice that it is possible, by logic, to "triangulate" from what is known to what's in the hole. (See Fig. 1-20.)
FIGURE 1-19 Imaginary map of all the physics formulas.
FIG URE 1-20 Forgotten facts can be recreated by triangulating from known facts.
And you can re-create the things that you've forgotten perpetually-if you don't forget too much, and if you know enough. In other words, there comes a time-which you haven't quite got to, yet-where you'll know so many things that as you forget them, you can reconstruct them from the pieces that you can still remember. It is therefore of first-rate importance that you know how to "triangulate" -that is, to know how to figure something out from what you already know. It is absolutely necessary. You might say, "Ah, I don't care; I'm a good memorizer! I know how to really memorize! In fact, I took a course in memory!"
That still doesn't work! Because the real utility of physicists-both to discover new laws of nature, and to develop new things in industry, and so
FIGURE 1-21 New discoveries are made by physicists triangulating from the known to the previously unknown.
on-is not to talk about what's already known, but to do something newand so they triangulate out from the known things: they make a "triangulation" that no one has ever made before. (See Fig. 1-21.)
In order to learn how to do that, you've got to forget the memorizing of formulas, and to try to learn to understand the interrelationships of nature. That's very much more difficult at the beginning, but it's the only successful way.
Laws and Intuition
REVIEW LECTURE B
Last time we discussed the mathematics that you need to know to do the physics, and I pointed out that equations should be memorized as a tool, but that it isn't a good idea to memorize everything. In fact, it's impossible in the long run to do everything by memory. That doesn't mean to do nothing by memory-the more you remember, in a certain sense, the better it isbut you should be able to re-create anything that you forgot.
Incidentally, on the subject of suddenly finding yourself below average when you come to Caltech, which we also discussed last time, if you somehow escape from being in the bottom half of the class, you're just making it miserable for somebody else, because now you're forcing somebody else to go down to the bottom half! But there is a way you can do it without disturbing anybody: find and pursue something interesting that delights you especially, so you become a kind of temporary expert in some phenomenon that you heard about. It's the way to save your soul—then you can always say, "Well, at least the other guys don't know anything about this!"
2-1 The physical laws
Now, in this review, I'm going to talk about the physical laws, and the first thing to do is to state what they are. We stated them in words a lot during the lectures so far, and it's hard to say it all again without using the same amount of time, but the physical laws can also be summarized by some equations, which I'll write down here. (By this time I'll suppose that your mathematics is developed to a point that you can understand the notation right away.) The following are all the physical laws that you should know.
First:
That is, the force, , is equal to the rate of change, with respect to time, of the momentum, . ( and are vectors. You're supposed to know what the symbols mean by this time.)
I'd like to emphasize that in any physical equation it is necessary to understand what the letters stand for. That doesn't mean to say, "Oh, I know that's , which stands for the mass in motion times the velocity, or the mass at rest times the velocity over the square root of 1 minus squared over c squared": 1
Instead, to understand physically what the stands for, you have to know that is not just "the momentum"; it's the momentum of something-the momentum of a particle whose mass is and whose velocity is . And, in Eq. 2.1, F is the total force-the vector sum of all the forces that are acting on that particle. Only then can you have an understanding of these equations.
Now, here's another physical law that you should know, called the conservation of momentum:
The law of conservation of momentum says that the total momentum is a constant in any situation. What does that mean, physically? For instance in a collision, it's the same as saying that the sum of the momenta of all the particles before a collision is the same as the sum of the momenta of all the particles after the collision. In the relativistic world, the particles can be different after the collision-you can create new particles and destroy old ones-but it's still true that the vector sum of the total momenta of everything before and after is the same.
The next physical law you should know, called the conservation of energy, takes the same form:
That is, the sum of the energies of all the particles before a collision is equal to the sum of the energies of all the particles after the collision. In order to use this formula, you have to know what the energy of a particle is. The energy of a particle with rest mass and speed is
2-2 The nonrelativistic approximation
Now, those are the laws that are correct in the relativistic world. In the nonrelativistic approximation-that is, if we look at particles at low velocity compared to the speed of light-then there are some special cases of the above laws.
To begin with, the momentum at low velocities is easy: is almost 1, so Eq. (2.2) becomes
That means the formula for the force, , can also be written . Then, by moving the constant, , out in front, we see that for low velocities, the force equals the mass times the acceleration:
The conservation of momentum for particles at low velocities has the same form as Eq. (2.3), except that the formula for the momenta is (and the masses are all constant):
However, the conservation of energy at low velocities becomes two laws: first, that the mass of each particle is constant-you can't create or destroy any material-and second, that the sum of the s (the total kinetic energy, or K.E.) of all the particles is constant:
If we think of large, everyday objects as particles with low velocitieslike an ashtray is a particle, approximately-then the law that the sum of the kinetic energies before equals the sum after is not true, because there can be some s of the particles all mixed up on the inside of the objects, in the form of internal motion-heat, for example. So in a collision between large objects, this law appears to fail. It's only true for fundamental particles. Of course with large objects, in can happen that not much energy goes into the internal motion, so the conservation of energy appears to be nearly true, and that's called a nearly elastic collision-which is sometimes idealized as a perfectly elastic collision. So energy is much more difficult to keep track of than momentum, because the conservation of energy needn't be true when the objects involved are large, like weights and so on.
2-3 Motion with forces
Now, if we look not at a collision, but at motion when forces act-then we get first a theorem that tells us that the change in kinetic energy of a particle is equal to the work done on it by the forces:
Remember, this means something-you have to know what all the letters mean: it means that if a particle is moving on some curve, , from point to point , and it's moving under the influence of a force , where is the total force acting on the particle, then if you knew what the of the particle is at point , and what it is over at point , they differ by the integral, from to , of , where is an increment of displacement along . (See Fig. 2-1).
and
In certain cases, that integral can be calculated easily ahead of time, because the force on the particle depends only on its position in a simple way. Under those circumstances we can write that the work done on the particle is equal to the change in another quantity called its potential energy, or P.E. Such forces are said to be "conservative":
FIGURE 2-1 .
Incidentally, the words that we use in physics are terrible: "conservative forces" doesn't mean that the forces are conserved, but rather that the forces are such that the energy of the things that the forces work on can be conserved. It's very confusing, I admit, and I can't help it.
The total energy of a particle is its kinetic energy plus its potential energy:
When only conservative forces act, a particle's total energy does not change:
But when nonconservative forces act-forces not included in any potential-then the change in a particle's energy is equal to the work done on it by those forces.
FIGURE 2-2 Velocity and acceleration vectors for circular motion.
Now, the end of this part of the review comes when we give all the rules that are known for the various forces.
But before I do that, there's a formula for acceleration that is very useful: if, at a given instant, a thing is moving on a circle of radius at velocity , then its acceleration is directed toward the center, and is equal in magnitude to . (See Fig. 2-2.) That's sort of at "right angles" to everything else I've been talking about, but it's good to remember that formula, because it's a pain in the neck to derive it:
TABLE 2-1
False in general
True always
(true only at low velocities)
Force
Momentum
Energy
TABLE
True with conservative forces
. is undefined.
Definitions: Kinetic Energy, K.E. Work, .
2-4 Forces and their potentials
Now, to get back on the track, I will list a series of laws of force, and the formulas for their potentials.
TABLE 2-3
Force
Potential
Gravity, near the earth's surface
Gravity, between particles
Electric Charge
Electric Field
Ideal Spring
Friction
First is surface gravity on the earth. The force is down, but never mind the sign; just remember which direction the force is, because who knows what your axes are-maybe you're making the axis down! (You're allowed to.) So the force is , and potential energy is , where is the mass of an object, is a constant (the acceleration of gravity at the surface of the earth-otherwise, the formula is no good!), and is the height above the ground, or any other level. That means the value of the potential energy can be zero any place you want. The way we're going to use potential energy is to talk about its changes-and then, of course, it doesn't make any difference if you add a constant.
The problem of how to deduce new things from old, and how to solve problems, is really very difficult to teach, and I don't really know how to do it. I don't know how to tell you something that will transform you from a person who can't analyze new situations or solve problems, to a person who can. In the case of the mathematics, I can transform you from somebody who can't differentiate to somebody who can, by giving you all the rules. But in the case of the physics, I can't transform you from somebody who can't to somebody who can, so I don't know what to do.
Because I intuitively understand what's going on physically, I find it difficult to communicate: I can only do it by showing you examples. Therefore, the rest of this lecture, as well as the next one, will consist of doing a whole lot of little examples-of applications, of phenomena in the physical world or in the industrial world, of applications of physics in different places-to show you how what you already know will permit you to understand or to analyze what's going on. Only from the examples will you be able to catch on.
We have found many old texts of ancient Babylonian mathematics. Among them is a great library full of mathematics workbooks for students. And it's very interesting: the Babylonians could solve quadratic equations; they even had tables for solving cubic equations. They could do triangles (See Fig. 2-3); they could do all kinds of things, but they never wrote down
FIG U RE 2-3 Pythagorean triples in the Plimpton 322 tablet from about 1700 B.C.
an algebraic formula. The ancient Babylonians had no way of writing formulas; instead, they did one example after the other-that's all. The idea was you're supposed to look at examples until you get the idea. That's because the ancient Babylonians didn't have the power of expression in mathematical form.
Today we do not have the power of expression to tell a student how to understand physics physically! We can write the laws, but we still can't say how to understand them physically. The only way you can understand physics physically, because of our lack of machinery for expressing this, is to follow the dull, Babylonian way of doing a whole lot of problems until you get the idea. That's all I can do for you. And the students who didn't get the idea in Babylonia flunked, and the guys who did get the idea died, so it's all the same!
So, now we try.
2-6 Understanding physics physically
The first problem that I mentioned in Chapter 1 involved a lot of physical things. There were two rods, a roller, a pivot, and a weight-it was , I believe. The geometrical relation of the rods was , and 0.5 , and
FIGURE 2-4 The simple machine of Chapter 1.
the problem was, what is the horizontal force required at the roller to hold the weight up, as shown in Figure 2-4? It took a little fiddling around (in fact, I had to do it twice before I got it right), but we found that the horizontal force on the roller corresponded to a weight of , as shown in Figure 2-5.
Now, if you just let yourself loose of the equations and think about it a while, and you pull back your sleeves and wave your arms, you can almost understand what the answer's going to be-at least can. Now, I have to teach you how to do that.
FIG URE 2-5 Distribution of force from the weight, through the rods, to the roller and pivot.
You could say, "Well, the force from the weight comes straight down, and it corresponds to , and the weight is balanced equally on two legs. So the vertical force from each leg must be enough to hold up . Now, the corresponding horizontal force on each leg must be the fraction of the vertical force that is merely the horizontal to vertical ratio in this right triangle, which is 3 to 4 . Therefore, the horizontal force on the roller corresponds to weight—period."
Now, let's see if it makes sense: according to that idea, if the roller were shoved much closer to the pivot, so that the distance between the legs was much smaller, I would expect much less force on the roller. Is it true, that when the weight is waaaay up there, the force on the roller should be low? Yeah! (See Fig. 2-6.)
If you can't see it, it's hard to explain why-but if you try to hold something up with a ladder, say, and you get the ladder directly under the thing, it's easy to keep the ladder from sliding out. But if the ladder is leaning way out at an angle, it's damn hard to keep the thing up! In fact, if you go waaaaay out, so that the far end of the ladder is only a very tiny distance from the ground, you'll find a nearly infinite horizontal force is required to hold the thing up at a very slight angle.
Now, all these things you can feel. You don't have to feel them; you can work them out by making diagrams and calculations, but as problems get more and more difficult, and as you try to understand nature in more and more complicated situations, the more you can guess at, feel, and understand without actually calculating, the much better off you are! So that's what you should practice doing on the various problems: when you have time somewhere, and you're not worried about getting the answer for a quiz or something, look the problem over and see if you can understand the way it behaves, roughly, when you change some of the numbers.
FIGURE 2-6 The force on the roller varies with the height of the weight.
Now, how to explain how to do that, I don't know. I remember once trying to teach somebody who was having a great deal of trouble taking the physics course, even though he did well in mathematics. A good example of a problem that he found impossible to solve was this: "There's a round table on three legs. Where should you lean on it, so the table will be the most unstable?"
The student's solution was, "Probably on top of one of the legs, but let me see: I'll calculate how much force will produce what lift, and so on, at different places."
Then I said, "Never mind calculating. Can you imagine a real table?"
"But that's not the way you're supposed to do it!"
"Never mind how you're supposed to do it; you've got a real table here with the various legs, you see? Now, where do you think you'd lean? What would happen if you pushed down directly over a leg?"
"Nothin'!"
I say, "That's right; and what happens if you push down near the edge, halfway between two of the legs?"
"It flips over!"
I say, "OK! That's better!"
The point is that the student had not realized that these were not just mathematical problems; they described a real table with legs. Actually, it wasn't a real table, because it was perfectly circular, the legs were straight up and down, and so on. But it nearly described, roughly speaking, a real table, and from knowing what a real table does, you can get a very good idea of what this table does without having to calculate anything-you know darn well where you have to lean to make the table flip over.
So, how to explain that, I don't know! But once you get the idea that the problems are not mathematical problems but physical problems, it helps a lot.
Now I'm going to apply this approach to a series of problems: first, in machine design; second, to motions of satellites; third, to the propulsion of rockets; fourth, to beam analyzers, and then, if I still have time, to the disintegration of pi mesons, and a couple of other things. All these problems are pretty difficult, but they illustrate various points as we go along. So, let's see what happens.
2-7 A problem in machine design
First, machine design. Here's the problem: there are two pivoted rods, each a half a meter long, which carry a weight of -sound familiar?-and at the left a roller is being driven back or forth by some machinery at a constant
velocity of 2 meters per second, OK? And the question for you is, what is the force required to do that when the height of the weight is 0.4 meters? (See Fig. 2-7.)
You might be thinking, "We did that already! The horizontal force required to balance the weight was of a weight."
But I argue, "The force is not , because the weight is moving."
You might counter, "When an object is moving, is a force required to keep it moving? No!"
"But a force is required to change the object's motion."
"Yes, but the roller is moving at a constant velocity!"
"Ah, yes, that's true: the roller is moving at a constant velocity of 2 meters per second. But what about the weight: is that moving at a constant velocity? Let's feel it: does the weight move slowly sometimes, and fast sometimes?"
"Yes . . ."
"Then its motion is changing-and that's the problem we have: to figure out the force required to keep the roller moving constantly at 2 meters per second when the weight is at a height of 0.4 meters."
Let's see if we can understand how the weight's motion is changing.
Well, if the weight is near the top and the roller is almost directly underneath it, the weight hardly moves up and down. In this position the weight is not moving very fast. But if the weight is down low, like we had before, and you push the roller just a shade to the right-boy, that weight has to move way up to get out of the way! So, as we push the roller, the weight starts moving up very fast, and then slows down, correct? If it's going up very fast and it gets slower, which way is the acceleration, then? The acceleration must be down: it's like I threw it up fast and it slowed down-like it's falling, sort of, so that the force must be reduced. That is, the horizontal force I'm going to
FIGURE 2-7 The simple machine, in motion.
get on the roller is going to be less than it would be if it weren't moving. So we have to figure out how much less. (The reason I went through all this is that I couldn't keep the signs right in the equations, so I had at the end to figure out which way the sign was by this physical argument.)
Incidentally, I have done this problem about four times-making a mistake every time-but I have, at last, got it right. I appreciate that when you do a problem the first time, there are many, many things that get confused: I got the numbers mixed up, I forgot to square, I put the sign of the time wrong, and I did a lot of other things wrong, but anyway, now I have it right, and I can show you how it can be done correctly-but I must admit, frankly, that it took me quite a while to get it right. (Boy, I'm glad I've still got my notes!)
Now, in order to calculate the force, we need the acceleration. It's impossible to find the acceleration by just looking at the diagram, with all dimensions fixed at the time of interest. To find the rate of change, we can't leave it fixed-I mean, we can't say, "Well, this is 0.3 , this is 0.4 , this is 0.5 , this is 2 meters per second, what's the acceleration?" There's no easy way to get at that. The only way to find the acceleration is to find the general motion and differentiate it with respect to time. Then we can put in the value of the time that corresponds to this particular diagram.
So I need, therefore, to analyze this thing in a more general circumstance, when the weight is at some arbitrary position. Let's say the pivot and the roller are together at time , and that the distance between them is , because the roller is moving at 2 meters per second. The time when we want to make the analysis is 0.3 seconds before they're together, which is , and so the distance between them is actually negative -but it'll be all right if we use and let the distance be . There will be a lot of signs wrong at the end, but because of my little fishing around at the beginning as to what the right sign was for the force, I'll be all right-I'd rather leave the mathematics alone and get the sign right from physics, than the other way around. Anyhow, here we are. (Don't you do this; it's too difficult-it takes practice!)
(Remember what the means: is the time before the pivots are together, which is sort of a negative time, which will make everybody crazy, but I can't help it-this is the way I did it.)
Now, the geometry is such that the weight is always (horizontally) halfway between the roller and the pivot. So, if we put the origin of our coordinate system at the pivot, then the coordinate of the weight is . The length of the rods is 0.5 , so for the height of the weight, its coordinate, I got , by the Pythagorean theorem. (See Fig. 2-8.) Can you imagine, the first time I worked this problem out, very carefully, I got ?
Now we need the acceleration, and the acceleration has two components: one is the horizontal acceleration, and the other is the vertical acceleration. If there's a horizontal acceleration, then there's a horizontal force, and we've got to chase that down through the rod and figure out what it is on the roller. This problem is a little easier than it looks because there is no horizontal acceleration-the coordinate of the weight is always half that of the roller; it moves in the same direction, but at half its speed. So, the weight moves horizontally at a constant 1 meter per second. There's no acceleration sideways, thank god! That makes the problem a little easier; we only have to worry about the up and down acceleration.
Therefore to get the acceleration, I must differentiate the height of the weight twice: once to get the velocity in the direction, and again, to get the acceleration. The height is . You should be able to differentiate this fast, and the answer is
It's negative, even though the weight is moving up. But I got my signs all bungled up, so I'll leave it this way; anyway, I know the speed is up, so this would be wrong if were positive, but should really be negative-so it's right anyway.
Now, we calculate the acceleration. There are several ways you can do this: You can do it using ordinary methods, but I'll use the new "super" method I showed you in Chapter 1: you write down again; then you say,
FIGURE 2-8 Using the Pythagorean Theorem to find the height of the weight.
"The first term that I want to differentiate is to the first power, . Derivative of is -1 . The next term that I want to differentiate is to the minus one-half power; the term is . The derivative is . Done!"
Now we have the acceleration at any time. In order to find the force, we need to multiply it by the mass. So, the force-that is, the extra force besides gravity that's involved because of the acceleration-is the mass, which is 2 kilograms, times this acceleration. Let's put the numbers into this thing: is 0.3 . The square root of is the square root of 0.25 minus 0.09 , which is 0.16 , the square root of which is 0.4 -well, how convenient! Is that right? Yes indeed, sir; this square root is the same as itself, and when is 0.3 , according to our diagram, is 0.4 . , no mistake.
(I'm always checking things while I calculate because I make so many mistakes. One way to check it is to do the mathematics very carefully; the other way to check it is to keep seeing whether the numbers that come out are sensible, whether they describe what's really happening.)
Now we calculate. (The first time I did this I put instead of 0.16 -it took me a while to find that one!) We get some number or other, which I have worked out; it's about 3.9.
So, the acceleration is 3.9 , and now for the force: the vertical force that this acceleration corresponds to is 3.9 times 2 kilograms times . No, that's not right! I forgot there's no now; 3.9 is the true acceleration. The vertical force of gravity is times the acceleration due to gravity, 9.8-that's -and the vertical component of the force of the rod on the weight is the sum of these two, with a minus sign for one; the relative signs are opposite. So, you subtract, and you get
But remember, now, this is the vertical force on the weight. How much is the horizontal force on the roller? The answer is, the horizontal force on the roller is three-quarters of one-half of the vertical force on the weight. We noticed that before: the force pulling down is balanced by the two legs, which divides it by two, and then the geometry is such that the ratio of the horizontal component to the vertical component is -and so the answer is that the horizontal force on the roller is three-eighths of the vertical force
on the weight. I worked out the three-eighths of each of these things, and I got 7.35 for gravity, and 2.925 for the inertial force, and the difference is 4.425 newtons-about 3 newtons less than the force required to support the weight in the same position when it was not moving. (See Fig. 2-9.)
Anyway, that's how you design machines; you know how much force you need to drive that thing forward.
Now, you say, is that the correct way do to it?
There is no such thing! There is no "correct" way to do anything. A particular way of doing it may be correct, but it is not the correct way. You can do it any damn way you want! (Well, excuse me: there are incorrect ways to do things . . . )
Now, if I were sufficiently smart, I could just look at this thing and tell you what the force is, but I'm not sufficiently smart, so I had to do it some way or other-but there are many ways of doing it. I will illustrate one other way, which is very useful, especially if you are involved in designing real machines. This problem is somewhat simplified by having the legs equal, and so on, because I didn't want to complicate the arithmetic. But the physical ideas are such that you can figure the whole thing out another way, even when the geometry is not so simple. And that is the following, interesting, other way.
When you have a whole lot of levers moving a lot of weights, you can do this: as you drive the thing along, and all the weights begin to move
FIGURE 2-9 Using similar triangles to find the force on the roller.
because of all the levers, you're doing a certain amount of work, . At any given time there's a certain power going in, which is the rate at which you are working, . At the same time, the energy of all the weights, , is changing at some rate, , and those should match each other; that is, the rate at which you put work in should match the rate of change of the total energy of all of the weights:
As you may recall from the lectures, power is equal to force times velocity:
And so, we have
The idea, then, is that at a given instant the weights have some kind of a speed, and thus they have a kinetic energy. They also have a certain height above the ground, and so they have a potential energy. So if we can figure out how fast the weights are moving and where they are, in order to get their total energy, and then we differentiate that with respect to time, that would be equal to the product of the component of force in the direction that the thing being worked on is moving, times its speed.
Let's see if we can apply that to our problem.
Now, when I push on the roller with a force while moving it at a velocity , the rate of change of the energy of the whole darn thing, with respect to time, should equal the magnitude of the force times the speed, , because in this case the force and the velocity are both in the same direction. It's not a general formula; if I had asked you for the force in some other direction, I couldn't have gotten it by this argument directly because this method only gives you the component of the force that does the work! (Of course, you can get it indirectly because you can know the force is going along the rod. If there were several more rods connected, this method would still work, provided you took the force in a direction of motion.)
What about all the work done by all the forces of the constraints-the roller, the pivots, and all the other machinery that holds this stuff in the right motion? No work is done by them, provided they aren't worked on by other forces as they go along. For example, if somebody else is sitting over there, pulling one leg out while I'm pushing the other one in, I've got to take the work done by the other guy into account! But nobody's doing that, so, with , we have
So I'm all set if I can calculate -divide by two, and lo and behold: the force!
Ready? Let's go!
Now, we have the total energy of the weight in two pieces: kinetic energy plus potential energy. Well, the potential energy is easy: it's mgy (see Table 2-3). We already know that is 0.4 meters, is , and is 9.8 meters per second squared. So the potential energy is joules. And now the kinetic energy: well, after a lot of fiddling around, I'll get the velocity of the weight, and I'll write in the kinetic energy for that; we'll do that in just a second. Then I'm all set because I'll have the total energy.
I'm not all set: unfortunately, I don't want the energy! I need the derivative of the energy with respect to time, and you cannot find how fast something changes by figuring out how much it is right now! You've either got to figure it out at two adjacent times-now, and an instant later-or, if you want to use the mathematical form, you figure it out for an arbitrary time, , and differentiate with respect to . It depends on which is the easiest to do: it may be numerically much easier to figure out the geometry for two positions than it is to figure out the geometry in general, and to differentiate.
(Most people immediately try to put a problem in mathematical form and differentiate it because they don't have enough experience with arithmetic to appreciate the tremendous power and ease of doing calculations with numbers instead of letters. Nevertheless, we'll do it with letters.)
Again, we have to solve this problem, where , and , so that we will be able to calculate the derivative.
Now, we need the potential energy. That we can get very easily: it's times the height, , and that comes out to
But more interesting, and harder to figure out, is the kinetic energy. The kinetic energy is . To figure out the kinetic energy, I need to figure out the velocity squared, and that takes a lot of fooling around: the velocity squared is its component squared plus its component squared. I could figure out the component just like I did before; the component, I've already pointed out, is 1 , and I could have squared those and added them together. But supposing I hadn't already done that, and I wanted to think of still another way to get the velocity.
Well, after thinking about it, a good machine designer usually can figure it out from the principles of geometry and the layout of the machinery. For example, since the pivot is stationary, the weight must move around it in a circle. So, in which direction must the velocity of the weight be? It can have no velocity parallel to the rod, because that would change the length of the rod, right? Therefore, the velocity vector is perpendicular to the rod. (See Fig. 2-10.)
You might say to yourself, "Ooh! I have to learn that trick!"
No. That trick is only good for a special kind of problem; it doesn't work most of the time. Very rarely do you happen to need the velocity of something that is rotating around a fixed point; there's no rule that says, "velocities are perpendicular to rods," or anything like that. You have to use common sense as often as possible. It's the general idea of analyzing the machine geometrically that's important here-not any specific rule.
So, now we know the direction of the velocity. The horizontal component of the velocity, we already know, is 1 , because it's half the speed of the roller. But look! The velocity is the hypotenuse of a right triangle that is
FIGURE 2-10 The weight moves in a circle, so its velocity is perpendicular to the rod.
similar to a triangle having the rod as its hypotenuse! To obtain the magnitude of the velocity is no harder than finding its proportion to its horizontal component, and we can get that proportion from the other triangle, which we already know all about. (See Fig. 2-11.)
Finally, for the kinetic energy we get
joules.
Now, for the signs: the kinetic energy is certainly positive, and the potential energy is positive because I measured the distance from the floor. So now I'm all right with the signs. So, the energy at any time is
Now, in order to find the force using this trick, we need to differentiate the energy and then we can divide by two and everything will be ready. (The apparent ease with which I do this is false: I swear I did it more than once before I got it right!)
Now, we differentiate the energy with respect to time. I'm not going to stall around with this; you're supposed to know how to differentiate by now. So there we are, with the answer for (which, incidentally, is twice the force required):
FIGURE 2-11 Using similar triangles to find the velocity of the weight.
So I'm all finished; I need merely put 0.3 in for , and I'm all done:
Now, let's see whether this makes sense. If there were no motion, and I didn't have to worry about the kinetic energy, then the total energy of the weight would just be its potential energy, and its derivative should be the force due to the weight. And sure enough, it comes out here the same as we calculated in Chapter 1, 2 times 9.8 times .
The sign of is negative, which must mean that the direction of the gravitational part of the force is opposite the direction of the kinetic part of the force. Anyhow, one is positive and the other is negative, which is all I want to know. I know which way the gravitational part of the force is: I've got to push on the roller to support the weight, so the kinetic part must reduce the force. You can put the numbers in, and sure enough, the force comes out to be the same as before:
In fact, this is why I had to do it so many times: after doing it the first time, and being completely satisfied with my wrong answer, I decided to try to do it another, completely different, way. After I did it the other way, I was satisfied with a completely different answer! When you work hard, there are moments when you think, "At last, I've discovered that mathematics is inconsistent!" But pretty soon you discover the error, as I finally did.
Anyway, that's just two ways of solving this problem. There's no unique way of doing any specific problem. By greater and greater ingenuity, you can find ways that require less and less work, but that takes experience.
2-8 Earth's escape velocity
I don't have much time left, but the next problem we'll talk about is something involving the motion of planets. I'll have to come back to it because I certainly can't tell you everything about it this time. The first problem is, what is the velocity required to leave the earth's surface? How fast does something have to move so that it can just escape from Earth's gravity?
Now, one way to work that out would be to calculate the motion under the force of gravity, but another way is by the conservation of energy. When the thing reaches way out there, infinitely far away, the kinetic energy will be zero, and the potential energy will be whatever it comes out for infinite distance. The formula for the gravitational potential is in Table 2-3; and it tells us that the potential energy, for particles that are infinitely distant, equals zero.
So, the total energy of something when it leaves Earth at escape velocity must be the same after the thing has gone an infinite distance and Earth's gravity has slowed it down to zero velocity (assuming there are no other forces involved). If is the mass of the earth, is the radius of the earth, and is the universal gravitational constant, we find that the square of the escape velocity must be .
Incidentally, the gravity constant, (the acceleration of gravity near the earth's surface) is because the law of force, for a mass, , is . In terms of the easier-to-remember gravity constant I can write . Now, is , and the radius of the earth is , so the earth's escape velocity is
So you have to go 11 kilometers per second to get out-which is pretty fast.
Next, I would talk about what happens if you are going 15 kilometers per second, and you're shooting past the earth at some distance.
Now, at 15 kilometers per second, the thing has enough energy to get out, going straight up. But is it obviously necessary that it gets out if it's not going straight up? Is it possible that the thing will go around and come back? That's not self-evident; it takes some thought. You say, "It has enough energy to get out," but how do you know? We didn't calculate the escape velocity for that direction. Could it be that the sideways acceleration due to Earth's gravity is enough to make it turn around? (See Fig. 2-12.)
It is possible, in principle. You know the law that you sweep out equal areas in equal times, so you know that when you get far out, you have to be moving sideways somehow or other. It's not clear that some of the motion that you need to escape isn't going sideways, so that even at 15 kilometers per second you don't escape.
Actually, it turns out that at 15 kilometers per second it does escape-it escapes as long as the velocity is greater than the escape velocity we computed above. As long as it can escape, it does escape-although that's not self-evident-and the next time, I'm going to try to show it. But to give you a hint as to how I'm going to show it, so you can play around with it yourself, it's the following.
We'll use the conservation of energy at two points, A and , at its shortest distance from Earth, , and at its longest distance from Earth, , as shown in Figure 2-13; the problem is to calculate . We know the total
This one gets out all right!
FIGURE 2-12 Does having the escape velocity guarantee escape?
energy of the thing at , and it's the same at because the energy is conserved, so if we knew the velocity at , we could calculate its potential energy, and thus . But we don't know the velocity at B!
Yet we do: from the law that equal areas are swept out in equal times, we know that the speed at must be lower than the speed at , in a certain proportion-in fact, it's to . Using that fact to get the speed at , we're able to find this distance in terms of , and we'll do that next time.
FIGURE 2-13 Satellite distance and velocity at perihelion and aphelion.
Alternate Solutions By Michael A. Gottlieb
Here are three more approaches to solving the machine design problem presented earlier in this chapter (Section 2-7), beginning on p. 39 .
A Finding the acceleration of the weight using geometry
But if we remember Kepler's law of equal areas, we know that in a given time the same area is swept out at the aphelion as is swept out at the perihelion: in a short time the particle at the perihelion moves a distance so the area swept out is about , while at the aphelion, where the particle moves , the area swept out is about . And so "equal areas" means that equals -which means that the velocities vary inversely as the radii. (See Fig. 3-4.)
That gives us, then, a formula for in terms of , which we can substitute in Eq. (3.2). Then we will have an equation to determine :
FIGURE 3-4 Using Kepler's law of equal areas to find the velocity of a satellite at aphelion.
Dividing by , and rearranging, we get
If you look at Eq. (3.5) a while, you could say, "Well, I can multiply by , and then it'll be a quadratic equation in ," or, if you prefer, you could look at it just the way it is, and solve the quadratic equation for -either way. The solution for is
I'm not going to discuss the algebra from here on; you know how to solve a quadratic equation, and there are two solutions for : one of them is equals , it turns out-and that's happy, because if you look at Eq. (3.2) you see it's obvious that if equals , the equation will match. (Of course, that doesn't mean that is .) With the other solution, we get a formula for in terms of , which is given here:
The question is whether we can write the formula in such a way that the relationship of to the escape velocity at the distance can readily be seen. Notice that by Eq. (3.1) 2 GM/a is the square of the escape velocity, and therefore we can write the formula this way:
That's the final result, and it is rather interesting. Suppose, first, that is less than the escape velocity. Under those circumstances, we'd expect the particle not to escape, so we should get a sensible value for . And sure enough, if is less than , then is greater than 1 , and the square is also greater than 1 ; taking away 1 , you get some nice positive number, and divided by that number tells us .
To check roughly how accurate our analysis is, a good thing to play around with is the numerical calculation we made of the orbit in the ninth lecture, to see how close the that we calculated then agrees with the we get from Eq. (3.8). Why should they not agree perfectly? Because, of course, the numerical method of integration treats time as little blobs instead of continuous, and therefore it isn't perfect.
Anyway, that's how we get when is less than . (Incidentally, knowing and knowing , we know the semi-major axis of the ellipse, and thus we could figure out the period of the orbit from Eq. (3.2), if we wanted to.)
But the interesting thing is this: suppose, first, that is exactly the velocity of escape. Then is 1 , and Eq. (3.8) says that then is infinite. That means that the orbit is not an ellipse; it means that the orbit goes off to infinity. (It can be shown that it is a parabola, in this special case.) So, it turns out, that if you're anywhere near a star or a planet, and no matter what direction you're moving, if you have the velocity of escape, you'll escape, all right-you won't get caught, even though you're not pointed in the right direction.
Still another question is, what happens if exceeds the velocity of escape? Then is less than 1 , and turns out negative-and that doesn't mean anything; there is no real . Physically, that solution looks more like this: with a very high velocity, much higher than the velocity of escape, a particle coming in is deflected-but its orbit is not an ellipse. It is, in fact, a hyperbola. So the orbits of objects moving around the sun are not only ellipses, as Kepler thought, but the generalization to higher speeds includes ellipses, parabolas, and hyperbolas. (We didn't prove here that they are ellipses, parabolas, or hyperbolas, but that's the answer to the problem.)
3-2 Discovery of the atomic nucleus
This hyperbolic orbit business is interesting, and has a very interesting historical application, which I'd like to show you; it is illustrated in Figure 3-5. We take the limiting case of an enormously high speed, and a relatively small force. That is, the object is going by so fast that in the first approximation it goes in a straight line. (See Fig. 3-5.)
Suppose we have a nucleus with charge (where is the electron charge), and a charged particle that is moving past it at a distance an ion of some kind (it was originally done with an alpha particle), it doesn't make any difference; you can put in your own case-let's take a proton of mass , velocity , and charge (for an alpha particle, it would be ). The proton doesn't go quite in a straight line, but is deflected through a very small angle. The question is, what's the angle? Now, I'm not going to do it exactly, but roughly-to get some idea of how the angle varies with . (I'll do it nonrelativistically, although it's just as easy to take relativity into account-just a minor change that you can figure out for yourself.) Of course, the bigger is, the smaller the angle ought to be. And the question is, does the angle decrease as the square of , the cube of , as , or what? We want to get some idea about this.
(This is, as a matter of fact, the way you start on any complicated or unfamiliar problem: you first get a rough idea; then you go back when you understand it better and do it more carefully.)
So the first rough analysis will run something like this: as the proton flies by, there are sideways forces on it from the nucleus-of course, there
nucleus
FIGURE 3-5 A high-speed proton is deflected by the electric field as it passes near the nucleus of an atom.
'Our stamp appears in the liner notes of Back TUVA Future, a CD featuring the Tuvan throat-singing master Ondar and a cameo appearance by Richard Feynman (Warner Bros. 9 47131-2), released in 1999.
Feynman Lectures on Computation, by Richard P. Feynman, edited by Anthony J.G. Hey and Robin W. Allen. 1996. Addison-Wesley, ISBN 0-201-48991-0.
The academic year at Caltech is divided into three terms; the first runs from late September to early December, the second from early January to early March, and the third from late March to early June.
The exercises in Chapter 5 of this volume include more than a dozen problems from Foster Strong's collection that were reproduced with permission in Exercises in Introductory Physics by Robert B. Leighton and Rochus E. Vogt.
Principles of Modern Physics, by Robert B. Leighton, 1959, McGraw-Hill, Library of Congress Catalog Card Number 58-8847.
All footnotes are comments from the authors (other than Feynman), editors, or contributors.
No one went out.
Only men were admitted to Caltech in 1961.
See Vol. I, Sec. 11-7.
is the speed of the particle; is the speed of light.
The relationship between the kinetic energy of a particle and its total (relativistic) energy can readily be seen by substituting the first two terms of the Taylor series expansion of into Eq. (2.5):
force is defined to be conservative when the total work it does on a particle that moves from one place to another is the same regardless of the path the particle moves on-the total work done depends only on the endpoints of the path. In particular, the work done by a conservative force on a particle that goes around a closed path, ending where it began, is always zero. See Vol. I, Section 14-3.
See Alternate Solutions on page 67 for a way to find the acceleration of the weight without differentiating.
See Vol. I, Chapter 13 .
The derivative of the total energy with respect to is the magnitude of the force due to the weight (in the direction). However, because happens to equal in this particular problem, the derivative of the weight's energy with respect to equals its derivative with respect to .
See Alternate Solutions, beginning on p. 67, for three other approaches to solving this problem.
