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2Laws and Intuition—Review Lecture B

(There was no summary for this lecture.)

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 is—but 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!

2–1The 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: \begin{equation} \label{Eq:TIPS:2:1} \FLPF = \ddt{\FLPp}{t}. \end{equation} That is, the force, $\FLPF$, is equal to the rate of change, with respect to time, of the momentum, $\FLPp$. ($\FLPF$ and $\FLPp$ 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 $\FLPp$, 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 $v$ squared over $c$ squared:”1

\begin{equation} \label{Eq:TIPS:2:2} \FLPp = \frac{m\FLPv}{\sqrt{1-{v^2}/{c^2}}}. \end{equation}

Instead, to understand physically what the $\FLPp$ stands for, you have to know that $\FLPp$ is not just “the momentum”; it’s the momentum of something—the momentum of a particle whose mass is $m$ and whose velocity is $\FLPv$. And, in Eq. (2.1), $\FLPF$ 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: \begin{equation} \label{Eq:TIPS:2:3} \sum_\text{particles} \FLPp_\text{after} = \sum_\text{particles} \FLPp_\text{before}. \end{equation}

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: \begin{equation} \label{Eq:TIPS:2:4} \sum_\text{particles} E_\text{after} = \sum_\text{particles} E_\text{before}. \end{equation}

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 $m$ and speed $v$ is \begin{equation} \label{Eq:TIPS:2:5} E = \frac{mc^2}{\sqrt{1-{v^2}/{c^2}}}. \end{equation}

2–2The 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: $\sqrt{1-{v^2}/{c^2}}$ is almost $1$, so Eq. (2.2) becomes \begin{equation} \label{Eq:TIPS:2:6} \FLPp = m\FLPv. \end{equation}

That means the formula for the force, $\FLPF = d\FLPp/dt$, can also be written $\FLPF = d(m\FLPv)/dt$ Then, by moving the constant, $m$, out in front, we see that for low velocities, the force equals the mass times the acceleration: \begin{equation} \label{Eq:TIPS:2:7} \FLPF = \ddt{\FLPp}{t} = \ddt{(m\FLPv)}{t} = m\ddt{\FLPv}{t} = m\FLPa. \end{equation}

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 $\FLPp = m\FLPv$ (and the masses are all constant): \begin{equation} \label{Eq:TIPS:2:8} \sum_\text{particles} (m\FLPv)_\text{after} = \sum_\text{particles} (m\FLPv)_\text{before}. \end{equation}

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 $\frac12mv^2\text{s}$ (the total kinetic energy, or K.E.) of all the particles is constant:2 \begin{equation} \begin{aligned} m_\text{after} &= m_\text{before}\\[1ex] \sum_\text{particles} ({\textstyle\frac12}mv^2)_\text{after} &= \sum_\text{particles} ({\textstyle\frac12}mv^2)_\text{before}\\ \end{aligned} \label{Eq:TIPS:2:9} \end{equation}

If we think of large, everyday objects as particles with low velocities—like 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 $\frac12mv^2\text{s}$ 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, it 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 kinetic energy needn’t be true when the objects involved are large, like weights and so on, that have inelastic collisions.

Table 2–1

True Always
False in general
(true only at low velocities)
Force $\displaystyle \FLPF = \ddt{\FLPp}{t}$ $\FLPF = m\FLPa$
Momentum $\displaystyle \FLPp = \frac{m\FLPv}{\sqrt{1-v^2/c^2}}$ $\FLPp = m\FLPv$
Energy $\displaystyle E = \frac{mc^2}{\sqrt{1-v^2/c^2}}$ $E = \frac12mv^2 (+mc^2)$

2–3Motion 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: \begin{equation} \label{Eq:TIPS:2:10} \Delta\,K\!.\!E. = \Delta W. \end{equation}

Remember, this means something—you have to know what all the letters mean: it means that if a particle is moving on some curve, $S$, from point $A$ to point $B$, and it’s moving under the influence of a force $\FLPF$, where $\FLPF$ is the total force acting on the particle, then if you knew what the $\frac12mv^2$ of the particle is at point $A$, and what it is over at point $B$, they differ by the integral, from $A$ to $B$, of $\FLPF\cdot d\FLPs$ where $d\FLPs$ is an increment of displacement along $S$. (See Fig. 2-1). \begin{equation} \label{Eq:TIPS:2:11} \Delta\,K\!.\!E. = {\textstyle\frac12}mv^2_\text{B} - {\textstyle\frac12}mv^2_\text{A} \end{equation} and \begin{equation} \label{Eq:TIPS:2:12} \Delta W = \int_\text{A}^\text{B} \FLPF \cdot d\FLPs. \end{equation}

Fig. 2–1.$\frac12mv_\text{B}^2 - \frac12mv_\text{A}^2 = \int_\text{A}^\text{B} \FLPF \cdot d\FLPs.$

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 in magnitude and opposite to the change in another quantity called its potential energy, or $P\!.\!E$. Such forces are said to be “conservative”: \begin{equation} \label{Eq:TIPS:2:13} \Delta W = -\Delta\,P\!.\!E.\hskip1ex(\text{with a }\textit{conservative}\text{ force, $\FLPF$)}. \end{equation} \begin{equation} \label{Eq:TIPS:2:13} \Delta W = -\Delta\,P\!.\!E.\\ (\text{with a }\textit{conservative}\text{ force, $\FLPF$)}. \end{equation}

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.3 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: \begin{equation} \label{Eq:TIPS:2:14} E = K\!.\!E+P\!.\!E. \end{equation}

When only conservative forces act, a particle’s total energy does not change: (with conservative forces). \begin{equation} \label{Eq:TIPS:2:15} \Delta E = \Delta\,K\!.\!E.+\,\Delta\,P\!.\!E.=0\hskip1ex(\text{with }\textit{conservative}\text{ forces}). \end{equation} \begin{equation} \label{Eq:TIPS:2:15} \Delta E = \Delta\,K\!.\!E.+\,\Delta\,P\!.\!E.=0\\ (\text{with }\textit{conservative}\text{ forces}). \end{equation}

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. \begin{equation} \label{Eq:TIPS:2:16} \Delta E = \Delta W\hskip1ex(\text{with }\textit{nonconservative}\text{ forces}). \end{equation} \begin{equation} \label{Eq:TIPS:2:16} \Delta E = \Delta W\\ (\text{with }\textit{nonconservative}\text{ forces}). \end{equation}

Table 2–2
True with conservative forces True with nonconservative forces
$\Delta\textit{P.E.} = -\Delta W$ $\textit{P.E. is undefined}$
$\Delta E = \textit{K.E.} + \textit{P.E.} = 0$ $\Delta E = \Delta W$
Definitions: $\textit{Kinetic Energy, K.E.}=\frac12mv^2$; $\textit{Work, } W=\int \FLPF\cdot d\FLPs$.

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 $r$ at constant speed $v$, then its acceleration is directed towards the center, and is equal in magnitude to $v^2/r$ (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:4 \begin{equation} \label{Eq:TIPS:2:17} |\FLPa| = \frac{v^2}r. \end{equation}

Fig. 2–2.Velocity and acceleration vectors for constant-speed circular motion.

2–4Forces 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 $-mg$ $mgz$
Gravity, between particles $-Gm_1m_2/r^2$ $-Gm_1m_2/r$
Electric Charge $q_1q_2/4\pi\epsilon_0r^2$ $q_1q_2/4\pi\epsilon_0r$
Electric Field $q\FLPE$ $q\phi$
Ideal Spring $-kx$ $\frac12kx^2$
Friction $-\mu N$ $\textit{No!}$

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 $z$ axis down! (You’re allowed to.) So the force is $-mg$, and potential energy is $mgz$, where $m$ is the mass of an object, $g$ is a constant (the acceleration of gravity at the surface of the earth—otherwise, the formula is no good!), and $z$ 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.

Next is gravity in space between particles; this force is directed centrally, and is proportional to the product of the one mass by the other mass divided by the distance between the two squared, $-mm'/r^2$, or $-m_1m_2/r^2$, or any other way you want to write it. It’s better to just remember which direction the force is, than to worry about the sign. But this part you’ve got to remember: the force of gravity goes as the inverse square of the distance between the particles. (So which way is the sign? Well, likes attract in gravity, so the force is in the opposite direction to the radius vector. That shows you that I don’t remember the sign; I just remember physically which way the sign is: the particles attract—that’s all I have to remember.)

Now, the potential energy between two particles is $-Gm_1m_2/r$. It’s hard for me to remember which way the potential energy goes. Let’s see: the particles lose potential energy when they come together, so that means when r is smaller, the potential energy should be less, so it’s negative—I hope that’s right! I have a great deal of difficulty with signs.

For electricity, the force is proportional to the product of the charges, $q_1$ and $q_2$, divided by the distance between them squared. But the constant of proportionality, instead of being written in the numerator (as with gravity), is written as $4\pi\epsilon_0$ in the denominator. The electrical force is directed radially, just like gravitation is, but with the opposite law of sign: likes repel, electrically, and therefore the sign of electrical potential energy is opposite that of gravitational potential energy, but then the constant of proportionality is different: $1/4\pi\epsilon_0$ instead of $G$.

Some technical points from the laws of electricity: the force on $q$ units of charge can be written as $q$ times the electric field, $q\FLPE$, and the energy can be written as $q$ times the electrical potential, $q\phi$. Here, $\FLPE$ is a vector field and $\phi$ is a scalar field. $q$ is measured in coulombs, and $\phi$ is measured in volts—when the energy is in the usual units of joules.

To continue this table of formulas, we have next an ideal spring. The force to pull out an ideal spring to a distance $x$ is a constant, $k$, times $x$. Now, you have to know what the letters mean again: $x$ is the distance that you pull the spring away from the equilibrium position, and the force pulls it back an amount $-kx$. I put the sign in just to say the spring pulls backwards; you know damn well a spring pulls a thing back, and doesn’t push it out further when you pull on it. Now, the potential energy is $\frac12kx^2$ In order to pull out a spring you do work on it, so after it’s pulled out, the potential energy is plus. So this sign business is easy—for the spring.

You see, details like the signs that I can’t remember, I try to reconstruct by arguments—that’s how I remember all the things I don’t remember.

Friction: the force of friction against a dry surface is $-\mu N$, and again you have to know what the symbols mean: when an object is pushed against another surface with a force whose component perpendicular to the surface is $N$, then in order to keep it sliding along the surface, the force required is $\mu$ times $N$. You can easily figure out which direction the force is; it’s opposite to the direction you slide it.

Now, under the potential energy for friction in Table 2-3, the answer is No: friction does not conserve energy, and therefore we have no formula for the potential energy for friction. If you push an object along a surface one way, you do work; then, when you drag it back, you do work again. So after you’ve gone through a complete cycle, you haven’t come out with no energy change; you’ve done work—and so friction has no potential energy.

2–5Learning physics by example

Those are all the rules I can remember as being necessary. So you say, “Well, that’s very easy: I’ll just memorize the whole damn table, and then I’ll know all the physics.” Well, it won’t work.

Actually, it might work fairly well at the beginning, but it gets harder and harder, as I pointed out in Chapter 1. Therefore, what we have to learn next is how to apply the mathematics to the physics in order to understand the world. The equations keep track of things for us, so we use them as tools—but to do that, we have to know what objects the equations are talking about.

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.

Fig. 2–3.Pythagorean triples in the Plimpton 322 tablet from about 1700 B.C.

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 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.

2–6Understanding physics physically

Fig. 2–4.The simple machine of Chapter 1.

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 $2$ kg, I believe. The geometrical relation of the rods was $0.3$, $0.4$, and $0.5$, and the problem was, what is the horizontal force $\FLPP$ 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 $\frac34$ kg, as shown in Figure 2-5.

Fig. 2–5.Distribution of force from the weight, through the rods, to the roller and pivot.

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 I can. Now, I have to teach you how to do that.

You could say, “Well, the force from the weight comes straight down, and it corresponds to $2$ kg, and the weight is balanced equally on two legs. So the vertical force from each leg must be enough to hold up $1$ kg. 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 $\frac34$ kg 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 waaaaay up there, the force on the roller should be low? Yeah! (See Fig. 2-6.)

Fig. 2–6.The force on the roller varies with the height of the weight.

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.

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.

2–7A 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 $2$ kg—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 $\textit{0.4}$ meters? (See Fig. 2-7.)

Fig. 2–7.The simple machine, in motion.

You might be thinking, “We did that already! The horizontal force required to balance the weight was $\frac34$ of a $1$ kg weight.”

But I argue, “The force is not $\frac34$ kg, 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 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.5 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 $t=0$, and that the distance between them is $2t$, 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 $t=-0.3$, and so the distance between them is actually negative $2t$—but it’ll be all right if we use $t=0.3$ and let the distance be $2t$. 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 $t$ means: $t$ 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 $x$ coordinate of the weight is $x=\frac12(2t)=t$. The length of the rods is $0.5$, so for the height of the weight, its $y$ coordinate, I got $y=\sqrt{0.25-t^2}$, by the Pythagorean theorem. (See Fig. 2-8.) Can you imagine, the first time I worked this problem out, very carefully, I got $y=\sqrt{0.25+t^2}$?

Fig. 2–8.Using the Pythagorean Theorem to find the height of the weight.

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 $x$ 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 $y$ direction, and again, to get the acceleration. The height is $y=\sqrt{0.25-t^2}$. You should be able to differentiate this fast, and the answer is \begin{equation} \label{Eq:TIPS:2:18} y' = \frac{-t}{\sqrt{0.25-t^2}} \end{equation}

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 $t$ were positive, but $t$ 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 $y'$ again; then you say, “The first term that I want to differentiate is to the first power, $-t$. Derivative of $-t$ is $-1$. The next term that I want to differentiate is to the minus one-half power; the term is $0.25-t^2$. The derivative is $-2t$. Done! \begin{equation} \begin{aligned} y' &= -t(0.25-t^2)^{-1/2}\\[1ex] y'' &= -t(0.25-t^2)^{-1/2}\bigg[1\cdot\frac{-1}{(-t)}-\frac12\cdot\frac{-2t}{(0.25-t^2)}\bigg]\\ \end{aligned} \label{Eq:TIPS:2:19} \end{equation} \begin{equation} \begin{aligned} y' &= -t(0.25-t^2)^{-1/2}\\[1ex] y'' &= -t(0.25-t^2)^{-1/2}\;\large\times\\ &\hspace2em\bigg[1\cdot\frac{-1}{(-t)}-\frac12\cdot\frac{-2t}{(0.25-t^2)}\bigg]\\ \end{aligned} \label{Eq:TIPS:2:19} \end{equation}

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: $t$ is $0.3$. The square root of $0.25-t^2$ 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 $y$ itself, and when $t$ is $0.3$, according to our diagram, $y$ is $0.4$. OK, 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 $0.25-t^2 = 0.4$ instead of $0.16$—it took me a while to find that one!) We get some number6 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 $g$. No, that’s not right! I forgot there’s no $g$ now; $3.9$ is the true acceleration. The vertical force of gravity is $2$ kg times the acceleration due to gravity, $9.8$—that’s $g$—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 \begin{equation} \label{Eq:TIPS:2:20} F_\text{w} = ma-mg = 7.8-19.6 = -11.8 \text{ newtons.} \end{equation} \begin{equation} \begin{aligned} F_\text{w} &= ma-mg = 7.8-19.6\\[1ex] &= -11.8 \text{ newtons.} \end{aligned} \label{Eq:TIPS:2:20} \end{equation}

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 $\frac34$—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 term due to the acceleration, 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.)

Fig. 2–9.Using similar triangles to find the force on the roller.

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 because of all the levers, you’re doing a certain amount of work, $W$. At any given time there’s a certain power going in, which is the rate at which you are working, $dW/dt$. At the same time, the energy of all the weights, $E$, is changing at some rate, $dE/dt$, 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: \begin{equation} \label{Eq:TIPS:2:21} \ddt{E}{t} = \ddt{W}{t}. \end{equation}

As you may recall from the lectures, power is equal to force times velocity:7 \begin{equation} \label{Eq:TIPS:2:22} \ddt{W}{t} =\frac{\FLPF\cdot d\FLPs}{dt} = \FLPF\cdot\ddt{\FLPs}{t} = \FLPF\cdot\FLPv. \end{equation}

And so, we have \begin{equation} \label{Eq:TIPS:2:23} \ddt{E}{t} = \FLPF\cdot\FLPv. \end{equation}

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 $\FLPP = -\FLPF_\text{R}$ while moving it at a velocity $\FLPv_\text{R}$, 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, $F_\text{R}v_\text{R}$ 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 $v_\text{R} = 2$, we have \begin{equation} \label{Eq:TIPS:2:24} \ddt{E}{t} = 2F_\text{R}. \end{equation} So I’m all set if I can calculate $dE/dt$—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 $y$ is $0.4$ meters, $m$ is $2$ kg, and $g$ is $9.8$ meters per second squared. So the potential energy is $2 \times 9.8 \times 0.4 = 7.84$ 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, $t$, and differentiate with respect to $t$. 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 $x=t$, and $y=\sqrt{0.25-t^2}$, so that we will be able to calculate the derivative.

Now, we need the potential energy. That we can get very easily: it’s $mg$ times the height, $y$, and that comes out to \begin{equation} \begin{aligned} \textit{P.E.} &= mgy = 2\text{ kg} \times 9.8 \text{ m}/\text{s}^2 \times \sqrt{0.25-t^2} \text{ m}\\[1ex] &= 19.6 \text{ newtons} \times \sqrt{0.25-t^2} \text{ m}\\[1ex] &= 19.6\sqrt{0.25-t^2} \text{ joules}. \end{aligned} \label{Eq:TIPS:2:25} \end{equation} \begin{equation} \begin{aligned} &\textit{P.E.} = mgy\\[1ex] &= 2\text{ kg} \times 9.8 \text{ m}/\text{s}^2\!\times\!\sqrt{0.25-t^2} \text{ m}\\[1ex] &= 19.6 \text{ newtons}\times\!\sqrt{0.25-t^2} \text{ m}\\[1ex] &= 19.6\sqrt{0.25-t^2} \text{ joules}. \end{aligned} \label{Eq:TIPS:2:25} \end{equation}

But more interesting, and harder to figure out, is the kinetic energy. The kinetic energy is $\frac12 mv^2$. 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 $x$ component squared plus its $y$ component squared. I could figure out the $y$ component just like I did before; the $x$ 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.)

Fig. 2–10.The weight moves in a circle, so its velocity is perpendicular to the rod.

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 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.)

Fig. 2–11.Using similar triangles to find the velocity of the weight.

Finally, for the kinetic energy we get \begin{equation} \label{Eq:TIPS:2:26} \textit{K.E.} = {\textstyle\frac12}mv^2 = \frac12 \times 2\text{ kg} \times \bigg(\frac{0.5}{\sqrt{0.25-t^2}}\text{ m/s}\bigg)^2 = \frac{1}{1-4t^2}\text{ joules}. \end{equation} \begin{equation} \begin{aligned} &\textit{K.E.} = {\textstyle\frac12}mv^2\\ &= \frac12 \times 2\text{ kg} \times \bigg(\frac{0.5}{\sqrt{0.25-t^2}}\text{ m/s}\bigg)^2\\ &= \frac{1}{1-4t^2}\text{ joules}. \end{aligned} \label{Eq:TIPS:2:26} \end{equation}

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 \begin{equation} \label{Eq:TIPS:2:27} E = \textit{K.E.} + \textit{P.E.} = \frac{1}{1-4t^2} + 19.6\sqrt{0.25-t^2}. \end{equation} \begin{equation} \begin{aligned} E &= \textit{K.E.} + \textit{P.E.}\\ & = \frac{1}{1-4t^2} + 19.6\sqrt{0.25-t^2}. \end{aligned} \label{Eq:TIPS:2:27} \end{equation}

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): \begin{equation} \label{Eq:TIPS:2:28} \ddt{E}{t} = \frac{8t}{(1-4t^2)^2} - \frac{19.6t}{\sqrt{0.25-t^2}}. \end{equation}

So I’m all finished; I need merely put $0.3$ in for $t$, and I’m all done. Well, not quite—to make the signs come out right, I have to use $t = -0.3$: \begin{equation} \begin{aligned} \ddt{E}{t}(-0.3) &= -\frac{2.4}{0.4096}\!+ 19.6 \times\!\frac{0.3}{0.4}.\\[1ex] &\approx 8.84 \text{ watts}. \end{aligned} \label{Eq:TIPS:2:29} \end{equation}

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.8 And sure enough, it comes out here the same as we calculated in Chapter 1, $2$ times $9.8$ times $\frac{3}{4}$.

The first term on the right-hand side of Eq. (2.29) is negative because the weight is decelerating, so it’s losing kinetic energy; the second term is positive because the weight is going up, so it’s gaining potential energy. Anyhow, they’re opposite each other, which is all I want to know, and you can put the numbers in, and sure enough, the force comes out to be the same as before: \begin{equation} \begin{aligned} 2F_\text{R} &= \ddt{E}{t} \approx 8.84\\[1ex] F_\text{R} &\approx 4.42 \text{ newtons}. \end{aligned} \label{Eq:TIPS:2:30} \end{equation}

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.9

2–8Earth’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 $M$ is the mass of the earth, $R$ is the radius of the earth, and $G$ is the universal gravitational constant, we find that the square of the escape velocity must be $2GM/R$.

$(\textit{K.E.}+\textit{P.E.})\text{ at }\infty,\;v\!=\!0$ $(\textit{K.E.}+\textit{P.E.})\text{ at }\infty$,
$v\!=\!0$
$=$ $(\textit{K.E.}+\textit{P.E.})\text{ at }R,\;v\!=\!v_\text{escape}$ $(\textit{K.E.}+\textit{P.E.})\text{ at }R$,
$v\!=\!v_\text{escape}$
(conservation of energy)
\begin{equation*} \begin{alignedat}{3} &\textit{P.E.}\text{ at }\infty\hskip1em &= -\frac{GMm}{\infty} &=0\\[1ex] &\textit{K.E.}\text{ at }v\!=\!0\hskip1em &=-\frac{~~m0^2~}{2} &=0 \end{alignedat} \end{equation*} \begin{equation*} \begin{alignedat}{3} &\textit{P.E.}\text{ at }\infty\\ &\hskip1em= -\frac{GMm}{\infty} &=0\\[1.5ex] &\textit{K.E.}\text{ at }v\!=\!0\\ &\hskip1em=-\frac{~~m0^2~}{2} &=0 \end{alignedat} \end{equation*} \begin{equation*} \begin{alignedat}{2} &\textit{P.E.}\text{ at }R &\hskip1em=-\frac{GMm}{R}\\[1ex] &\textit{K.E.}\text{ at }v\!=\!v_\text{escape}&\hskip1em=\frac{mv_\text{escape}^2}{2} \end{alignedat} \end{equation*} \begin{equation*} \begin{alignedat}{2} &\textit{P.E.}\text{ at }R\\ &\hskip1em=-\frac{GMm}{R}\\[1.5ex] &\textit{K.E.}\text{ at }v\!=\!v_\text{escape}\\ &\hskip1em=\frac{mv_\text{escape}^2}{2} \end{alignedat} \end{equation*}
$+$ $+$
$0$ $=$ \begin{equation*} \biggl(\!-\frac{GMm}{R}\!+\!\frac{mv_\text{escape}^2}{2}\!\biggr) \end{equation*}
\begin{equation} \label{Eq:TIPS:2:31} \therefore v^2_\text{escape} = \frac{2GM}{R}\:\:\: \end{equation}

Incidentally, the gravity constant, $g$ (the acceleration of gravity near the earth’s surface) is $GM/R^2$ because the law of force, for a mass, $m$, is $mg=GMm/R^2$. In terms of the easier-to-remember gravity constant I can write $v^2 = 2gR$. Now, $g$ is $9.8\text{ m}/\text{s}^2$, and the radius of the earth is $6400$ km, so the earth’s escape velocity is \begin{equation} \label{Eq:TIPS:2:32} v_\text{escape} = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 1000} = 11,200 \text{ m/s}. \end{equation} \begin{equation} \begin{aligned} v_\text{escape} &= \sqrt{2gR}\\[1ex] &= \sqrt{2 \times 9.8 \times 6400 \times 1000}\\[1ex] &= 11,200 \text{ m/s}. \end{aligned} \label{Eq:TIPS:2:32} \end{equation} 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.)

Fig. 2–12.Does having the escape velocity guarantee escape?

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.

Fig. 2–13.Satellite distance and velocity at perihelion and aphelion.

We’ll use the conservation of energy at two points, $A$ and $B$, at its shortest distance from Earth, $a$, and at its longest distance from Earth, $b$, as shown in Figure 2-13; the problem is to calculate $b$. We know the total energy of the thing at $A$, and it’s the same at $B$ because the energy is conserved, so if we knew the velocity at $B$, we could calculate its potential energy, and thus $b$. 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 $B$ must be lower than the speed at $A$, in a certain proportion—in fact, it’s $a$ to $b$. Using that fact to get the speed at $B$, we’re able to find this distance $b$ in terms of $a$, and we’ll do that next time.

2–9Alternate Solutions By Michael A. Gottlieb

Here are three more approaches to solving the machine design problem presented earlier in this chapter (Section 2-7).

AFinding the acceleration of the weight using geometry

The weight is always horizontally halfway between the roller and the pivot, so its horizontal speed is $1$ m/s, half the speed of the roller. The weight moves on a circle (centered at the pivot), so its velocity is perpendicular to the rod. By similar triangles we obtain the velocity of the weight. (See Fig. 2-14a.)

Fig. 2–14.

Because the weight moves on a circle, the radial component of its acceleration is \begin{equation*} a_\text{rad} = \frac{v^2}{r} = \frac{(1.25)^2}{0.5} = 3.125. \end{equation*} as per Eq. (2.17). The vertical acceleration of the weight is the sum of its radial and transverse components. (See Fig. 2-14b.)

Using similar triangles again, we obtain the vertical acceleration: \begin{equation*} a_y = \frac{a_y}{a_\text{rad}} \times a_\text{rad} = \frac{0.5}{0.4} \times 3.125 = 3.90625. \end{equation*}

BFinding the acceleration of the weight using trigonometry

The weight moves on a circular arc of radius $\frac12$, so its equations of motion can be expressed in terms of the angle the rods make with the ground. (See Fig. 2-15.) \begin{equation*} \begin{aligned} x &= {\textstyle\frac12}\cos\theta,\\[1ex] y &= {\textstyle\frac12}\sin\theta. \end{aligned} \end{equation*}

Fig. 2–15.

The horizontal speed of the weight is $1$ m/s (half the speed of the roller). So $x = t$, $dx/dt = 1$, and $d^2x/dt^2 = 0$. The vertical acceleration can be calculated by differentiating $y$ with respect to $t$ twice. But first, since $t = \frac12\cos\theta$, \begin{equation*} \ddt{\theta}{t} = -\frac{2}{\sin\theta}. \end{equation*}

Therefore, \begin{equation*} \begin{aligned} \ddt{y}{t} &= {\textstyle\frac12}\cos\theta \cdot \ddt{\theta}{t} = {\textstyle\frac12}\cos\theta \cdot\biggl(-\frac{2}{\sin\theta}\biggr) = -\cot\theta,\\[2ex] \frac{d^2y}{dt^2} &= \frac{1}{\sin^2\theta}\cdot \ddt{\theta}{t} = \frac{1}{\sin^2\theta}\cdot\biggl(-\frac{2}{\sin\theta}\biggr) = -\frac{2}{\sin^3\theta}. \end{aligned} \end{equation*} \begin{equation*} \begin{aligned} \ddt{y}{t} &= {\textstyle\frac12}\cos\theta \cdot \ddt{\theta}{t}\\[1ex] &= {\textstyle\frac12}\cos\theta \cdot\biggl(-\frac{2}{\sin\theta}\biggr) = -\cot\theta,\\[2ex] \frac{d^2y}{dt^2} &= \frac{1}{\sin^2\theta}\cdot \ddt{\theta}{t}\\[1ex] &= \frac{1}{\sin^2\theta}\cdot\biggl(-\frac{2}{\sin\theta}\biggr) = -\frac{2}{\sin^3\theta}. \end{aligned} \end{equation*}

When $x=t=0.3$, we have $y=0.4$ and $\sin\theta=0.8$ (since $y=\frac12\sin\theta$). The magnitude of the vertical acceleration is thus \begin{equation*} a_y = \biggl|\frac{d^2y}{dt^2}\biggr| = \frac{2}{(0.8)^3} = 3.90625. \end{equation*}

CFinding the force on the weight using torque and angular momentum

The torque on the weight is $\tau=xF_y-yF_x$. The weight moves horizontally at $1$ m/s, so there is no horizontal force on it: $F_x=0$. Letting $x=t$, the torque reduces to $\tau=tF_y$. Torque is the time derivative of angular momentum, so if we can find the angular momentum $L$ of the weight, we can differentiate it and divide by $t$ to get $F_y$: \begin{equation*} F_y = \frac{\tau}{t}=\frac{1}{t}\ddt{L}{t}. \end{equation*}

The angular momentum of the weight is easy to find because the weight moves in a circle. Its angular momentum is simply the length of the rod $r$, times the momentum of the weight, which is its mass $m$, times its speed $v$. The speed can be found using Feynman’s geometric method (see Fig. 2-16) or by differentiating the weight’s equations of motion.

Fig. 2–16.

Putting this all together we have: \begin{equation*} \begin{aligned} F_y &= \frac{1}{t}\ddt{L}{t} = \frac{1}{t}\ddt{}{t}(rmv) = \frac{rm}{t}\cdot\ddt{}{t}\biggl(\frac{0.5}{\sqrt{0.25-t^2}}\biggr)\\[1ex] &= \frac{0.5\cdot2}{t}\cdot\frac{0.5t}{(0.25-t^2)^{3/2}} = \frac{4}{(1-4t^2)^{3/2}}. \end{aligned} \end{equation*} \begin{equation*} \begin{aligned} F_y &= \frac{1}{t}\ddt{L}{t} = \frac{1}{t}\ddt{}{t}(rmv)\\[1ex] &= \frac{rm}{t}\cdot\ddt{}{t}\biggl(\frac{0.5}{\sqrt{0.25-t^2}}\biggr)\\[1ex] &= \frac{0.5\cdot2}{t}\cdot\frac{0.5t}{(0.25-t^2)^{3/2}} = \frac{4}{(1-4t^2)^{3/2}}. \end{aligned} \end{equation*}

At time $t=0.3$ we have $F_y=7.8125$. Dividing by $2$ kg gives the vertical acceleration we found before: $3.90625$.

  1. $v = |\FLPv|$ is the speed of the particle; $c$ is the speed of light.
  2. 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 $1/\sqrt{1-v^2/c^2}$ into Eq. (2.5): \begin{equation*} \begin{aligned} \frac{1}{\sqrt{1-x^2}} &= 1 + \frac12x^2 + \frac{1\cdot3}{2\cdot4}x^4 + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^6 + \dots\\[2ex] E &= \frac{mc^2}{\sqrt{1-v^2/c^2}} = mc^2(1+v^2/2c^2+\dots)\\[2ex] &\approx mc^2 + {\textstyle\frac12}mv^2 = \text{rest energy} + \textit{K.E. } \text{(for $v \ll c$)}. \end{aligned} \end{equation*}
  3. A 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 FLP Vol. I, Section 14–3.
  4. See FLP Vol. I, Section 11–6.
  5. See Subsection 2-9A Alternate Solution A for a way to find the acceleration of the weight without differentiating.
  6. The exact number is $3.90625$.
  7. See FLP Vol. I, Chapter 13.
  8. The derivative of the energy with respect to the position of the roller is the magnitude of the force on the roller. However, because the position of the roller happens to equal $2t$ in this particular problem, the derivative of the energy with respect to $t$ equals twice the force on the roller.
  9. See Section 2–9 Alternate Solutions for three other approaches to solving this problem.