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Forces transfer momentum to the girl.

# Chapter 3. Conservation of Momentum

I think, therefore I am.\par{}I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery. -- René Descartes

## 3.1 Momentum In One Dimension

a / Systems consisting of material particles that interact through an energy $$U(r)$$. Top: The galaxy M100. Here the “particles” are stars. Middle: The pool balls don't interact until they come together and become compressed; the energy $$U(r)$$ has a sharp upturn when the center-to-center distance $$r$$ gets small enough for the balls to be in contact. Bottom: A uranium nucleus undergoing fission. The energy $$U(r)$$ has a repulsive contribution from the electrical interactions of the protons, plus an attractive one due to the strong nuclear interaction. (M100: Hubble Space Telescope image.)

b / A collision between two pool balls is seen in two different frames of reference. The solid ball catches up with the striped ball. Velocities are shown with arrows. The second observer is moving to the left at velocity $$u$$ compared to the first observer, so all the velocities in the second frame have $$u$$ added onto them. The two observers must agree on conservation of energy.

c / The ion drive engine of the NASA Deep Space 1 probe, shown under construction (top) and being tested in a vacuum chamber (bottom) prior to its October 1998 launch. Intended mainly as a test vehicle for new technologies, the craft nevertheless also carried out a scientific program that included a rendezvous with a comet in 2004. (NASA)

### 3.1.1 Mechanical momentum

In the martial arts movie Crouching Tiger, Hidden Dragon, those who had received mystical enlightenment are able to violate the laws of physics. Some of the violations are obvious, such as their ability to fly, but others are a little more subtle. The rebellious young heroine/antiheroine Jen Yu gets into an argument while sitting at a table in a restaurant. A young tough, Iron Arm Lu, comes running toward her at full speed, and she puts up one arm and effortlessly makes him bounce back, without even getting out of her seat or bracing herself against anything. She does all this between bites.

Although kinetic energy doesn't depend on the direction of motion, we've already seen on page 89 how conservation of energy combined with Galilean relativity allows us to make some predictions about the direction of motion. One of the examples was a demonstration that it isn't possible for a hockey puck to spontaneously reverse its direction of motion. In the scene from the movie, however, the woman's assailant isn't just gliding through space. He's interacting with her, so the previous argument doesn't apply here, and we need to generalize it to more than one object. We consider the case of a physical system composed of pointlike material particles, in which every particle interacts with every other particle through an energy $$U(r)$$ that depends only on the distance $$r$$ between them. This still allows for a fairly general mechanical system, by which I mean roughly a system made of matter, not light. The characters in the movie are made of protons, neutrons, and electrons, so they would constitute such a system if the interactions among all these particles were of the form $$U(r)$$.1 We might even be able to get away with thinking of each person as one big particle, if it's a good approximation to say that every part of each person's whole body moves in the same direction at the same speed.

The basic insight can be extracted from the special case where there are only two particles interacting, and they only move in one dimension, as in the example shown in figure b. Conservation of energy says

$\begin{equation*} K_{1i}+K_{2i}+U_i = K_{1f}+K_{2f}+U_f . \end{equation*}$

For simplicity, let's assume that the interactions start after the time we're calling initial, and end before the instant we choose as final. This is true in figure b, for example. Then $$U_i=U_f$$, and we can subtract the interaction energies from both sides, giving.

\begin{align*} K_{1i}+K_{2i} &= K_{1f}+K_{2f} \\ \frac{1}{2}m_1v_{1i}^2+\frac{1}{2}m_2v_{2i}^2 &= \frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{2f}^2 . \end{align*}

As in the one-particle argument on page 89, the trick is to require conservation of energy not just in one particular frame of reference, but in every frame of reference. In a frame of reference moving at velocity $$u$$ relative to the first one, the velocities all have $$u$$ added onto them:2

$\begin{multline*} \frac{1}{2}m_1(v_{1i}+u)^2+\frac{1}{2}m_2(v_{2i}+u)^2 = \frac{1}{2}m_1(v_{1f}+u)^2+\frac{1}{2}m_2(v_{2f}+u)^2 \end{multline*}$

When we square a quantity like $$(v_{1i}+u)^2$$, we get the same $$v_{1i}^2$$ that occurred in the original frame of reference, plus two $$u$$-dependent terms, $$2v_{1i}u+u^2$$. Subtracting the original conservation of energy equation from the version in the new frame of reference, we have

\begin{align*} m_1v_{1i}u+m_2v_{2i}u = m_1v_{1f}u+m_2v_{2f}u , \\ \text{or, dividing by u,} m_1v_{1i}+m_2v_{2i} = m_1v_{1f}+m_2v_{2f} . \end{align*}

This is a statement that when you add up $$mv$$ for the whole system, that total remains constant over time. In other words, this is a conservation law. The quantity $$mv$$ is called momentum, notated $$p$$ for obscure historical reasons. Its units are $$\text{kg}\cdot\text{m}/\text{s}$$.

Unlike kinetic energy, momentum depends on the direction of motion, since the velocity is not squared. In one dimension, motion in the same direction as the positive $$x$$ axis is represented with positive values of $$v$$ and $$p$$. Motion in the opposite direction has negative $$v$$ and $$p$$.

##### Example 1: Jen Yu meets Iron Arm Lu

$$\triangleright$$ Initially, Jen Yu is at rest, and Iron Arm Lu is charging to the left, toward her, at 5 m/s. Jen Yu's mass is 50 kg, and Lu's is 100 kg. After the collision, the movie shows Jen Yu still at rest, and Lu rebounding at 5 m/s to the right. Is this consistent with the laws of physics, or would it be impossible in real life?

$$\triangleright$$ This is perfectly consistent with conservation of mass (50 kg+100 kg=50 kg+100 kg), and also with conservation of energy, since neither person's kinetic energy changes, and there is therefore no change in the total energy. (We don't have to worry about interaction energies, because the two points in time we're considering are ones at which the two people aren't interacting.) To analyze whether the scene violates conservation of momentum, we have to pick a coordinate system. Let's define positive as being to the right. The initial momentum is (50 kg)(0 m/s)+(100 kg)($$-$$5 m/s)=$$-$$500 $$\text{kg}\cdot\text{m}/\text{s}$$, and the final momentum is (50 kg)(0 m/s)+(100 kg)(5 m/s)=500 $$\text{kg}\cdot\text{m}/\text{s}$$. This is a change of 1000 $$\text{kg}\cdot\text{m}/\text{s}$$, which is impossible if the two people constitute a closed system.

One could argue that they're not a closed system, since Lu might be exchanging momentum with the floor, and Jen Yu might be exchanging momentum with the seat of her chair. This is a reasonable objection, but in the following section we'll see that there are physical reasons why, in this situation, the force of friction would be relatively weak, and would not be able to transfer that much momentum in a fraction of a second.

This example points to an intuitive interpretation of conservation of momentum, which is that interactions are always mutual. That is, Jen Yu can't change Lu's momentum without having her own momentum changed as well.

##### Example 2: A cannon

$$\triangleright$$ A cannon of mass 1000 kg fires a 10-kg shell at a velocity of 200 m/s. At what speed does the cannon recoil?

$$\triangleright$$ The law of conservation of momentum tells us that

$\begin{equation*} p_{cannon,i} + p_{shell,i} = p_{cannon,f} + p_{shell,f} . \end{equation*}$

Choosing a coordinate system in which the cannon points in the positive direction, the given information is

\begin{align*} p_{cannon,i} &= 0 \\ p_shell,i &= 0 \\ p_{shell,f} &= 2000\ \text{kg}\cdot\text{m/s} . \end{align*}

We must have $$p_{cannon,f}=-\text{2000 kg}\cdot\text{m/s}$$, so the recoil velocity of the cannon is 2 m/s.

##### Example 3: Ion drive

$$\triangleright$$ The experimental solar-powered ion drive of the Deep Space 1 space probe expels its xenon gas exhaust at a speed of 30,000 m/s, ten times faster than the exhaust velocity for a typical chemical-fuel rocket engine. Roughly how many times greater is the maximum speed this spacecraft can reach, compared with a chemical-fueled probe with the same mass of fuel (“reaction mass”) available for pushing out the back as exhaust?

$$\triangleright$$ Momentum equals mass multiplied by velocity. Both spacecraft are assumed to have the same amount of reaction mass, and the ion drive's exhaust has a velocity ten times greater, so the momentum of its exhaust is ten times greater. Before the engine starts firing, neither the probe nor the exhaust has any momentum, so the total momentum of the system is zero. By conservation of momentum, the total momentum must also be zero after all the exhaust has been expelled. If we define the positive direction as the direction the spacecraft is going, then the negative momentum of the exhaust is canceled by the positive momentum of the spacecraft. The ion drive allows a final speed that is ten times greater. (This simplified analysis ignores the fact that the reaction mass expelled later in the burn is not moving backward as fast, because of the forward speed of the already-moving spacecraft.)

d / Halley's comet. Top: A photograph made from earth. Bottom: A view of the nucleus from the Giotto space probe. (W. Liller and European Space Agency)

### 3.1.2 Nonmechanical momentum

So far, it sounds as though conservation of momentum can be proved mathematically, unlike conservation of mass and energy, which are entirely based on observations. The proof, however, was only for a mechanical system, with interactions of the form $$U(r)$$. Conservation of momentum can be extended to other systems as well, but this generalization is based on experiments, not mathematical proof. Light is the most important example of momentum that doesn't equal $$mv$$ --- light doesn't have mass at all, but it does have momentum. For example, a flashlight left on for an hour would absorb about $$10^{-5}\ \text{kg}\cdot\text{m}/\text{s}$$ of momentum as it recoiled.

##### Example 4: Halley's comet

Momentum is not always equal to $$mv$$. Halley's comet, shown in figure d, has a very elongated elliptical orbit, like those of many other comets. About once per century, its orbit brings it close to the sun. The comet's head, or nucleus, is composed of dirty ice, so the energy deposited by the intense sunlight gradually removes ice from the surface and turns it into water vapor. The bottom photo shows a view of the water coming off of the nucleus from the European Giotto space probe, which passed within 596 km of the comet's head on March 13, 1986.

The sunlight does not just carry energy, however. It also carries momentum. Once the steam comes off, the momentum of the sunlight impacting on it pushes it away from the sun, forming a tail as shown in in the top image. The tail always points away from the sun, so when the comet is receding from the sun, the tail is in front. By analogy with matter, for which momentum equals $$mv$$, you would expect that massless light would have zero momentum, but the equation $$p= mv$$ is not the correct one for light, and light does have momentum. (Some comets also have a second tail, which is propelled by electrical forces rather than by the momentum of sunlight.)

The reason for bringing this up is not so that you can plug numbers into formulas in these exotic situations. The point is that the conservation laws have proven so sturdy exactly because they can easily be amended to fit new circumstances. The momentum of light will be a natural consequence of the discussion of the theory of relativity in chapter 7.

e / Examples 5 and 6. The momenta cancel, but the energies don't.

f / Example 9.

### 3.1.3 Momentum compared to kinetic energy

Momentum and kinetic energy are both measures of the quantity of motion, and a sideshow in the Newton-Leibniz controversy over who invented calculus was an argument over whether $$mv$$ (i.e., momentum) or $$mv^2$$ (i.e., kinetic energy without the 1/2 in front) was the “true” measure of motion. The modern student can certainly be excused for wondering why we need both quantities, when their complementary nature was not evident to the greatest minds of the 1700s. The following table highlights their differences.

Kinetic energy... & Momentum...
\hline\hline doesn't depend on direction. & depends on direction.
\hline

is always positive, and cannot cancel out. & cancels with momentum in the opposite direction.
\hline

can be traded for forms of energy that do not involve motion. Kinetic energy is not a conserved quantity by itself. & is always conserved in a closed system.
\hline

is quadrupled if the velocity is doubled. & is doubled if the velocity is doubled.
\hline

Here are some examples that show the different behaviors of the two quantities.

##### Example 5: A spinning top
A spinning top has zero total momentum, because for every moving point, there is another point on the opposite side that cancels its momentum. It does, however, have kinetic energy.
##### Example 6: Why a tuning fork has two prongs
A tuning fork is made with two prongs so that they can vibrate in opposite directions, canceling their momenta. In a hypothetical version with only one prong, the momentum would have to oscillate, and this momentum would have to come from somewhere, such as the hand holding the fork. The result would be that vibrations would be transmitted to the hand and rapidly die out. In a two-prong fork, the two momenta cancel, but the energies don't.
##### Example 7: Momentum and kinetic energy in firing a rifle

The rifle and bullet have zero momentum and zero kinetic energy to start with. When the trigger is pulled, the bullet gains some momentum in the forward direction, but this is canceled by the rifle's backward momentum, so the total momentum is still zero. The kinetic energies of the gun and bullet are both positive numbers, however, and do not cancel. The total kinetic energy is allowed to increase, because kinetic energy is being traded for other forms of energy. Initially there is chemical energy in the gunpowder. This chemical energy is converted into heat, sound, and kinetic energy. The gun's “backward” kinetic energy does not refrigerate the shooter's shoulder!

##### Example 8: The wobbly earth

As the moon completes half a circle around the earth, its motion reverses direction. This does not involve any change in kinetic energy. The reversed velocity does, however, imply a reversed momentum, so conservation of momentum in the closed earth-moon system tells us that the earth must also reverse its momentum. In fact, the earth wobbles in a little “orbit” about a point below its surface on the line connecting it and the moon. The two bodies' momenta always point in opposite directions and cancel each other out.

##### Example 9: The earth and moon get a divorce
Why can't the moon suddenly decide to fly off one way and the earth the other way? It is not forbidden by conservation of momentum, because the moon's newly acquired momentum in one direction could be canceled out by the change in the momentum of the earth, supposing the earth headed off in the opposite direction at the appropriate, slower speed. The catastrophe is forbidden by conservation of energy, because their energies would have to increase greatly.
##### Example 10: Momentum and kinetic energy of a glacier

A cubic-kilometer glacier would have a mass of about $$10^{12}$$ kg. If it moves at a speed of $$10^{-5}$$ m/s, then its momentum is $$10^7\ \text{kg}\cdot\text{m}/\text{s}$$. This is the kind of heroic-scale result we expect, perhaps the equivalent of the space shuttle taking off, or all the cars in LA driving in the same direction at freeway speed. Its kinetic energy, however, is only 50 J, the equivalent of the calories contained in a poppy seed or the energy in a drop of gasoline too small to be seen without a microscope. The surprisingly small kinetic energy is because kinetic energy is proportional to the square of the velocity, and the square of a small number is an even smaller number.

##### Discussion Questions

If all the air molecules in the room settled down in a thin film on the floor, would that violate conservation of momentum as well as conservation of energy?

A refrigerator has coils in back that get hot, and heat is molecular motion. These moving molecules have both energy and momentum. Why doesn't the refrigerator need to be tied to the wall to keep it from recoiling from the momentum it loses out the back?

g / This Hubble Space Telescope photo shows a small galaxy (yellow blob in the lower right) that has collided with a larger galaxy (spiral near the center), producing a wave of star formation (blue track) due to the shock waves passing through the galaxies' clouds of gas. This is considered a collision in the physics sense, even though it is statistically certain that no star in either galaxy ever struck a star in the other --- the stars are very small compared to the distances between them. (NASA)

Gory Details of the Proof in Example 12 The equation $$A+B = C+D$$ says that the change in one ball's velocity is equal and opposite to the change in the other's. We invent a symbol $$x=C-A$$ for the change in ball 1's velocity. The second equation can then be rewritten as $$A^2+B^2 = (A+x)^2+(B-x)^2$$. Squaring out the quantities in parentheses and then simplifying, we get $$0 = Ax-Bx+x^2$$. The equation has the trivial solution $$x=0$$, i.e., neither ball's velocity is changed, but this is physically impossible because the balls can't travel through each other like ghosts. Assuming $$x\ne 0$$, we can divide by $$x$$ and solve for $$x=B-A$$. This means that ball 1 has gained an amount of velocity exactly sufficient to match ball 2's initial velocity, and vice-versa. The balls must have swapped velocities.}

### 3.1.4 Collisions in one dimension

Physicists employ the term “collision” in a broader sense than in ordinary usage, applying it to any situation where objects interact for a certain period of time. A bat hitting a baseball, a cosmic ray damaging DNA, and a gun and a bullet going their separate ways are all examples of collisions in this sense. Physical contact is not even required. A comet swinging past the sun on a hyperbolic orbit is considered to undergo a collision, even though it never touches the sun. All that matters is that the comet and the sun interacted gravitationally with each other.

The reason for broadening the term “collision” in this way is that all of these situations can be attacked mathematically using the same conservation laws in similar ways. In our first example, conservation of momentum is all that is required.

##### Example 11: Getting rear-ended

$$\triangleright$$ Ms.\ Chang is rear-ended at a stop light by Mr.\ Nelson, and sues to make him pay her medical bills. He testifies that he was only going 55 km per hour when he hit Ms.\ Chang. She thinks he was going much faster than that. The cars skidded together after the impact, and measurements of the length of the skid marks show that their joint velocity immediately after the impact was 30 km per hour. Mr.\ Nelson's Nissan has a mass of 1400 kg, and Ms.\ Chang 's Cadillac is 2400 kg. Is Mr.\ Nelson telling the truth?

$$\triangleright$$ Since the cars skidded together, we can write down the equation for conservation of momentum using only two velocities, $$v$$ for Mr.\ Nelson's velocity before the crash, and $$v'$$ for their joint velocity afterward:

$\begin{equation*} m_{N} v = m_{N} v' + m_{C} v' . \end{equation*}$

Solving for the unknown, $$v$$, we find

\begin{align*} v &= \left(1+\frac{ m_C}{ m_{N}}\right) v' \\ &= \text{80\ km/hr} . \end{align*}

He is lying.

The above example was simple because both cars had the same velocity afterward. In many one-dimensional collisions, however, the two objects do not stick. If we wish to predict the result of such a collision, conservation of momentum does not suffice, because both velocities after the collision are unknown, so we have one equation in two unknowns.

Conservation of energy can provide a second equation, but its application is not as straightforward, because kinetic energy is only the particular form of energy that has to do with motion. In many collisions, part of the kinetic energy that was present before the collision is used to create heat or sound, or to break the objects or permanently bend them. Cars, in fact, are carefully designed to crumple in a collision. Crumpling the car uses up energy, and that's good because the goal is to get rid of all that kinetic energy in a relatively safe and controlled way. At the opposite extreme, a superball is “super” because it emerges from a collision with almost all its original kinetic energy, having only stored it briefly as interatomic electrical energy while it was being squashed by the impact.

Collisions of the superball type, in which almost no kinetic energy is converted to other forms of energy, can thus be analyzed more thoroughly, because they have $$K_f=K_i$$, as opposed to the less useful inequality $$K_f\ltK_i$$ for a case like a tennis ball bouncing on grass.

##### Example 12: Pool balls colliding head-on
$$\triangleright$$ Two pool balls collide head-on, so that the collision is restricted to one dimension. Pool balls are constructed so as to lose as little kinetic energy as possible in a collision, so under the assumption that no kinetic energy is converted to any other form of energy, what can we predict about the results of such a collision?

$$\triangleright$$ Pool balls have identical masses, so we use the same symbol $$m$$ for both. Conservation of energy and no loss of kinetic energy give us the two equations

\begin{align*} mv_{1i}+mv_{2i} &= mv_{1f}+mv_{2f} \\ \frac{1}{2}mv_{1i}^2+\frac{1}{2}mv_{2i}^2 &= \frac{1}{2}mv_{1f}^2+\frac{1}{2}mv_{2f}^2 \end{align*}

The masses and the factors of 1/2 can be divided out, and we eliminate the cumbersome subscripts by replacing the symbols $$v_{1i}$$,... with the symbols $$A$$, $$B$$, $$C$$, and $$D$$:

\begin{align*} A+B &= C+D \\ A^2+B^2 &= C^2+D^2 . \end{align*}

A little experimentation with numbers shows that given values of $$A$$ and $$B$$, it is impossible to find $$C$$ and $$D$$ that satisfy these equations unless $$C$$ and $$D$$ equal $$A$$ and $$B$$, or $$C$$ and $$D$$ are the same as $$A$$ and $$B$$ but swapped around. A formal proof of this fact is given in the sidebar. In the special case where ball 2 is initially at rest, this tells us that ball 1 is stopped dead by the collision, and ball 2 heads off at the velocity originally possessed by ball 1. This behavior will be familiar to players of pool.

Often, as in example 12, the details of the algebra are the least interesting part of the problem, and considerable physical insight can be gained simply by counting the number of unknowns and comparing to the number of equations. Suppose a beginner at pool notices a case where her cue ball hits an initially stationary ball and stops dead. “Wow, what a good trick,” she thinks. “I bet I could never do that again in a million years.” But she tries again, and finds that she can't help doing it even if she doesn't want to. Luckily she has just learned about collisions in her physics course. Once she has written down the equations for conservation of energy and no loss of kinetic energy, she really doesn't have to complete the algebra. She knows that she has two equations in two unknowns, so there must be a well-defined solution. Once she has seen the result of one such collision, she knows that the same thing must happen every time. The same thing would happen with colliding marbles or croquet balls. It doesn't matter if the masses or velocities are different, because that just multiplies both equations by some constant factor.

#### The discovery of the neutron

This was the type of reasoning employed by James Chadwick in his 1932 discovery of the neutron. At the time, the atom was imagined to be made out of two types of fundamental particles, protons and electrons. The protons were far more massive, and clustered together in the atom's core, or nucleus. Electrical attraction caused the electrons to orbit the nucleus in circles, in much the same way that gravity kept the planets from cruising out of the solar system. Experiments showed, for example, that twice as much energy was required to strip the last electron off of a helium atom as was needed to remove the single electron from a hydrogen atom, and this was explained by saying that helium had two protons to hydrogen's one. The trouble was that according to this model, helium would have two electrons and two protons, giving it precisely twice the mass of a hydrogen atom with one of each. In fact, helium has about four times the mass of hydrogen.

Chadwick suspected that the helium nucleus possessed two additional particles of a new type, which did not participate in electrical interactions at all, i.e., were electrically neutral. If these particles had very nearly the same mass as protons, then the four-to-one mass ratio of helium and hydrogen could be explained. In 1930, a new type of radiation was discovered that seemed to fit this description. It was electrically neutral, and seemed to be coming from the nuclei of light elements that had been exposed to other types of radiation. At this time, however, reports of new types of particles were a dime a dozen, and most of them turned out to be either clusters made of previously known particles or else previously known particles with higher energies. Many physicists believed that the “new” particle that had attracted Chadwick's interest was really a previously known particle called a gamma ray, which was electrically neutral. Since gamma rays have no mass, Chadwick decided to try to determine the new particle's mass and see if it was nonzero and approximately equal to the mass of a proton.

h / Chadwick's subatomic pool table. A disk of the naturally occurring metal polonium provides a source of radiation capable of kicking neutrons out of the beryllium nuclei. The type of radiation emitted by the polonium is easily absorbed by a few mm of air, so the air has to be pumped out of the left-hand chamber. The neutrons, Chadwick's mystery particles, penetrate matter far more readily, and fly out through the wall and into the chamber on the right, which is filled with nitrogen or hydrogen gas. When a neutron collides with a nitrogen or hydrogen nucleus, it kicks it out of its atom at high speed, and this recoiling nucleus then rips apart thousands of other atoms of the gas. The result is an electrical pulse that can be detected in the wire on the right. Physicists had already calibrated this type of apparatus so that they could translate the strength of the electrical pulse into the velocity of the recoiling nucleus. The whole apparatus shown in the figure would fit in the palm of your hand, in dramatic contrast to today's giant particle accelerators.

Unfortunately a subatomic particle is not something you can just put on a scale and weigh. Chadwick came up with an ingenious solution. The masses of the nuclei of the various chemical elements were already known, and techniques had already been developed for measuring the speed of a rapidly moving nucleus. He therefore set out to bombard samples of selected elements with the mysterious new particles. When a direct, head-on collision occurred between a mystery particle and the nucleus of one of the target atoms, the nucleus would be knocked out of the atom, and he would measure its velocity.

Suppose, for instance, that we bombard a sample of hydrogen atoms with the mystery particles. Since the participants in the collision are fundamental particles, there is no way for kinetic energy to be converted into heat or any other form of energy, and Chadwick thus had two equations in three unknowns:

equation #1: conservation of momentum

equation #2: no loss of kinetic energy

unknown #1: mass of the mystery particle

unknown #2: initial velocity of the mystery particle

unknown #3: final velocity of the mystery particle

The number of unknowns is greater than the number of equations, so there is no unique solution. But by creating collisions with nuclei of another element, nitrogen, he gained two more equations at the expense of only one more unknown:

equation #3: conservation of momentum in the new collision

equation #4: no loss of kinetic energy in the new collision

unknown #4: final velocity of the mystery particle in the new collision

He was thus able to solve for all the unknowns, including the mass of the mystery particle, which was indeed within 1% of the mass of a proton. He named the new particle the neutron, since it is electrically neutral.

##### Discussion Questions

Good pool players learn to make the cue ball spin, which can cause it not to stop dead in a head-on collision with a stationary ball. If this does not violate the laws of physics, what hidden assumption was there in the example in the text where it was proved that the cue ball must stop?

i / The highjumper's body passes over the bar, but his center of mass passes under it. (Dunia Young)

j / Two pool balls collide.

l / No matter what point you hang the pear from, the string lines up with the pear's center of mass. The center of mass can therefore be defined as the intersection of all the lines made by hanging the pear in this way. Note that the X in the figure should not be interpreted as implying that the center of mass is on the surface --- it is actually inside the pear.

m / The circus performers hang with the ropes passing through their centers of mass.

n / An improperly balanced wheel has a center of mass that is not at its geometric center. When you get a new tire, the mechanic clamps little weights to the rim to balance the wheel.

o / This toy was intentionally designed so that the mushroom-shaped piece of metal on top would throw off the center of mass. When you wind it up, the mushroom spins, but the center of mass doesn't want to move, so the rest of the toy tends to counter the mushroom's motion, causing the whole thing to jump around.

q / Self-check A.

### 3.1.5 The center of mass

Figures i and k show two examples where a motion that appears complicated actually has a very simple feature. In both cases, there is a particular point, called the center of mass, whose motion is surprisingly simple. The highjumper flexes his body as he passes over the bar, so his motion is intrinsically very complicated, and yet his center of mass's motion is a simple parabola, just like the parabolic arc of a pointlike particle. The wrench's center of mass travels in a straight line as seen from above, which is what we'd expect for a pointlike particle flying through the air.

k / In this multiple-flash photograph, we see the wrench from above as it flies through the air, rotating as it goes. Its center of mass, marked with the black cross, travels along a straight line, unlike the other points on the wrench, which execute loops. (PSSC Physics)

The highjumper and the wrench are both complicated systems, each consisting of zillions of subatomic particles. To understand what's going on, let's instead look at a nice simple system, two pool balls colliding. We assume the balls are a closed system (i.e., their interaction with the felt surface is not important) and that their rotation is unimportant, so that we'll be able to treat each one as a single particle. By symmetry, the only place their center of mass can be is half-way in between, at an $$x$$ coordinate equal to the average of the two balls' positions, $$x_{cm}=(x_1+x_2)/2$$.

Figure j makes it appear that the center of mass, marked with an $$\times$$, moves with constant velocity to the right, regardless of the collision, and we can easily prove this using conservation of momentum:

\begin{align*} v_{cm} &= d{}x_{cm}/d{}t \\ &= \frac{1}{2}(v_1+v_2) \\ & = \frac{1}{2m}(mv_1+mv_2) \\ & = \frac{p_{total}}{m_{total}} \end{align*}

Since momentum is conserved, the last expression is constant, which proves that $$v_{cm}$$ is constant.

Rearranging this a little, we have $$p_{total}=m_{total}v_{cm}$$. In other words, the total momentum of the system is the same as if all its mass was concentrated at the center of mass point.

#### Sigma notation

When there is a large, potentially unknown number of particles, we can write sums like the ones occurring above using symbols like “$$+...$$,” but that gets awkward. It's more convenient to use the Greek uppercase sigma, $$\Sigma$$, to indicate addition. For example, the sum $$1^2+2^2+3^2+4^2=30$$ could be written as

$\begin{equation*} \sum_{j=1}^n{j^2} = 30 , \end{equation*}$

read “the sum from $$j=1$$ to $$n$$ of $$j^2$$.” The variable $$j$$ is a dummy variable, just like the $$d{}x$$ in an integral that tells you you're integrating with respect to $$x$$, but has no significance outside the integral. The $$j$$ below the sigma tells you what variable is changing from one term of the sum to the next, but $$j$$ has no significance outside the sum.

As an example, let's generalize the proof of $$p_{total}=m_{total}v_{cm}$$ to the case of an arbitrary number $$n$$ of identical particles moving in one dimension, rather than just two particles. The center of mass is at

\begin{align*} x_{cm} &= \frac{1}{n}\sum_{j=1}^{n}{x_j} , \end{align*}

where $$x_1$$ is the mass of the first particle, and so on. The velocity of the center of mass is

\begin{align*} v_{cm} &= d{}x_{cm}/d{}t \\ &= \frac{1}{n}\sum_{j=1}^{n}{v_j} \\ &= \frac{1}{nm}\sum_{j=1}^{n}{mv_j} \\ & = \frac{p_{total}}{m_{total}} \end{align*}

What about a system containing objects with unequal masses, or containing more than two objects? The reasoning above can be generalized to a weighted average:

\begin{align*} x_{cm} &= \frac{\sum_{j=1}^{n}{m_jx_j}}{\sum_{j=1}^{n}{m_j}} \end{align*}

##### Example 13: The solar system's center of mass

In the discussion of the sun's gravitational field on page 99, I mentioned in a footnote that the sun doesn't really stay in one place while the planets orbit around it. Actually, motion is relative, so it's meaningless to ask whether the sun is absolutely at rest, but it is meaningful to ask whether it moves in a straight line at constant velocity. We can now see that since the solar system is a closed system, its total momentum must be constant, and $$p_{total}= m_{total}v_{cm}$$ then tells us that it's the solar system's center of mass that has constant velocity, not the sun. The sun wobbles around this point irregularly due to its interactions with the planets, Jupiter in particular.

##### Example 14: The earth-moon system
The earth-moon system is much simpler than the solar system because it contains only two objects. Where is the center of mass of this system? Let $$x$$=0 be the earth's center, so that the moon lies at $$x=3.8\times10^5\ \text{km}$$. Then
\begin{align*} x_{cm} &= \frac{\sum_{ j=1}^{2}{ m_{j} x_{j}}} {\sum_{ j=1}^{2}{ m_j}} \\ &= \frac{ m_{1} x_{1}+ m_2 x_{2}} { m_{1}+ m_2} , \\ \text{and letting 1 be the earth and 2 the moon, we have} x_{cm} &= \frac{ m_{earth}\times0+ m_{moon} x_{moon}} { m_{earth}+ m_{moon}} \\ &= 4600\ \text{km} , \end{align*}
or about three quarters of the way from the earth's center to its surface.
##### Example 15: The center of mass as an average
$$\triangleright$$ Explain how we know that the center of mass of each object is at the location shown in figure p.

p / Example 15.

$$\triangleright$$ The center of mass is a sort of average, so the height of the centers of mass in 1 and 2 has to be midway between the two squares, because that height is the average of the heights of the two squares. Example 3 is a combination of examples 1 and 2, so we can find its center of mass by averaging the horizontal positions of their centers of mass. In example 4, each square has been skewed a little, but just as much mass has been moved up as down, so the average vertical position of the mass hasn't changed. Example 5 is clearly not all that different from example 4, the main difference being a slight clockwise rotation, so just as in example 4, the center of mass must be hanging in empty space, where there isn't actually any mass. Horizontally, the center of mass must be between the heels and toes, or else it wouldn't be possible to stand without tipping over.

##### Example 16: Momentum and Galilean relativity

The principle of Galilean relativity states that the laws of physics are supposed to be equally valid in all inertial frames of reference. If we first calculate some momenta in one frame of reference and find that momentum is conserved, and then rework the whole problem in some other frame of reference that is moving with respect to the first, the numerical values of the momenta will all be different. Even so, momentum will still be conserved. All that matters is that we work a single problem in one consistent frame of reference.

One way of proving this is to apply the equation $$p_{total}=m_{total}v_{cm}$$. If the velocity of one frame relative to the other is $$u$$, then the only effect of changing frames of reference is to change $$v_{cm}$$ from its original value to $$v_{cm}+u$$. This adds a constant onto the momentum, which has no effect on conservation of momentum.

self-check:

The figure shows a gymnast holding onto the inside of a big wheel. From inside the wheel, how could he make it roll one way or the other?

r / The same collision of two pools balls, but now seen in the center of mass frame of reference.

s / The sun's frame of reference.

t / The c.m. frame.

### 3.1.6 The center of mass frame of reference

A particularly useful frame of reference in many cases is the frame that moves along with the center of mass, called the center of mass (c.m.) frame. In this frame, the total momentum is zero. The following examples show how the center of mass frame can be a powerful tool for simplifying our understanding of collisions.

##### Example 17: A collision of pool balls viewed in the c.m. frame
If you move your head so that your eye is always above the point halfway in between the two pool balls, as in figure r, you are viewing things in the center of mass frame. In this frame, the balls come toward the center of mass at equal speeds. By symmetry, they must therefore recoil at equal speeds along the lines on which they entered. Since the balls have essentially swapped paths in the center of mass frame, the same must also be true in any other frame. This is the same result that required laborious algebra to prove previously without the concept of the center of mass frame.
##### Example 18: The slingshot effect
It is a counterintuitive fact that a spacecraft can pick up speed by swinging around a planet, if it arrives in the opposite direction compared to the planet's motion. Although there is no physical contact, we treat the encounter as a one-dimensional collision, and analyze it in the center of mass frame. Since Jupiter is so much more massive than the spacecraft, the center of mass is essentially fixed at Jupiter's center, and Jupiter has zero velocity in the center of mass frame, as shown in figure 3.1.6. The c.m. frame is moving to the left compared to the sun-fixed frame used in figure 3.1.6, so the spacecraft's initial velocity is greater in this frame than in the sun's frame.

Things are simpler in the center of mass frame, because it is more symmetric. In the sun-fixed frame, the incoming leg of the encounter is rapid, because the two bodies are rushing toward each other, while their separation on the outbound leg is more gradual, because Jupiter is trying to catch up. In the c.m. frame, Jupiter is sitting still, and there is perfect symmetry between the incoming and outgoing legs, so by symmetry we have $$v_{1f}=- v_{1i}$$. Going back to the sun-fixed frame, the spacecraft's final velocity is increased by the frames' motion relative to each other. In the sun-fixed frame, the spacecraft's velocity has increased greatly.

##### Example 19: Einstein's motorcycle

We've assumed we were dealing with a system of material objects, for which the equation $$p=mv$$ was true. What if our system contains only light rays, or a mixture of light and matter? As a college student, Einstein kept worrying about was what a beam of light would look like if you could ride alongside it on a motorcycle. In other words, he imagined putting himself in the light beam's center of mass frame. Chapter 7 discusses Einstein's resolution of this problem, but the basic point is that you can't ride the motorcycle alongside the light beam, because material objects can't go as fast as the speed of light. A beam of light has no center of mass frame of reference.

##### Discussion Questions

Make up a numerical example of two unequal masses moving in one dimension at constant velocity, and verify the equation $$p_{total}=m_{total}v_{cm}$$ over a time interval of one second.

A more massive tennis racquet or baseball bat makes the ball fly off faster. Explain why this is true, using the center of mass frame. For simplicity, assume that the racquet or bat is simply sitting still before the collision, and that the hitter's hands do not make any force large enough to have a significant effect over the short duration of the impact.

## 3.2 Force In One Dimension

a / Power and force are the rates at which energy and momentum are transferred.

b / The airbag increases $$\Delta t$$ so as to reduce $$F=\Delta p/\Delta t$$.

### 3.2.1 Momentum transfer

For every conserved quantity, we can define an associated rate of flow. An open system can have mass transferred in or out of it, and we can measure the rate of mass flow, $$d{}m/d{}t$$ in units of kg/s. Energy can flow in or out, and the rate of energy transfer is the power, $$P=d{}E/d{}t$$, measured in watts.3 The rate of momentum transfer is called force,

$\begin{equation*} F = \frac{d{}p}{d{}t} \text{[definition of force]} . \end{equation*}$

The units of force are $$\text{kg}\!\cdot\!\text{m}/\text{s}^2$$, which can be abbreviated as newtons, $$1\ \text{N}=\text{kg}\!\cdot\!\text{m}/\text{s}^2$$. Newtons are unfortunately not as familiar as watts. A newton is about how much force you'd use to pet a dog. The most powerful rocket engine ever built, the first stage of the Saturn V that sent astronauts to the moon, had a thrust of about 30 million newtons. In one dimension, positive and negative signs indicate the direction of the force --- a positive force is one that pushes or pulls in the direction of the positive $$x$$ axis.

##### Example 20: Walking into a lamppost

$$\triangleright$$ Starting from rest, you begin walking, bringing your momentum up to 100 $$\text{kg}\!\cdot\!\text{m}/\text{s}$$. You walk straight into a lamppost. Why is the momentum change of $$-100\ \text{kg}\!\cdot\!\text{m}/\text{s}$$ so much more painful than the change of $$+100\ \text{kg}\!\cdot\!\text{m}/\text{s}$$ when you started walking?

$$\triangleright$$ The forces are not really constant, but for this type of qualitative discussion we can pretend they are, and approximate $$d{} p/d{} t$$ as $$\Delta{} p/\Delta{} t$$. It probably takes you about 1 s to speed up initially, so the ground's force on you is $$F=\Delta{} p/\Delta{} t\approx100\ \text{N}$$. Your impact with the lamppost, however, is over in the blink of an eye, say 1/10 s or less. Dividing by this much smaller $$\Delta{} t$$ gives a much larger force, perhaps thousands of newtons (with a negative sign).

This is also the principle of airbags in cars. The time required for the airbag to decelerate your head is fairly long: the time it takes your face to travel 20 or 30 cm. Without an airbag, your face would have hit the dashboard, and the time interval would have been the much shorter time taken by your skull to move a couple of centimeters while your face compressed. Note that either way, the same amount of momentum is transferred: the entire momentum of your head.

Force is defined as a derivative, and the derivative of a sum is the sum of the derivatives. Therefore force is additive: when more than one force acts on an object, you add the forces to find out what happens. An important special case is that forces can cancel. Consider your body sitting in a chair as you read this book. Let the positive $$x$$ axis be upward. The chair's upward force on you is represented with a positive number, which cancels out with the earth's downward gravitational force, which is negative. The total rate of momentum transfer into your body is zero, and your body doesn't change its momentum.

##### Example 21: Finding momentum from force

$$\triangleright$$ An object of mass $$m$$ starts at rest at $$t=t_\text{o}$$. A force varying as $$F=bt^{-2}$$, where $$b$$ is a constant, begins acting on it. Find the greatest speed it will ever have.

$$\triangleright$$

\begin{align*} F &= \frac{dp}{dt} \\ dp &= F dt \\ p &= \int F dt + p_\text{o}\\ &= -\frac{b}{t}+p_\text{o} , \end{align*}

where $$p_\text{o}$$ is a constant of integration. The given initial condition is that $$p=0$$ at $$t=t_\text{o}$$, so we find that $$p_\text{o}=b/t_\text{o}$$. The negative term gets closer to zero with increasing time, so the maximum momentum is achieved by letting $$t$$ approach infinity. That is, the object will never stop speeding up, but it will also never surpass a certain speed. In the limit $$t\rightarrow\infty$$, we identify $$p_\text{o}$$ as the momentum that the object will approach asymptotically. The maximum velocity is $$v=p_\text{o}/m=b/mt_\text{o}$$.

##### Discussion Question

Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities:

(1) The collision is instantaneous.
(2) The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
(3) The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.

How can two of these be ruled out based on energy or momentum considerations?

c / Isaac Newton (1643-1727).

d / Two magnets exert forces on each other.

e / It doesn't make sense for the man to talk about the woman's money canceling out his bar tab, because there is no good reason to combine his debts and her assets.

f / Newton's third law does not mean that forces always cancel out so that nothing can ever move. If these two ice skaters, initially at rest, push against each other, they will both move.

g / A swimmer doing the breast stroke pushes backward against the water. By Newton's third law, the water pushes forward on him.

h / A coin slides across a table. Even for motion in one dimension, some of the forces may not lie along the line of the motion.

 x (m) t (s) 10 1.84 20 2.86 30 3.80 40 4.67 50 5.53 60 6.38 70 7.23 80 8.10 90 8.96 100 9.83

Discussion question B. }

### 3.2.2 Newton's laws

Although momentum is the third conserved quantity we've encountered, historically it was the first to be discovered. Isaac Newton formulated a complete treatment of mechanical systems in terms of force and momentum. Newton's theory was based on three laws of motion, which we now think of as consequences of conservation of mass, energy, and momentum.

 Newton’s laws in one dimension: Newton’s first law: If there is no force acting on an object, it stays in the same state of motion. Newton’s second law: The sum of all the forces acting on an object determines the rate at which its momentum changes,Ftotalderpdert. Newton’s third law: Forces occur in opposite pairs. If object A interacts with object B, then A’s force on B and B’s force on A are related byFAB = − FBA.

{}The second law is the definition of force, which we've already encountered.4 The first law is a special case of the second law --- if $$d{}p/d{}t$$ is zero, then $$p=mv$$ is a constant, and since mass is conserved, constant $$p$$ implies constant $$v$$. The third law is a restatement of conservation of momentum: for two objects interacting, we have constant total momentum, so $$0=\frac{d}{d{}t}(p_A+p_B) =F_{BA}+F_{AB}$$.

##### Example 22: a=F/m

Many modern textbooks restate Newton's second law as $$a= F/ m$$, i.e., as an equation that predicts an object's acceleration based on the force exerted on it. This is easily derived from Newton's original form as follows: $$a=\text{d} v/\text{d} t=(\text{d} p/\text{d} t)/ m= F/ m$$.

##### Example 23: Gravitational force related to g

As a special case of the previous example, consider an object in free fall, and let the $$x$$ axis point down. Then $$a=+g$$, and $$F= ma= mg$$. For example, the gravitational force on a 1 kg mass at the earth's surface is about 9.8 N. Even if other forces act on the object, and it isn't in free fall, the gravitational force on it is still the same, and can still be calculated as $$mg$$.

##### Example 24: Changing frames of reference

Suppose we change from one frame reference into another, which is moving relative to the first one at a constant velocity $$u$$. If an object of mass $$m$$ is moving at velocity $$v$$ (which need not be constant), then the effect is to change its momentum from $$mv$$ in one frame to $$mv+mu$$ in the other. Force is defined as the derivative of momentum with respect to time, and the derivative of a constant is zero, so adding the constant $$mu$$ has no effect on the result. We therefore conclude that observers in different inertial frames of reference agree on forces.

#### Using the third law correctly

If you've already accepted Galilean relativity in your heart, then there is nothing really difficult about the first and second laws. The third law, however, is more of a conceptual challenge. The first hurdle is that it is counterintuitive. Is it really true that if a fighter jet collides with a mosquito, the mosquito's force on the jet is just as strong as the jet's force on the mosquito? Yes, it is true, but it is hard to believe at first. That amount of force simply has more of an effect on the mosquito, because it has less mass.

A more humane and practical experiment is shown in figure d. A large magnet and a small magnet are weighed separately, and then one magnet is hung from the pan of the top balance so that it is directly above the other magnet. There is an attraction between the two magnets, causing the reading on the top scale to increase and the reading on the bottom scale to decrease. The large magnet is more “powerful” in the sense that it can pick up a heavier paperclip from the same distance, so many people have a strong expectation that one scale's reading will change by a far different amount than the other. Instead, we find that the two changes are equal in magnitude but opposite in direction, so the upward force of the top magnet on the bottom magnet is of the same magnitude as the downward force of the bottom magnet on the top magnet.

To students, it often sounds as though Newton's third law implies nothing could ever change its motion, since the two equal and opposite forces would always cancel. As illustrated in figure e, the fallacy arises from assuming that we can add things that it doesn't make sense to add. It only makes sense to add up forces that are acting on the same object, whereas two forces related to each other by Newton's third law are always acting on two different objects. If two objects are interacting via a force and no other forces are involved, then both objects will accelerate --- in opposite directions, as shown in figure f!

Here are some suggestions for avoiding misapplication of Newton's third law:

1. It always relates exactly two forces, not more.
2. The two forces involve exactly two objects, in the pattern A on B, B on A.
3. The two forces are always of the same type, e.g., friction and friction, or gravity and gravity.

#### Directions of forces

We've already seen that momentum, unlike energy, has a direction in space. Since force is defined in terms of momentum, force also has a direction in space. For motion in one dimension, we have to pick a coordinate system, and given that choice, forces and momenta will be positive or negative. We've already used signs to represent directions of forces in Newton's third law, $$F_{AB}=-F_{BA}$$.

There is, however, a complication with force that we were able to avoid with momentum. If an object is moving on a line, we're guaranteed that its momentum is in one of two directions: the two directions along the line. But even an object that stays on a line may still be subject to forces that act perpendicularly to the line. For example, suppose a coin is sliding to the right across a table, h, and let's choose a positive $$x$$ axis that points to the right. The coin's motion is along a horizontal line, and its momentum is positive and decreasing. Because the momentum is decreasing, its time derivative $$dp/dt$$ is negative. This derivative equals the horizontal force of friction $$F_1$$, and its negative sign tells us that this force on the coin is to the left.

But there are also vertical forces on the coin. The Earth exerts a downward gravitational force $$F_2$$ on it, and the table makes an upward force $$F_3$$ that prevents the coin from sinking into the wood. In fact, without these vertical forces the horizontal frictional force wouldn't exist: surfaces don't exert friction against one another unless they are being pressed together.

To avoid mathematical complication, we want to postpone the full three-dimensional treatment of force and momentum until section 3.4. For now, we'll limit ourselves to examples like the coin, in which the motion is confined to a line, and any forces perpendicular to the line cancel each other out.

##### Discussion Questions

Criticize the following incorrect statement:
“If an object is at rest and the total force on it is zero, it stays at rest. There can also be cases where an object is moving and keeps on moving without having any total force on it, but that can only happen when there's no friction, like in outer space.”

The table gives laser timing data for Ben Johnson's 100 m dash at the 1987 World Championship in Rome. (His world record was later revoked because he tested positive for steroids.) How does the total force on him change over the duration of the race?

You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “According to Newton's third law, there has to be a force opposite to your force on the ball. The opposite force is the ball's mass, which resists acceleration, and also air resistance.”

Tam Anh grabs Sarah by the hand and tries to pull her. She tries to remain standing without moving. A student analyzes the situation as follows. “If Tam Anh's force on Sarah is greater than her force on him, he can get her to move. Otherwise, she'll be able to stay where she is.” What's wrong with this analysis?

### 3.2.3 What force is not

Violin teachers have to endure their beginning students' screeching. A frown appears on the woodwind teacher's face as she watches her student take a breath with an expansion of his ribcage but none in his belly. What makes physics teachers cringe is their students' verbal statements about forces. Below I have listed several dicta about what force is not.

#### Force is not a property of one object.

A great many of students' incorrect descriptions of forces could be cured by keeping in mind that a force is an interaction of two objects, not a property of one object.

Incorrect statement: “That magnet has a lot of force.”

If the magnet is one millimeter away from a steel ball bearing, they may exert a very strong attraction on each other, but if they were a meter apart, the force would be virtually undetectable. The magnet's strength can be rated using certain electrical units $$(\text{ampere}-\text{meters}^2)$$, but not in units of force.

#### Force is not a measure of an object's motion.

If force is not a property of a single object, then it cannot be used as a measure of the object's motion.

Incorrect statement: “The freight train rumbled down the tracks with awesome force.”

Force is not a measure of motion. If the freight train collides with a stalled cement truck, then some awesome forces will occur, but if it hits a fly the force will be small.

#### Force is not energy.

Incorrect statement: “How can my chair be making an upward force on my rear end? It has no power!”

Power is a concept related to energy, e.g., a 100-watt lightbulb uses up 100 joules per second of energy. When you sit in a chair, no energy is used up, so forces can exist between you and the chair without any need for a source of power.

#### Force is not stored or used up.

Because energy can be stored and used up, people think force also can be stored or used up.

Incorrect statement: “If you don't fill up your tank with gas, you'll run out of force.”

Energy is what you'll run out of, not force.

#### Forces need not be exerted by living things or machines.

Transforming energy from one form into another usually requires some kind of living or mechanical mechanism. The concept is not applicable to forces, which are an interaction between objects, not a thing to be transferred or transformed.

Incorrect statement: “How can a wooden bench be making an upward force on my rear end? It doesn't have any springs or anything inside it.”

No springs or other internal mechanisms are required. If the bench didn't make any force on you, you would obey Newton's second law and fall through it. Evidently it does make a force on you!

#### A force is the direct cause of a change in motion.

I can click a remote control to make my garage door change from being at rest to being in motion. My finger's force on the button, however, was not the force that acted on the door. When we speak of a force on an object in physics, we are talking about a force that acts directly. Similarly, when you pull a reluctant dog along by its leash, the leash and the dog are making forces on each other, not your hand and the dog. The dog is not even touching your hand.

self-check:

Which of the following things can be correctly described in terms of force?

(1) A nuclear submarine is charging ahead at full steam.

(2) A nuclear submarine's propellers spin in the water.

(3) A nuclear submarine needs to refuel its reactor periodically.

##### Discussion Questions

Criticize the following incorrect statement: “If you shove a book across a table, friction takes away more and more of its force, until finally it stops.”

You hit a tennis ball against a wall. Explain any and all incorrect ideas in the following description of the physics involved: “The ball gets some force from you when you hit it, and when it hits the wall, it loses part of that force, so it doesn't bounce back as fast. The muscles in your arm are the only things that a force can come from.”

i / Static friction: the tray doesn't slip on the waiter's fingers.

j / Kinetic friction: the car skids.

### 3.2.4 Forces between solids

Conservation laws are more fundamental than Newton's laws, and they apply where Newton's laws don't, e.g., to light and to the internal structure of atoms. However, there are certain problems that are much easier to solve using Newton's laws. As a trivial example, if you drop a rock, it could conserve momentum and energy by levitating, or by falling in the usual manner.5 With Newton's laws, however, we can reason that $$a=F/m$$, so the rock must respond to the gravitational force by accelerating.

Less trivially, suppose a person is hanging onto a rope, and we want to know if she will slip. Unlike the case of the levitating rock, here the no-motion solution could be perfectly reasonable if her grip is strong enough. We know that her hand's interaction with the rope is fundamentally an electrical interaction between the atoms in the surface of her palm and the nearby atoms in the surface of the rope. For practical problem-solving, however, this is a case where we're better off forgetting the fundamental classification of interactions at the atomic level and working with a more practical, everyday classification of forces. In this practical scheme, we have three types of forces that can occur between solid objects in contact:

 A normal force,Fn, is perpendicular to the surface of contact, and prevents objects from passing through each other by becoming as strong as necessary (up to the point where the objects break). “Normal” means perpendicular. Static friction,Fs, is parallel to the surface of contact, and prevents the surfaces from starting to slip by becoming as strong as necessary, up to a maximum value ofFs,max. “Static” means not moving, i.e., not slipping. Kinetic friction,Fk, is parallel to the surface of contact, and tends to slow down any slippage once it starts. “Kinetic” means moving, i.e., slipping.
self-check:

Can a frictionless surface exert a normal force? Can a frictional force exist without a normal force?

If you put a coin on this page, which is horizontal, gravity pulls down on the coin, but the atoms in the paper and the coin repel each other electrically, and the atoms are compressed until the repulsion becomes strong enough to stop the downward motion of the coin. We describe this complicated and invisible atomic process by saying that the paper makes an upward normal force on the coin, and the coin makes a downward normal force on the paper. (The two normal forces are related by Newton's third law. In fact, Newton's third law only relates forces that are of the same type.)

If you now tilt the book a little, static friction keeps the coin from slipping. The picture at the microscopic level is even more complicated than the previous description of the normal force. One model is to think of the tiny bumps and depressions in the coin as settling into the similar irregularities in the paper. This model predicts that rougher surfaces should have more friction, which is sometimes true but not always. Two very smooth, clean glass surfaces or very well finished machined metal surfaces may actually stick better than rougher surfaces would, the probable explanation being that there is some kind of chemical bonding going on, and the smoother surfaces allow more atoms to be in contact.

Finally, as you tilt the book more and more, there comes a point where static friction reaches its maximum value. The surfaces become unstuck, and the coin begins to slide over the paper. Kinetic friction slows down this slipping motion significantly. In terms of energy, kinetic friction is converting mechanical energy into heat, just like when you rub your hands together to keep warm. One model of kinetic friction is that the tiny irregularities in the two surfaces bump against each other, causing vibrations whose energy rapidly converts to heat and sound --- you can hear this sound if you rub your fingers together near your ear.

For dry surfaces, experiments show that the following equations usually work fairly well:

$\begin{equation*} F_{s,max} \approx \mu_sF_n , \end{equation*}$

and

$\begin{equation*} F_k \approx \mu_kF_n , \end{equation*}$

where $$\mu_s$$, the coefficient of static friction, and $$\mu_k$$, the coefficient of kinetic friction, are constants that depend on the properties of the two surfaces, such as what they're made of and how rough they are.

self-check:

1. When a baseball player slides in to a base, is the friction static, or kinetic?

2. A mattress stays on the roof of a slowly accelerating car. Is the friction static, or kinetic?

3. Does static friction create heat? Kinetic friction?

##### Example 25: Maximum acceleration of a car

$$\triangleright$$ Rubber on asphalt gives $$\mu_k\approx0.4$$ and $$\mu_s\approx 0.6$$. What is the upper limit on a car's acceleration on a flat road, assuming that the engine has plenty of power and that air friction is negligible?

$$\triangleright$$ This isn't a flying car, so we don't expect it to accelerate vertically. The vertical forces acting on the car should cancel out. The earth makes a downward gravitational force on the car whose absolute value is $$mg$$, so the road apparently makes an upward normal force of the same magnitude, $$F_n= mg$$.

Now what about the horizontal motion? As is always true, the coefficient of static friction is greater than the coefficient of kinetic friction, so the maximum acceleration is obtained with static friction, i.e., the driver should try not to burn rubber. The maximum force of static friction is $$F_{s,max}=\mu_{s} F_{n} =\mu_s mg$$. The maximum acceleration is $$a= F_s/ m=\mu_{s} g\approx6\ \text{m}/\text{s}^2$$. This is true regardless of how big the tires are, since the experimentally determined relationship $$F_{s,max}=\mu_{s} F_n$$ is independent of surface area.

self-check:

Find the direction of each of the forces in figure k.

k / 1. The cliff's normal force on the climber's feet. 2. The track's static frictional force on the wheel of the accelerating dragster. 3. The ball's normal force on the bat.

##### Example 26: Locomotives
Looking at a picture of a locomotive, l, we notice two obvious things that are different from an automobile. Where a car typically has two drive wheels, a locomotive normally has many --- ten in this example. (Some also have smaller, unpowered wheels in front of and behind the drive wheels, but this example doesn't.) Also, cars these days are generally built to be as light as possible for their size, whereas locomotives are very massive, and no effort seems to be made to keep their weight low. (The steam locomotive in the photo is from about 1900, but this is true even for modern diesel and electric trains.)

l / Example 26.

The reason locomotives are built to be so heavy is for traction. The upward normal force of the rails on the wheels, $$F_N$$, cancels the downward force of gravity, $$F_W$$, so ignoring plus and minus signs, these two forces are equal in absolute value, $$F_N=F_W$$. Given this amount of normal force, the maximum force of static friction is $$F_s=\mu_s F_N=\mu_s F_W$$. This static frictional force, of the rails pushing forward on the wheels, is the only force that can accelerate the train, pull it uphill, or cancel out the force of air resistance while cruising at constant speed. The coefficient of static friction for steel on steel is about 1/4, so no locomotive can pull with a force greater than about 1/4 of its own weight. If the engine is capable of supplying more than that amount of force, the result will be simply to break static friction and spin the wheels.

The reason this is all so different from the situation with a car is that a car isn't pulling something else. If you put extra weight in a car, you improve the traction, but you also increase the inertia of the car, and make it just as hard to accelerate. In a train, the inertia is almost all in the cars being pulled, not in the locomotive.

The other fact we have to explain is the large number of driving wheels. First, we have to realize that increasing the number of driving wheels neither increases nor decreases the total amount of static friction, because static friction is independent of the amount of surface area in contact. (The reason four-wheel-drive is good in a car is that if one or more of the wheels is slipping on ice or in mud, the other wheels may still have traction. This isn't typically an issue for a train, since all the wheels experience the same conditions.) The advantage of having more driving wheels on a train is that it allows us to increase the weight of the locomotive without crushing the rails, or damaging bridges.

m / The wheelbases of the Hummer H3 and the Toyota Prius are surprisingly similar, differing by only 10%. The main difference in shape is that the Hummer is much taller and wider. It presents a much greater cross-sectional area to the wind, and this is the main reason that it uses about 2.5 times more gas on the freeway.

### 3.2.5 Fluid friction

Try to drive a nail into a waterfall and you will be confronted with the main difference between solid friction and fluid friction. Fluid friction is purely kinetic; there is no static fluid friction. The nail in the waterfall may tend to get dragged along by the water flowing past it, but it does not stick in the water. The same is true for gases such as air: recall that we are using the word “fluid” to include both gases and liquids.

Unlike kinetic friction between solids, fluid friction increases rapidly with velocity. It also depends on the shape of the object, which is why a fighter jet is more streamlined than a Model T. For objects of the same shape but different sizes, fluid friction typically scales up with the cross-sectional area of the object, which is one of the main reasons that an SUV gets worse mileage on the freeway than a compact car.

##### Discussion Question

Criticize the following analysis: “A book is sitting on a table. I shove it, overcoming static friction. Then it slows down until it has less force than static friction, and it stops.”

### 3.2.6 Analysis of forces

Newton's first and second laws deal with the total of all the forces exerted on a specific object, so it is very important to be able to figure out what forces there are. Once you have focused your attention on one object and listed the forces on it, it is also helpful to describe all the corresponding forces that must exist according to Newton's third law. We refer to this as “analyzing the forces” in which the object participates.

##### Example 27: A barge

A barge is being pulled along a canal by teams of horses on the shores. Analyze all the forces in which the barge participates.

 force acting on barge force related to it by Newton’s third law ropes’ forward normal forces on barge barge’s backward normal force on ropes water’s backward fluid friction force on barge barge’s forward fluid friction force on water planet earth’s downward gravitational force on barge barge’s upward gravitational force on earth water’s upward “floating” force on barge barge’s downward “floating” force on water

Here I've used the word “floating” force as an example of a sensible invented term for a type of force not classified on the tree in the previous section. A more formal technical term would be “hydrostatic force.”

Note how the pairs of forces are all structured as “A's force on B, B's force on A”: ropes on barge and barge on ropes; water on barge and barge on water. Because all the forces in the left column are forces acting on the barge, all the forces in the right column are forces being exerted by the barge, which is why each entry in the column begins with “barge.”

Often you may be unsure whether you have missed one of the forces. Here are three strategies for checking your list:

1. See what physical result would come from the forces you've found so far. Suppose, for instance, that you'd forgotten the “floating” force on the barge in the example above. Looking at the forces you'd found, you would have found that there was a downward gravitational force on the barge which was not canceled by any upward force. The barge isn't supposed to sink, so you know you need to find a fourth, upward force.

2. Whenever one solid object touches another, there will be a normal force, and possibly also a frictional force; check for both.

3. Make a drawing of the object, and draw a dashed boundary line around it that separates it from its environment. Look for points on the boundary where other objects come in contact with your object. This strategy guarantees that you'll find every contact force that acts on the object, although it won't help you to find non-contact forces.

The following is another example in which we can profit by checking against our physical intuition for what should be happening.

##### Example 28: Rappelling

As shown in the figure below, Cindy is rappelling down a cliff. Her downward motion is at constant speed, and she takes little hops off of the cliff, as shown by the dashed line. Analyze the forces in which she participates at a moment when her feet are on the cliff and she is pushing off.

 force acting on Cindy force related to it by Newton’s third law planet earth’s downward gravitational force on Cindy Cindy’s upward gravitational force on earth ropes upward frictional force on Cindy (her hand) Cindy’s downward frictional force on the rope cliff’s rightward normal force on Cindy Cindy’s leftward normal force on the cliff

The two vertical forces cancel, which is what they should be doing if she is to go down at a constant rate. The only horizontal force on her is the cliff's force, which is not canceled by any other force, and which therefore will produce an acceleration of Cindy to the right. This makes sense, since she is hopping off. (This solution is a little oversimplified, because the rope is slanting, so it also applies a small leftward force to Cindy. As she flies out to the right, the slant of the rope will increase, pulling her back in more strongly.)

I believe that constructing the type of table described in this section is the best method for beginning students. Most textbooks, however, prescribe a pictorial way of showing all the forces acting on an object. Such a picture is called a free-body diagram. It should not be a big problem if a future physics professor expects you to be able to draw such diagrams, because the conceptual reasoning is the same. You simply draw a picture of the object, with arrows representing the forces that are acting on it. Arrows representing contact forces are drawn from the point of contact, noncontact forces from the center of mass. Free-body diagrams do not show the equal and opposite forces exerted by the object itself.

##### Discussion Questions

When you fire a gun, the exploding gases push outward in all directions, causing the bullet to accelerate down the barrel. What Newton's-third-law pairs are involved? [Hint: Remember that the gases themselves are an object.]

In the example of the barge going down the canal, I referred to a “floating” or “hydrostatic” force that keeps the boat from sinking. If you were adding a new branch on the force-classification tree to represent this force, where would it go?

A pool ball is rebounding from the side of the pool table. Analyze the forces in which the ball participates during the short time when it is in contact with the side of the table.

The earth's gravitational force on you, i.e., your weight, is always equal to $$mg$$, where $$m$$ is your mass. So why can you get a shovel to go deeper into the ground by jumping onto it? Just because you're jumping, that doesn't mean your mass or weight is any greater, does it?

o / Example 29. The forces between the rope and other objects are normal and frictional forces.

### 3.2.7 Transmission of forces by low-mass objects

You're walking your dog. The dog wants to go faster than you do, and the leash is taut. Does Newton's third law guarantee that your force on your end of the leash is equal and opposite to the dog's force on its end? If they're not exactly equal, is there any reason why they should be approximately equal?

If there was no leash between you, and you were in direct contact with the dog, then Newton's third law would apply, but Newton's third law cannot relate your force on the leash to the dog's force on the leash, because that would involve three separate objects. Newton's third law only says that your force on the leash is equal and opposite to the leash's force on you,

$\begin{equation*} F_{yL} = - F_{Ly} , \end{equation*}$

and that the dog's force on the leash is equal and opposite to its force on the dog

$\begin{equation*} F_{dL} = - F_{Ld} . \end{equation*}$

Still, we have a strong intuitive expectation that whatever force we make on our end of the leash is transmitted to the dog, and vice-versa. We can analyze the situation by concentrating on the forces that act on the leash, $$F_{dL}$$ and $$F_{yL}$$. According to Newton's second law, these relate to the leash's mass and acceleration:

$\begin{equation*} F_{dL} + F_{yL} = m_La_L . \end{equation*}$

The leash is far less massive then any of the other objects involved, and if $$m_L$$ is very small, then apparently the total force on the leash is also very small, $$F_{dL}$$ + $$F_{yL}\approx 0$$, and therefore

$\begin{equation*} F_{dL}\approx - F_{yL} . \end{equation*}$

Thus even though Newton's third law does not apply directly to these two forces, we can approximate the low-mass leash as if it was not intervening between you and the dog. It's at least approximately as if you and the dog were acting directly on each other, in which case Newton's third law would have applied.

In general, low-mass objects can be treated approximately as if they simply transmitted forces from one object to another. This can be true for strings, ropes, and cords, and also for rigid objects such as rods and sticks.

n / If we imagine dividing a taut rope up into small segments, then any segment has forces pulling outward on it at each end. If the rope is of negligible mass, then all the forces equal $$+T$$ or $$-T$$, where $$T$$, the tension, is a single number.

If you look at a piece of string under a magnifying glass as you pull on the ends more and more strongly, you will see the fibers straightening and becoming taut. Different parts of the string are apparently exerting forces on each other. For instance, if we think of the two halves of the string as two objects, then each half is exerting a force on the other half. If we imagine the string as consisting of many small parts, then each segment is transmitting a force to the next segment, and if the string has very little mass, then all the forces are equal in magnitude. We refer to the magnitude of the forces as the tension in the string, $$T$$.

The term “tension” refers only to internal forces within the string. If the string makes forces on objects at its ends, then those forces are typically normal or frictional forces (example 29).

##### Example 29: Types of force made by ropes
$$\triangleright$$ Analyze the forces in figures o/1 and o/2.

$$\triangleright$$ In all cases, a rope can only make “pulling” forces, i.e., forces that are parallel to its own length and that are toward itself, not away from itself. You can't push with a rope!

In o/1, the rope passes through a type of hook, called a carabiner, used in rock climbing and mountaineering. Since the rope can only pull along its own length, the direction of its force on the carabiner must be down and to the right. This is perpendicular to the surface of contact, so the force is a normal force.

 force acting on Cindy force related to it by Newton’s third law planet earth’s downward gravitational force on Cindy Cindy’s upward gravitational force on earth ropes upward frictional force on Cindy (her hand) Cindy’s downward frictional force on the rope cliff’s rightward normal force on Cindy Cindy’s leftward normal force on the cliff

(There are presumably other forces acting on the carabiner from other hardware above it.)

In figure o/2, the rope can only exert a net force at its end that is parallel to itself and in the pulling direction, so its force on the hand is down and to the left. This is parallel to the surface of contact, so it must be a frictional force. If the rope isn't slipping through the hand, we have static friction. Friction can't exist without normal forces. These forces are perpendicular to the surface of contact. For simplicity, we show only two pairs of these normal forces, as if the hand were a pair of pliers.

 force acting on Cindy force related to it by Newton’s third law planet earth’s downward gravitational force on Cindy Cindy’s upward gravitational force on earth ropes upward frictional force on Cindy (her hand) Cindy’s downward frictional force on the rope cliff’s rightward normal force on Cindy Cindy’s leftward normal force on the cliff

(There are presumably other forces acting on the person as well, such as gravity.)

If a rope goes over a pulley or around some other object, then the tension throughout the rope is approximately equal so long as the pulley has negligible mass and there is not too much friction. A rod or stick can be treated in much the same way as a string, but it is possible to have either compression or tension.

##### Discussion Question

When you step on the gas pedal, is your foot's force being transmitted in the sense of the word used in this section?

p / A simplified drawing of an airgun.

q / The black box does work by reeling in its cable.

r / The wheel spinning in the air has $$K_{cm}=0$$. The space shuttle has all its kinetic energy in the form of center of mass motion, $$K=K_{cm}$$. The rolling ball has some, but not all, of its energy in the form of center of mass motion, $$K_{cm}\ltK$$. (Space Shuttle photo by NASA)

### 3.2.8 Work

#### Energy transferred to a particle

To change the kinetic energy, $$K=(1/2)mv^2$$, of a particle moving in one dimension, we must change its velocity. That will entail a change in its momentum, $$p=mv$$, as well, and since force is the rate of transfer of momentum, we conclude that the only way to change a particle's kinetic energy is to apply a force.6 A force in the same direction as the motion speeds it up, increasing the kinetic energy, while a force in the opposite direction slows it down.

Consider an infinitesimal time interval during which the particle moves an infinitesimal distance $$d{}x$$, and its kinetic energy changes by $$d{}K$$. In one dimension, we represent the direction of the force and the direction of the motion with positive and negative signs for $$F$$ and $$d{}x$$, so the relationship among the signs can be summarized as follows: \begin{center}

 F > 0 derx0 derK0 F < 0 derx0 derK0 F > 0 derx0 derK0 F < 0 derx0 derK0

\end{center} This looks exactly like the rule for determining the sign of a product, and we can easily show using the chain rule that this is indeed a multiplicative relationship:

\begin{align*} d{}K &= \frac{d{}K}{d{}v}\frac{d{}v}{d{}t}\frac{d{}t}{d{}x}d{}x \text{[chain rule]} \\ &= (mv)(a)(1/v)d{}x \\ & = m\,a\,d{}x \\ & = F\,d{}x \text{[Newton's second law]} \text{We can verify that force multiplied by distance has units of energy:} \text{N}\!\cdot\!\text{m} &= \frac{\text{kg}\!\cdot\!\text{m}/\text{s}}{\text{s}}\times\text{m} \\ &= \text{kg}\!\cdot\!\text{m}^2/\text{s}^2 \\ &= \text{J} \end{align*}

##### Example 30: A TV picture tube

$$\triangleright$$ At the back of a typical TV's picture tube, electrical forces accelerate each electron to an energy of $$5\times10^{-16}\ \text{J}$$ over a distance of about 1 cm. How much force is applied to a single electron? (Assume the force is constant.) What is the corresponding acceleration?

$$\triangleright$$ Integrating

\begin{align*} d{} K &= Fd{} x , \text{we find} K_{f}- K_{i} &= F( x_{f}- x_{i}) \text{or} \Delta K &= F\Delta x . \text{The force is} F &= \Delta K/\Delta x \\ &= \frac{5\times10^{-16}\ \text{J}}{ 0.01\ \text{m}} \\ & = 5\times10^{-14}\ \text{N} . \text{This may not sound like an impressive force, but it's enough to supply an electron with a spectacular acceleration. Looking up the mass of an electron on p. 943, we find} a &= F/ m \\ &= 5\times10^{16}\ \text{m}/\text{s}^2 . \end{align*}

##### Example 31: An air gun

$$\triangleright$$ An airgun, figure p, uses compressed air to accelerate a pellet. As the pellet moves from $$x_1$$ to $$x_2$$, the air decompresses, so the force is not constant. Using methods from chapter 5, one can show that the air's force on the pellet is given by $$F= bx^\text{-7/5}$$. A typical high-end airgun used for competitive target shooting has

\begin{align*} x_1 &= 0.046\ \text{m} , \\ x_2 &= 0.41\ \text{m} , \\ \text{and} b &= 4.4\ \text{N}\!\cdot\!\text{m}^\text{7/5} . \end{align*}

What is the kinetic energy of the pellet when it leaves the muzzle? (Assume friction is negligible.)

$$\triangleright$$ Since the force isn't constant, it would be incorrect to do $$F = \Delta K/\Delta x$$. Integrating both sides of the equation $$d{} K= Fd{} x$$, we have

\begin{align*} \Delta K &= \int_{ x_{1}}^{ x_{2}} Fd{} x \\ &= -\frac{5 b}{2}\left( x_2^\text{-2/5} - x_1^\text{-2/5}\right) \\ &= 22\ \text{J} \end{align*}

In general, when energy is transferred by a force,7 we use the term work to refer to the amount of energy transferred. This is different from the way the word is used in ordinary speech. If you stand for a long time holding a bag of cement, you get tired, and everyone will agree that you've worked hard, but you haven't changed the energy of the cement, so according to the definition of the physics term, you haven't done any work on the bag. There has been an energy transformation inside your body, of chemical energy into heat, but this just means that one part of your body did positive work (lost energy) while another part did a corresponding amount of negative work (gained energy).

#### Work in general

I derived the expression $$Fd{}x$$ for one particular type of kinetic-energy transfer, the work done in accelerating a particle, and then defined work as a more general term. Is the equation correct for other types of work as well? For example, if a force lifts a mass $$m$$ against the resistance of gravity at constant velocity, the increase in the mass's gravitational energy is $$d{}(mgy)=mgd{}y=Fd{}y$$, so again the equation works, but this still doesn't prove that the equation is always correct as a way of calculating energy transfers.

Imagine a black box8, containing a gasoline-powered engine, which is designed to reel in a steel cable, exerting a certain force $$F$$. For simplicity, we imagine that this force is always constant, so we can talk about $$\Delta{}x$$ rather than an infinitesimal $$d{}x$$. If this black box is used to accelerate a particle (or any mass without internal structure), and no other forces act on the particle, then the original derivation applies, and the work done by the box is $$W=F\Delta{}x$$. Since $$F$$ is constant, the box will run out of gas after reeling in a certain amount of cable $$\Delta{}x$$. The chemical energy inside the box has decreased by $$-W$$, while the mass being accelerated has gained $$W$$ worth of kinetic energy.9

Now what if we use the black box to pull a plow? The energy increase in the outside world is of a different type than before; it takes the forms of (1) the gravitational energy of the dirt that has been lifted out to the sides of the furrow, (2) frictional heating of the dirt and the plowshare, and (3) the energy needed to break up the dirt clods (a form of electrical energy involving the attractions among the atoms in the clod). The box, however, only communicates with the outside world via the hole through which its cable passes. The amount of chemical energy lost by the gasoline can therefore only depend on $$F$$ and $$\Delta{}x$$, so it is the same $$-W$$ as when the box was being used to accelerate a mass, and thus by conservation of energy, the work done on the outside world is again $$W$$.

This is starting to sound like a proof that the force-times-distance method is always correct, but there was one subtle assumption involved, which was that the force was exerted at one point (the end of the cable, in the black box example). Real life often isn't like that. For example, a cyclist exerts forces on both pedals at once. Serious cyclists use toe-clips, and the conventional wisdom is that one should use equal amounts of force on the upstroke and downstroke, to make full use of both sets of muscles. This is a two-dimensional example, since the pedals go in circles. We're only discussing one-dimensional motion right now, so let's just pretend that the upstroke and downstroke are both executed in straight lines. Since the forces are in opposite directions, one is positive and one is negative. The cyclist's total force on the crank set is zero, but the work done isn't zero. We have to add the work done by each stroke, $$W=F_1\Delta{}x_1+F_2\Delta{}x_2$$. (I'm pretending that both forces are constant, so we don't have to do integrals.) Both terms are positive; one is a positive number multiplied by a positive number, while the other is a negative times a negative.

This might not seem like a big deal --- just remember not to use the total force --- but there are many situations where the total force is all we can measure. The ultimate example is heat conduction. Heat conduction is not supposed to be counted as a form of work, since it occurs without a force. But at the atomic level, there are forces, and work is done by one atom on another. When you hold a hot potato in your hand, the transfer of heat energy through your skin takes place with a total force that's extremely close to zero. At the atomic level, atoms in your skin are interacting electrically with atoms in the potato, but the attractions and repulsions add up to zero total force. It's just like the cyclist's feet acting on the pedals, but with zillions of forces involved instead of two. There is no practical way to measure all the individual forces, and therefore we can't calculate the total energy transferred.

To summarize, $$\sum{F_jd{}x_j}$$ is a correct way of calculating work, where $$F_j$$ is the individual force acting on particle $$j$$, which moves a distance $$d{}x_j$$. However, this is only useful if you can identify all the individual forces and determine the distance moved at each point of contact. For convenience, I'll refer to this as the work theorem. (It doesn't have a standard name.)

There is, however, something useful we can do with the total force. We can use it to calculate the part of the work done on an object that consists of a change in the kinetic energy it has due to the motion of its center of mass. The proof is essentially the same as the proof on p. 161, except that now we don't assume the force is acting on a single particle, so we have to be a little more delicate. Let the object consist of $$n$$ particles. Its total kinetic energy is $$K=\sum_{j=1}^n{(1/2)m_jv_j^2}$$, but this is what we've already realized can't be calculated using the total force. The kinetic energy it has due to motion of its center of mass is

\begin{align*} K_{cm} &= \frac{1}{2}m_{total}v_{cm}^2 . \end{align*}

Figure r shows some examples of the distinction between $$K_{cm}$$ and $$K$$. Differentiating $$K_{cm}$$, we have

\begin{align*} d{}K_{cm} &= m_{total}v_{cm}d{}v_{cm} \\ &= m_{total}v_{cm} \frac{d{}v_{cm}}{d{}t}\frac{d{}t}{d{}x_{cm}}d{}x_{cm} \text{[chain rule]} \\ &= m_{total}\frac{d{}v_{cm}}{d{}t}d{}x_{cm} \text{[d{}t/d{}x_{cm}=1/v_{cm}]} \\ &= \frac{d{}p_{total}}{d{}t}d{}x_{cm} \text{[p_{total}=m_{total}v_{cm}]} \\ &= F_{total}d{}x_{cm} \end{align*}

I'll call this the kinetic energy theorem --- like the work theorem, it has no standard name.

##### Example 32: An ice skater pushing off from a wall

The kinetic energy theorem tells us how to calculate the skater's kinetic energy if we know the amount of force and the distance her center of mass travels while she is pushing off.

The work theorem tells us that the wall does no work on the skater, since the point of contact isn't moving. This makes sense, because the wall does not have any source of energy.

##### Example 33: Absorbing an impact without recoiling?

$$\triangleright$$ Is it possible to absorb an impact without recoiling? For instance, if a ping-pong ball hits a brick wall, does the wall “give” at all?

$$\triangleright$$ There will always be a recoil. In the example proposed, the wall will surely have some energy transferred to it in the form of heat and vibration. The work theorem tells us that we can only have an energy transfer if the distance traveled by the point of contact is nonzero.

##### Example 34: Dragging a refrigerator at constant velocity

The fridge's momentum is constant, so there is no net momentum transfer, and the total force on it must be zero: your force is canceling the floor's kinetic frictional force. The kinetic energy theorem is therefore true but useless. It tells us that there is zero total force on the refrigerator, and that the refrigerator's kinetic energy doesn't change.

The work theorem tells us that the work you do equals your hand's force on the refrigerator multiplied by the distance traveled. Since we know the floor has no source of energy, the only way for the floor and refrigerator to gain energy is from the work you do. We can thus calculate the total heat dissipated by friction in the refrigerator and the floor.

Note that there is no way to find how much of the heat is dissipated in the floor and how much in the refrigerator.

##### Example 35: Accelerating a cart

If you push on a cart and accelerate it, there are two forces acting on the cart: your hand's force, and the static frictional force of the ground pushing on the wheels in the opposite direction.

Applying the work theorem to your force tells us how to calculate the work you do.

Applying the work theorem to the floor's force tells us that the floor does no work on the cart. There is no motion at the point of contact, because the atoms in the floor are not moving. (The atoms in the surface of the wheel are also momentarily at rest when they touch the floor.) This makes sense, because the floor does not have any source of energy.

The kinetic energy theorem refers to the total force, and because the floor's backward force cancels part of your force, the total force is less than your force. This tells us that only part of your work goes into the kinetic energy associated with the forward motion of the cart's center of mass. The rest goes into rotation of the wheels.

##### Discussion Questions

Criticize the following incorrect statement: “A force doesn't do any work unless it's causing the object to move.”

To stop your car, you must first have time to react, and then it takes some time for the car to slow down. Both of these times contribute to the distance you will travel before you can stop. The figure shows how the average stopping distance increases with speed. Because the stopping distance increases more and more rapidly as you go faster, the rule of one car length per 10 m.p.h. of speed is not conservative enough at high speeds. In terms of work and kinetic energy, what is the reason for the more rapid increase at high speeds?

s / Discussion question B.

t / The force is transmitted to the block.

u / A mechanical advantage of 2.

v / An inclined plane.

w / A wedge.

x / Archimedes' screw

### 3.2.9 Simple Machines

Conservation of energy provided the necessary tools for analyzing some mechanical systems, such as the seesaw on page 85 and the pulley arrangements of the homework problems on page 120, but we could only analyze those machines by computing the total energy of the system. That approach wouldn't work for systems like the biceps/forearm machine on page 85, or the one in figure t, where the energy content of the person's body is impossible to compute directly. Even though the seesaw and the biceps/forearm system were clearly just two different forms of the lever, we had no way to treat them both on the same footing. We can now successfully attack such problems using the work and kinetic energy theorems.

##### Example 36: Constant tension around a pulley

$$\triangleright$$ In figure t, what is the relationship between the force applied by the person's hand and the force exerted on the block?

$$\triangleright$$ If we assume the rope and the pulley are ideal, i.e., frictionless and massless, then there is no way for them to absorb or release energy, so the work done by the hand must be the same as the work done on the block. Since the hand and the block move the same distance, the work theorem tells us the two forces are the same.

Similar arguments provide an alternative justification for the statement made in section 3.2.7 that show that an idealized rope exerts the same force, the tension, anywhere it's attached to something, and the same amount of force is also exerted by each segment of the rope on the neighboring segments. Going around an ideal pulley also has no effect on the tension.

This is an example of a simple machine, which is any mechanical system that manipulates forces to do work. This particular machine reverses the direction of the motion, but doesn't change the force or the speed of motion.

##### Example 37: A mechanical advantage

The idealized pulley in figure u has negligible mass, so its kinetic energy is zero, and the kinetic energy theorem tells us that the total force on it is zero. We know, as in the preceding example, that the two forces pulling it to the right are equal to each other, so the force on the left must be twice as strong. This simple machine doubles the applied force, and we refer to this ratio as a mechanical advantage (M.A.) of 2. There's no such thing as a free lunch, however; the distance traveled by the load is cut in half, and there is no increase in the amount of work done.

##### Example 38: Inclined plane and wedge
In figure v, the force applied by the hand is equal to the one applied to the load, but there is a mechanical advantage compared to the force that would have been required to lift the load straight up. The distance traveled up the inclined plane is greater by a factor of 1/sin $$\theta$$, so by the work theorem, the force is smaller by a factor of sin $$\theta$$, and we have M.A.=1/sin $$\theta$$. The wedge, w, is similar.
##### Example 39: Archimedes' screw

In one revolution, the crank travels a distance $$2\pi{} b$$, and the water rises by a height $$h$$. The mechanical advantage is $$2\pi{} b/ h$$.

### 3.2.10 Force related to interaction energy

In section 2.3, we saw that there were two equivalent ways of looking at gravity, the gravitational field and the gravitational energy. They were related by the equation $$d{}U=mgd{}r$$, so if we knew the field, we could find the energy by integration, $$U=\int{mgd{}r}$$, and if we knew the energy, we could find the field by differentiation, $$g=(1/m)d{}U/d{}r$$.

The same approach can be applied to other interactions, for example a mass on a spring. The main difference is that only in gravitational interactions does the strength of the interaction depend on the mass of the object, so in general, it doesn't make sense to separate out the factor of $$m$$ as in the equation $$d{}U=mgd{}r$$. Since $$F=mg$$ is the gravitational force, we can rewrite the equation in the more suggestive form $$d{}U=Fd{}r$$. This form no longer refers to gravity specifically, and can be applied much more generally. The only remaining detail is that I've been fairly cavalier about positive and negative signs up until now. That wasn't such a big problem for gravitational interactions, since gravity is always attractive, but it requires more careful treatment for nongravitational forces, where we don't necessarily know the direction of the force in advance, and we need to use positive and negative signs carefully for the direction of the force.

In general, suppose that forces are acting on a particle --- we can think of them as coming from other objects that are “off stage” --- and that the interaction between the particle and the off-stage objects can be characterized by an interaction energy, $$U$$, which depends only on the particle's position, $$x$$. Using the kinetic energy theorem, we have $$d{}K=Fd{}x$$. (It's not necessary to write $$K_{cm}$$, since a particle can't have any other kind of kinetic energy.) Conservation of energy tells us $$d{}K+d{}U=0$$, so the relationship between force and interaction energy is $$d{}U=-Fd{}x$$, or

$\begin{equation*} F = -\frac{d{}U}{d{}x} \text{[relationship between force and interaction energy]} . \end{equation*}$

##### Example 40: Force exerted by a spring
$$\triangleright$$ A mass is attached to the end of a spring, and the energy of the spring is $$U=(1/2) kx^2$$, where $$x$$ is the position of the mass, and $$x=0$$ is defined to be the equilibrium position. What is the force the spring exerts on the mass? Interpret the sign of the result.

$$\triangleright$$ Differentiating, we find

\begin{align*} F &= -\frac{d{} U}{d{} x} \\ &= - kx . \end{align*}

If $$x$$ is positive, then the force is negative, i.e., it acts so as to bring the mass back to equilibrium, and similarly for $$x\lt0$$ we have $$F>0$$.

Most books do the $$F=- kx$$ form before the $$U=(1/2) kx^2$$ form, and call it Hooke's law. Neither form is really more fundamental than the other --- we can always get from one to the other by integrating or differentiating.

##### Example 41: Newton's law of gravity

$$\triangleright$$ Given the equation $$U=- Gm_1 m_{2}/ r$$ for the energy of gravitational interactions, find the corresponding equation for the gravitational force on mass $$m_2$$. Interpret the positive and negative signs.

$$\triangleright$$ We have to be a little careful here, because we've been taking $$r$$ to be positive by definition, whereas the position, $$x$$, of mass $$m_2$$ could be positive or negative, depending on which side of $$m_1$$ it's on.

For positive $$x$$, we have $$r= x$$, and differentiation gives

\begin{align*} F &= -\frac{d{} U}{d{} x} \\ &= - Gm_1 m_{2}/ x^2 . \end{align*}

As in the preceding example, we have $$F\lt0$$ when $$x$$ is positive, because the object is being attracted back toward $$x=0$$.

When $$x$$ is negative, the relationship between $$r$$ and $$x$$ becomes $$r=- x$$, and the result for the force is the same as before, but with a minus sign. We can combine the two equations by writing

$\begin{equation*} | F| = Gm_1 m_{2}/ r^2 , \end{equation*}$

and this is the form traditionally known as Newton's law of gravity. As in the preceding example, the $$U$$ and $$F$$ equations contain equivalent information, and neither is more fundamental than the other.

##### Example 42: Equilibrium

I previously described the condition for equilibrium as a local maximum or minimum of $$U$$. A differentiable function has a zero derivative at its extrema, and we can now relate this directly to force: zero force acts on an object when it is at equilibrium.

## 3.3 Resonance

a / An $$x$$-versus-$$t$$ graph for a swing pushed at resonance.

b / A swing pushed at twice its resonant frequency.

c / The $$F$$-versus-$$t$$ graph for an impulsive driving force.

d / A sinusoidal driving force.

Resonance is a phenomenon in which an oscillator responds most strongly to a driving force that matches its own natural frequency of vibration. For example, suppose a child is on a playground swing with a natural frequency of 1 Hz. That is, if you pull the child away from equilibrium, release her, and then stop doing anything for a while, she'll oscillate at 1 Hz. If there was no friction, as we assumed in section 2.5, then the sum of her gravitational and kinetic energy would remain constant, and the amplitude would be exactly the same from one oscillation to the next. However, friction is going to convert these forms of energy into heat, so her oscillations would gradually die out. To keep this from happening, you might give her a push once per cycle, i.e., the frequency of your pushes would be 1 Hz, which is the same as the swing's natural frequency. As long as you stay in rhythm, the swing responds quite well. If you start the swing from rest, and then give pushes at 1 Hz, the swing's amplitude rapidly builds up, as in figure a, until after a while it reaches a steady state in which friction removes just as much energy as you put in over the course of one cycle.

self-check:

In figure a, compare the amplitude of the cycle immediately following the first push to the amplitude after the second. Compare the energies as well.

What will happen if you try pushing at 2 Hz? Your first push puts in some momentum, $$p$$, but your second push happens after only half a cycle, when the swing is coming right back at you, with momentum $$-p$$! The momentum transfer from the second push is exactly enough to stop the swing. The result is a very weak, and not very sinusoidal, motion, b.

\formatlikesubsubsection{Making the math easy}
This is a simple and physically transparent example of resonance: the swing responds most strongly if you match its natural rhythm. However, it has some characteristics that are mathematically ugly and possibly unrealistic. The quick, hard pushes are known as impulse forces, c, and they lead to an $$x$$-$$t$$ graph that has nondifferentiable kinks. Impulsive forces like this are not only badly behaved mathematically, they are usually undesirable in practical terms. In a car engine, for example, the engineers work very hard to make the force on the pistons change smoothly, to avoid excessive vibration. Throughout the rest of this section, we'll assume a driving force that is sinusoidal, d, i.e., one whose $$F$$-$$t$$ graph is either a sine function or a function that differs from a sine wave in phase, such as a cosine. The force is positive for half of each cycle and negative for the other half, i.e., there is both pushing and pulling. Sinusoidal functions have many nice mathematical characteristics (we can differentiate and integrate them, and the sum of sinusoidal functions that have the same frequency is a sinusoidal function), and they are also used in many practical situations. For instance, my garage door zapper sends out a sinusoidal radio wave, and the receiver is tuned to resonance with it.

A second mathematical issue that I glossed over in the swing example was how friction behaves. In section 3.2.4, about forces between solids, the empirical equation for kinetic friction was independent of velocity. Fluid friction, on the other hand, is velocity-dependent. For a child on a swing, fluid friction is the most important form of friction, and is approximately proportional to $$v^2$$. In still other situations, e.g., with a low-density gas or friction between solid surfaces that have been lubricated with a fluid such as oil, we may find that the frictional force has some other dependence on velocity, perhaps being proportional to $$v$$, or having some other complicated velocity dependence that can't even be expressed with a simple equation. It would be extremely complicated to have to treat all of these different possibilities in complete generality, so for the rest of this section, we'll assume friction proportional to velocity

$\begin{equation*} F = -bv , \end{equation*}$

simply because the resulting equations happen to be the easiest to solve. Even when the friction doesn't behave in exactly this way, many of our results may still be at least qualitatively correct.

e / A damped sine wave, of the form $$x = Ae^{- ct}\text{sin} (\omega_{f} t+\delta)$$.

f / Self-check G.

g / A damped sine wave is compared with an undamped one, with $$m$$ and $$k$$ kept the same and only $$b$$ changed.

### 3.3.1 Damped, free motion

#### Numerical treatment

An oscillator that has friction is referred to as damped. Let's use numerical techniques to find the motion of a damped oscillator that is released away from equilibrium, but experiences no driving force after that. We can expect that the motion will consist of oscillations that gradually die out.

In section 2.5, we simulated the undamped case using our tried and true Python function based on conservation of energy. Now, however, that approach becomes a little awkward, because it involves splitting up the path to be traveled into $$n$$ tiny segments, but in the presence of damping, each swing is a little shorter than the last one, and we don't know in advance exactly how far the oscillation will get before turning around. An easier technique here is to use force rather than energy. Newton's second law, $$a=F/m$$, gives $$a=(-kx-bv)/m$$, where we've made use of the result of example 40 for the force exerted by the spring. This becomes a little prettier if we rewrite it in the form

$\begin{equation*} ma+bv+kx = 0 , \end{equation*}$

which gives symmetric treatment to three terms involving $$x$$ and its first and second derivatives, $$v$$ and $$a$$. Now instead of calculating the time $$\Delta{}t=\Delta{}x/v$$ required to move a predetermined distance $$\Delta{}x$$, we pick $$\Delta{}t$$ and determine the distance traveled in that time, $$\Delta{}x=v\Delta{}t$$. Also, we can no longer update $$v$$ based on conservation of energy, since we don't have any easy way to keep track of how much mechanical energy has been changed into heat energy. Instead, we recalculate the velocity using $$\Delta{}v=a\Delta{}t$$.

import math
k=39.4784 # chosen to give a period of 1 second
m=1.
b=0.211 # chosen to make the results simple
x=1.
v=0.
t=0.
dt=.01
n=1000
for j in range(n):
x=x+v*dt
a=(-k*x-b*v)/m
if (v>0) and (v+a*dt<0) :
print("turnaround at t=",t,", x=",x)
v=v+a*dt
t=t+dt

turnaround at t= 0.99 , x= 0.899919262445
turnaround at t= 1.99 , x= 0.809844934046
turnaround at t= 2.99 , x= 0.728777519477
turnaround at t= 3.99 , x= 0.655817260033
turnaround at t= 4.99 , x= 0.590154191135
turnaround at t= 5.99 , x= 0.531059189965
turnaround at t= 6.99 , x= 0.477875914756
turnaround at t= 7.99 , x= 0.430013546991
turnaround at t= 8.99 , x= 0.386940256644
turnaround at t= 9.99 , x= 0.348177318484


The spring constant, $$k=4\pi{}=39.4784$$ N/m, is designed so that if the undamped equation $$f=(1/2\pi)\sqrt{k/m}$$ was still true, the frequency would be 1 Hz. We start by noting that the addition of a small amount of damping doesn't seem to have changed the period at all, or at least not to within the accuracy of the calculation.10 You can check for yourself, however, that a large value of $$b$$, say 5 $$\text{N}\!\cdot\!\text{s}/\text{m}$$, does change the period significantly.

We release the mass from $$x=1\ \text{m}$$, and after one cycle, it only comes back to about $$x=0.9\ \text{m}$$. I chose $$b=0.211\ \text{N}\!\cdot\!\text{s}/\text{m}$$ by fiddling around until I got this result, since a decrease of exactly 10% is easy to discuss. Notice how the amplitude after two cycles is about $$0.81\ \text{m}$$, i.e., $$1\ \text{m}$$ times $$0.9^2$$: the amplitude has again dropped by exactly 10%. This pattern continues for as long as the simulation runs, e.g., for the last two cycles, we have 0.34818/0.38694=0.89982, or almost exactly 0.9 again. It might have seemed capricious when I chose to use the unrealistic equation $$F=-bv$$, but this is the payoff. Only with $$-bv$$ friction do we get this kind of mathematically simple exponential decay.

Because the decay is exponential, it never dies out completely; this is different from the behavior we would have had with Coulomb friction, which does make objects grind completely to a stop at some point. With friction that acts like $$F=-bv$$, $$v$$ gets smaller as the oscillations get smaller. The smaller and smaller force then causes them to die out at a rate that is slower and slower.

#### Analytic treatment

Taking advantage of this unexpectedly simple result, let's find an analytic solution for the motion. The numerical output suggests that we assume a solution of the form

$\begin{equation*} x = Ae^{-ct}\sin (\omega{}_f t+\delta) , \end{equation*}$

where the unknown constants $$\omega_f$$ and $$c$$ will presumably be related to $$m$$, $$b$$, and $$k$$. The constant $$c$$ indicates how quickly the oscillations die out. The constant $$\omega_f$$ is, as before, defined as $$2\pi$$ times the frequency, with the subscript $$f$$ to indicate a free (undriven) solution. All our equations will come out much simpler if we use $$\omega$$s everywhere instead of $$f$$s from now on, and, as physicists often do, I'll generally use the word “frequency” to refer to $$\omega$$ when the context makes it clear what I'm talking about. The phase angle $$\delta$$ has no real physical significance, since we can define $$t=0$$ to be any moment in time we like.

self-check:

In figure f, which graph has the greater value of $$c$$?

The factor $$A$$ for the initial amplitude can also be omitted without loss of generality, since the equation we're trying to solve, $$ma+bv+kx = 0$$, is linear. That is, $$v$$ and $$a$$ are the first and second derivatives of $$x$$, and the derivative of $$Ax$$ is simply $$A$$ times the derivative of $$x$$. Thus, if $$x(t)$$ is a solution of the equation, then multiplying it by a constant gives an equally valid solution. This is another place where we see that a damping force proportional to $$v$$ is the easiest to handle mathematically. For a damping force proportional to $$v^2$$, for example, we would have had to solve the equation $$ma+bv^2+kx = 0$$, which is nonlinear.

For the purpose of determining $$\omega_f$$ and $$c$$, the most general form we need to consider is therefore $$x = e^{-ct}\sin \omega_f t$$ , whose first and second derivatives are $$v = e^{-ct}\left(-c \sin \omega_f t + \omega\cos \omega_f t\right)$$ and $$a = e^{-ct}\left(c^2 \sin \omega_f t -2\omega_f c \cos \omega_f t -\omega_f^2\sin \omega_f t\right)$$. Plugging these into the equation $$ma+bv+kx = 0$$ and setting the sine and cosine parts equal to zero gives, after some tedious algebra,

\begin{align*} c &= \frac{b}{2m} \\ \text{and} \omega_f &= \sqrt{\frac{k}{m}-\frac{b^2}{4m^2}} . \end{align*}

Intuitively, we expect friction to “slow down” the motion, as when we ride a bike into a big patch of mud. “Slow down,” however, could have more than one meaning here. It could mean that the oscillator would take more time to complete each cycle, or it could mean that as time went on, the oscillations would die out, thus giving smaller velocities.

Our mathematical results show that both of these things happen. The first equation says that $$c$$, which indicates how quickly the oscillations damp out, is directly related to $$b$$, the strength of the damping.

The second equation, for the frequency, can be compared with the result from page 116 of $$\sqrt{k/m}$$ for the undamped system. Let's refer to this now as $$\omega_\text{o}$$, to distinguish it from the actual frequency $$\omega_f$$ of the free oscillations when damping is present. The result for $$\omega_f$$ will be less than $$\omega_\text{o}$$, due to the presence of the $$b^2/4m^2$$ term. This tells us that the addition of friction to the system does increase the time required for each cycle. However, it is very common for the $$b^2/4m^2$$ term to be negligible, so that $$\omega_f\approx\omega_\text{o}$$.

Figure g shows an example. The damping here is quite strong: after only one cycle of oscillation, the amplitude has already been reduced by a factor of 2, corresponding to a factor of 4 in energy. However, the frequency of the damped oscillator is only about 1% lower than that of the undamped one; after five periods, the accumulated lag is just barely visible in the offsetting of the arrows. We can see that extremely strong damping --- even stronger than this --- would have been necessary in order to make $$\omega_f\approx\omega_\text{o}$$ a poor approximation.

### 3.3.2 The quality factor

It's usually impractical to measure $$b$$ directly and determine $$c$$ from the equation $$c=b/2m$$. For a child on a swing, measuring $$b$$ would require putting the child in a wind tunnel! It's usually much easier to characterize the amount of damping by observing the actual damped oscillations and seeing how many cycles it takes for the mechanical energy to decrease by a certain factor. The unitless quality factor, $$Q$$, is defined as $$Q=\omega_\text{o}/2c$$, and in the limit of weak damping, where $$\omega\approx\omega_\text{o}$$, this can be interpreted as the number of cycles required for the mechanical energy to fall off by a factor of $$e^{2\pi}=535.49...$$ Using this new quantity, we can rewrite the equation for the frequency of damped oscillations in the slightly more elegant form $$\omega_f = \omega_\text{o}\sqrt{1-1/4Q^2}$$.

self-check:

What if we wanted to make a simpler definition of $$Q$$, as the number of oscillations required for the vibrations to die out completely, rather than the number required for the energy to fall off by this obscure factor?

##### Example 43: A graph

The damped motion in figure g has $$Q\approx 4.5$$, giving $$\sqrt{1-1/4Q^2}\approx 0.99$$, as claimed at the end of the preceding subsection.

##### Example 44: Exponential decay in a trumpet

$$\triangleright$$ The vibrations of the air column inside a trumpet have a $$Q$$ of about 10. This means that even after the trumpet player stops blowing, the note will keep sounding for a short time. If the player suddenly stops blowing, how will the sound intensity 20 cycles later compare with the sound intensity while she was still blowing?

$$\triangleright$$ The trumpet's $$Q$$ is 10, so after 10 cycles the energy will have fallen off by a factor of 535. After another 10 cycles we lose another factor of 535, so the sound intensity is reduced by a factor of $$535\times535= 2.9\times10^5$$.

The decay of a musical sound is part of what gives it its character, and a good musical instrument should have the right $$Q$$, but the $$Q$$ that is considered desirable is different for different instruments. A guitar is meant to keep on sounding for a long time after a string has been plucked, and might have a $$Q$$ of 1000 or 10000. One of the reasons why a cheap synthesizer sounds so bad is that the sound suddenly cuts off after a key is released.

{Summary of Notation}
 k spring constant m mass of the oscillator b sets the amount of damping,F = − bv T period f frequency,1 / T ω (Greek letter omega), angular frequency,2πf, often referred to simply as “frequency” omegatextupo frequency the oscillator would have without damping,sqrtkm ωf frequency of the free vibrations c sets the time scale for the exponential decay envelopee − ct of the free vibrations Fm strength of the driving force, which is assumed to vary sinusoidally with frequencyω A amplitude of the steady-state response δ phase angle of the steady-state response
}

h / Dependence of the amplitude and phase angle on the driving frequency, for an undamped oscillator. The amplitudes were calculated with $$F_m$$, $$m$$, and $$\omega_\text{o}$$, all set to 1.

k / The definition of $$\Delta\omega$$, the full width at half maximum.

l / The collapsed section of the Nimitz Freeway

m / An $$x$$-versus-$$t$$ graph of the steady-state motion of a swing being pushed at twice its resonant frequency by an impulsive force.

### 3.3.3 Driven motion

The driven case is extremely important in science, technology, and engineering. We have an external driving force $$F=F_m \sin \omega t$$, where the constant $$F_m$$ indicates the maximum strength of the force in either direction. The equation of motion is now

$\begin{multline*} ma+bv+kx = F_m \sin \omega t \\ \text{[equation of motion for a driven oscillator]} . \end{multline*}$

After the driving force has been applied for a while, we expect that the amplitude of the oscillations will approach some constant value. This motion is known as the steady state, and it's the most interesting thing to find out; as we'll see later, the most general type of motion is only a minor variation on the steady-state motion. For the steady-state motion, we're going to look for a solution of the form

$\begin{equation*} x = A \sin (\omega{}t+\delta) . \end{equation*}$

In contrast to the undriven case, here it's not possible to sweep $$A$$ and $$\delta$$ under the rug. The amplitude of the steady-state motion, $$A$$, is actually the most interesting thing to know about the steady-state motion, and it's not true that we still have a solution no matter how we fiddle with $$A$$; if we have a solution for a certain value of $$A$$, then multiplying $$A$$ by some constant would break the equality between the two sides of the equation of motion. It's also no longer true that we can get rid of $$\delta$$ simply be redefining when we start the clock; here $$\delta$$ represents a difference in time between the start of one cycle of the driving force and the start of the corresponding cycle of the motion.

The velocity and acceleration are $$v=\omega{}A\cos(\omega t+\delta)$$ and $$a=-\omega^2A\sin(\omega t+\delta)$$, and if we plug these into the equation of motion, \eqref{eqn:resonancemotion}, and simplify a little, we find

$\begin{equation*} (k-m\omega^2)\sin (\omega t+\delta) +\omega b \cos (\omega t+\delta) = \frac{F_m}{A} \sin \omega t . \end{equation*}$

The sum of any two sinusoidal functions with the same frequency is also a sinusoidal, so the whole left side adds up to a sinusoidal. By fiddling with $$A$$ and $$\delta$$ we can make the amplitudes and phases of the two sides of the equation match up.

$$A$$ and $$\delta$$ are easy to find in the case where there is no damping at all. There are now no cosines in equation \eqref{eqn:steadystate} above, only sines, so if we wish we can set $$\delta$$ to zero, and we find $$A=F_m/(k-m\omega^2)=F_m/m(\omega_\text{o}^2-\omega^2)$$. This, however, makes $$A$$ negative for $$\omega>\omega_\text{o}$$. The variable $$\delta$$ was designed to represent this kind of phase relationship, so we prefer to keep $$A$$ positive and set $$\delta=\pi$$ for $$\omega>\omega_\text{o}$$. Our results are then

\begin{align*} A &= \frac{F_m}{m\left|\omega^2-\omega_\text{o}^2\right|} \\ \text{and} \delta &= \left\{\begin{array}{ll} 0, & \omega\lt\omega_\text{o}\\ \pi, & \omega>\omega_\text{o}\end{array}\right. . \end{align*}

The most important feature of the result is that there is a resonance: the amplitude becomes greater and greater, and approaches infinity, as $$\omega$$ approaches the resonant frequency $$\omega_\text{o}$$. This is the physical behavior we anticipated on page 171 in the example of pushing a child on a swing. If the driving frequency matches the frequency of the free vibrations, then the driving force will always be in the right direction to add energy to the swing. At a driving frequency very different from the resonant frequency, we might get lucky and push at the right time during one cycle, but our next push would come at some random point in the next cycle, possibly having the effect of slowing the swing down rather than speeding it up.

The interpretation of the infinite amplitude at $$\omega=\omega_\text{o}$$ is that there really isn't any steady state if we drive the system exactly at resonance --- the amplitude will just keep on increasing indefinitely. In real life, the amplitude can't be infinite both because there is always some damping and because there will always be some difference, however small, between $$\omega$$ and $$\omega_\text{o}$$. Even though the infinity is unphysical, it has entered into the popular consciousness, starting with the eccentric Serbian-American inventor and physicist Nikola Tesla. Around 1912, the tabloid newspaper The World Today credulously reported a story which Tesla probably fabricated --- or wildly exaggerated --- for the sake of publicity. Supposedly he created a steam-powered device “no larger than an alarm clock,” containing a piston that could be made to vibrate at a tunable and precisely controlled frequency. “He put his little vibrator in his coat-pocket and went out to hunt a half-erected steel building. Down in the Wall Street district, he found one --- ten stories of steel framework without a brick or a stone laid around it. He clamped the vibrator to one of the beams, and fussed with the adjustment [presumably hunting for the building's resonant frequency] until he got it. Tesla said finally the structure began to creak and weave and the steel-workers came to the ground panic-stricken, believing that there had been an earthquake. Police were called out. Tesla put the vibrator in his pocket and went away. Ten minutes more and he could have laid the building in the street. And, with the same vibrator he could have dropped the Brooklyn Bridge into the East River in less than an hour.”

The phase angle $$\delta$$ also exhibits surprising behavior. As the frequency is tuned upward past resonance, the phase abruptly shifts so that the phase of the response is opposite to that of the driving force. There is a simple interpretation for this. The system's mechanical energy can only change due to work done by the driving force, since there is no damping to convert mechanical energy to heat. In the steady state, then, the power transmitted by the driving force over a full cycle of motion must average out to zero. In general, the work theorem $$d{}E=Fd{}x$$ can always be divided by $$d{}t$$ on both sides to give the useful relation $$P=Fv$$. If $$Fv$$ is to average out to zero, then $$F$$ and $$v$$ must be out of phase by $$\pm\pi/2$$, and since $$v$$ is ahead of $$x$$ by a phase angle of $$\pi/2$$, the phase angle between $$x$$ and $$F$$ must be zero or $$\pi$$.

Given that these are the two possible phases, why is there a difference in behavior between $$\omega\lt\omega_\text{o}$$ and $$\omega>\omega_\text{o}$$? At the low-frequency limit, consider $$\omega=0$$, i.e., a constant force. A constant force will simply displace the oscillator to one side, reaching an equilibrium that is offset from the usual one. The force and the response are in phase, e.g., if the force is to the right, the equilibrium will be offset to the right. This is the situation depicted in the amplitude graph of figure h at $$\omega=0$$. The response, which is not zero, is simply this static displacement of the oscillator to one side.

At high frequencies, on the other hand, imagine shaking the poor child on the swing back and forth with a force that oscillates at 10 Hz. This is so fast that there is essentially no time for the force $$F=-kx$$ from gravity and the chain to act from one cycle to the next. The problem becomes equivalent to the oscillation of a free object. If the driving force varies like $$\sin(\omega t)$$, with $$\delta=0$$, then the acceleration is also proportional to the sine. Integrating, we find that the velocity goes like minus a cosine, and a second integration gives a position that varies as minus the sine --- opposite in phase to the driving force. Intuitively, this mathematical result corresponds to the fact that at the moment when the object has reached its maximum displacement to the right, that is the time when the greatest force is being applied to the left, in order to turn it around and bring it back toward the center.

##### Example 45: A practice mute for a violin
The amplitude of the driven vibrations, $$A=F_m/(m|\omega^2-\omega_\text{o}^2|)$$, contains an inverse proportionality to the mass of the vibrating object. This is simply because a given force will produce less acceleration when applied to a more massive object. An application is shown in figure 45.

In a stringed instrument, the strings themselves don't have enough surface area to excite sound waves very efficiently. In instruments of the violin family, as the strings vibrate from left to right, they cause the bridge (the piece of wood they pass over) to wiggle clockwise and counterclockwise, and this motion is transmitted to the top panel of the instrument, which vibrates and creates sound waves in the air.

A string player who wants to practice at night without bothering the neighbors can add some mass to the bridge. Adding mass to the bridge causes the amplitude of the vibrations to be smaller, and the sound to be much softer. A similar effect is seen when an electric guitar is used without an amp. The body of an electric guitar is so much more massive than the body of an acoustic guitar that the amplitude of its vibrations is very small.

i / Example 45: a viola without a mute (left), and with a mute (right). The mute doesn't touch the strings themselves.

The extension of the analysis to the damped case involves some lengthy algebra, which I've outlined on page 912 in appendix 2. The results are shown in figure j. It's not surprising that the steady state response is weaker when there is more damping, since the steady state is reached when the power extracted by damping matches the power input by the driving force. The maximum amplitude, at the peak of the resonance curve, is approximately proportional to $$Q$$.

j / Dependence of the amplitude and phase angle on the driving frequency. The undamped case is $$Q=\infty$$, and the other curves represent $$Q$$=1, 3, and 10. $$F_m$$, $$m$$, and $$\omega_\text{o}$$ are all set to 1.

self-check:

From the final result of the analysis on page 912, substitute $$\omega=\omega_\text{o}$$, and satisfy yourself that the result is proportional to $$Q$$. Why is $$A_{res}\propto Q$$ only an approximation?

What is surprising is that the amplitude is strongly affected by damping close to resonance, but only weakly affected far from it. In other words, the shape of the resonance curve is broader with more damping, and even if we were to scale up a high-damping curve so that its maximum was the same as that of a low-damping curve, it would still have a different shape. The standard way of describing the shape numerically is to give the quantity $$\Delta\omega$$, called the full width at half-maximum, or FWHM, which is defined in figure k. Note that the $$y$$ axis is energy, which is proportional to the square of the amplitude. Our previous observations amount to a statement that $$\Delta\omega$$ is greater when the damping is stronger, i.e., when the $$Q$$ is lower. It's not hard to show from the equations on page 912 that for large $$Q$$, the FWHM is given approximately by

$\begin{equation*} \Delta\omega \approx \omega_\text{o}/Q . \end{equation*}$

Another thing we notice in figure j is that for small values of $$Q$$ the frequency $$\omega_{res}$$ of the maximum $$A$$ is less than $$\omega_\text{o}$$.11 At even lower values of $$Q$$, like $$Q=1$$, the $$A-\omega$$ curve doesn't even have a maximum near $$\omega>0$$.

##### Example 46: An opera singer breaking a wineglass

In order to break a wineglass by singing, an opera singer must first tap the glass to find its natural frequency of vibration, and then sing the same note back, so that her driving force will produce a response with the greatest possible amplitude. If she's shopping for the right glass to use for this display of her prowess, she should look for one that has the greatest possible $$Q$$, since the resonance curve has a higher maximum for higher values of $$Q$$.

##### Example 47: Collapse of the Nimitz Freeway

Figure l shows a section of the Nimitz Freeway in Oakland, CA, that collapsed during an earthquake in 1989. An earthquake consists of many low-frequency vibrations that occur simultaneously, which is why it sounds like a rumble of indeterminate pitch rather than a low hum. The frequencies that we can hear are not even the strongest ones; most of the energy is in the form of vibrations in the range of frequencies from about 1 Hz to 10 Hz.

All the structures we build are resting on geological layers of dirt, mud, sand, or rock. When an earthquake wave comes along, the topmost layer acts like a system with a certain natural frequency of vibration, sort of like a cube of jello on a plate being shaken from side to side. The resonant frequency of the layer depends on how stiff it is and also on how deep it is. The ill-fated section of the Nimitz freeway was built on a layer of mud, and analysis by geologist Susan E. Hough of the U.S. Geological Survey shows that the mud layer's resonance was centered on about 2.5 Hz, and had a width covering a range from about 1 Hz to 4 Hz.

When the earthquake wave came along with its mixture of frequencies, the mud responded strongly to those that were close to its own natural 2.5 Hz frequency. Unfortunately, an engineering analysis after the quake showed that the overpass itself had a resonant frequency of 2.5 Hz as well! The mud responded strongly to the earthquake waves with frequencies close to 2.5 Hz, and the bridge responded strongly to the 2.5 Hz vibrations of the mud, causing sections of it to collapse.

#### Physical reason for the relationship between Q and the FWHM

What is the reason for this surprising relationship between the damping and the width of the resonance? Fundamentally, it has to do with the fact that friction causes a system to lose its “memory” of its previous state. If the Pioneer 10 space probe, coasting through the frictionless vacuum of interplanetary space, is detected by aliens a million years from now, they will be able to trace its trajectory backwards and infer that it came from our solar system. On the other hand, imagine that I shove a book along a tabletop, it comes to rest, and then someone else walks into the room. There will be no clue as to which direction the book was moving before it stopped --- friction has erased its memory of its motion. Now consider the playground swing driven at twice its natural frequency, figure m, where the undamped case is repeated from figure b on page 171. In the undamped case, the first push starts the swing moving with momentum $$p$$, but when the second push comes, if there is no friction at all, it now has a momentum of exactly $$-p$$, and the momentum transfer from the second push is exactly enough to stop it dead. With moderate damping, however, the momentum on the rebound is not quite $$-p$$, and the second push's effect isn't quite as disastrous. With very strong damping, the swing comes essentially to rest long before the second push. It has lost all its memory, and the second push puts energy into the system rather than taking it out. Although the detailed mathematical results with this kind of impulsive driving force are different,12 the general results are the same as for sinusoidal driving: the less damping there is, the greater the penalty you pay for driving the system off of resonance.

##### Example 48: High-Q speakers
Most good audio speakers have $$Q\approx1$$, but the resonance curve for a higher-$$Q$$ oscillator always lies above the corresponding curve for one with a lower $$Q$$, so people who want their car stereos to be able to rattle the windows of the neighboring cars will often choose speakers that have a high $$Q$$. Of course they could just use speakers with stronger driving magnets to increase $$F_m$$, but the speakers might be more expensive, and a high-$$Q$$ speaker also has less friction, so it wastes less energy as heat.

One problem with this is that whereas the resonance curve of a low-$$Q$$ speaker (its “response curve” or “frequency response” in audiophile lingo) is fairly flat, a higher-$$Q$$ speaker tends to emphasize the frequencies that are close to its natural resonance. In audio, a flat response curve gives more realistic reproduction of sound, so a higher quality factor, $$Q$$, really corresponds to a lower -quality speaker.

Another problem with high-$$Q$$ speakers is discussed in example 51 on page 185 .

##### Example 49: Changing the pitch of a wind instrument

$$\triangleright$$ A saxophone player normally selects which note to play by choosing a certain fingering, which gives the saxophone a certain resonant frequency. The musician can also, however, change the pitch significantly by altering the tightness of her lips. This corresponds to driving the horn slightly off of resonance. If the pitch can be altered by about 5% up or down (about one musical half-step) without too much effort, roughly what is the $$Q$$ of a saxophone?

$$\triangleright$$ Five percent is the width on one side of the resonance, so the full width is about 10%, $$\Delta f/f_\text{o}\approx 0.1$$. The equation $$\Delta\omega=\omega_\text{o}/ Q$$ is defined in terms of angular frequency, $$\omega=2\pi\ f$$, and we've been given our data in terms of ordinary frequency, $$f$$. The factors of $$2\pi$$ end up canceling out, however:

\begin{align*} Q &= \frac{\omega_\text{o}}{\Delta\omega} \\ & = \frac{2\pi f_\text{o}}{2\pi\Delta f} \\ & = \frac{f_\text{o}}{f} \\ & \approx 10 \end{align*}

In other words, once the musician stops blowing, the horn will continue sounding for about 10 cycles before its energy falls off by a factor of 535. (Blues and jazz saxophone players will typically choose a mouthpiece that gives a low $$Q$$, so that they can produce the bluesy pitch-slides typical of their style. “Legit,” i.e., classically oriented players, use a higher-$$Q$$ setup because their style only calls for enough pitch variation to produce a vibrato, and the higher $$Q$$ makes it easier to play in tune.)

$$\triangleright$$ A radio receiver used in the FM band needs to be tuned in to within about 0.1 MHz for signals at about 100 MHz. What is its $$Q$$?

$$\triangleright$$ As in the last example, we're given data in terms of $$f$$s, not $$\omega$$s, but the factors of $$2\pi$$ cancel. The resulting $$Q$$ is about 1000, which is extremely high compared to the $$Q$$ values of most mechanical systems.

#### Transients

What about the motion before the steady state is achieved? When we computed the undriven motion numerically on page 172, the program had to initialize the position and velocity. By changing these two variables, we could have gotten any of an infinite number of simulations.13 The same is true when we have an equation of motion with a driving term, $$ma+bv+kx = F_m \sin \omega{}t$$ (p. , equation \eqref{eqn:resonancemotion}). The steady-state solutions, however, have no adjustable parameters at all --- $$A$$ and $$\delta$$ are uniquely determined by the parameters of the driving force and the oscillator itself. If the oscillator isn't initially in the steady state, then it will not have the steady-state motion at first. What kind of motion will it have?

The answer comes from realizing that if we start with the solution to the driven equation of motion, and then add to it any solution to the free equation of motion, the result,

$\begin{equation*} x = A \sin (\omega t+\delta) + A' e^{-ct}\sin (\omega_f t+\delta') , \end{equation*}$

is also a solution of the driven equation. Here, as before, $$\omega_f$$ is the frequency of the free oscillations ($$\omega_f\approx\omega_\text{o}$$ for small $$Q$$), $$\omega$$ is the frequency of the driving force, $$A$$ and $$\delta$$ are related as usual to the parameters of the driving force, and $$A'$$ and $$\delta'$$ can have any values at all. Given the initial position and velocity, we can always choose $$A'$$ and $$\delta'$$ to reproduce them, but this is not something one often has to do in real life. What's more important is to realize that the second term dies out exponentially over time, decaying at the same rate at which a free vibration would. For this reason, the $$A'$$ term is called a transient. A high-$$Q$$ oscillator's transients take a long time to die out, while a low-$$Q$$ oscillator always settles down to its steady state very quickly.

##### Example 51: Boomy bass
In example 48 on page 183, I've already discussed one of the drawbacks of a high-$$Q$$ speaker, which is an uneven response curve. Another problem is that in a high-$$Q$$ speaker, transients take a long time to die out. The bleeding-eardrums crowd tend to focus mostly on making their bass loud, so it's usually their woofers that have high $$Q$$s. The result is that bass notes, “ring” after the onset of the note, a phenomenon referred to as “boomy bass.”

#### Overdamped motion

The treatment of free, damped motion on page 174 skipped over a subtle point: in the equation $$\omega_f = \sqrt{k/m-b^2/4m^2} = \omega_\text{o}\sqrt{1-1/4Q^2}$$, $$Q\lt1/2$$ results in an answer that is the square root of a negative number. For example, suppose we had $$k=0$$, which corresponds to a neutral equilibrium. A physical example would be a mass sitting in a tub of syrup. If we set it in motion, it won't oscillate --- it will simply slow to a stop. This system has $$Q=0$$. The equation of motion in this case is $$ma+bv=0$$, or, more suggestively,

$\begin{equation*} m\frac{d{}v}{d{}t}+bv=0 . \end{equation*}$

One can easily verify that this has the solution $$v=\text{(constant)}e^{-bt/m}$$, and integrating, we find $$x=\text{(constant)}e^{-bt/m}+\text{(constant)}$$. In other words, the reason $$\omega_f$$ comes out to be mathematical nonsense14 is that we were incorrect in assuming a solution that oscillated at a frequency $$\omega_f$$. The actual motion is not oscillatory at all.

In general, systems with $$Q\lt1/2$$, called overdamped systems, do not display oscillatory motion. Most cars' shock absorbers are designed with $$Q\approx1/2$$, since it's undesirable for the car to undulate up and down for a while after you go over a bump. (Shocks with extremely low values of $$Q$$ are not good either, because such a system takes a very long time to come back to equilibrium.) It's not particularly important for our purposes, but for completeness I'll note, as you can easily verify, that the general solution to the equation of motion for $$0\ltQ\lt1/2$$ is of the form $$x=Ae^{-ct}+Be^{-dt}$$, while $$Q=1/2$$, called the critically dampedcase, gives $$x=(A+Bt)e^{-ct}$$.

## 3.4 Motion In Three Dimensions

a / The car can change its $$x$$ and $$y$$ motions by one square every turn.

d / Two surfaces that could be used to extract energy from a stream of water.

e / An asteroid absorbs visible light from the sun, and gets rid of the energy by radiating infrared light.

### 3.4.1 The Cartesian perspective

When my friends and I were bored in high school, we used to play a paper-and-pencil game which, although we never knew it, was Very Educational --- in fact, it pretty much embodies the entire world-view of classical physics. To play the game, you draw a racetrack on graph paper, and try to get your car around the track before anyone else. The default is for your car to continue at constant speed in a straight line, so if it moved three squares to the right and one square up on your last turn, it will do the same this turn. You can also control the car's motion by changing its $$\Delta x$$ and $$\Delta y$$ by up to one unit. If it moved three squares to the right last turn, you can have it move anywhere from two to four squares to the right this turn.

b / French mathematician René Descartes invented analytic geometry; Cartesian ($$xyz$$) coordinates are named after him. He did work in philosophy, and was particularly interested in the mind-body problem. He was a skeptic and an antiaristotelian, and, probably for fear of religious persecution, spent his adult life in the Netherlands, where he fathered a daughter with a Protestant peasant whom he could not marry. He kept his daughter's existence secret from his enemies in France to avoid giving them ammunition, but he was crushed when she died of scarlatina at age 5. A pious Catholic, he was widely expected to be sainted. His body was buried in Sweden but then reburied several times in France, and along the way everything but a few fingerbones was stolen by peasants who expected the body parts to become holy relics.

The fundamental way of dealing with the direction of an object's motion in physics is to use conservation of momentum, since momentum depends on direction. Up until now, we've only done momentum in one dimension. How does this relate to the racetrack game? In the game, the motion of a car from one turn to the next is represented by its $$\Delta x$$ and $$\Delta y$$. In one dimension, we would only need $$\Delta x$$, which could be related to the velocity, $$\Delta x/\Delta t$$, and the momentum, $$m\Delta x/\Delta t$$. In two dimensions, the rules of the game amount to a statement that if there is no momentum transfer, then both $$m\Delta x/\Delta t$$ and $$m\Delta y/\Delta t$$ stay the same. In other words, there are two flavors of momentum, and they are separately conserved. All of this so far has been done with an artificial division of time into “turns,” but we can fix that by redefining everything in terms of derivatives, and for motion in three dimensions rather than two, we augment $$x$$ and $$y$$ with $$z$$:

\begin{align*} v_x &= dx/dt & v_y &= dy/dt & v_z &= dz/dt \\ \text{and} p_x &= mv_x & p_y &= mv_y &p_z &= mv_z \end{align*}

We call these the $$x$$, $$y$$, and $$z$$ components of the velocity and the momentum.

There is both experimental and theoretical evidence that the $$x$$, $$y$$, and $$z$$ momentum components are separately conserved, and that a momentum transfer (force) along one axis has no effect on the momentum components along the other two axes. On page 89, for example, I argued that it was impossible for an air hockey puck to make a 180-degree turn spontaneously, because then in the frame moving along with the puck, it would have begun moving after starting from rest. Now that we're working in two dimensions, we might wonder whether the puck could spontaneously make a 90-degree turn, but exactly the same line of reasoning shows that this would be impossible as well, which proves that the puck can't trade $$x$$-momentum for $$y$$-momentum. A more general proof of separate conservation will be given on page 214, after some of the appropriate mathematical techniques have been introduced.

c / Bullets are dropped and shot at the same time.

As an example of the experimental evidence for separate conservation of the momentum components, figure c shows correct and incorrect predictions of what happens if you shoot a rifle and arrange for a second bullet to be dropped from the same height at exactly the same moment when the first one left the barrel. Nearly everyone expects that the dropped bullet will reach the dirt first, and Aristotle would have agreed, since he believed that the bullet had to lose its horizontal motion before it could start moving vertically. In reality, we find that the vertical momentum transfer between the earth and the bullet is completely unrelated to the horizontal momentum. The bullet ends up with $$p_y\lt0$$, while the planet picks up an upward momentum $$p_y>0$$, and the total momentum in the $$y$$ direction remains zero. Both bullets hit the ground at the same time. This is much simpler than the Aristotelian version!

##### Example 52: The Pelton waterwheel

$$\triangleright$$ There is a general class of machines that either do work on a gas or liquid, like a boat's propeller, or have work done on them by a gas or liquid, like the turbine in a hydroelectric power plant. Figure d shows two types of surfaces that could be attached to the circumference of an old-fashioned waterwheel. Compare the force exerted by the water in the two cases.

$$\triangleright$$ Let the $$x$$ axis point to the right, and the $$y$$ axis up. In both cases, the stream of water rushes down onto the surface with momentum $$p_{y,i}=- p_\text{o}$$, where the subscript $$i$$ stands for “initial,” i.e., before the collision.

In the case of surface 1, the streams of water leaving the surface have no momentum in the $$y$$ direction, and their momenta in the $$x$$ direction cancel. The final momentum of the water is zero along both axes, so its entire momentum, $$- p_\text{o}$$, has been transferred to the waterwheel.

When the water leaves surface 2, however, its momentum isn't zero. If we assume there is no friction, it's $$p_{y,f}= +p_\text{o}$$, with the positive sign indicating upward momentum. The change in the water's momentum is $$p_{y,f}- p_{y,i}=2 p_\text{o}$$, and the momentum transferred to the waterwheel is $$-2 p_\text{o}$$.

Force is defined as the rate of transfer of momentum, so surface 2 experiences double the force. A waterwheel constructed in this way is known as a Pelton waterwheel.

##### Example 53: The Yarkovsky effect
We think of the planets and asteroids as inhabiting their orbits permanently, but it is possible for an orbit to change over periods of millions or billions of years, due to a variety of effects. For asteroids with diameters of a few meters or less, an important mechanism is the Yarkovsky effect, which is easiest to understand if we consider an asteroid spinning about an axis that is exactly perpendicular to its orbital plane.

The illuminated side of the asteroid is relatively hot, and radiates more infrared light than the dark (night) side. Light has momentum, and a total force away from the sun is produced by combined effect of the sunlight hitting the asteroid and the imbalance between the momentum radiated away on the two sides. This force, however, doesn't cause the asteroid's orbit to change over time, since it simply cancels a tiny fraction of the sun's gravitational attraction. The result is merely a tiny, undetectable violation of Kepler's law of periods.

Consider the sideways momentum transfers, however. In figure e, the part of the asteroid on the right has been illuminated for half a spin-period (half a “day”) by the sun, and is hot. It radiates more light than the morning side on the left. This imbalance produces a total force in the $$x$$ direction which points to the left. If the asteroid's orbital motion is to the left, then this is a force in the same direction as the motion, which will do positive work, increasing the asteroid's energy and boosting it into an orbit with a greater radius. On the other hand, if the asteroid's spin and orbital motion are in opposite directions, the Yarkovsky push brings the asteroid spiraling in closer to the sun.

Calculations show that it takes on the order of $$10^7$$ to $$10^8$$ years for the Yarkovsky effect to move an asteroid out of the asteroid belt and into the vicinity of earth's orbit, and this is about the same as the typical age of a meteorite as estimated by its exposure to cosmic rays. The Yarkovsky effect doesn't remove all the asteroids from the asteroid belt, because many of them have orbits that are stabilized by gravitational interactions with Jupiter. However, when collisions occur, the fragments can end up in orbits which are not stabilized in this way, and they may then end up reaching the earth due to the Yarkovsky effect. The cosmic-ray technique is really telling us how long it has been since the fragment was broken out of its parent.

##### Discussion Questions

The following is an incorrect explanation of a fact about target shooting:

“Shooting a high-powered rifle with a high muzzle velocity is different from shooting a less powerful gun. With a less powerful gun, you have to aim quite a bit above your target, but with a more powerful one you don't have to aim so high because the bullet doesn't drop as fast.”

What is the correct explanation?

f / Discussion question A.

You have thrown a rock, and it is flying through the air in an arc. If the earth's gravitational force on it is always straight down, why doesn't it just go straight down once it leaves your hand?

Consider the example of the bullet that is dropped at the same moment another bullet is fired from a gun. What would the motion of the two bullets look like to a jet pilot flying alongside in the same direction as the shot bullet and at the same horizontal speed?

g / Two balls roll down a cone and onto a plane.

### 3.4.2 Rotational invariance

The Cartesian approach requires that we choose $$x$$, $$y$$, and $$z$$ axes. How do we choose them correctly? The answer is that it had better not matter which directions the axes point (provided they're perpendicular to each other), or where we put the origin, because if it did matter, it would mean that space was asymmetric. If there was a certain point in the universe that was the right place to put the origin, where would it be? The top of Mount Olympus? The United Nations headquarters? We find that experiments come out the same no matter where we do them, and regardless of which way the laboratory is oriented, which indicates that no location in space or direction in space is special in any way.15

This is closely related to the idea of Galilean relativity stated on page 62, from which we already know that the absolute motion of a frame of reference is irrelevant and undetectable. Observers using frames of reference that are in motion relative to each other will not even agree on the permanent identity of a particular point in space, so it's not possible for the laws of physics to depend on where you are in space. For instance, if gravitational energies were proportional to $$m_1m_2$$ in one location but to $$(m_1m_2)^{1.00001}$$ in another, then it would be possible to determine when you were in a state of absolute motion, because the behavior of gravitational interactions would change as you moved from one region to the other.

Because of this close relationship, we restate the principle of Galilean relativity in a more general form. This extended principle of Galilean relativity states that the laws of physics are no different in one time and place than in another, and that they also don't depend on your orientation or your motion, provided that your motion is in a straight line and at constant speed.

The irrelevance of time and place could have been stated in chapter 1, but since this section is the first one in which we're dealing with three-dimensional physics in full generality, the irrelevance of orientation is what we really care about right now. This property of the laws of physics is called rotational invariance. The word “invariance” means a lack of change, i.e., the laws of physics don't change when we reorient our frame of reference.

##### Example 54: Rotational invariance of gravitational interactions

Gravitational energies depend on the quantity 1/$$r$$, which by the Pythagorean theorem equals

$\begin{equation*} \frac{1}{\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}} . \end{equation*}$

Rotating a line segment doesn't change its length, so this expression comes out the same regardless of which way we orient our coordinate axes. Even though $$\Delta x$$, $$\Delta y$$, and $$\Delta z$$ are different in differently oriented coordinate systems, $$r$$ is the same.

##### Example 55: Kinetic energy
Kinetic energy equals $$\text{(1/2)} mv^2$$, but what does that mean in three dimensions, where we have $$v_x$$, $$v_y$$, and $$v_z$$? If you were tempted to add the components and calculate $$K=(1/2) m( v_x+ v_y+ v_z)^2$$, figure g should convince you otherwise. Using that method, we'd have to assign a kinetic energy of zero to ball number 1, since its negative $$v_y$$ would exactly cancel its positive $$v_x$$, whereas ball number 2's kinetic energy wouldn't be zero. This would violate rotational invariance, since the balls would behave differently.

The only possible way to generalize kinetic energy to three dimensions, without violating rotational invariance, is to use an expression that resembles the Pythagorean theorem,

$\begin{equation*} v=\sqrt{ v_{x}^2+ v_{y}^2+ v_z^2} , \end{equation*}$

which results in

$\begin{equation*} K=\frac{1}{2} m \left( v_{x}^2+ v_{y}^2+ v_z^2\right) . \end{equation*}$

Since the velocity components are squared, the positive and negative signs don't matter, and the two balls in the example behave the same way.

h / Example 56.

i / Example 57.

j / The geometric interpretation of a vector's components.

k / Two vectors, 1, to which we apply the same operation in two different frames of reference, 2 and 3.

l / Example 59.

m / Example 61.

n / Example 63.

p / Adding vectors graphically by placing them tip-to-tail, like a train.

q / Example 65

r / Example 68.

s / Archimedes' principle works regardless of whether the object is a cube. The fluid makes a force on every square millimeter of the object's surface.

t / Example 70.

u / An artist's rendering of what Cosmos 1 would have looked like in orbit.

w / Discussion question E.

x / Discussion question H.

### 3.4.3 Vectors

Remember the title of this book? It would have been possible to obtain the result of example 55 by applying the Pythagorean theorem to $$dx$$, $$dy$$, and $$dz$$, and then dividing by $$dt$$, but the rotational invariance approach is simpler, and is useful in a much broader context. Even with a quantity you presently know nothing about, say the magnetic field, you can infer that if the components of the magnetic field are $$B_x$$, $$B_y$$, and $$B_z$$, then the physically useful way to talk about the strength of the magnetic field is to define it as $$\sqrt{B_x^2+B_y^2+B_z^2}$$. Nature knows your brain cells are precious, and doesn't want you to have to waste them by memorizing mathematical rules that are different for magnetic fields than for velocities.

When mathematicians see that the same set of techniques is useful in many different contexts, that's when they start making definitions that allow them to stop reinventing the wheel. The ancient Greeks, for example, had no general concept of fractions. They couldn't say that a circle's radius divided by its diameter was equal to the number 1/2. They had to say that the radius and the diameter were in the ratio of one to two. With this limited number concept, they couldn't have said that water was dripping out of a tank at a rate of 3/4 of a barrel per day; instead, they would have had to say that over four days, three barrels worth of water would be lost. Once enough of these situations came up, some clever mathematician finally realized that it would make sense to define something called a fraction, and that one could think of these fraction thingies as numbers that lay in the gaps between the traditionally recognized numbers like zero, one, and two. Later generations of mathematicians introduced further subversive generalizations of the number concepts, inventing mathematical creatures like negative numbers, and the square root of two, which can't be expressed as a fraction.

In this spirit, we define a vector as any quantity that has both an amount and a direction in space. In contradistinction, a scalar has an amount, but no direction. Time and temperature are scalars. Velocity, acceleration, momentum, and force are vectors. In one dimension, there are only two possible directions, and we can use positive and negative numbers to indicate the two directions. In more than one dimension, there are infinitely many possible directions, so we can't use the two symbols $$+$$ and $$-$$ to indicate the direction of a vector. Instead, we can specify the three components of the vector, each of which can be either negative or positive. We represent vector quantities in handwriting by writing an arrow above them, so for example the momentum vector looks like this, $$\vec{p}$$, but the arrow looks ugly in print, so in books vectors are usually shown in bold-face type: p. A straightforward way of thinking about vectors is that a vector equation really represents three different equations. For instance, conservation of momentum could be written in terms of the three components,

\begin{align*} \Delta p_x &= 0 \\ \Delta p_y &= 0 \\ \Delta p_z &= 0 , \end{align*}

or as a single vector equation,16

$\begin{equation*} \Delta \mathbf{p} = 0 . \end{equation*}$

The following table summarizes some vector operations.

 operation definition mathbfvector sqrtvectorx2vectory2vectorz2 mathbfvectormathbfvector Add component by component. mathbfvectormathbfvector Subtract component by component. mathbfvector cdot scalar Multiply each component by the scalar. mathbfvector scalar Divide each component by the scalar.

{}The first of these is called the magnitude of the vector; in one dimension, where a vector only has one component, it amounts to taking the absolute value, hence the similar notation.

self-check:

Translate the equations $$F_{x}\text{=} ma_x$$, $$F_{y}\text{=} ma_y$$, and $$F_{z}\text{=} ma_z$$ into a single equation in vector notation.

##### Example 56: An explosion
$$\triangleright$$ Astronomers observe the planet Mars as the Martians fight a nuclear war. The Martian bombs are so powerful that they rip the planet into three separate pieces of liquefied rock, all having the same mass. If one fragment flies off with velocity components $$v_{1 x}=0$$, $$v_{1 y}\text{=1.0x10}^4$$ km/hr, and the second with $$v_{2 x}\text{=1.0x10}^4$$ km/hr, $$v_{2 y}=0$$, what is the magnitude of the third one's velocity?

$$\triangleright$$ We work the problem in the center of mass frame, in which the planet initially had zero momentum. After the explosion, the vector sum of the momenta must still be zero. Vector addition can be done by adding components, so

\begin{align*} mv_{1 x} + mv_{2 x} + mv_{3 x} &= 0 \\ \text{and} mv_{1 y} + mv_{2 y} + mv_{3 y} &= 0 , \end{align*}

where we have used the same symbol $$m$$ for all the terms, because the fragments all have the same mass. The masses can be eliminated by dividing each equation by $$m$$, and we find

\begin{align*} v_{3 x} &= -\text{1.0x10}^4\ \text{km/hr} \\ v_{3 y} &= -\text{1.0x10}^4\ \text{km/hr} , \end{align*}

which gives a magnitude of

\begin{align*} |\mathbf{v}_3| &= \sqrt{ v_{3 x}^2+ v_{3 y}^2} \\ &= \text{1.4x10}^4\ \text{km/hr} . \end{align*}

##### Example 57: A toppling box
If you place a box on a frictionless surface, it will fall over with a very complicated motion that is hard to predict in detail. We know, however, that its center of mass's motion is related to its momentum, and the rate at which momentum is transferred is the force. Moreover, we know that these relationships apply separately to each component. Let $$x$$ and $$y$$ be horizontal, and $$z$$ vertical. There are two forces on the box, an upward force from the table and a downward gravitational force. Since both of these are along the $$z$$ axis, $$p_z$$ is the only component of the box's momentum that can change. We conclude that the center of mass travels vertically. This is true even if the box bounces and tumbles. [Based on an example by Kleppner and Kolenkow.]

#### Geometric representation of vectors

A vector in two dimensions can be easily visualized by drawing an arrow whose length represents its magnitude and whose direction represents its direction. The $$x$$ component of a vector can then be visualized, j, as the length of the shadow it would cast in a beam of light projected onto the $$x$$ axis, and similarly for the $$y$$ component. Shadows with arrowheads pointing back against the direction of the positive axis correspond to negative components.

In this type of diagram, the negative of a vector is the vector with the same magnitude but in the opposite direction. Multiplying a vector by a scalar is represented by lengthening the arrow by that factor, and similarly for division.

self-check:

Given vector Q represented by an arrow below, draw arrows representing the vectors 1.5Q and $$-\mathbf{Q}$$.\ \begin{center}\end{center}

[4]{A useless vector operation} The way I've defined the various vector operations above aren't as arbitrary as they seem. There are many different vector operations that we could define, but only some of the possible definitions are mathematically useful. Consider the operation of multiplying two vectors component by component to produce a third vector:

\begin{align*} R_{x} & = P_{x} Q_x \\ R_{y} & = P_{y} Q_y \\ R_{z} & = P_{z} Q_z \end{align*}

As a simple example, we choose vectors P and Q to have length 1, and make them perpendicular to each other, as shown in figure k/1. If we compute the result of our new vector operation using the coordinate system shown in k/2, we find:

\begin{align*} R_{x} &= 0 \\ R_{y} &= 0 \\ R_z &= 0 \end{align*}

The $$x$$ component is zero because $$P_x =0$$, the $$y$$ component is zero because $$Q_y=0$$, and the $$z$$ component is of course zero because both vectors are in the $$x$$-$$y$$ plane. However, if we carry out the same operations in coordinate system k/3, rotated 45 degrees with respect to the previous one, we find

\begin{align*} R_{x} &= -\text{1/2} \\ R_{y} &= \text{1/2} \\ R_z &= 0 \end{align*}

The operation's result depends on what coordinate system we use, and since the two versions of R have different lengths (one being zero and the other nonzero), they don't just represent the same answer expressed in two different coordinate systems. Such an operation will never be useful in physics, because experiments show physics works the same regardless of which way we orient the laboratory building! The useful vector operations, such as addition and scalar multiplication, are rotationally invariant, i.e., come out the same regardless of the orientation of the coordinate system.

All the vector techniques can be applied to any kind of vector, but the graphical representation of vectors as arrows is particularly natural for vectors that represent lengths and distances. We define a vector called $$\mathbf{r}$$ whose components are the coordinates of a particular point in space, $$x$$, $$y$$, and $$z$$. The $$\Delta\mathbf{r}$$ vector, whose components are $$\Delta x$$, $$\Delta y$$, and $$\Delta z$$, can then be used to represent motion that starts at one point and ends at another. Adding two $$\Delta \mathbf{r}$$ vectors is interpreted as a trip with two legs: by computing the $$\Delta \mathbf{r}$$ vector going from point A to point B plus the vector from B to C, we find the vector that would have taken us directly from A to C.

#### Calculations with magnitude and direction

If you ask someone where Las Vegas is compared to Los Angeles, she is unlikely to say that the $$\Delta x$$ is 290 km and the $$\Delta y$$ is 230 km, in a coordinate system where the positive $$x$$ axis is east and the $$y$$ axis points north. She will probably say instead that it's 370 km to the northeast. If she was being precise, she might specify the direction as $$38°$$ counterclockwise from east. In two dimensions, we can always specify a vector's direction like this, using a single angle. A magnitude plus an angle suffice to specify everything about the vector. The following two examples show how we use trigonometry and the Pythagorean theorem to go back and forth between the $$x$$-$$y$$ and magnitude-angle descriptions of vectors.

##### Example 58: Finding magnitude and angle from components
$$\triangleright$$ Given that the $$\Delta\mathbf{r}$$ vector from LA to Las Vegas has $$\Delta x$$=290 km and $$\Delta y$$=230 km, how would we find the magnitude and direction of $$\Delta\mathbf{r}$$?

$$\triangleright$$ We find the magnitude of $$\Delta\mathbf{r}$$ from the Pythagorean theorem:

\begin{align*} |\Delta\mathbf{r}| &= \sqrt{\Delta x^2+\Delta y^2} \\ &= 370\ \text{km} \end{align*}

We know all three sides of the triangle, so the angle $$\theta$$ can be found using any of the inverse trig functions. For example, we know the opposite and adjacent sides, so

\begin{align*} \theta &= \text{tan}^{-1} \frac{\Delta y}{\Delta x}\\ &= 38° . \end{align*}

##### Example 59: Finding the components from the magnitude and angle
$$\triangleright$$ Given that the straight-line distance from Los Angeles to Las Vegas is 370 \textup{km}, and that the angle $$\theta$$ in the figure is 38°, how can the $$x$$ and $$y$$ components of the $$\Delta\mathbf{r}$$ vector be found?

$$\triangleright$$ The sine and cosine of $$\theta$$ relate the given information to the information we wish to find:

\begin{align*} \text{cos}\ \theta &= \frac{\Delta x}{|\Delta\mathbf{r}|}\\ \text{sin}\ \theta &= \frac{\Delta y}{|\Delta\mathbf{r}|} \end{align*}

Solving for the unknowns gives

\begin{align*} \Delta x &= |\Delta\mathbf{r}|\:\text{cos}\ \theta\\ &= 290\ \text{km}\\ \Delta y &= |\Delta\mathbf{r}|\:\text{sin}\ \theta\\ &= 230\ \text{km} \end{align*}

The following example shows the correct handling of the plus and minus signs, which is usually the main cause of mistakes by students.

##### Example 60: Negative components
$$\triangleright$$ San Diego is 120 km east and 150 km south of Los Angeles. An airplane pilot is setting course from San Diego to Los Angeles. At what angle should she set her course, measured counterclockwise from east, as shown in the figure?

$$\triangleright$$ If we make the traditional choice of coordinate axes, with $$x$$ pointing to the right and $$y$$ pointing up on the map, then her $$\Delta x$$ is negative, because her final $$x$$ value is less than her initial $$x$$ value. Her $$\Delta y$$ is positive, so we have

\begin{align*} \Delta x &= -120\ \text{km}\\ \Delta y &= 150\ \text{km} . \end{align*}

If we work by analogy with the example 59, we get

\begin{align*} \theta &= \text{tan}^{-1} \frac{\Delta y}{\Delta x}\\ &= \text{tan}^{-1}\left(- 1.25\right) \\ &= -51° . \end{align*}

According to the usual way of defining angles in trigonometry, a negative result means an angle that lies clockwise from the $$x$$ axis, which would have her heading for the Baja California. What went wrong? The answer is that when you ask your calculator to take the arctangent of a number, there are always two valid possibilities differing by 180°. That is, there are two possible angles whose tangents equal -1.25:

\begin{align*} \text{tan}\ 129° &= - 1.25 \\ \text{tan}\left(-51°\right) &= - 1.25 \end{align*}

You calculator doesn't know which is the correct one, so it just picks one. In this case, the one it picked was the wrong one, and it was up to you to add $$180°$$ to it to find the right answer.

##### Example 61: A shortcut
$$\triangleright$$ A split second after nine o'clock, the hour hand on a clock dial has moved clockwise past the nine-o'clock position by some imperceptibly small angle $$\phi$$. Let positive $$x$$ be to the right and positive $$y$$ up. If the hand, with length $$\ell$$, is represented by a $$\Delta\mathbf{r}$$ vector going from the dial's center to the tip of the hand, find this vector's $$\Delta x$$.

$$\triangleright$$ The following shortcut is the easiest way to work out examples like these, in which a vector's direction is known relative to one of the axes. We can tell that $$\Delta\mathbf{r}$$ will have a large, negative $$x$$ component and a small, positive $$y$$. Since $$\Delta x\lt0$$, there are really only two logical possibilities: either $$\Delta x = -\ell \cos\phi$$, or $$\Delta x = -\ell \sin\phi$$. Because $$\phi$$ is small, $$\cos\phi$$ is large and $$\sin\phi$$ is small. We conclude that $$\Delta x = -\ell \cos\phi$$.

A typical application of this technique to force vectors is given in example 71 on p. 205.

#### Addition of vectors given their components

The easiest type of vector addition is when you are in possession of the components, and want to find the components of their sum.

##### Example 62: San Diego to Las Vegas
$$\triangleright$$ Given the $$\Delta x$$ and $$\Delta y$$ values from the previous examples, find the $$\Delta x$$ and $$\Delta y$$ from San Diego to Las Vegas.

$$\triangleright$$

\begin{align*} \Delta x_{total} &= \Delta x_1 + \Delta x_2 \\ &= -120\ \text{km} + 290\ \text{km} \\ &= 170\ \text{km} \\ \Delta y_{total} &= \Delta y_1 + \Delta y_2 \\ &= 150\ \text{km} + 230\ \text{km} \\ &= 380 \end{align*}

#### Addition of vectors given their magnitudes and directions

In this case, you must first translate the magnitudes and directions into components, and the add the components.

Often the easiest way to add vectors is by making a scale drawing on a piece of paper. This is known as graphical addition, as opposed to the analytic techniques discussed previously.

o / Example 64.

##### Example 63: From San Diego to Las Vegas, graphically
$$\triangleright$$ Given the magnitudes and angles of the $$\Delta\mathbf{r}$$ vectors from San Diego to Los Angeles and from Los Angeles to Las Vegas, find the magnitude and angle of the $$\Delta\mathbf{r}$$ vector from San Diego to Las Vegas.

$$\triangleright$$ Using a protractor and a ruler, we make a careful scale drawing, as shown in figure o on page 200. A scale of 1 cm$$\leftrightarrow$$10 \textup{km} was chosen for this solution. With a ruler, we measure the distance from San Diego to Las Vegas to be 3.8 cm, which corresponds to 380 \textup{km}. With a protractor, we measure the angle $$\theta$$ to be 71°.

Even when we don't intend to do an actual graphical calculation with a ruler and protractor, it can be convenient to diagram the addition of vectors in this way, as shown in figure p. With $$\Delta\mathbf{r}$$ vectors, it intuitively makes sense to lay the vectors tip-to-tail and draw the sum vector from the tail of the first vector to the tip of the second vector. We can do the same when adding other vectors such as force vectors.

#### Unit vector notation

When we want to specify a vector by its components, it can be cumbersome to have to write the algebra symbol for each component:

$\begin{equation*} \Delta x = 290\ \text{km}, \Delta y = 230\ \text{km} \end{equation*}$

A more compact notation is to write

$\begin{equation*} \Delta\mathbf{r} = (290\ \text{km})\hat{\mathbf{x}} + (230\ \text{km})\hat{\mathbf{y}} , \end{equation*}$

where the vectors $$\hat{\mathbf{x}}$$, $$\hat{\mathbf{y}}$$, and $$\hat{\mathbf{z}}$$, called the unit vectors, are defined as the vectors that have magnitude equal to 1 and directions lying along the $$x$$, $$y$$, and $$z$$ axes. In speech, they are referred to as “x-hat,” “y-hat,” and “z-hat.”

A slightly different, and harder to remember, version of this notation is unfortunately more prevalent. In this version, the unit vectors are called $$\hat{\mathbf{i}}$$, $$\hat{\mathbf{j}}$$, and $$\hat{\mathbf{k}}$$:

$\begin{equation*} \Delta\mathbf{r} = (290\ \text{km})\hat{\mathbf{i}} + (230\ \text{km})\hat{\mathbf{j}} . \end{equation*}$

#### Applications to relative motion, momentum, and force

Vector addition is the correct way to generalize the one-dimensional concept of adding velocities in relative motion, as shown in the following example:

##### Example 64: Velocity vectors in relative motion
$$\triangleright$$ You wish to cross a river and arrive at a dock that is directly across from you, but the river's current will tend to carry you downstream. To compensate, you must steer the boat at an angle. Find the angle $$\theta$$, given the magnitude, $$|\mathbf{v}_{WL}|$$, of the water's velocity relative to the land, and the maximum speed, $$|\mathbf{v}_{BW}|$$, of which the boat is capable relative to the water.

$$\triangleright$$ The boat's velocity relative to the land equals the vector sum of its velocity with respect to the water and the water's velocity with respect to the land,

$\begin{equation*} \mathbf{v}_{BL} = \mathbf{v}_{BW}+ \mathbf{v}_{WL} . \end{equation*}$

If the boat is to travel straight across the river, i.e., along the $$y$$ axis, then we need to have $$\mathbf{v}_{BL,x}=0$$. This $$x$$ component equals the sum of the $$x$$ components of the other two vectors,

$\begin{equation*} \mathbf{v}_BL,x = \mathbf{v}_BW,x + \mathbf{v}_WL,x , \end{equation*}$

or

$\begin{equation*} 0 = -|\mathbf{v}_{BW}| \sin \theta + |\mathbf{v}_{WL}| . \end{equation*}$

Solving for $$\theta$$, we find

\begin{align*} \sin \theta &= |\mathbf{v}_{WL}|/|\mathbf{v}_{BW}| ,\\ \text{so} \theta &= \sin^{-1}\frac{|\mathbf{v}_{WL}|}{\mathbf{v}_{BW}} . \end{align*}

##### Example 65: How to generalize one-dimensional equations

$$\triangleright$$ How can the one-dimensional relationships

$\begin{gather*} p_{total} = m_{total} v_{cm} \text{and} x_{cm} = \frac{\sum_{j}{ m_{j} x_j}}{\sum_{j}{ m_j}} \end{gather*}$

be generalized to three dimensions?

$$\triangleright$$ Momentum and velocity are vectors, since they have directions in space. Mass is a scalar. If we rewrite the first equation to show the appropriate quantities notated as vectors,

$\begin{equation*} \mathbf{p}_{total} = m_{total} \mathbf{v}_{cm} , \end{equation*}$

we get a valid mathematical operation, the multiplication of a vector by a scalar. Similarly, the second equation becomes

$\begin{equation*} \mathbf{r}_{cm} = \frac{\sum_j{ m_{j}\mathbf{r}_j}}{\sum_{j}{ m_j}} , \end{equation*}$

which is also valid. Each term in the sum on top contains a vector multiplied by a scalar, which gives a vector. Adding up all these vectors gives a vector, and dividing by the scalar sum on the bottom gives another vector.

This kind of wave-the-magic-wand-and-write-it-all-in-bold-face technique will always give the right generalization from one dimension to three, provided that the result makes sense mathematically --- if you find yourself doing something nonsensical, such as adding a scalar to a vector, then you haven't found the generalization correctly.

##### Example 66: Colliding coins

$$\triangleright$$ Take two identical coins, put one down on a piece of paper, and slide the other across the paper, shooting it fairly rapidly so that it hits the target coin off-center. If you trace the initial and final positions of the coins, you can determine the directions of their momentum vectors after the collision. The angle between these vectors is always fairly close to, but a little less than, 90 degrees. Why is this?

$$\triangleright$$ Let the velocity vector of the incoming coin be $$\mathbf{a}$$, and let the two outgoing velocity vectors be $$\mathbf{b}$$ and $$\mathbf{c}$$. Since the masses are the same, conservation of momentum amounts to $$\mathbf{a}=\mathbf{b}+\mathbf{c}$$, which means that it has to be possible to assemble the three vectors into a triangle. If we assume that no energy is converted into heat and sound, then conservation of energy gives (discarding the common factor of $$m/2$$) $$a^2=b^2+c^2$$ for the magnitudes of the three vectors. This is the Pythagorean theorem, which will hold only if the three vectors form a right triangle.

The fact that we observe the angle to be somewhat less than 90 degrees shows that the assumption used in the proof is only approximately valid: a little energy is converted into heat and sound. The opposite case would be a collision between two blobs of putty, where the maximum possible amount of energy is converted into heat and sound, the two blobs fly off together, giving an angle of zero between their momentum vectors. The real-life experiment interpolates between the ideal extremes of 0 and 90 degrees, but comes much closer to 90.

Force is a vector, and we add force vectors when more than one force acts on the same object.

##### Example 67: Pushing a block up a ramp
$$\triangleright$$ Figure r/1 shows a block being pushed up a frictionless ramp at constant speed by an applied force $$F_a$$. How much force is required, in terms of the block's mass, $$m$$, and the angle of the ramp, $$\theta$$?

$$\triangleright$$ We analyzed this simple machine in example 38 on page 168 using the concept of work. Here we'll do it using vector addition of forces. Figure r/2 shows the other two forces acting on the block: a normal force, $$F_n$$, created by the ramp, and the gravitational force, $$F_g$$. Because the block is being pushed up at constant speed, it has zero acceleration, and the total force on it must be zero. In figure r/3, we position all the force vectors tip-to-tail for addition. Since they have to add up to zero, they must join up without leaving a gap, so they form a triangle. Using trigonometry we find

\begin{align*} F_{a} &= F_g\ \text{sin}\:\theta \\ &= mg\ \text{sin}\:\theta . \end{align*}

##### Example 68: Buoyancy, again

In example 10 on page 85, we found that the energy required to raise a cube immersed in a fluid is as if the cube's mass had been reduced by an amount equal to the mass of the fluid that otherwise would have been in the volume it occupies (Archimedes' principle). From the energy perspective, this effect occurs because raising the cube allows a certain amount of fluid to move downward, and the decreased gravitational energy of the fluid tends to offset the increased gravitational energy of the cube. The proof given there, however, could not easily be extended to other shapes.

Thinking in terms of force rather than energy, it becomes easier to give a proof that works for any shape. A certain upward force is needed to support the object in figure s. If this force was applied, then the object would be in equilibrium: the vector sum of all the forces acting on it would be zero. These forces are $$\mathbf{F}_a$$, the upward force just mentioned, $$\mathbf{F}_g$$, the downward force of gravity, and $$\mathbf{F}_f$$, the total force from the fluid:

$\begin{equation*} \mathbf{F}_a+\mathbf{F}_{g}+\mathbf{F}_f = 0 \end{equation*}$

Since the fluid is under more pressure at a greater depth, the part of the fluid underneath the object tends to make more force than the part above, so the fluid tends to help support the object.

Now suppose the object was removed, and instantly replaced with an equal volume of fluid. The new fluid would be in equilibrium without any force applied to hold it up, so

$\begin{equation*} \mathbf{F}_{gf}+\mathbf{F}_f = 0 , \end{equation*}$

where $$\mathbf{F}_{gf}$$, the weight of the fluid, is not the same as $$\mathbf{F}_g$$, the weight of the object, but $$\mathbf{F}_f$$ is the same as before, since the pressure of the surrounding fluid is the same as before at any particular depth. We therefore have

$\begin{equation*} \mathbf{F}_a=-\left(\mathbf{F}_g-\mathbf{F}_{gf}\right) , \end{equation*}$

which is Archimedes' principle in terms of force: the force required to support the object is lessened by an amount equal to the weight of the fluid that would have occupied its volume.

By the way, the word “pressure” that I threw around casually in the preceding example has a precise technical definition: force per unit area. The SI units of pressure are $$\text{N}/\text{m}^2$$, which can be abbreviated as pascals, 1 Pa = 1 $$\text{N}/\text{m}^2$$. Atmospheric pressure is about 100 kPa. By applying the equation $$\mathbf{F}_{g}+\mathbf{F}_f = 0$$ to the top and bottom surfaces of a cubical volume of fluid, one can easily prove that the difference in pressure between two different depths is $$\Delta P=\rho g\Delta y$$. (In physics, “fluid” can refer to either a gas or a liquid.)

Pressure is discussed in more detail in chapter 5.

##### Example 69: A solar sail
A solar sail, figure t/1, allows a spacecraft to get its thrust without using internal stores of energy or having to carry along mass that it can shove out the back like a rocket. Sunlight strikes the sail and bounces off, transferring momentum to the sail. A working 30-meter-diameter solar sail, Cosmos 1, was built by an American company, and was supposed to be launched into orbit aboard a Russian booster launched from a submarine, but launch attempts in 2001 and 2005 both failed.

In this example, we will calculate the optimal orientation of the sail, assuming that “optimal” means changing the vehicle's energy as rapidly as possible. For simplicity, we model the complicated shape of the sail's surface as a disk, seen edge-on in figure t/2, and we assume that the craft is in a nearly circular orbit around the sun, hence the 90-degree angle between the direction of motion and the incoming sunlight. We assume that the sail is 100% reflective. The orientation of the sail is specified using the angle $$\theta$$ between the incoming rays of sunlight and the perpendicular to the sail. In other words, $$\theta\text{=0}$$ if the sail is catching the sunlight full-on, while $$\theta\text{=90}°$$ means that the sail is edge-on to the sun.

Conservation of momentum gives

\begin{align*} \mathbf{p}_{light,i} &= \mathbf{p}_{light,f}+\Delta\mathbf{p}_{sail} , \\ \text{where \Delta\mathbf{p}_{sail} is the change in momentum picked up by the sail. Breaking this down into components, we have} 0 &= p_{light,f,x}+\Delta p_{sail,x} \text{and} \\ p_{light,i,y} &= p_{light,f,y}+\Delta p_{sail,y} . \end{align*}

As in example 53 on page 189, the component of the force that is directly away from the sun (up in figure t/2) doesn't change the energy of the craft, so we only care about $$\Delta p_{sail,x}$$, which equals $$- p_{light,f,x}$$. The outgoing light ray forms an angle of 2$$\theta$$ with the negative $$y$$ axis, or 270$$°-2\theta$$ measured counterclockwise from the $$x$$ axis, so the useful thrust depends on $$-\text{cos}(270°-2\theta) =\text{sin}\:2\theta$$.

However, this is all assuming a given amount of light strikes the sail. During a certain time period, the amount of sunlight striking the sail depends on the cross-sectional area the sail presents to the sun, which is proportional to cos $$\theta$$. For $$\theta\text{=90}°$$, cos $$\theta$$ equals zero, since the sail is edge-on to the sun.

Putting together these two factors, the useful thrust is proportional to $$\text{sin}\ 2\theta\ \text{cos}\:\theta$$, and this quantity is maximized for $$\theta\approx35°$$. A counterintuitive fact about this maneuver is that as the spacecraft spirals outward, its total energy (kinetic plus gravitational) increases, but its kinetic energy actually decreases!

##### Example 70: A layback
The figure shows a rock climber using a technique called a layback. He can make the normal forces $$\mathbf{F}_{N1}$$ and $$\mathbf{F}_{N2}$$ large, which has the side-effect of increasing the frictional forces $$\mathbf{F}_{F1}$$ and $$\mathbf{F}_{F2}$$, so that he doesn't slip down due to the gravitational (weight) force $$\mathbf{F}_W$$. The purpose of the problem is not to analyze all of this in detail, but simply to practice finding the components of the forces based on their magnitudes. To keep the notation simple, let's write $$F_{N1}$$ for $$|\mathbf{F}_{N1}|$$, etc. The crack overhangs by a small, positive angle $$\theta\approx9°$$.

In this example, we determine the $$x$$ component of $$\mathbf{F}_{N1}$$. The other nine components are left as an exercise to the reader (problem 81, p. 236).

The easiest method is the one demonstrated in example 62 on p. 199. Casting vector $$\mathbf{F}_{N1}$$'s shadow on the ground, we can tell that it would point to the left, so its $$x$$ component is negative. The only two possibilities for its $$x$$ component are therefore $$-F_{N1}\cos\theta$$ or $$-F_{N1}\sin\theta$$. We expect this force to have a large $$x$$ component and a much smaller $$y$$. Since $$\theta$$ is small, $$\cos\theta\approx 1$$, while $$\sin\theta$$ is small. Therefore the $$x$$ component must be $$-F_{N1}\cos\theta$$.

v / Example 71 and problem 81 on p. 236.

##### Discussion Questions

An object goes from one point in space to another. After it arrives at its destination, how does the magnitude of its $$\Deltar$$ vector compare with the distance it traveled?

In several examples, I've dealt with vectors having negative components. Does it make sense as well to talk about negative and positive vectors?

If you're doing graphical addition of vectors, does it matter which vector you start with and which vector you start from the other vector's tip?

If you add a vector with magnitude 1 to a vector of magnitude 2, what magnitudes are possible for the vector sum?

Which of these examples of vector addition are correct, and which are incorrect?

Is it possible for an airplane to maintain a constant velocity vector but not a constant $$|v|$$? How about the opposite -- a constant $$|v|$$ but not a constant velocity vector? Explain.

New York and Rome are at about the same latitude, so the earth's rotation carries them both around nearly the same circle. Do the two cities have the same velocity vector (relative to the center of the earth)? If not, is there any way for two cities to have the same velocity vector?

The figure shows a roller coaster car rolling down and then up under the influence of gravity. Sketch the car's velocity vectors and acceleration vectors. Pick an interesting point in the motion and sketch a set of force vectors acting on the car whose vector sum could have resulted in the right acceleration vector.

The following is a question commonly asked by students:

“Why does the force vector always have to point in the same direction as the acceleration vector? What if you suddenly decide to change your force on an object, so that your force is no longer pointing in the same direction that the object is accelerating?”

What misunderstanding is demonstrated by this question? Suppose, for example, a spacecraft is blasting its rear main engines while moving forward, then suddenly begins firing its sideways maneuvering rocket as well. What does the student think Newton's laws are predicting?

Debug the following incorrect solutions to this vector addition problem.

Problem: Freddi Fish$$^\text{TM}$$ swims 5.0 km northeast, and then 12.0 km in the direction 55 degrees west of south. How far does she end up from her starting point, and in what direction is she from her starting point?

Incorrect solution #1:
5.0 km+12.0 km=17.0 km

Incorrect solution #2:
$$\sqrt{(5.0\ \text{km})^2+(12.0\ \text{km})^2}$$=13.0 km

Incorrect solution #3:
Let A and B be her two $$\Deltar$$ vectors, and let $$C=A+B$$. Then

\begin{align*} A_x &= (5.0 \ \text{km}) \cos 45° = 3.5 \ \text{km}\\ B_x &= (12.0 \ \text{km}) \cos 55° = 6.9 \ \text{km} \\ A_y &= (5.0 \ \text{km}) \sin 45° = 3.5 \ \text{km} \\ B_y &= (12.0 \ \text{km}) \sin 55° = 9.8 \ \text{km} \\ C_x &= A_x+B_x \\ &= 10.4 \ \text{km} \\ C_y &= A_y+B_y \\ &= 13.3 \ \text{km} \\ |C| &= \sqrt{C_x^2+C_y^2} \\ &= 16.9 \ \text{km} \\ \text{direction} &= \tan^{-1} (13.3/10.4) \\ &= 52 ° \ \text{north of east} \end{align*}

Incorrect solution #4:
(same notation as above)

\begin{align*} A_x &= (5.0 \ \text{km}) \cos 45° = 3.5 \ \text{km}\\ B_x &= -(12.0 \ \text{km}) \cos 55° = -6.9 \ \text{km} \\ A_y &= (5.0 \ \text{km}) \sin 45° = 3.5 \ \text{km} \\ B_y &= -(12.0 \ \text{km}) \sin 55° = -9.8 \ \text{km} \\ C_x &= A_x+B_x \\ &= -3.4 \ \text{km} \\ C_y &= A_y+B_y \\ &= -6.3 \ \text{km} \\ |C| &= \sqrt{C_x^2+C_y^2} \\ &= 7.2 \ \text{km} \\ \text{direction} &= \tan^{-1} (-6.3/-3.4) \\ &= 62 °\ \text{north of east} \end{align*}

Incorrect solution #5:
(same notation as above)

\begin{align*} A_x &= (5.0 \ \text{km}) \cos 45° = 3.5 \ \text{km}\\ B_x &= -(12.0 \ \text{km}) \sin 55° = -9.8 \ \text{km} \\ A_y &= (5.0 \ \text{km}) \sin 45° = 3.5 \ \text{km} \\ B_y &= -(12.0 \ \text{km}) \cos 55° = -6.9 \ \text{km} \\ C_x &= A_x+B_x \\ &= -6.3 \ \text{km} \\ C_y &= A_y+B_y \\ &= -3.4 \ \text{km} \\ |C| &= \sqrt{C_x^2+C_y^2} \\ &= 7.2 \ \text{km} \\ \text{direction} &= \tan^{-1} (-3.4/-6.3) \\ &= 28 °\ \text{north of east} \end{align*}

y / Visualizing the acceleration vector.

z / This figure shows an intuitive justification for the fact proved mathematically in the example, that the direction of the force and acceleration in circular motion is inward. The heptagon, 2, is a better approximation to a circle than the triangle, 1. To make an infinitely good approximation to circular motion, we would need to use an infinitely large number of infinitesimal taps, which would amount to a steady inward force.

aa / The total force in the forward-backward direction is zero in both cases.

ab / There is no outward force on the bowling ball, but in the noninertial frame it seems like one exists.

ac / Discussion question A.

### 3.4.4 Calculus with vectors

#### Differentiation

In one dimension, we define the velocity as the derivative of the position with respect to time, and we can think of the derivative as what we get when we calculate $$\Delta x/\Delta t$$ for very short time intervals. The quantity $$\Delta x=x_f-x_i$$ is calculated by subtraction. In three dimensions, $$x$$ becomes r, and the $$\Delta\mathbf{r}$$ vector is calculated by vector subtraction, $$\Delta\mathbf{r}=\mathbf{r}_f-\mathbf{r}_i$$. Vector subtraction is defined component by component, so when we take the derivative of a vector, this means we end up taking the derivative component by component,

$\begin{equation*} v_x = \frac{dx}{dt} , v_y = \frac{dy}{dt}, v_z = \frac{dz}{dt} \end{equation*}$

or

$\begin{equation*} \frac{d\mathbf{r}}{dt} = \frac{dx}{dt}\hat{\mathbf{x}} +\frac{dy}{dt}\hat{\mathbf{y}}+\frac{dz}{dt}\hat{\mathbf{z}} . \end{equation*}$

All of this reasoning applies equally well to any derivative of a vector, so for instance we can take the second derivative,

$\begin{equation*} a_x = \frac{dv_x}{dt} , a_y = \frac{dv_y}{dt} , a_z = \frac{dv_z}{dt} \end{equation*}$

or

$\begin{equation*} \frac{d\mathbf{v}}{dt} = \frac{dv_x}{dt}\hat{\mathbf{x}}+\frac{dv_y}{dt}\hat{\mathbf{y}} +\frac{dv_z}{dt}\hat{\mathbf{z}} . \end{equation*}$

A counterintuitive consequence of this is that the acceleration vector does not need to be in the same direction as the motion. The velocity vector points in the direction of motion, but by Newton's second law, $$\mathbf{a}=\mathbf{F}/m$$, the acceleration vector points in the same direction as the force, not the motion. This is easiest to understand if we take velocity vectors from two different moments in the motion, and visualize subtracting them graphically to make a $$\Delta\mathbf{v}$$ vector. The direction of the $$\Delta\mathbf{v}$$ vector tells us the direction of the acceleration vector as well, since the derivative $$d\mathbf{v}/dt$$ can be approximated as $$\Delta\mathbf{v}/\Delta t$$. As shown in figure y/1, a change in the magnitude of the velocity vector implies an acceleration that is in the direction of motion. A change in the direction of the velocity vector produces an acceleration perpendicular to the motion, y/2.

##### Example 71: Circular motion
$$\triangleright$$ An object moving in a circle of radius $$r$$ in the $$x$$-$$y$$ plane has
\begin{align*} x &= r\ \text{cos}\ \omega t \text{and}\\ y &= r\ \text{sin}\ \omega t , \end{align*}
where $$\omega$$ is the number of radians traveled per second, and the positive or negative sign indicates whether the motion is clockwise or counterclockwise. What is its acceleration?

$$\triangleright$$ The components of the velocity are

\begin{align*} v_{x} &= -\omega r\ \text{sin}\ \omega t \text{and}\\ v_{y} &= \ \ \ \omega r\ \text{cos}\ \omega t , \end{align*}

and for the acceleration we have

\begin{align*} a_{x} &= -\omega^2 r\ \text{cos}\ \omega t \text{and}\\ a_{y} &= -\omega^2 r\ \text{sin}\ \omega t . \end{align*}

The acceleration vector has cosines and sines in the same places as the r vector, but with minus signs in front, so it points in the opposite direction, i.e., toward the center of the circle. By Newton's second law, a=F/$$m$$, this shows that the force must be inward as well; without this force, the object would fly off straight.

The magnitude of the acceleration is

\begin{align*} |\mathbf{a}| &= \sqrt{ a_x^2+ a_{y}^2}\\ &= \omega^2 r . \end{align*}

It makes sense that $$\omega$$ is squared, since reversing the sign of $$\omega$$ corresponds to reversing the direction of motion, but the acceleration is toward the center of the circle, regardless of whether the motion is clockwise or counterclockwise. This result can also be rewritten in the form

$\begin{equation*} |\mathbf{a}| = \frac{|\mathbf{v}|^2}{r} . \end{equation*}$

Although I've relegated the results $$a=\omega^2 r=|\mathbf{v}|^2/r$$ to an example because they are a straightforward corollary of more general principles already developed, they are important and useful enough to record for later use.

These results are counterintuitive as well. Until Newton, physicists and laypeople alike had assumed that the planets would need a force to push them forward in their orbits. Figure z may help to make it more plausible that only an inward force is required. A forward force might be needed in order to cancel out a backward force such as friction, aa, but the total force in the forward-backward direction needs to be exactly zero for constant-speed motion.

When you are in a car undergoing circular motion, there is also a strong illusion of an outward force. But what object could be making such a force? The car's seat makes an inward force on you, not an outward one. There is no object that could be exerting an outward force on your body. In reality, this force is an illusion that comes from our brain's intuitive efforts to interpret the situation within a noninertial frame of reference. As shown in figure ab, we can describe everything perfectly well in an inertial frame of reference, such as the frame attached to the sidewalk. In such a frame, the bowling ball goes straight because there is no force on it. The wall of the truck's bed hits the ball, not the other way around.

#### Integration

An integral is really just a sum of many infinitesimally small terms. Since vector addition is defined in terms of addition of the components, an integral of a vector quantity is found by doing integrals component by component.

##### Example 72: Projectile motion

$$\triangleright$$ Find the motion of an object whose acceleration vector is constant, for instance a projectile moving under the influence of gravity.

$$\triangleright$$ We integrate the acceleration to get the velocity, and then integrate the velocity to get the position as a function of time. Doing this to the $$x$$ component of the acceleration, we find

\begin{align*} x &= \int{\left(\int{a_x\ \text{d}t}\right)\ \text{d}t} \\ &= \int{\left(a_xt+v_{x\text{o}}\right)\text{d}t} ,\\ \text{where v_{ x\text{o}} is a constant of integration, and} x &= \frac{1}{2}a_xt^2 + v_{x\text{o}}t + x_\text{o} . \end{align*}

Similarly, $$y\text{=(1/2)} a_{y} t^2+ v_{ y\text{o}} t + y_\text{o}$$ and $$z=\text{(1/2)} a_{z} t^2+ v_{ z\text{o}} t + z_\text{o}$$. Once one has gained a little confidence, it becomes natural to do the whole thing as a single vector integral,

\begin{align*} \mathbf{r} &= \int{\left(\int{\mathbf{a}\ \text{d} t}\right)\ \text{d} t} \\ &= \int{\left(\mathbf{a} t+\mathbf{v}_\text{o}\right)\text{d} t} \\ &= \frac{1}{2}\mathbf{a} t^2+\mathbf{v}_\text{o} t+\mathbf{r}_\text{o} , \end{align*}

where now the constants of integration are vectors.

##### Discussion Questions

In the game of crack the whip, a line of people stand holding hands, and then they start sweeping out a circle. One person is at the center, and rotates without changing location. At the opposite end is the person who is running the fastest, in a wide circle. In this game, someone always ends up losing their grip and flying off. Suppose the person on the end loses her grip. What path does she follow as she goes flying off? (Assume she is going so fast that she is really just trying to put one foot in front of the other fast enough to keep from falling; she is not able to get any significant horizontal force between her feet and the ground.)

Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What force or forces are acting on her, and in what directions are they? (We are not interested in the vertical forces, which are the earth's gravitational force pulling down, and the ground's normal force pushing up.) Make a table in the format shown in subsection 3.2.6.

Suppose the person on the outside is still holding on, but feels that she may loose her grip at any moment. What is wrong with the following analysis of the situation? “The person whose hand she's holding exerts an inward force on her, and because of Newton's third law, there's an equal and opposite force acting outward. That outward force is the one she feels throwing her outward, and the outward force is what might make her go flying off, if it's strong enough.”

If the only force felt by the person on the outside is an inward force, why doesn't she go straight in?

In the amusement park ride shown in the figure, the cylinder spins faster and faster until the customer can pick her feet up off the floor without falling. In the old Coney Island version of the ride, the floor actually dropped out like a trap door, showing the ocean below. (There is also a version in which the whole thing tilts up diagonally, but we're discussing the version that stays flat.) If there is no outward force acting on her, why does she stick to the wall? Analyze all the forces on her.

What is an example of circular motion where the inward force is a normal force? What is an example of circular motion where the inward force is friction? What is an example of circular motion where the inward force is the sum of more than one force?

Does the acceleration vector always change continuously in circular motion? The velocity vector?

A certain amount of force is needed to provide the acceleration of circular motion. What if were are exerting a force perpendicular to the direction of motion in an attempt to make an object trace a circle of radius $$r$$, but the force isn't as big as $$m|v|^2/r$$?

Suppose a rotating space station is built that gives its occupants the illusion of ordinary gravity. What happens when a person in the station lets go of a ball? What happens when she throws a ball straight “up” in the air (i.e., towards the center)?

ae / The geometric interpretation of the dot product.

af / Breaking trail, by Walter E. Bohl. The pack horse is not doing any work on the pack, because the pack is moving in a horizontal line at constant speed, and therefore there is no kinetic or gravitational energy being transferred into or out of it.

### 3.4.5 The dot product

How would we generalize the mechanical work equation $$dE=F dx$$ to three dimensions? Energy is a scalar, but force and distance are vectors, so it might seem at first that the kind of “magic-wand” generalization discussed on page 202 failed here, since we don't know of any way to multiply two vectors together to get a scalar. Actually, this is Nature giving us a hint that there is such a multiplication operation waiting for us to invent it, and since Nature is simple, we can be assured that this operation will work just fine in any situation where a similar generalization is required.

How should this operation be defined? Let's consider what we would get by performing this operation on various combinations of the unit vectors $$\hat{\mathbf{x}}$$, $$\hat{\mathbf{y}}$$, and $$\hat{\mathbf{z}}$$. The conventional notation for the operation is to put a dot, $$\cdot$$, between the two vectors, and the operation is therefore called the dot product. Rotational invariance requires that we handle the three coordinate axes in the same way, without giving special treatment to any of them, so we must have $$\hat{\mathbf{x}}\cdot\hat{\mathbf{x}}=\hat{\mathbf{y}}\cdot\hat{\mathbf{y}}=\hat{\mathbf{z}}\cdot\hat{\mathbf{z}}$$ and $$\hat{\mathbf{x}}\cdot\hat{\mathbf{y}}=\hat{\mathbf{y}}\cdot\hat{\mathbf{z}}=\hat{\mathbf{z}}\cdot\hat{\mathbf{x}}$$. This is supposed to be a way of generalizing ordinary multiplication, so for consistency with the property $$1\times1=1$$ of ordinary numbers, the result of multiplying a magnitude-one vector by itself had better be the scalar 1, so $$\hat{\mathbf{x}}\cdot\hat{\mathbf{x}}=\hat{\mathbf{y}}\cdot\hat{\mathbf{y}}=\hat{\mathbf{z}}\cdot\hat{\mathbf{z}}=1$$. Furthermore, there is no way to satisfy rotational invariance unless we define the mixed products to be zero, $$\hat{\mathbf{x}}\cdot\hat{\mathbf{y}}=\hat{\mathbf{y}}\cdot\hat{\mathbf{z}}=\hat{\mathbf{z}}\cdot\hat{\mathbf{x}}=0$$; for example, a 90-degree rotation of our frame of reference about the $$z$$ axis reverses the sign of $$\hat{\mathbf{x}}\cdot\hat{\mathbf{y}}$$, but rotational invariance requires that $$\hat{\mathbf{x}}\cdot\hat{\mathbf{y}}$$ produce the same result either way, and zero is the only number that stays the same when we reverse its sign. Establishing these six products of unit vectors suffices to define the operation in general, since any two vectors that we want to multiply can be broken down into components, e.g., $$(2\hat{\mathbf{x}}+3\hat{\mathbf{z}})\cdot\hat{\mathbf{z}} =2\hat{\mathbf{x}}\cdot\hat{\mathbf{z}}+3\hat{\mathbf{z}}\cdot\hat{\mathbf{z}}=0+3=3$$. Thus by requiring rotational invariance and consistency with multiplication of ordinary numbers, we find that there is only one possible way to define a multiplication operation on two vectors that gives a scalar as the result.17 The dot product has all of the properties we normally associate with multiplication, except that there is no “dot division.”

##### Example 73: Dot product in terms of components

If we know the components of any two vectors b and c, we can find their dot product:

\begin{align*} \mathbf{b}\cdot\mathbf{c} &= \left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right) \cdot \left( c_{x}\hat{\mathbf{x}}+ c_{y}\hat{\mathbf{y}}+ c_z\hat{\mathbf{z}}\right) \\ &= b_x c_x+ b_y c_y+ b_z c_z . \end{align*}

##### Example 74: Magnitude expressed with a dot product
If we take the dot product of any vector b with itself, we find
\begin{align*} \mathbf{b}\cdot\mathbf{b} &= \left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right) \cdot \left( b_{x}\hat{\mathbf{x}}+ b_{y}\hat{\mathbf{y}}+ b_z\hat{\mathbf{z}}\right) \\ &= b_{x}^2+ b_{y}^2+ b_z^2 ,\\ \text{so its magnitude can be expressed as} |\mathbf{b}| &= \sqrt{\mathbf{b}\cdot\mathbf{b}} . \end{align*}
We will often write $$b^2$$ to mean $$\mathbf{b}\cdot\mathbf{b}$$, when the context makes it clear what is intended. For example, we could express kinetic energy as $$\text{(1/2)} m|\mathbf{v}|^2$$, $$\text{(1/2)} m\mathbf{v}\cdot\mathbf{v}$$, or $$\text{(1/2)} m v^2$$. In the third version, nothing but context tells us that $$v$$ really stands for the magnitude of some vector $$\mathbf{v}$$.
##### Example 75: Geometric interpretation
In figure ae, vectors a, b, and c represent the sides of a triangle, and $$\mathbf{a}=\mathbf{b}+\mathbf{c}$$. The law of cosines gives
$\begin{equation*} |\mathbf{c}|^2 = |\mathbf{a}|^2+|\mathbf{b}|^2-2|\mathbf{a}||\mathbf{b}|\ \text{cos}\ \theta . \end{equation*}$
Using the result of example 75, we can also write this as
\begin{align*} |\mathbf{c}|^2 &= \mathbf{c}\cdot\mathbf{c} \\ &= (\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b}) \\ &= \mathbf{a}\cdot\mathbf{a}+\mathbf{b}\cdot\mathbf{b}-2\mathbf{a}\cdot\mathbf{b} . \end{align*}
Matching up terms in these two expressions, we find
$\begin{equation*} \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\ \text{cos}\ \theta , \end{equation*}$
which is a geometric interpretation for the dot product.

The result of example 76 is very useful. It gives us a way to find the angle between two vectors if we know their components. It can be used to show that the dot product of any two perpendicular vectors is zero. It also leads to a nifty proof that the dot product is rotationally invariant --- up until now I've only proved that if a rotationally invariant product exists, the dot product is it --- because angles and lengths aren't affected by a rotation, so the right side of the equation is rotationally invariant, and therefore so is the left side.

I introduced the whole discussion of the dot product by way of generalizing the equation $$dE=Fdx$$ to three dimensions. In terms of a dot product, we have

\begin{align*} dE &= \mathbf{F}\cdotd\mathbf{r} .\\ \text{If \mathbf{F} is a constant, integrating both sides gives} \Delta E &= \mathbf{F}\cdot\Delta\mathbf{r} . \end{align*}

(If that step seemed like black magic, try writing it out in terms of components.) If the force is perpendicular to the motion, as in figure af, then the work done is zero. The pack horse is doing work within its own body, but is not doing work on the pack.

##### Example 76: Pushing a lawnmower

$$\triangleright$$ I push a lawnmower with a force $$\mathbf{F}\text{=(110 N)}\hat{\mathbf{x}}-\text{(40 N)}\hat{\mathbf{y}}$$, and the total distance I travel is $$\text{(100 m)}\hat{\mathbf{x}}$$. How much work do I do?

$$\triangleright$$ The dot product is 11000 $$\text{N}\!\cdot\!\text{m}$$ = 11000 J.

A good application of the dot product is to allow us to write a simple, streamlined proof of separate conservation of the momentum components. (You can skip the proof without losing the continuity of the text.) The argument is a generalization of the one-dimensional proof on page 130, and makes the same assumption about the type of system of particles we're dealing with. The kinetic energy of one of the particles is $$(1/2)m\mathbf{v}\cdot\mathbf{v}$$, and when we transform into a different frame of reference moving with velocity u relative to the original frame, the one-dimensional rule $$v\rightarrow v+u$$ turns into vector addition, $$\mathbf{v}\rightarrow \mathbf{v}+\mathbf{u}$$. In the new frame of reference, the kinetic energy is $$(1/2)m(\mathbf{v}+\mathbf{u})\cdot(\mathbf{v}+\mathbf{u})$$. For a system of $$n$$ particles, we have

\begin{align*} K &= \sum_{j=1}^{n}{\frac{1}{2}m_j(\mathbf{v}_j+\mathbf{u})\cdot(\mathbf{v}_j+\mathbf{u})} \\ &= \frac{1}{2}\left[\sum_{j=1}^{n}{m_j\mathbf{v}_j\cdot\mathbf{v}_j} +2\sum_{j=1}^{n}{m_j\mathbf{v}_j\cdot\mathbf{u}} +\sum_{j=1}^{n}{m_j\mathbf{u}\cdot\mathbf{u}}\right] . \end{align*}

As in the proof on page 130, the first sum is simply the total kinetic energy in the original frame of reference, and the last sum is a constant, which has no effect on the validity of the conservation law. The middle sum can be rewritten as

\begin{align*} 2\sum_{j=1}^{n}{m_j\mathbf{v}_j\cdot\mathbf{u}} &= 2\ \mathbf{u}\cdot\sum_{j=1}^{n}{m_j\mathbf{v}_j}\\ &= 2\ \mathbf{u}\cdot\sum_{j=1}^{n}{\mathbf{p}_j} , \end{align*}

so the only way energy can be conserved for all values of u is if the vector sum of the momenta is conserved as well.

ag / An object moves through a field of force.

### 3.4.6 Gradients and line integrals (optional)

This subsection introduces a little bit of vector calculus. It can be omitted without loss of continuity, but the techniques will be needed in our study of electricity and magnetism, and it may be helpful to be exposed to them in easy-to-visualize mechanical contexts before applying them to invisible electrical and magnetic phenomena.

In physics we often deal with fields of force, meaning situations where the force on an object depends on its position. For instance, figure ag could represent a map of the trade winds affecting a sailing ship, or a chart of the gravitational forces experienced by a space probe entering a double-star system. An object moving under the influence of this force will not necessarily be moving in the same direction as the force at every moment. The sailing ship can tack against the wind, due to the force from the water on the keel. The space probe, if it entered from the top of the diagram at high speed, would start to curve around to the right, but its inertia would carry it forward, and it wouldn't instantly swerve to match the direction of the gravitational force. For convenience, we've defined the gravitational field, g, as the force per unit mass, but that trick only leads to a simplification because the gravitational force on an object is proportional to its mass. Since this subsection is meant to apply to any kind of force, we'll discuss everything in terms of the actual force vector, F, in units of newtons.

If an object moves through the field of force along some curved path from point $$\mathbf{r}_1$$ to point $$\mathbf{r}_2$$, the force will do a certain amount of work on it. To calculate this work, we can break the path up into infinitesimally short segments, find the work done along each segment, and add them all up. For an object traveling along a nice straight $$x$$ axis, we use the symbol $$dx$$ to indicate the length of any infinitesimally short segment. In three dimensions, moving along a curve, each segment is a tiny vector $$d\mathbf{r}=\hat{\mathbf{x}}dx+\hat{\mathbf{y}}dy+\hat{\mathbf{z}}dz$$. The work theorem can be expressed as a dot product, so the work done along a segment is $$\mathbf{F}\cdotd\mathbf{r}$$. We want to integrate this, but we don't know how to integrate with respect to a variable that's a vector, so let's define a variable $$s$$ that indicates the distance traveled so far along the curve, and integrate with respect to it instead. The expression $$\mathbf{F}\cdotd\mathbf{r}$$ can be rewritten as $$|\mathbf{F}|\:|d\mathbf{r}|\:\cos\theta$$, where $$\theta$$ is the angle between F and $$d\mathbf{r}$$. But $$|d\mathbf{r}|$$ is simply $$ds$$, so the amount of work done becomes

$\begin{equation*} \Delta E = \int_{\mathbf{r}_1}^{\mathbf{r}_2}{|\mathbf{F}| \cos\theta}\ ds . \end{equation*}$

Both F and $$\theta$$ are functions of $$s$$. As a matter of notation, it's cumbersome to have to write the integral like this. Vector notation was designed to eliminate this kind of drudgery. We therefore define the line integral

\begin{align*} \int_C{\mathbf{F}\cdotd\mathbf{r}} \end{align*}

as a way of notating this type of integral. The `C' refers to the curve along which the object travels. If we don't know this curve then we typically can't evaluate the line integral just by knowing the initial and final positions $$\mathbf{r}_1$$ and $$\mathbf{r}_2$$.

The basic idea of calculus is that integration undoes differentiation, and vice-versa. In one dimension, we could describe an interaction either in terms of a force or in terms of an interaction energy. We could integrate force with respect to position to find minus the energy, or we could find the force by taking minus the derivative of the energy. In the line integral, position is represented by a vector. What would it mean to take a derivative with respect to a vector? The correct way to generalize the derivative $$dU/dx$$ to three dimensions is to replace it with the following vector,

$\begin{equation*} \frac{dU}{dx}\hat{\mathbf{x}} +\frac{dU}{dy}\hat{\mathbf{y}} +\frac{dU}{dz}\hat{\mathbf{z}} , \end{equation*}$

called the gradient of $$U$$, and written with an upside-down delta18 like this, $$\nabla U$$. Each of these three derivatives is really what's known as a partial derivative. What that means is that when you're differentiating $$U$$ with respect to $$x$$, you're supposed to treat $$y$$ and $$z$$ and constants, and similarly when you do the other two derivatives. To emphasize that a derivative is a partial derivative, it's customary to write it using the symbol $$\partial$$ in place of the differential d's. Putting all this notation together, we have

$\begin{equation*} \nabla U = \frac{\partial U}{\partial x}\hat{\mathbf{x}} +\frac{\partial U}{\partial y}\hat{\mathbf{y}} +\frac{\partial U}{\partial z}\hat{\mathbf{z}} \text{[definition of the gradient]} . \end{equation*}$

The gradient looks scary, but it has a very simple physical interpretation. It's a vector that points in the direction in which $$U$$ is increasing most rapidly, and it tells you how rapidly $$U$$ is increasing in that direction. For instance, sperm cells in plants and animals find the egg cells by traveling in the direction of the gradient of the concentration of certain hormones. When they reach the location of the strongest hormone concentration, they find their destiny. In terms of the gradient, the force corresponding to a given interaction energy is $$\mathbf{F}=-\nabla U$$.

##### Example 77: Force exerted by a spring

In one dimension, Hooke's law is $$U=\text{(1/2)} kx^2$$. Suppose we tether one end of a spring to a post, but it's free to stretch and swing around in a plane. Let's say its equilibrium length is zero, and let's choose the origin of our coordinate system to be at the post. Rotational invariance requires that its energy only depend on the magnitude of the r vector, not its direction, so in two dimensions we have $$U=\text{(1/2)} k|\mathbf{r}|^2 =\text{(1/2)} k\left( x^2+ y^2\right)$$. The force exerted by the spring is then

\begin{align*} \mathbf{F} &= -\nabla U \\ &= -\frac{\partial U}{\partial x}\hat{\mathbf{x}} -\frac{\partial U}{\partial y}\hat{\mathbf{y}}\\ &= - kx\hat{\mathbf{x}}- ky\hat{\mathbf{y}} . \end{align*}

The magnitude of this force vector is $$k|\mathbf{r}|$$, and its direction is toward the origin.

\backofchapterboilerplate{3}

## Homework Problems

a / Problem 16.

b / Problem 19.

c / Problem 23, part c.

d / Problem 29.

e / Problem 31.

f / Problem 32

g / Problem 33.

h / Problem 38

i / Problem 45

k / Problem 54.

m / Problem 59

n / Problem 63

o / Problem 64

p / Problem 66.

q / Problem 70.

s / Problem 78.

t / Problem 79.

u / Problem 80.

w / Problem 82.

x / Problem 83.

1. Derive a formula expressing the kinetic energy of an object in terms of its momentum and mass.(answer check available at lightandmatter.com)

2. Two people in a rowboat wish to move around without causing the boat to move. What should be true about their total momentum? Explain.

3. A bullet leaves the barrel of a gun with a kinetic energy of 90 J. The gun barrel is 50 cm long. The gun has a mass of 4 kg, the bullet 10 g.
(a) Find the bullet's final velocity. (answer check available at lightandmatter.com)
(b) Find the bullet's final momentum. (answer check available at lightandmatter.com)
(c) Find the momentum of the recoiling gun.
(d) Find the kinetic energy of the recoiling gun, and explain why the recoiling gun does not kill the shooter. (answer check available at lightandmatter.com)

4. (solution in the pdf version of the book) The big difference between the equations for momentum and kinetic energy is that one is proportional to $$v$$ and one to $$v^2$$. Both, however, are proportional to $$m$$. Suppose someone tells you that there's a third quantity, funkosity, defined as $$f=m^2v$$, and that funkosity is conserved. How do you know your leg is being pulled?

5. A ball of mass $$2m$$ collides head-on with an initially stationary ball of mass $$m$$. No kinetic energy is transformed into heat or sound. In what direction is the mass-$$2m$$ ball moving after the collision, and how fast is it going compared to its original velocity? \hwans{hwans:twotoonecollision}

6. A very massive object with velocity $$v$$ collides head-on with an object at rest whose mass is very small. No kinetic energy is converted into other forms. Prove that the low-mass object recoils with velocity $$2v$$. [Hint: Use the center-of-mass frame of reference.]

7. A mass $$m$$ moving at velocity $$v$$ collides with a stationary target having the same mass $$m$$. Find the maximum amount of energy that can be released as heat and sound. (answer check available at lightandmatter.com)

8. A rocket ejects exhaust with an exhaust velocity $$u$$. The rate at which the exhaust mass is used (mass per unit time) is $$b$$. We assume that the rocket accelerates in a straight line starting from rest, and that no external forces act on it. Let the rocket's initial mass (fuel plus the body and payload) be $$m_i$$, and $$m_f$$ be its final mass, after all the fuel is used up. (a) Find the rocket's final velocity, $$v$$, in terms of $$u$$, $$m_i$$, and $$m_f$$. Neglect the effects of special relativity. (b) A typical exhaust velocity for chemical rocket engines is 4000 m/s. Estimate the initial mass of a rocket that could accelerate a one-ton payload to 10% of the speed of light, and show that this design won't work. (For the sake of the estimate, ignore the mass of the fuel tanks. The speed is fairly small compared to $$c$$, so it's not an unreasonable approximation to ignore relativity.) (answer check available at lightandmatter.com)

9. An object is observed to be moving at constant speed along a line. Can you conclude that no forces are acting on it? Explain. [Based on a problem by Serway and Faughn.]

10. At low speeds, every car's acceleration is limited by traction, not by the engine's power. Suppose that at low speeds, a certain car is normally capable of an acceleration of $$3\ \text{m}/\text{s}^2$$. If it is towing a trailer with half as much mass as the car itself, what acceleration can it achieve? [Based on a problem from PSSC Physics.]

11. (a) Let $$T$$ be the maximum tension that an elevator's cable can withstand without breaking, i.e., the maximum force it can exert. If the motor is programmed to give the car an acceleration $$a$$ ($$a>0$$ is upward), what is the maximum mass that the car can have, including passengers, if the cable is not to break?(answer check available at lightandmatter.com)
(b) Interpret the equation you derived in the special cases of $$a=0$$ and of a downward acceleration of magnitude $$g$$.

12. (solution in the pdf version of the book) When the contents of a refrigerator cool down, the changed molecular speeds imply changes in both momentum and energy. Why, then, does a fridge transfer power through its radiator coils, but not force?

13. A helicopter of mass $$m$$ is taking off vertically. The only forces acting on it are the earth's gravitational force and the force, $$F_{air}$$, of the air pushing up on the propeller blades.
(a) If the helicopter lifts off at $$t=0$$, what is its vertical speed at time $$t$$?
(b) Check that the units of your answer to part a make sense.
(c) Discuss how your answer to part a depends on all three variables, and show that it makes sense. That is, for each variable, discuss what would happen to the result if you changed it while keeping the other two variables constant. Would a bigger value give a smaller result, or a bigger result? Once you've figured out this mathematical relationship, show that it makes sense physically.
(d) Plug numbers into your equation from part a, using $$m=2300$$ kg, $$F_{air}=27000\ \text{N}$$, and $$t=4.0\ \text{s}$$. (answer check available at lightandmatter.com)

14. A blimp is initially at rest, hovering, when at $$t=0$$ the pilot turns on the motor of the propeller. The motor cannot instantly get the propeller going, but the propeller speeds up steadily. The steadily increasing force between the air and the propeller is given by the equation $$F=kt$$, where $$k$$ is a constant. If the mass of the blimp is $$m$$, find its position as a function of time. (Assume that during the period of time you're dealing with, the blimp is not yet moving fast enough to cause a significant backward force due to air resistance.)(answer check available at lightandmatter.com)

15. (solution in the pdf version of the book) A car is accelerating forward along a straight road. If the force of the road on the car's wheels, pushing it forward, is a constant 3.0 kN, and the car's mass is 1000 kg, then how long will the car take to go from 20 m/s to 50 m/s?

16. A little old lady and a pro football player collide head-on. Compare their forces on each other, and compare their accelerations. Explain.

17. The earth is attracted to an object with a force equal and opposite to the force of the earth on the object. If this is true, why is it that when you drop an object, the earth does not have an acceleration equal and opposite to that of the object?

18. When you stand still, there are two forces acting on you, the force of gravity (your weight) and the normal force of the floor pushing up on your feet. Are these forces equal and opposite? Does Newton's third law relate them to each other? Explain.

19. Today's tallest buildings are really not that much taller than the tallest buildings of the 1940's. One big problem with making an even taller skyscraper is that every elevator needs its own shaft running the whole height of the building. So many elevators are needed to serve the building's thousands of occupants that the elevator shafts start taking up too much of the space within the building. An alternative is to have elevators that can move both horizontally and vertically: with such a design, many elevator cars can share a few shafts, and they don't get in each other's way too much because they can detour around each other. In this design, it becomes impossible to hang the cars from cables, so they would instead have to ride on rails which they grab onto with wheels. Friction would keep them from slipping. The figure shows such a frictional elevator in its vertical travel mode. (The wheels on the bottom are for when it needs to switch to horizontal motion.)
(a) If the coefficient of static friction between rubber and steel is $$\mu_s$$, and the maximum mass of the car plus its passengers is $$M$$, how much force must there be pressing each wheel against the rail in order to keep the car from slipping? (Assume the car is not accelerating.)(answer check available at lightandmatter.com)
(b) Show that your result has physically reasonable behavior with respect to $$\mu_s$$. In other words, if there was less friction, would the wheels need to be pressed more firmly or less firmly? Does your equation behave that way?

20. A tugboat of mass $$m$$ pulls a ship of mass $$M$$, accelerating it. Ignore fluid friction acting on their hulls, although there will of course need to be fluid friction acting on the tug's propellers.
(a) If the force acting on the tug's propeller is $$F$$, what is the tension, $$T$$, in the cable connecting the two ships? \hwhint{hwhint:tugboat}(answer check available at lightandmatter.com)
(b) Interpret your answer in the special cases of $$M=0$$ and $$M=\infty$$.

21. Someone tells you she knows of a certain type of Central American earthworm whose skin, when rubbed on polished diamond, has $$\mu_k>\mu_s$$. Why is this not just empirically unlikely but logically suspect?

22. A uranium atom deep in the earth spits out an alpha particle. An alpha particle is a fragment of an atom. This alpha particle has initial speed $$v$$, and travels a distance $$d$$ before stopping in the earth.
(a) Find the force, $$F$$, from the dirt that stopped the particle, in terms of $$v,d$$, and its mass, $$m$$. Don't plug in any numbers yet. Assume that the force was constant.(answer check available at lightandmatter.com)
(c) Discuss how your answer to part a depends on all three variables, and show that it makes sense. That is, for each variable, discuss what would happen to the result if you changed it while keeping the other two variables constant. Would a bigger value give a smaller result, or a bigger result? Once you've figured out this mathematical relationship, show that it makes sense physically.
(d) Evaluate your result for $$m=6.7\times10^{-27}$$ kg, $$v=2.0\times10^4$$ km/s, and $$d=0.71$$ mm.(answer check available at lightandmatter.com)

23. You are given a large sealed box, and are not allowed to open it. Which of the following experiments measure its mass, and which measure its weight? [Hint: Which experiments would give different results on the moon?]
(a) Put it on a frozen lake, throw a rock at it, and see how fast it scoots away after being hit.
(b) Drop it from a third-floor balcony, and measure how loud the sound is when it hits the ground.
(c) As shown in the figure, connect it with a spring to the wall, and watch it vibrate.

24. While escaping from the palace of the evil Martian emperor, Sally Spacehound jumps from a tower of height $$h$$ down to the ground. Ordinarily the fall would be fatal, but she fires her blaster rifle straight down, producing an upward force of magnitude $$F_B$$. This force is insufficient to levitate her, but it does cancel out some of the force of gravity. During the time $$t$$ that she is falling, Sally is unfortunately exposed to fire from the emperor's minions, and can't dodge their shots. Let $$m$$ be her mass, and $$g$$ the strength of gravity on Mars.
(a) Find the time $$t$$ in terms of the other variables.
(c) For sufficiently large values of $$F_B$$, your answer to part a becomes nonsense --- explain what's going on.(answer check available at lightandmatter.com)

25. When I cook rice, some of the dry grains always stick to the measuring cup. To get them out, I turn the measuring cup upside-down and hit the “roof” with my hand so that the grains come off of the “ceiling.” (a) Explain why static friction is irrelevant here. (b) Explain why gravity is negligible. (c) Explain why hitting the cup works, and why its success depends on hitting the cup hard enough.

26. A flexible rope of mass $$m$$ and length $$L$$ slides without friction over the edge of a table. Let $$x$$ be the length of the rope that is hanging over the edge at a given moment in time.
(a) Show that $$x$$ satisfies the equation of motion $$d^2x/dt^2=gx/L$$. [Hint: Use $$F=dp/dt$$, which allows you to handle the two parts of the rope separately even though mass is moving out of one part and into the other.]
(b) Give a physical explanation for the fact that a larger value of $$x$$ on the right-hand side of the equation leads to a greater value of the acceleration on the left side.
(c) When we take the second derivative of the function $$x(t)$$ we are supposed to get essentially the same function back again, except for a constant out in front. The function $$e^x$$ has the property that it is unchanged by differentiation, so it is reasonable to look for solutions to this problem that are of the form $$x=be^{ct}$$, where $$b$$ and $$c$$ are constants. Show that this does indeed provide a solution for two specific values of $$c$$ (and for any value of $$b)$$.
(d) Show that the sum of any two solutions to the equation of motion is also a solution.
(e) Find the solution for the case where the rope starts at rest at $$t=0$$ with some nonzero value of $$x$$.

In problems 27-31, analyze the forces using a table in the format shown in section 3.2.6. Analyze the forces in which the italicized object participates.

27. Some people put a spare car key in a little magnetic box that they stick under the chassis of their car. Let's say that the box is stuck directly underneath a horizontal surface, and the car is parked. (See instructions above.)

28. Analyze two examples of objects at rest relative to the earth that are being kept from falling by forces other than the normal force. Do not use objects in outer space, and do not duplicate problem 27 or 31. (See instructions above.)

29. A person is rowing a boat, with her feet braced. She is doing the part of the stroke that propels the boat, with the ends of the oars in the water (not the part where the oars are out of the water). (See instructions above.)

30. A farmer is in a stall with a cow when the cow decides to press him against the wall, pinning him with his feet off the ground. Analyze the forces in which the farmer participates. (See instructions above.)

31. A propeller plane is cruising east at constant speed and altitude. (See instructions above.)

32. The figure shows a stack of two blocks, sitting on top of a table that is bolted to the floor. All three objects are made from identical wood, with their surfaces finished identically using the same sandpaper. We tap the middle block, giving it an initial velocity $$v$$ to the right. The tap is executed so rapidly that almost no initial velocity is imparted to the top block.
(a) Find the time that will elapse until the slipping between the top and middle blocks stops. Express your answer in terms of $$v$$, $$m$$, $$M$$, $$g$$, and the relevant coefficient of friction.(answer check available at lightandmatter.com)
(c) Check that your result has the correct behavior when you make $$m$$ bigger or smaller. Explain. This means that you should discuss the mathematical behavior of the result, and then explain how this corresponds to what would really happen physically.
(d) Similarly, discuss what happens when you make $$M$$ bigger or smaller.
(e) Similarly, discuss what happens when you make $$g$$ bigger or smaller.

33. Ginny has a plan. She is going to ride her sled while her dog Foo pulls her, and she holds on to his leash. However, Ginny hasn't taken physics, so there may be a problem: she may slide right off the sled when Foo starts pulling.
(a) Analyze all the forces in which Ginny participates, making a table as in subsection 3.2.6.
(b) Analyze all the forces in which the sled participates.
(c) The sled has mass $$m$$, and Ginny has mass $$M$$. The coefficient of static friction between the sled and the snow is $$\mu_1$$, and $$\mu_2$$ is the corresponding quantity for static friction between the sled and her snow pants. Ginny must have a certain minimum mass so that she will not slip off the sled. Find this in terms of the other three variables.(answer check available at lightandmatter.com)
(d) Interpreting your equation from part c, under what conditions will there be no physically realistic solution for $$M$$? Discuss what this means physically.

34. (solution in the pdf version of the book) In each case, identify the force that causes the acceleration, and give its Newton's-third-law partner. Describe the effect of the partner force. (a) A swimmer speeds up. (b) A golfer hits the ball off of the tee. (c) An archer fires an arrow. (d) A locomotive slows down.

35. A cop investigating the scene of an accident measures the length $$L$$ of a car's skid marks in order to find out its speed $$v$$ at the beginning of the skid. Express $$v$$ in terms of $$L$$ and any other relevant variables.(answer check available at lightandmatter.com)

36. An ice skater builds up some speed, and then coasts across the ice passively in a straight line. (a) Analyze the forces, using a table the format shown in subsection 3.2.6.
(b) If his initial speed is $$v$$, and the coefficient of kinetic friction is $$\mu_k$$, find the maximum theoretical distance he can glide before coming to a stop. Ignore air resistance.(answer check available at lightandmatter.com)
(c) Show that your answer to part b has the right units.
(d) Show that your answer to part b depends on the variables in a way that makes sense physically.
(e) Evaluate your answer numerically for $$\mu_k=0.0046$$, and a world-record speed of $$14.58$$ m/s. (The coefficient of friction was measured by De Koning et al., using special skates worn by real speed skaters.)(answer check available at lightandmatter.com)
(f) Comment on whether your answer in part e seems realistic. If it doesn't, suggest possible reasons why.

37. (a) Using the solution of problem 37 on page 124, predict how the spring constant of a fiber will depend on its length and cross-sectional area.
(b) The constant of proportionality is called the Young's modulus, $$E$$, and typical values of the Young's modulus are about $$10^{10}$$ to $$10^{11}$$. What units would the Young's modulus have in the SI system? (solution in the pdf version of the book)

38. This problem depends on the results of problems problem 37 on page 124 and problem 37 from this chapter. When atoms form chemical bonds, it makes sense to talk about the spring constant of the bond as a measure of how “stiff” it is. Of course, there aren't really little springs --- this is just a mechanical model. The purpose of this problem is to estimate the spring constant, $$k$$, for a single bond in a typical piece of solid matter. Suppose we have a fiber, like a hair or a piece of fishing line, and imagine for simplicity that it is made of atoms of a single element stacked in a cubical manner, as shown in the figure, with a center-to-center spacing $$b$$. A typical value for $$b$$ would be about $$10^{-10}\ \text{m}$$.
(a) Find an equation for $$k$$ in terms of $$b$$, and in terms of the Young's modulus, $$E$$, defined in problem 37 and its solution.
(b) Estimate $$k$$ using the numerical data given in problem 37.
(c) Suppose you could grab one of the atoms in a diatomic molecule like $$\text{H}_2$$ or $$\text{O}_2$$, and let the other atom hang vertically below it. Does the bond stretch by any appreciable fraction due to gravity?

39. This problem has been deleted.

40. Many fish have an organ known as a swim bladder, an air-filled cavity whose main purpose is to control the fish's buoyancy an allow it to keep from rising or sinking without having to use its muscles. In some fish, however, the swim bladder (or a small extension of it) is linked to the ear and serves the additional purpose of amplifying sound waves. For a typical fish having such an anatomy, the bladder has a resonant frequency of 300 Hz, the bladder's $$Q$$ is 3, and the maximum amplification is about a factor of 100 in energy. Over what range of frequencies would the amplification be at least a factor of 50?

41. An oscillator with sufficiently strong damping has its maximum response at $$\omega=0$$. Using the result derived on page 912 , find the value of $$Q$$ at which this behavior sets in. \hwhint{hwhint:maxampatdc}\hwans{hwans:maxampatdc}

42. An oscillator has $$Q$$=6.00, and, for convenience, let's assume $$F_m=1.00$$, $$\omega_\text{o}=1.00$$, and $$m=1.00$$. The usual approximations would give

$\begin{gather*} \omega_{res} =\omega_\text{o} ,\\ A_{res}=6.00 , \text{and}\\ \Delta\omega=1/6.00 . \end{gather*}$

Determine these three quantities numerically using the result derived on page 912 , and compare with the approximations.

43. The apparatus in figure d on page 61 had a natural period of oscillation of 5 hours and 20 minutes. The authors estimated, based on calculations of internal friction in the tungsten wire, that its $$Q$$ was on the order of $$10^6$$, but they were unable to measure it empirically because it would have taken years for the amplitude to die down by any measurable amount. Although each aluminum or platinum mass was really moving along an arc of a circle, any actual oscillations caused by a violation of the equivalence of gravitational and inertial mass would have been measured in millions of a degree, so it's a good approximation to say that each mass's motion was along a (very short!) straight line segment. We can also treat each mass as if it was oscillating separately from the others. If the principle of equivalence had been violated at the $$10^{-12}$$ level, the limit of their experiment's sensitivity, the sun's gravitational force on one of the 0.4-gram masses would have been about $$3\times10^{-19}$$ N, oscillating with a period of 24 hours due to the rotation of the earth. (We ignore the inertia of the arms, whose total mass was only about 25% of the total mass of the rotating assembly.)
(a) Find the amplitude of the resulting oscillations, and determine the angle to which they would have corresponded, given that the radius of the balance arms was 10 cm.\hwans{hwans:braginskii}
(b) Show that even if their estimate of $$Q$$ was wildly wrong, it wouldn't have affected this result.

44. (solution in the pdf version of the book) A firework shoots up into the air, and just before it explodes it has a certain momentum and kinetic energy. What can you say about the momenta and kinetic energies of the pieces immediately after the explosion? [Based on a problem from PSSC Physics.]

45. (solution in the pdf version of the book) The figure shows a view from above of a collision about to happen between two air hockey pucks sliding without friction. They have the same speed, $$v_i$$, before the collision, but the big puck is 2.3 times more massive than the small one. Their sides have sticky stuff on them, so when they collide, they will stick together. At what angle will they emerge from the collision? In addition to giving a numerical answer, please indicate by drawing on the figure how your angle is defined.

46. A learjet traveling due east at 300 mi/hr collides with a jumbo jet which was heading southwest at 150 mi/hr. The jumbo jet's mass is five times greater than that of the learjet. When they collide, the learjet sticks into the fuselage of the jumbo jet, and they fall to earth together. Their engines stop functioning immediately after the collision. On a map, what will be the direction from the location of the collision to the place where the wreckage hits the ground? (Give an angle.)(answer check available at lightandmatter.com)

47. (a) A ball is thrown straight up with velocity $$v$$. Find an equation for the height to which it rises.(answer check available at lightandmatter.com)
(b) Generalize your equation for a ball thrown at an angle $$\theta$$ above horizontal, in which case its initial velocity components are $$v_x=v \cos \theta$$ and $$v_y=v \sin \theta$$.(answer check available at lightandmatter.com)

48. At the 2010 Salinas Lettuce Festival Parade, the Lettuce Queen drops her bouquet while riding on a float moving toward the right. Sketch the shape of its trajectory in her frame of reference, and compare with the shape seen by one of her admirers standing on the sidewalk.

49. Two daredevils, Wendy and Bill, go over Niagara Falls. Wendy sits in an inner tube, and lets the 30 km/hr velocity of the river throw her out horizontally over the falls. Bill paddles a kayak, adding an extra 10 km/hr to his velocity. They go over the edge of the falls at the same moment, side by side. Ignore air friction. Explain your reasoning.
(a) Who hits the bottom first?
(b) What is the horizontal component of Wendy's velocity on impact?
(c) What is the horizontal component of Bill's velocity on impact?
(d) Who is going faster on impact?

50. A baseball pitcher throws a pitch clocked at $$v_x$$=73.3 mi/h. He throws horizontally. By what amount, $$d$$, does the ball drop by the time it reaches home plate, $$L$$=60.0 ft away?
(a) First find a symbolic answer in terms of $$L$$, $$v_x$$, and $$g$$.(answer check available at lightandmatter.com)
(b) Plug in and find a numerical answer. Express your answer in units of ft. (Note: 1 ft=12 in, 1 mi=5280 ft, and 1 in=2.54 cm) (answer check available at lightandmatter.com)

j / Problem 50.

51. A batter hits a baseball at speed $$v$$, at an angle $$\theta$$ above horizontal.
(a) Find an equation for the range (horizontal distance to where the ball falls), $$R$$, in terms of the relevant variables. Neglect air friction and the height of the ball above the ground when it is hit. \hwans{hwans:baseballrange}
(b) Interpret your equation in the cases of $$\theta$$=0 and $$\theta=90°$$.
(c) Find the angle that gives the maximum range.\hwans{hwans:baseballrange}

52. In this problem you'll extend the analysis in problem 51

to include air friction by writing a computer program. For a game played at sea level, the force due to air friction is approximately $$(7\times10^{-4}\ \text{N}\!\cdot\!\text{s}^2/\text{m}^2)v^2$$, in the direction opposite to the motion of the ball.The mass of a baseball is 0.146 kg.
(a) For a ball hit at a speed of 45.0 m/s from a height of 1.0 m, find the optimal angle and the resulting range. \hwans{hwans:baseballrangeair}
(b) How much farther would the ball fly at the Colorado Rockies' stadium, where the thinner air gives 18 percent less air friction? \hwans{hwans:baseballrangeair}

53. If you walk 35 km at an angle 25° counterclockwise from east, and then 22 km at 230° counterclockwise from east, find the distance and direction from your starting point to your destination. (answer check available at lightandmatter.com)

54. The figure shows vectors $$\mathbf{A}$$ and $$\mathbf{B}$$. As in figure p on p. 200, graphically calculate the following:

$$\mathbf{A}+\mathbf{B}$$, $$\mathbf{A}-\mathbf{B}$$, $$\mathbf{B}-\mathbf{A}$$, $$-2\mathbf{B}$$, $$\mathbf{A}-2\mathbf{B}$$

No numbers are involved.

55. Phnom Penh is 470 km east and 250 km south of Bangkok. Hanoi is 60 km east and 1030 km north of Phnom Penh.
(a) Choose a coordinate system, and translate these data into $$\Delta x$$ and $$\Delta y$$ values with the proper plus and minus signs.
(b) Find the components of the $$\Delta \mathbf{r}$$ vector pointing from Bangkok to Hanoi.(answer check available at lightandmatter.com)

56. Is it possible for a helicopter to have an acceleration due east and a velocity due west? If so, what would be going on? If not, why not?

l / Problem 57.

57. A dinosaur fossil is slowly moving down the slope of a glacier under the influence of wind, rain and gravity. At the same time, the glacier is moving relative to the continent underneath. The dashed lines represent the directions but not the magnitudes of the velocities. Pick a scale, and use graphical addition of vectors to find the magnitude and the direction of the fossil's velocity relative to the continent. You will need a ruler and protractor. (answer check available at lightandmatter.com)

58. A bird is initially flying horizontally east at 21.1 m/s, but one second later it has changed direction so that it is flying horizontally and $$7°$$ north of east, at the same speed. What are the magnitude and direction of its acceleration vector during that one second time interval? (Assume its acceleration was roughly constant.) (answer check available at lightandmatter.com)

59. Your hand presses a block of mass $$m$$ against a wall with a force $$\mathbf{F}_H$$ acting at an angle $$\theta$$, as shown in the figure. Find the minimum and maximum possible values of $$|\mathbf{F}_H|$$ that can keep the block stationary, in terms of $$m$$, $$g$$, $$\theta$$, and $$\mu_s$$, the coefficient of static friction between the block and the wall. Check both your answers in the case of $$\theta=90°$$, and interpret the case where the maximum force is infinite.(answer check available at lightandmatter.com)

60. A skier of mass $$m$$ is coasting down a slope inclined at an angle $$\theta$$ compared to horizontal. Assume for simplicity that the treatment of kinetic friction given in chapter 5 is appropriate here, although a soft and wet surface actually behaves a little differently. The coefficient of kinetic friction acting between the skis and the snow is $$\mu_k$$, and in addition the skier experiences an air friction force of magnitude $$bv^2$$, where $$b$$ is a constant.
(a) Find the maximum speed that the skier will attain, in terms of the variables $$m$$, $$g$$, $$\theta$$, $$\mu_k$$, and $$b$$.(answer check available at lightandmatter.com)
(b) For angles below a certain minimum angle $$\theta_{min}$$, the equation gives a result that is not mathematically meaningful. Find an equation for $$\theta_{min}$$, and give a physical explanation of what is happening for $$\theta \lt\theta_{min}$$.(answer check available at lightandmatter.com)

61. A gun is aimed horizontally to the west. The gun is fired, and the bullet leaves the muzzle at $$t=0$$. The bullet's position vector as a function of time is $$\mathbf{r}=b\hat{\mathbf{x}}+ct\hat{\mathbf{y}}+dt^2\hat{\mathbf{z}}$$, where $$b$$, $$c$$, and $$d$$ are positive constants.
(a) What units would $$b$$, $$c$$, and $$d$$ need to have for the equation to make sense?
(b) Find the bullet's velocity and acceleration as functions of time.
(c) Give physical interpretations of $$b$$, $$c$$, $$d$$, $$\hat{\mathbf{x}}$$, $$\hat{\mathbf{y}}$$, and $$\hat{\mathbf{z}}$$.

62. Annie Oakley, riding north on horseback at 30 mi/hr, shoots her rifle, aiming horizontally and to the northeast. The muzzle speed of the rifle is 140 mi/hr. When the bullet hits a defenseless fuzzy animal, what is its speed of impact? Neglect air resistance, and ignore the vertical motion of the bullet.(solution in the pdf version of the book)

63. A cargo plane has taken off from a tiny airstrip in the Andes, and is climbing at constant speed, at an angle of $$\theta=17°$$ with respect to horizontal. Its engines supply a thrust of $$F_{thrust}=200\ \text{kN}$$, and the lift from its wings is $$F_{lift}=654\ \text{kN}$$. Assume that air resistance (drag) is negligible, so the only forces acting are thrust, lift, and weight. What is its mass, in kg? (solution in the pdf version of the book)

64. A wagon is being pulled at constant speed up a slope $$\theta$$ by a rope that makes an angle $$\phi$$ with the vertical. (a) Assuming negligible friction, show that the tension in the rope is given by the equation

$\begin{equation*} T = \frac{\sin\theta}{\sin(\theta+\phi)}mg, \end{equation*}$

(b) Interpret this equation in the special cases of $$\phi=0$$ and $$\phi=180°-\theta$$.(solution in the pdf version of the book)

65. The angle of repose is the maximum slope on which an object will not slide. On airless, geologically inert bodies like the moon or an asteroid, the only thing that determines whether dust or rubble will stay on a slope is whether the slope is less steep than the angle of repose.
(a) Find an equation for the angle of repose, deciding for yourself what are the relevant variables.
(b) On an asteroid, where $$g$$ can be thousands of times lower than on Earth, would rubble be able to lie at a steeper angle of repose?(solution in the pdf version of the book)

66. When you're done using an electric mixer, you can get most of the batter off of the beaters by lifting them out of the batter with the motor running at a high enough speed. Let's imagine, to make things easier to visualize, that we instead have a piece of tape stuck to one of the beaters.
(a) Explain why static friction has no effect on whether or not the tape flies off.
(b) Analyze the forces in which the tape participates, using a table the format shown in subsection 3.2.6.
(c) Suppose you find that the tape doesn't fly off when the motor is on a low speed, but at a greater speed, the tape won't stay on. Why would the greater speed change things? [Hint: If you don't invoke any law of physics, you haven't explained it.]

67. Show that the expression $$|\mathbf{v}|^2/r$$ has the units of acceleration.

68. A plane is flown in a loop-the-loop of radius 1.00 km. The plane starts out flying upside-down, straight and level, then begins curving up along the circular loop, and is right-side up when it reaches the top. (The plane may slow down somewhat on the way up.) How fast must the plane be going at the top if the pilot is to experience no force from the seat or the seatbelt while at the top of the loop? (answer check available at lightandmatter.com)

69. Find the angle between the following two vectors:

$\begin{gather*} \hat{\mathbf{x}}+2\hat{\mathbf{y}}+3\hat{\mathbf{z}} \\ 4\hat{\mathbf{x}}+5\hat{\mathbf{y}}+6\hat{\mathbf{z}} \end{gather*}$

70. The two blocks shown in the figure have equal mass, $$m$$, and the surface is frictionless. (a) What is the tension in the massless rope?\hwhint{hwhint:ropeslopes}(answer check available at lightandmatter.com)
(c) Check the physical behavior of your answer in the special cases of $$\phi \le \theta$$ and $$\theta=0$$, $$\phi=90°$$.

71. (a) We observe that the amplitude of a certain free oscillation decreases from $$A_\text{o}$$ to $$A_\text{o}/Z$$ after $$n$$ oscillations. Find its $$Q$$.(answer check available at lightandmatter.com)
(b) The figure is from Shape memory in Spider draglines, Emile, Le Floch, and Vollrath, Nature 440:621 (2006). Panel 1 shows an electron microscope's image of a thread of spider silk. In 2, a spider is hanging from such a thread. From an evolutionary point of view, it's probably a bad thing for the spider if it twists back and forth while hanging like this. (We're referring to a back-and-forth rotation about the axis of the thread, not a swinging motion like a pendulum.) The authors speculate that such a vibration could make the spider easier for predators to see, and it also seems to me that it would be a bad thing just because the spider wouldn't be able to control its orientation and do what it was trying to do. Panel 3 shows a graph of such an oscillation, which the authors measured using a video camera and a computer, with a 0.1 g mass hung from it in place of a spider. Compared to human-made fibers such as kevlar or copper wire, the spider thread has an unusual set of properties:

1. It has a low $$Q$$, so the vibrations damp out quickly.
2. It doesn't become brittle with repeated twisting as a copper wire would.
3. When twisted, it tends to settle in to a new equilibrium angle, rather than insisting on returning to its original angle. You can see this in panel 2, because although the experimenters initially twisted the wire by 35 degrees, the thread only performed oscillations with an amplitude much smaller than $$\pm35$$ degrees, settling down to a new equilibrium at 27 degrees.
4. Over much longer time scales (hours), the thread eventually resets itself to its original equilbrium angle (shown as zero degrees on the graph). (The graph reproduced here only shows the motion over a much shorter time scale.) Some human-made materials have this “memory” property as well, but they typically need to be heated in order to make them go back to their original shapes.

Focusing on property number 1, estimate the $$Q$$ of spider silk from the graph.(answer check available at lightandmatter.com)

r / Problem 71.

72. A cross-country skier is gliding on a level trail, with negligible friction. Then, when he is at position $$x=0$$, the tip of his skis enters a patch of dirt. As he rides onto the dirt, more and more of his weight is being supported by the dirt. The skis have length $$\ell$$, so if he reached $$x=\ell$$ without stopping, his weight would be completely on the dirt. This problem deals with the motion for $$x\lt\ell$$. (a) Find the acceleration in terms of $$x$$, as well as any other relevant constants.
(b) This is a second-order differential equation. You should be able to find the solution simply by thinking about some commonly occuring functions that you know about, and finding two that have the right properties. If these functions are $$x=f(t)$$ and $$x=g(t)$$, then the most general solution to the equations of motion will be of the form $$x=af+bg$$, where $$a$$ and $$b$$ are constants to be determined from the initial conditions.
(c) Suppose that the initial velocity $$v_\text{o}$$ at $$x=0$$ is such that he stops at $$x\lt\ell$$. Find the time until he stops, and show that, counterintuitively, this time is independent of $$v_\text{o}$$. Explain physically why this is true.(answer check available at lightandmatter.com)

73. A microwave oven works by twisting molecules one way and then the other, counterclockwise and then clockwise about their own centers, millions of times a second. If you put an ice cube or a stick of butter in a microwave, you'll observe that the solid doesn't heat very quickly, although eventually melting begins in one small spot. Once this spot forms, it grows rapidly, while the rest of the solid remains solid; it appears that a microwave oven heats a liquid much more rapidly than a solid. Explain why this should happen, based on the atomic-level description of heat, solids, and liquids. (See, e.g., figure b on page 110.)

Don't repeat the following common mistakes:

In a solid, the atoms are packed more tightly and have less space between them. Not true. Ice floats because it's less dense than water.

In a liquid, the atoms are moving much faster. No, the difference in average speed between ice at $$-1°\text{C}$$ and water at $$1°\text{C}$$ is only 0.4%.

74. Problem 2-16 on page 120 was intended to be solved using conservation of energy. Solve the same problem using Newton's laws.

75. A bead slides down along a piece of wire that is in the shape of a helix. The helix lies on the surface of a vertical cylinder of radius $$r$$, and the vertical distance between turns is $$d$$.
(a) Ordinarily when an object slides downhill under the influence of kinetic friction, the velocity-independence of kinetic friction implies that the acceleration is constant, and therefore there is no limit to the object's velocity. Explain the physical reason why this argument fails here, so that the bead will in fact have some limiting velocity.
(b) Find the limiting velocity.
(c) Show that your result has the correct behavior in the limit of $$r \rightarrow \infty$$. [Problem by B. Korsunsky.](answer check available at lightandmatter.com)

76. A person on a bicycle is to coast down a ramp of height $$h$$ and then pass through a circular loop of radius $$r$$. What is the smallest value of $$h$$ for which the cyclist will complete the loop without falling? (Ignore the kinetic energy of the spinning wheels.)(answer check available at lightandmatter.com)

77. A car accelerates from rest. At low speeds, its acceleration is limited by static friction, so that if we press too hard on the gas, we will “burn rubber” (or, for many newer cars, a computerized traction-control system will override the gas pedal). At higher speeds, the limit on acceleration comes from the power of the engine, which puts a limit on how fast kinetic energy can be developed.
(a) Show that if a force $$F$$ is applied to an object moving at speed $$v$$, the power required is given by $$P=vF$$.
(b) Find the speed $$v$$ at which we cross over from the first regime described above to the second. At speeds higher than this, the engine does not have enough power to burn rubber. Express your result in terms of the car's power $$P$$, its mass $$m$$, the coefficient of static friction $$\mu_s$$, and $$g$$.(answer check available at lightandmatter.com)
(c) Show that your answer to part b has units that make sense.
(d) Show that the dependence of your answer on each of the four variables makes sense physically.
(e) The 2010 Maserati Gran Turismo Convertible has a maximum power of $$3.23\times10^5\ \text{W}$$ (433 horsepower) and a mass (including a 50-kg driver) of $$2.03\times10^3\ \text{kg}$$. (This power is the maximum the engine can supply at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available in which the engine will be running at very nearly this frequency when the car is at moving at $$v$$.) Rubber on asphalt has $$\mu_s\approx0.9$$. Find $$v$$ for this car. Answer: $$18\ \text{m}/\text{s}$$, or about 40 miles per hour.
(f) Our analysis has neglected air friction, which can probably be approximated as a force proportional to $$v^2$$. The existence of this force is the reason that the car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation it is to ignore air friction, find what fraction of the engine's maximum power is being used to overcome air resistance when the car is moving at the speed $$v$$ found in part e. Answer: 1%

78. Two wheels of radius $$r$$ rotate in the same vertical plane with angular velocities $$+\Omega$$ and $$-\Omega$$ about axes that are parallel and at the same height. The wheels touch one another at a point on their circumferences, so that their rotations mesh like gears in a gear train. A board is laid on top of the wheels, so that two friction forces act upon it, one from each wheel. Characterize the three qualitatively different types of motion that the board can exhibit, depending on the initial conditions.

79. For safety, mountain climbers often wear a climbing harness and tie in to other climbers on a rope team or to anchors such as pitons or snow anchors. When using anchors, the climber usually wants to tie in to more than one, both for extra strength and for redundancy in case one fails. The figure shows such an arrangement, with the climber hanging from a pair of anchors forming a “Y” at an angle $$\theta$$. The usual advice is to make $$\theta\lt90°$$; for large values of $$\theta$$, the stress placed on the anchors can be many times greater than the actual load $$L$$, so that two anchors are actually less safe than one.
(a) Find the force $$S$$ at each anchor in terms of $$L$$ and $$\theta$$. (answer check available at lightandmatter.com)
(b) Verify that your answer makes sense in the case of $$\theta=0$$.
(c) Interpret your answer in the case of $$\theta=180°$$.
(d) What is the smallest value of $$\theta$$ for which $$S$$ equals or exceeds $$L$$, so that for larger angles a failure of at least one anchor is more likely than it would have been with a single anchor?(answer check available at lightandmatter.com)

80. Mountain climbers with masses $$m$$ and $$M$$ are roped together while crossing a horizontal glacier when a vertical crevasse opens up under the climber with mass $$M$$. The climber with mass $$m$$ drops down on the snow and tries to stop by digging into the snow with the pick of an ice ax. Alas, this story does not have a happy ending, because this doesn't provide enough friction to stop. Both $$m$$ and $$M$$ continue accelerating, with $$M$$ dropping down into the crevasse and $$m$$ being dragged across the snow, slowed only by the kinetic friction with coefficient $$\mu_k$$ acting between the ax and the snow. There is no significant friction between the rope and the lip of the crevasse.
(a) Find the acceleration $$a$$.(answer check available at lightandmatter.com)
(b) Check the units of your result.
(c) Check the dependence of your equation on the variables. That means that for each variable, you should determine what its effect on $$a$$ should be physically, and then what your answer from part a says its effect would be mathematically.

v / Problem 81.

81. Complete example 71 on p. 205 by expressing the remaining nine $$x$$ and $$y$$ components of the forces in terms of the five magnitudes and the small, positive angle $$\theta\approx9°$$ by which the crack overhangs. (answer check available at lightandmatter.com)

82. In a well known stunt from circuses and carnivals, a motorcyclist rides around inside a big bowl, gradually speeding up and rising higher. Eventually the cyclist can get up to where the walls of the bowl are vertical. Let's estimate the conditions under which a running human could do the same thing.
(a) If the runner can run at speed $$v$$, and her shoes have a coefficient of static friction $$\mu_s$$, what is the maximum radius of the circle?(answer check available at lightandmatter.com)
(c) Check that its dependence on the variables makes sense.
(d) Evaluate your result numerically for $$v=10\ \text{m}/\text{s}$$ (the speed of an olympic sprinter) and $$\mu_s=5$$. (This is roughly the highest coefficient of static friction ever achieved for surfaces that are not sticky. The surface has an array of microscopic fibers like a hair brush, and is inspired by the hairs on the feet of a gecko. These assumptions are not necessarily realistic, since the person would have to run at an angle, which would be physically awkward.)(answer check available at lightandmatter.com)

83. Problem 79 discussed a possible correct way of setting up a redundant anchor for mountaineering. The figure for this problem shows an incorrect way of doing it, by arranging the rope in a triangle (which we'll take to be isoceles). One of the bad things about the triangular arrangement is that it requires more force from the anchors, making them more likely to fail. (a) Using the same notation as in problem 79, find $$S$$ in terms of $$L$$ and $$\theta$$. (answer check available at lightandmatter.com)
(b) Verify that your answer makes sense in the case of $$\theta=0$$, and compare with the correct setup.

84. At a picnic, someone hands you a can of beer. The ground is uneven, and you don't want to spill your drink. You reason that it will be more stable if you drink some of it first in order to lower its center of mass. How much should you drink in order to make the center of mass as low as possible? [Based on a problem by Walter van B. Roberts and Martin Gardner.]

## Exercises

### Exercise A: Force and Motion

Equipment:

2-meter pieces of butcher paper

wood blocks with hooks

string

masses to put on top of the blocks to increase friction

spring scales (preferably calibrated in Newtons)

Suppose a person pushes a crate, sliding it across the floor at a certain speed, and then repeats the same thing but at a higher speed. This is essentially the situation you will act out in this exercise. What do you think is different about her force on the crate in the two situations? Discuss this with your group and write down your hypothesis:

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

1. First you will measure the amount of friction between the wood block and the butcher paper when the wood and paper surfaces are slipping over each other. The idea is to attach a spring scale to the block and then slide the butcher paper under the block while using the scale to keep the block from moving with it. Depending on the amount of force your spring scale was designed to measure, you may need to put an extra mass on top of the block in order to increase the amount of friction. It is a good idea to use long piece of string to attach the block to the spring scale, since otherwise one tends to pull at an angle instead of directly horizontally.

First measure the amount of friction force when sliding the butcher paper as slowly as possible:\_\_\_\_\_\_\_\_\_

Now measure the amount of friction force at a significantly higher speed, say 1 meter per second. (If you try to go too fast, the motion is jerky, and it is impossible to get an accurate reading.) \_\_\_\_\_\_\_\_\_

Discuss your results. Why are we justified in assuming that the string's force on the block (i.e., the scale reading) is the same amount as the paper's frictional force on the block?

2. Now try the same thing but with the block moving and the paper standing still. Try two different speeds.

Do your results agree with your original hypothesis? If not, discuss what's going on. How does the block “know” how fast to go?

### Exercise B: Vibrations

Equipment:

• air track and carts of two different masses
• springs
• spring scales

Place the cart on the air track and attach springs so that it can vibrate.

1. Test whether the period of vibration depends on amplitude. Try at least two moderate amplitudes, for which the springs do not go slack, and at least one amplitude that is large enough so that they do go slack.

2. Try a cart with a different mass. Does the period change by the expected factor, based on the equation $$T=2\pi\sqrt{m/k}$$?

3. Use a spring scale to pull the cart away from equilibrium, and make a graph of force versus position. Is it linear? If so, what is its slope?

4. Test the equation $$T=2\pi\sqrt{m/k}$$ numerically.

### Exercise C: Worksheet on Resonance

1. Compare the oscillator's energies at A, B, C, and D.

\widefignocaptionnofloat{ex-resonance-1}

2. Compare the Q values of the two oscillators.

\widefignocaptionnofloat{ex-resonance-2}

3. Match the x-t graphs in #2 with the amplitude-frequency graphs below.

\widefignocaptionnofloat{ex-resonance-3}

Exercise D is on the following two pages.

### Exercise D: Vectors and Motion

Each diagram on page 243 shows the motion of an object in an $$x-y$$ plane. Each dot is one location of the object at one moment in time. The time interval from one dot to the next is always the same, so you can think of the vector that connects one dot to the next as a $$\mathbf{v}$$ vector, and subtract to find $$\Delta\mathbf{v}$$ vectors.

1. Suppose the object in diagram 1 is moving from the top left to the bottom right. Deduce whatever you can about the force acting on it. Does the force always have the same magnitude? The same direction?

Invent a physical situation that this diagram could represent.

What if you reinterpret the diagram, and reverse the object's direction of motion?

2. What can you deduce about the force that is acting in diagram 2?

Invent a physical situation that diagram 2 could represent.

3. What can you deduce about the force that is acting in diagram 3?

Invent a physical situation.

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

##### Footnotes
[1] Electrical and magnetic interactions don't quite behave like this, which is a point we'll take up later in the book.
[2] We can now see that the derivation would have been equally valid for $$U_i\ne U_f$$. The two observers agree on the distance between the particles, so they also agree on the interaction energies, even though they disagree on the kinetic energies.
[3] Recall that uppercase $$P$$ is power, while lowercase $$p$$ is momentum.
[4] This is with the benefit of hindsight. At the time, the word “force” already had certain connotations, and people thought they understood what it meant and how to measure it, e.g., by using a spring scale. From their point of view, $$F=d{}p/d{}t$$ was not a definition but a testable --- and controversial! --- statement.
[5] This pathological solution was first noted on page 83, and discussed in more detail on page 910.
[6] The converse isn't true, because kinetic energy doesn't depend on the direction of motion, but momentum does. We can change a particle's momentum without changing its energy, as when a pool ball bounces off a bumper, reversing the sign of $$p$$.
[7] The part of the definition about “by a force” is meant to exclude the transfer of energy by heat conduction, as when a stove heats soup.
[8] “Black box” is a traditional engineering term for a device whose inner workings we don't care about.
[9] For conceptual simplicity, we ignore the transfer of heat energy to the outside world via the exhaust and radiator. In reality, the sum of these energies plus the useful kinetic energy transferred would equal $$W$$.
[10] This subroutine isn't as accurate a way of calculating the period as the energy-based one we used in the undamped case, since it only checks whether the mass turned around at some point during the time interval $$\Delta{}t$$.
[11] The relationship is $$\omega_{max\ A}/\omega_\text{o}=\sqrt{1-1/2Q^2}$$, which is similar in form to the equation for the frequency of the free vibration, $$\omega_{f}/\omega_\text{o}=\sqrt{1-1/4Q^2}$$. A subtle point here is that although the maximum of $$A$$ and the maximum of $$A^2$$ must occur at the same frequency, the maximum energy does not occur, as we might expect, at the same frequency as the maximum of $$A^2$$. This is because the interaction energy is proportional to $$A^2$$ regardless of frequency, but the kinetic energy is proportional to $$A^2\omega^2$$. The maximum energy actually occurs are precisely $$\omega_\text{o}$$.
[12] For example, the graphs calculated for sinusoidal driving have resonances that are somewhat below the natural frequency, getting lower with increasing damping, until for $$Q\le1$$ the maximum response occurs at $$\omega=0$$. In figure m, however, we can see that impulsive driving at $$\omega=2\omega_\text{o}$$ produces a steady state with more energy than at $$\omega=\omega_\text{o}$$.
[13] If you've learned about differential equations, you'll know that any second-order differential equation requires the specification of two boundary conditions in order to specify solution uniquely.
[14] Actually, if you know about complex numbers and Euler's theorem, it's not quite so nonsensical.
[15] Of course, you could tell in a sealed laboratory which way was down, but that's because there happens to be a big planet nearby, and the planet's gravitational field reaches into the lab, not because space itself has a special down direction. Similarly, if your experiment was sensitive to magnetic fields, it might matter which way the building was oriented, but that's because the earth makes a magnetic field, not because space itself comes equipped with a north direction.
[16] The zero here is really a zero vector, i.e., a vector whose components are all zero, so we should really represent it with a boldface \vc{0}. There's usually not much danger of confusion, however, so most books, including this one, don't use boldface for the zero vector.
[17] There is, however, a different operation, discussed in the next chapter, which multiplies two vectors to give a vector.
[18] The symbol $$\nabla$$ is called a “nabla.” Cool word!