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Contents

Section 15.1 - Conservation of angular momentum

Section 15.2 - Angular momentum in planetary motion

Section 15.3 - Two theorems about angular momentum

Section 15.4 - Torque: the rate of transfer of angular momentum

Section 15.5 - Statics

Section 15.6 - Simple Machines: the lever

Section 15.7 - Proof of Kepler's elliptical orbit law (optional)

Section 15.8 - Summary

Section 15.1 - Conservation of angular momentum

Section 15.2 - Angular momentum in planetary motion

Section 15.3 - Two theorems about angular momentum

Section 15.4 - Torque: the rate of transfer of angular momentum

Section 15.5 - Statics

Section 15.6 - Simple Machines: the lever

Section 15.7 - Proof of Kepler's elliptical orbit law (optional)

Section 15.8 - Summary

“Sure, and maybe the sun won't come up tomorrow.” Of course, the sun only appears to go up and down because the earth spins, so the cliche should really refer to the unlikelihood of the earth's stopping its rotation abruptly during the night. Why can't it stop? It wouldn't violate conservation of momentum, because the earth's rotation doesn't add anything to its momentum. While California spins in one direction, some equally massive part of India goes the opposite way, canceling its momentum. A halt to Earth's rotation would entail a drop in kinetic energy, but that energy could simply be converted into some other form, such as heat.

Other examples along these lines are not hard to find. A hydrogen atom spins at the same rate for billions of years. A high-diver who is rotating when he comes off the board does not need to make any physical effort to continue rotating, and indeed would be unable to stop rotating before he hit the water.

These observations have the hallmarks of a conservation law:

**A closed system is involved.** Nothing is making an effort to
twist the earth, the hydrogen atom, or the high-diver. They
are isolated from rotation-changing influences, i.e.,
they are closed systems.

**Something remains unchanged.** There appears to be a numerical
quantity for measuring rotational motion such that the total
amount of that quantity remains constant in a closed system.

**Something can be transferred back and forth without changing
the total amount.** In figure a,
the jumper wants to get his feet out in front of
him so he can keep from doing a “face plant” when he
lands. Bringing his feet forward would involve a certain
quantity of counterclockwise rotation, but he didn't start
out with any rotation when he left the ground. Suppose we
consider counterclockwise as positive and clockwise as
negative. The only way his legs can acquire some positive
rotation is if some other part of his body picks up an equal
amount of negative rotation. This is why he swings his arms
up behind him, clockwise.

What numerical measure of rotational motion is conserved? Car engines and old-fashioned LP records have speeds of rotation measured in rotations per minute (r.p.m.), but the number of rotations per minute (or per second) is not a conserved quantity. A twirling figure skater, for instance, can pull her arms in to increase her r.p.m.'s. The first section of this chapter deals with the numerical definition of the quantity of rotation that results in a valid conservation law.

When most people think of rotation, they think of a solid object like a wheel rotating in a circle around a fixed point. Examples of this type of rotation, called rigid rotation or rigid-body rotation, include a spinning top, a seated child's swinging leg, and a helicopter's spinning propeller. Rotation, however, is a much more general phenomenon, and includes noncircular examples such as a comet in an elliptical orbit around the sun, or a cyclone, in which the core completes a circle more quickly than the outer parts.

If there is a numerical measure of rotational motion that is a conserved quantity, then it must include nonrigid cases like these, since nonrigid rotation can be traded back and forth with rigid rotation. For instance, there is a trick for finding out if an egg is raw or hardboiled. If you spin a hardboiled egg and then stop it briefly with your finger, it stops dead. But if you do the same with a raw egg, it springs back into rotation because the soft interior was still swirling around within the momentarily motionless shell. The pattern of flow of the liquid part is presumably very complex and nonuniform due to the asymmetric shape of the egg and the different consistencies of the yolk and the white, but there is apparently some way to describe the liquid's total amount of rotation with a single number, of which some percentage is given back to the shell when you release it.

The best strategy is to devise a way of defining the amount of rotation of a single small part of a system. The amount of rotation of a system such as a cyclone will then be defined as the total of all the contributions from its many small parts.

The quest for a conserved quantity of rotation even requires
us to broaden the rotation concept to include cases where
the motion doesn't repeat or even curve around. If you throw
a piece of putty at a door, the door will recoil and start
rotating. The putty was traveling straight, not in a circle,
but if there is to be a general conservation law that can
cover this situation, it appears that we must describe the
putty as having had some “rotation,” which it then gave up
to the door. The best way of thinking about it is to
attribute rotation to any moving object or part of an object
that changes its angle in relation to the axis of rotation.
In the putty-and-door example, the hinge of the door is the
natural point to think of as an axis, and the putty changes
its angle as seen by someone standing at the hinge. For this
reason, the conserved quantity we are investigating is
called *angular * momentum.
The symbol for angular momentum can't be \(a\) or \(m\),
since those are used for acceleration and mass, so the
symbol \(L\) is arbitrarily chosen instead.

Imagine a 1-kg blob of putty, thrown at the door at a speed
of 1 m/s, which hits the door at a distance of 1 \(m\) from
the hinge. We define this blob to have 1 unit of angular
momentum. When it hits the door, the door
will recoil and start rotating.
We can use the speed at which the door recoils as a measure
of the angular momentum the blob brought in.^{1}

Experiments show, not surprisingly, that a 2-kg blob thrown in the same way makes the door rotate twice as fast, so the angular momentum of the putty blob must be proportional to mass,

\[\begin{equation*}
L\propto m .
\end{equation*}\]

Similarly, experiments show that doubling the velocity of the blob will have a doubling effect on the result, so its angular momentum must be proportional to its velocity as well,

\[\begin{equation*}
L\propto mv .
\end{equation*}\]

You have undoubtedly had the experience of approaching a closed door with one of those bar-shaped handles on it and pushing on the wrong side, the side close to the hinges. You feel like an idiot, because you have so little leverage that you can hardly budge the door. The same would be true with the putty blob. Experiments would show that the amount of rotation the blob can give to the door is proportional to the distance, \(r\), from the axis of rotation, so angular momentum must also be proportional to \(r\),

\[\begin{equation*}
L\propto mvr .
\end{equation*}\]

We are almost done, but there is one missing ingredient. We know on grounds of symmetry that a putty ball thrown directly inward toward the hinge will have no angular momentum to give to the door. After all, there would not even be any way to decide whether the ball's rotation was clockwise or counterclockwise in this situation. It is therefore only the component of the blob's velocity vector perpendicular to the door that should be counted in its angular momentum,

\[\begin{equation*}
L = m v_{\perp}r .
\end{equation*}\]

More generally, \(v_{\perp}\) should be thought of as the component of the object's velocity vector that is perpendicular to the line joining the object to the axis of rotation.

We find that this equation agrees with the definition of the original putty blob as having one unit of angular momentum, and we can now see that the units of angular momentum are \((\text{kg}\!\cdot\!\text{m}/\text{s})\!\cdot\!\text{m}\), i.e., \(\text{kg}\!\cdot\!\text{m}^2/\text{s}\). This gives us a way of calculating the angular momentum of any material object or any system consisting of material objects:

The angular momentum of a moving particle is

\[\begin{equation*}
L = mv_{\perp}r ,
\end{equation*}\]

where \(m\) is its mass, \(v_{\perp}\) is the component of its velocity vector perpendicular to the line joining it to the axis of rotation, and \(r\) is its distance from the axis. Positive and negative signs are used to describe opposite directions of rotation.

The angular momentum of a finite-sized object or a system of many objects is found by dividing it up into many small parts, applying the equation to each part, and adding to find the total amount of angular momentum.

Note that \(r\) is not necessarily the radius of a circle. (As implied by the qualifiers, matter isn't the only thing that can have angular momentum. Light can also have angular momentum, and the above equation would not apply to light.)

Conservation of angular momentum has been verified over and over again by experiment, and is now believed to be one of the three most fundamental principles of physics, along with conservation of energy and momentum.

When a figure skater is twirling, there is very little friction between her and the ice, so she is essentially a closed system, and her angular momentum is conserved. If she pulls her arms in, she is decreasing \(r\) for all the atoms in her arms. It would violate conservation of angular momentum if she then continued rotating at the same speed, i.e., taking the same amount of time for each revolution, because her arms' contributions to her angular momentum would have decreased, and no other part of her would have increased its angular momentum. This is impossible because it would violate conservation of angular momentum. If her total angular momentum is to remain constant, the decrease in \(r\) for her arms must be compensated for by an overall increase in her rate of rotation. That is, by pulling her arms in, she substantially reduces the time for each rotation.

But what force is causing the moon to speed up, drawing it out into a larger orbit? It is the gravitational forces of the earth's tidal bulges. The effect is described qualitatively in the caption of the figure. The result would obviously be extremely difficult to calculate directly, and this is one of those situations where a conservation law allows us to make precise quantitative statements about the outcome of a process when the calculation of the process itself would be prohibitively complex.

Is angular momentum a vector, or a scalar? It does have a direction in space, but it's a direction of rotation, not a straight-line direction like the directions of vectors such as velocity or force. It turns out (see problem 29) that there is a way of defining angular momentum as a vector, but in this book the examples will be confined to a single plane of rotation, i.e., effectively two-dimensional situations. In this special case, we can choose to visualize the plane of rotation from one side or the other, and to define clockwise and counterclockwise rotation as having opposite signs of angular momentum.

◊

Conservation of plain old momentum, \(p\), can be thought of
as the greatly expanded and modified descendant of Galileo's
original principle of inertia, that no force is required to
keep an object in motion. The principle of inertia is
counterintuitive, and there are many situations in which it
appears superficially that a force *is* needed to
maintain motion, as maintained by Aristotle. Think of a
situation in which conservation of angular momentum, \(L\),
also seems to be violated, making it seem incorrectly that
something external must act on a closed system to keep its
angular momentum from “running down.”

We now discuss the application of conservation of angular momentum to planetary motion, both because of its intrinsic importance and because it is a good way to develop a visual intuition for angular momentum.

Kepler's law of equal areas states that the area swept out by a planet in a certain length of time is always the same. Angular momentum had not been invented in Kepler's time, and he did not even know the most basic physical facts about the forces at work. He thought of this law as an entirely empirical and unexpectedly simple way of summarizing his data, a rule that succeeded in describing and predicting how the planets sped up and slowed down in their elliptical paths. It is now fairly simple, however, to show that the equal area law amounts to a statement that the planet's angular momentum stays constant.

There is no simple geometrical rule for the area of a pie wedge cut out of an ellipse, but if we consider a very short time interval, as shown in figure i, the shaded shape swept out by the planet is very nearly a triangle. We do know how to compute the area of a triangle. It is one half the product of the base and the height:

\[\begin{equation*}
\text{area} = \frac{1}{2}bh .
\end{equation*}\]

We wish to relate this to angular momentum, which contains the variables \(r\) and \(v_{\perp}\) . If we consider the sun to be the axis of rotation, then the variable \(r\) is identical to the base of the triangle, \(r=b\). Referring to the magnified portion of the figure, \(v_{\perp}\) can be related to \(h\), because the two right triangles are similar:

\[\begin{equation*}
\frac{h}{\text{distance traveled}} = \frac{v_\perp}{|\mathbf{v}|}
\end{equation*}\]

The area can thus be rewritten as

\[\begin{equation*}
\text{area} = \frac{1}{2}r\frac{v_\perp(\text{distance traveled})}{|\mathbf{v}|} .
\end{equation*}\]

The distance traveled equals \(|\mathbf{v}|\Delta t\), so this simplifies to

\[\begin{equation*}
\text{area} = \frac{1}{2}rv_\perp \Delta t .
\end{equation*}\]

We have found the following relationship between angular momentum and the rate at which area is swept out:

\[\begin{equation*}
L = 2m\frac{\text{area}}{\Delta t} .
\end{equation*}\]

The factor of 2 in front is simply a matter of convention, since any conserved quantity would be an equally valid conserved quantity if you multiplied it by a constant. The factor of \(m\) was not relevant to Kepler, who did not know the planets' masses, and who was only describing the motion of one planet at a time.

We thus find that Kepler's equal-area law is equivalent to a statement that the planet's angular momentum remains constant. But wait, why should it remain constant? --- the planet is not a closed system, since it is being acted on by the sun's gravitational force. There are two valid answers. The first is that it is actually the total angular momentum of the sun plus the planet that is conserved. The sun, however, is millions of times more massive than the typical planet, so it accelerates very little in response to the planet's gravitational force. It is thus a good approximation to say that the sun doesn't move at all, so that no angular momentum is transferred between it and the planet.

The second answer is that to change the planet's angular momentum requires not just a force but a force applied in a certain way. In section 15.4 we discuss the transfer of angular momentum by a force, but the basic idea here is that a force directly in toward the axis does not change the angular momentum.

◊

Suppose an object is simply traveling in a straight line at constant speed. If we pick some point not on the line and call it the axis of rotation, is area swept out by the object at a constant rate? Would it matter if we chose a different axis?

◊

The figure is a strobe photo of a pendulum bob, taken from underneath the pendulum looking straight up. The black string can't be seen in the photograph. The bob was given a slight sideways push when it was released, so it did not swing in a plane. The bright spot marks the center, i.e., the position the bob would have if it hung straight down at us. Does the bob's angular momentum appear to remain constant if we consider the center to be the axis of rotation? What if we choose a different axis?

With plain old momentum, \(p\), we had the freedom to work in any inertial frame of reference we liked. The same object could have different values of momentum in two different frames, if the frames were not at rest with respect to each other. Conservation of momentum, however, would be true in either frame. As long as we employed a single frame consistently throughout a calculation, everything would work.

The same is true for angular momentum, and in addition there is an ambiguity that arises from the definition of an axis of rotation. For a wheel, the natural choice of an axis of rotation is obviously the axle, but what about an egg rotating on its side? The egg has an asymmetric shape, and thus no clearly defined geometric center. A similar issue arises for a cyclone, which does not even have a sharply defined shape, or for a complicated machine with many gears. The following theorem, the first of two presented in this section without proof, explains how to deal with this issue. Although I have put descriptive titles above both theorems, they have no generally accepted names.

It is entirely arbitrary what point one defines as the axis for purposes of calculating angular momentum. If a closed system's angular momentum is conserved when calculated with one choice of axis, then it will also be conserved for any other choice. Likewise, any inertial frame of reference may be used.

With planet A as the axis, the two asteroids have the same amount of angular momentum, but one has positive angular momentum and the other has negative. Before the collision, the total angular momentum is therefore zero. After the collision, the two asteroids will have stopped moving, and again the total angular momentum is zero. The total angular momentum both before and after the collision is zero, so angular momentum is conserved if you choose planet A as the axis.

The only difference with planet B as axis is that \(r\) is smaller by a factor of two, so all the angular momenta are halved. Even though the angular momenta are different than the ones calculated by planet A, angular momentum is still conserved.

The earth spins on its own axis once a day, but simultaneously travels in its circular one-year orbit around the sun, so any given part of it traces out a complicated loopy path. It would seem difficult to calculate the earth's angular momentum, but it turns out that there is an intuitively appealing shortcut: we can simply add up the angular momentum due to its spin plus that arising from its center of mass's circular motion around the sun. This is a special case of the following general theorem:

An object's angular momentum with respect to some outside
axis A can be found by adding up two parts:

(1) The first part is the object's angular momentum
found by using its own center of mass as the axis, i.e., the
angular momentum the object has because it is spinning.

(2) The other part equals the angular momentum that the object
would have with respect to the axis A if it had all its mass
concentrated at and moving with its center of mass.

In the special case of an object whose center of mass is at rest, the spin theorem implies that the object's angular momentum is the same regardless of what axis we choose. (This is an even stronger statement than the choice of axis theorem, which only guarantees that angular momentum is conserved for any given choice of axis, without specifying that it is the same for all such choices.)

\(\triangleright\) A motorcycle wheel has almost all its mass concentrated at the outside. If the wheel has mass \(m\) and radius \(r\), and the time required for one revolution is \(T\), what is the spin part of its angular momentum?

\(\triangleright\) This is an example of the commonly encountered special case of rigid motion, as opposed to the rotation of a system like a hurricane in which the different parts take different amounts of time to go around. We don't really have to go through a laborious process of adding up contributions from all the many parts of a wheel, because they are all at about the same distance from the axis, and are all moving around the axis at about the same speed. The velocity is all perpendicular to the spokes,

\[\begin{align*}
v_\perp &= v \\
&= (\text{circumference})/T \\
&= 2\pi r/T ,
\end{align*}\]

and the angular momentum of the wheel about its center is

\[\begin{align*}
L &= mv_{\perp}r \\
&= m(2\pi r/T)r \\
&= 2\pi mr^2/T .
\end{align*}\]

Note that although the factors of \(2\pi \) in this expression is peculiar to a wheel with its mass concentrated on the rim, the proportionality to \(m/T\) would have been the same for any other rigidly rotating object. Although an object with a noncircular shape does not have a radius, it is also true in general that angular momentum is proportional to the square of the object's size for fixed values of \(m\) and \(T\). For instance doubling an object's size doubles both the \(v_{\perp}\) and \(r\) factors in the contribution of each of its parts to the total angular momentum, resulting in an overall factor of four increase.

The figure shows some examples of angular momenta of various shapes rotating about their centers of mass. The equations for their angular momenta were derived using calculus, as described in my calculus-based book Simple Nature. Do not memorize these equations!

\(\triangleright\) In the men's Olympic hammer throw, a steel ball of radius 6.1 cm is swung on the end of a wire of length 1.22 m. What fraction of the ball's angular momentum comes from its rotation, as opposed to its motion through space?

\(\triangleright\) It's always important to solve problems symbolically first, and plug in numbers only at the end, so let the radius of the ball be \(b\), and the length of the wire \(\ell\). If the time the ball takes to go once around the circle is \(T\), then this is also the time it takes to revolve once around its own axis. Its speed is \(v=2\pi\ell/T\), so its angular momentum due to its motion through space is \(mv\ell=2\pi m\ell^2/T\). Its angular momentum due to its rotation around its own center is \((4\pi/5)mb^2/T\). The ratio of these two angular momenta is \((2/5)(b/\ell)^2=1.0\times10^{-3}\). The angular momentum due to the ball's spin is extremely small.

\(\triangleright\) This is a nice example of a question that can very nearly be answered based only on units. Since the three variables, \(m\), \(b\), and \(g\), all have different units, they can't be added or subtracted. The only way to combine them mathematically is by multiplication or division. Multiplying one of them by itself is exponentiation, so in general we expect that the answer must be of the form

\[\begin{equation*}
p = A m^j b^k g^l ,
\end{equation*}\]

where \(A\), \(j\), \(k\), and \(l\) are unitless constants. The result has to have units of \(\text{kg}\!\cdot\!\text{m}/\text{s}\). To get kilograms to the first power, we need

\[\begin{equation*}
j=1 ,
\end{equation*}\]

meters to the first power requires

\[\begin{equation*}
k+l=1 ,
\end{equation*}\]

and seconds to the power \(-1\) implies

\[\begin{equation*}
l=1/2 .
\end{equation*}\]

We find \(j=1\), \(k=1/2\), and \(l=1/2\), so the solution must be of the form

\[\begin{equation*}
p = A m\sqrt{bg} .
\end{equation*}\]

Note that no physics was required!

Consideration of units, however, won't help us to find the unitless constant \(A\). Let \(t\) be the time the rod takes to fall, so that \((1/2)gt^2=b/2\). If the rod is going to land exactly on its side, then the number of revolutions it completes while in the air must be 1/4, or 3/4, or 5/4, ..., but all the possibilities greater than 1/4 would cause the head of the rod to collide with the floor prematurely. The rod must therefore rotate at a rate that would cause it to complete a full rotation in a time \(T=4t\), and it has angular momentum \(L=(\pi/6)mb^2/T\).

The momentum lost by the object striking the rod is \(p\), and by conservation of momentum, this is the amount of momentum, in the horizontal direction, that the rod acquires. In other words, the rod will fly forward a little. However, this has no effect on the solution to the problem. More importantly, the object striking the rod loses angular momentum \(bp/2\), which is also transferred to the rod. Equating this to the expression above for \(L\), we find \(p=(\pi/12)m\sqrt{bg}\).

Finally, we need to know whether this can really be done without having the foot of the rod scrape on the floor. The figure shows that the answer is no for this rod of finite width, but it appears that the answer would be yes for a sufficiently thin rod. This is analyzed further in homework problem 28 on page 413.

◊

In the example of the colliding asteroids, suppose planet A was moving toward the top of the page, at the same speed as the bottom asteroid. How would planet A's astronomers describe the angular momenta of the asteroids? Would angular momentum still be conserved?

Force can be interpreted as the rate of transfer of
momentum. The equivalent in the case of angular momentum is
called *torque* (rhymes with
“fork”). Where force tells us how hard we are pushing or
pulling on something, torque indicates how hard we are
twisting on it. Torque is represented by the Greek letter
tau, \(\tau \), and the rate of change of an object's angular
momentum equals the total torque acting on it,

\[\begin{equation*}
\tau_{total} = \frac{\Delta L}{\Delta t} .
\end{equation*}\]

(If the angular momentum does not change at a constant rate, the total torque equals the slope of the tangent line on a graph of \(L\) versus \(t\).)

As with force and momentum, it often happens that angular momentum recedes into the background and we focus our interest on the torques. The torque-focused point of view is exemplified by the fact that many scientifically untrained but mechanically apt people know all about torque, but none of them have heard of angular momentum. Car enthusiasts eagerly compare engines' torques, and there is a tool called a torque wrench which allows one to apply a desired amount of torque to a screw and avoid overtightening it.

Of course a force is necessary in order to create a torque --- you can't twist a screw without pushing on the wrench --- but force and torque are two different things. One distinction between them is direction. We use positive and negative signs to represent forces in the two possible directions along a line. The direction of a torque, however, is clockwise or counterclockwise, not a linear direction.

The other difference between torque and force is a matter of leverage. A given force applied at a door's knob will change the door's angular momentum twice as rapidly as the same force applied halfway between the knob and the hinge. The same amount of force produces different amounts of torque in these two cases.

It is possible to have a zero total torque with a nonzero total force. An airplane with four jet engines, o, would be designed so that their forces are balanced on the left and right. Their forces are all in the same direction, but the clockwise torques of two of the engines are canceled by the counterclockwise torques of the other two, giving zero total torque.

Conversely we can have zero total force and nonzero total torque. A merry-go-round's engine needs to supply a nonzero torque on it to bring it up to speed, but there is zero total force on it. If there was not zero total force on it, its center of mass would accelerate!

How do we calculate the amount of torque produced by a given force? Since it depends on leverage, we should expect it to depend on the distance between the axis and the point of application of the force. We'll derive an equation relating torque to force for a particular very simple situation, and state without proof that the equation actually applies to all situations.

To try to pin down this relationship more precisely, let's imagine hitting a tetherball, figure p. The boy applies a force \(F\) to the ball for a short time \(\Delta t\), accelerating the ball from rest to a velocity \(v\). Since force is the rate of transfer of momentum, we have

\[\begin{align*}
F & = \frac{m\Delta v}{\Delta t} . \\
\text{Since the initial velocity is zero, $\Delta v$ is the same as the final
velocity $v$. Multiplying both sides by $r$ gives}
Fr & = \frac{mvr}{\Delta t} . \\
\end{align*}\]

But \(mvr\) is simply the amount of angular momentum he's given the ball, so \(mvr/\Delta t\) also equals the amount of torque he applied. The result of this example is

\[\begin{equation*}
\tau = Fr .
\end{equation*}\]

Figure p was drawn so that the force \(F\) was in the direction
tangent to the circle, i.e., perpendicular to the radius \(r\).
If the boy had applied a force *parallel* to the radius line, either directly
inward or outward, then the ball would not have picked up any clockwise or counterclockwise
angular momentum.

If a force acts at an angle other than 0 or 90° with respect to the line joining the object and the axis, it would be only the component of the force perpendicular to the line that would produce a torque,

\[\begin{equation*}
\tau = F_{\perp}r .
\end{equation*}\]

Although this result was proved under a simplified set of circumstances, it is more generally valid:

The rate at which a force transfers angular momentum to an object, i.e., the torque produced by the force, is given by

\[\begin{equation*}
|\boldsymbol{\tau}| = r|F_\perp| ,
\end{equation*}\]

where \(r\) is the distance from the axis to the point of application of the force, and \(F_\perp\) is the component of the force that is perpendicular to the line joining the axis to the point of application.

The equation is stated with absolute value signs because the positive and negative signs of force and torque indicate different things, so there is no useful relationship between them. The sign of the torque must be found by physical inspection of the case at hand.

From the equation, we see that the units of torque can be written as newtons multiplied by meters. Metric torque wrenches are calibrated in \(\text{N}\!\cdot\!\text{m}\), but American ones use foot-pounds, which is also a unit of distance multiplied by a unit of force. We know from our study of mechanical work that newtons multiplied by meters equal joules, but torque is a completely different quantity from work, and nobody writes torques with units of joules, even though it would be technically correct.

Compare the magnitudes and signs of the four torques shown in the figure.

(answer in the back of the PDF version of the book)\(\triangleright\) How can the torque applied to the wrench in the figure be expressed in terms of \(r\), \(|F|\), and the angle \(\theta \) between these two vectors?

\(\triangleright\) The force vector and its \(F_{\perp}\) component form the hypotenuse and one leg of a right triangle,

and the interior angle opposite to \(F_{\perp}\) equals \(\theta \). The absolute value of \(F_{\perp}\) can thus be expressed as

\[\begin{equation*}
F_{\perp} = |\mathbf{F}| \sin \theta ,
\end{equation*}\]

leading to

\[\begin{equation*}
|\boldsymbol{\tau}| = r |\mathbf{F}| \sin \theta .
\end{equation*}\]

Sometimes torque can be more neatly visualized in terms of the quantity \(r_{\perp}\) shown in figure r, which gives us a third way of expressing the relationship between torque and force:

\[\begin{equation*}
|\boldsymbol{\tau} | = r_{\perp} |\mathbf{F}| .
\end{equation*}\]

Of course you would not want to go and memorize all three equations for torque. Starting from any one of them you could easily derive the other two using trigonometry. Familiarizing yourself with them can however clue you in to easier avenues of attack on certain problems.

Up until now we've been thinking in terms of a force that acts at a single point on an object, such as the force of your hand on the wrench. This is of course an approximation, and for an extremely realistic calculation of your hand's torque on the wrench you might need to add up the torques exerted by each square millimeter where your skin touches the wrench. This is seldom necessary. But in the case of a gravitational force, there is never any single point at which the force is applied. Our planet is exerting a separate tug on every brick in the Leaning Tower of Pisa, and the total gravitational torque on the tower is the sum of the torques contributed by all the little forces. Luckily there is a trick that allows us to avoid such a massive calculation. It turns out that for purposes of computing the total gravitational torque on an object, you can get the right answer by just pretending that the whole gravitational force acts at the object's center of mass.

\(\triangleright\) The total gravitational force acting on your arm is

\[\begin{equation*}
|F|= (3.0\ \text{kg})(9.8\ \text{m}/\text{s}^2)= 29\ \text{N} .
\end{equation*}\]

For the purpose of calculating the gravitational torque, we can treat the force as if it acted at the arm's center of mass. The force is straight down, which is perpendicular to the line connecting the shoulder to the center of mass, so

\[\begin{equation*}
F_{\perp} =|F|= 29\ \text{N} .
\end{equation*}\]

Continuing to pretend that the force acts at the center of the arm, \(r\) equals 30 cm = 0.30 m, so the torque is

\[\begin{equation*}
\tau =r F_{\perp} = 9\ \text{N}\!\cdot\!\text{m} .
\end{equation*}\]

There are three forces on the cow: the force of gravity \(\mathbf{F}_W\), the ground's normal force \(\mathbf{F}_N\), and the tippers' force \(\mathbf{F}_A\).

As soon as the cow's left hooves (on the right from our point of view) break contact with the ground, the ground's force is being applied only to hooves on the other side. We don't know the ground's force, and we don't want to find it. Therefore we take the axis to be at its point of application, so that its torque is zero.

For the purpose of computing torques, we can pretend that gravity acts at the cow's center of mass, which I've placed a little lower than the center of its torso, since its legs and head also have some mass, and the legs are more massive than the head and stick out farther, so they lower the c.m. more than the head raises it. The angle \(\theta_W\) between the vertical gravitational force and the line \(r_{W}\) is about \(14°\). (An estimate by Matt Semke at the University of Nebraska-Lincoln gives \(20°\), which is in the same ballpark.)

To generate the maximum possible torque with the least possible force, the tippers want to push at a point as far as possible from the axis, which will be the shoulder on the other side, and they want to push at a 90 degree angle with respect to the radius line \(r_{A}\).

When the tippers are just barely applying enough force to raise the cow's hooves on one side, the total torque has to be just slightly more than zero. (In reality, they want to push a lot harder than this --- hard enough to impart a lot of angular momentum to the cow fair in a short time, before it gets mad and hurts them. We're just trying to calculate the bare minimum force they can possibly use, which is the question that science can answer.) Setting the total torque equal to zero,

\[\begin{gather*}
\tau_{N}+\tau_{W}+\tau_{A} = 0 , \\
\text{and letting counterclockwise torques be positive, we have}
0-mgr_W\sin\theta_W+F_Ar_A\sin 90° = 0 \\
\end{gather*}\]

\[\begin{align*}
F_A &= \frac{r_W}{r_A} mg \sin\theta_W \\
&\approx \frac{1}{1.5} (680\ \text{kg})(9.8\ \text{m}/\text{s}^2) \sin 14° \\
&=1100\ \text{N} .
\end{align*}\]

The 680 kg figure for the typical mass of a cow is due to Lillie and Boechler, who are veterinarians, so I assume it's fairly accurate. My estimate of 1100 N comes out significantly lower than their 1400 N figure, mainly because their incorrect placement of the center of mass gives \(\theta_W=24°\). I don't think 1100 N is an impossible amount of force to require of one big, strong person (it's equivalent to lifting about 110 kg, or 240 pounds), but given that the tippers need to impart a large angular momentum fairly quickly, it's probably true that several people would be required.

The main practical issue with cow tipping is that cows generally sleep lying down. Falling on its side can also seriously injure a cow.

◊

This series of discussion questions deals with past students' incorrect reasoning about the following problem.

Suppose a comet is at the point in its orbit shown in the figure. The only force on the comet is the sun's gravitational force.

Throughout the question, define all torques and angular momenta using the sun as the axis.

(1) Is the sun producing a nonzero torque on the comet? Explain.

(2) Is the comet's angular momentum increasing, decreasing,
or staying the same? Explain.

Explain what is wrong with the following answers. In some
cases, the answer is correct, but the reasoning leading up to it is wrong.
(a) Incorrect answer to part (1): “Yes, because the sun is
exerting a force on the comet, and the comet is a certain
distance from the sun.”

(b) Incorrect answer to part (1): “No, because the
torques cancel out.”

(c) Incorrect answer to part (2): “Increasing, because the
comet is speeding up.”

◊

Which claw hammer would make it easier to get the nail out of the wood if the same force was applied in the same direction?

◊

You whirl a rock over your head on the end of a string, and gradually pull in the string, eventually cutting the radius in half. What happens to the rock's angular momentum? What changes occur in its speed, the time required for one revolution, and its acceleration? Why might the string break?

◊

A helicopter has, in addition to the huge fan blades on top, a smaller propeller mounted on the tail that rotates in a vertical plane. Why?

◊

The photo shows an amusement park ride whose two cars rotate in opposite directions. Why is this a good design?

There are many cases where a system is not closed but maintains constant angular momentum. When a merry-go-round is running at constant angular momentum, the engine's torque is being canceled by the torque due to friction.

When an object has constant momentum and constant angular momentum, we say that it is in equilibrium. This is a scientific redefinition of the common English word, since in ordinary speech nobody would describe a car spinning out on an icy road as being in equilibrium.

Very commonly, however, we are interested in cases where an object is not only in equilibrium but also at rest, and this corresponds more closely to the usual meaning of the word. Trees and bridges have been designed by evolution and engineers to stay at rest, and to do so they must have not just zero total force acting on them but zero total torque. It is not enough that they don't fall down, they also must not tip over. Statics is the branch of physics concerned with problems such as these.

Solving statics problems is now simply a matter of applying and combining some things you already know:

- You know the behaviors of the various types of forces, for example that a frictional force is always parallel to the surface of contact.
- You know about vector addition of forces. It is the vector sum of the forces that must equal zero to produce equilibrium.
- You know about torque. The total torque acting on an object must be zero if it is to be in equilibrium.
- You know that the choice of axis is arbitrary, so you can make a choice of axis that makes the problem easy to solve.

In general, this type of problem could involve four equations in four unknowns: three equations that say the force components add up to zero, and one equation that says the total torque is zero. Most cases you'll encounter will not be this complicated. In the following example, only the equation for zero total torque is required in order to get an answer.

\(\triangleright\) All three objects in the figure are supposed to be in equilibrium: the pole, the cable, and the wall. Whichever of the three objects we pick to investigate, all the forces and torques on it have to cancel out. It is not particularly helpful to analyze the forces and torques on the wall, since it has forces on it from the ground that are not given and that we don't want to find. We could study the forces and torques on the cable, but that doesn't let us use the given information about the pole. The object we need to analyze is the pole.

The pole has three forces on it, each of which may also result in a torque: (1) the gravitational force, (2) the cable's force, and (3) the wall's force.

We are free to define an axis of rotation at any point we wish, and it is helpful to define it to lie at the bottom end of the pole, since by that definition the wall's force on the pole is applied at \(r=0\) and thus makes no torque on the pole. This is good, because we don't know what the wall's force on the pole is, and we are not trying to find it.

With this choice of axis, there are two nonzero torques on the pole, a counterclockwise torque from the cable and a clockwise torque from gravity. Choosing to represent counterclockwise torques as positive numbers, and using the equation \(|\boldsymbol{\tau}| =r|F| \sin \theta\), we have

\[\begin{equation*}
r_{cable} |F_{cable}| \sin \theta_{cable} - r_{grav}|F_{grav}|\sin \theta_{grav} = 0 .
\end{equation*}\]

A little geometry gives \(\theta_{cable}=90°-\alpha\) and \(\theta_{grav}=\alpha\), so

\[\begin{equation*}
r_{cable} |F_{cable}| \sin (90°-\alpha) - r_{grav}|F_{grav}| \sin \alpha = 0 .
\end{equation*}\]

The gravitational force can be considered as acting at the pole's center of mass, i.e., at its geometrical center, so \(r_{cable}\) is twice \(r_{grav}\), and we can simplify the equation to read

\[\begin{equation*}
2 |F_{cable}| \sin (90°-\alpha) - |F_{grav}| \sin \alpha = 0 .
\end{equation*}\]

These are all quantities we were given, except for \(\alpha\), which is the angle we want to find. To solve for \(\alpha\) we need to use the trig identity \(\sin (90°-x)= \cos x\),

\[\begin{equation*}
2 |F_{cable}| \cos \alpha - |F_{grav}| \sin \alpha = 0 ,
\end{equation*}\]

which allows us to find

\[\begin{align*}
\tan\alpha &= 2\frac{|\mathbf{F}_{cable}|}{|\mathbf{F}_{grav}|}\\
\alpha &= \tan^{-1}\left(2\frac{|\mathbf{F}_{cable}|}{|\mathbf{F}_{grav}|}\right)\\
&= \tan^{-1}\left(2\times\frac{70\ \text{N}}{98\ \text{N}}\right)\\
&= 55° .
\end{align*}\]

\(\triangleright\) There are four forces on the cube: a gravitational force \(mg\), the force \(F_T\) from the cable, the upward normal force from the cylinder, \(F_N\), and the horizontal static frictional force from the cylinder, \(F_s\).

The total force on the cube in the vertical direction is zero:

\[\begin{equation*}
F_N-mg = 0 .
\end{equation*}\]

As our axis for defining torques, it's convenient to choose the point of contact between the cube and the cylinder, because then neither \(F_s\) nor \(F_N\) makes any torque. The cable's torque is counterclockwise, the torque due to gravity is clockwise. Letting counterclockwise torques be positive, and using the convenient equation \(\tau=r_\perp F\), we find the equation for the total torque:

\[\begin{equation*}
b F_T - mga = 0 .
\end{equation*}\]

We could also write down the equation saying that the total horizontal force is zero, but that would bring in the cylinder's frictional force on the cube, which we don't know and don't need to find. We already have two equations in the two unknowns \(F_T\) and \(F_N\), so there's no need to make it into three equations in three unknowns. Solving the first equation for \(F_N=mg\), we then substitute into the second equation to eliminate \(F_N\), and solve for \(F_T=(a/b)mg\).

As a check, our result makes sense when \(a=0\); the cube is balanced on the cylinder, so the cable goes slack.

A pencil balanced upright on its tip could theoretically be in equilibrium, but even if it was initially perfectly balanced, it would topple in response to the first air current or vibration from a passing truck. The pencil can be put in equilibrium, but not in stable equilibrium. The things around us that we really do see staying still are all in stable equilibrium.

Why is one equilibrium stable and another unstable? Try pushing your own nose to the left or the right. If you push it a millimeter to the left, your head responds with a gentle force to the right, which keeps your nose from flying off of your face. If you push your nose a centimeter to the left, your face's force on your nose becomes much stronger. The defining characteristic of a stable equilibrium is that the farther the object is moved away from equilibrium, the stronger the force is that tries to bring it back.

The opposite is true for an unstable equilibrium. In the top figure, the ball resting on the round hill theoretically has zero total force on it when it is exactly at the top. But in reality the total force will not be exactly zero, and the ball will begin to move off to one side. Once it has moved, the net force on the ball is greater than it was, and it accelerates more rapidly. In an unstable equilibrium, the farther the object gets from equilibrium, the stronger the force that pushes it farther from equilibrium.

This idea can be rephrased in terms of energy. The difference between the stable and unstable equilibria shown in figure x is that in the stable equilibrium, the potential energy is at a minimum, and moving to either side of equilibrium will increase it, whereas the unstable equilibrium represents a maximum.

Note that we are using the term “stable” in a weaker sense
than in ordinary speech. A domino standing upright is stable
in the sense we are using, since it will not spontaneously
fall over in response to a sneeze from across the room or
the vibration from a passing truck. We would only call it
unstable in the technical sense if it could be toppled by
*any* force, no matter how small. In everyday usage, of
course, it would be considered unstable, since the force
required to topple it is so small.

\(\triangleright\) The equilibrium points occur where the PE is at a minimum or maximum, and minima and maxima occur where the derivative (which equals minus the force on Nancy) is zero. This derivative is \(dPE/dx=4x^3-2x\), and setting it equal to zero, we have \(x=0, \pm1/\sqrt{2}\). Minima occur where the second derivative is positive, and maxima where it is negative. The second derivative is \(12x^2-2\), which is negative at \(x=0\) (unstable) and positive at \(x=\pm1/\sqrt{2}\) (stable). Interpretation: the graph of the PE is shaped like a rounded letter `W,' with the two troughs representing the two halves of the splitting nucleus. Nancy is going to have to decide which half she wants to go with.

Although we have discussed some simple machines such as the pulley, without the concept of torque we were not yet ready to address the lever, which is the machine nature used in designing living things, almost to the exclusion of all others. (We can speculate what life on our planet might have been like if living things had evolved wheels, gears, pulleys, and screws.) The figures show two examples of levers within your arm. Different muscles are used to flex and extend the arm, because muscles work only by contraction.

Analyzing example aa physically, there are two forces that do work. When we lift a load with our biceps muscle, the muscle does positive work, because it brings the bone in the forearm in the direction it is moving. The load's force on the arm does negative work, because the arm moves in the direction opposite to the load's force. This makes sense, because we expect our arm to do positive work on the load, so the load must do an equal amount of negative work on the arm. (If the biceps was lowering a load, the signs of the works would be reversed. Any muscle is capable of doing either positive or negative work.)

There is also a third force on the forearm: the force of the upper arm's bone exerted on the forearm at the elbow joint (not shown with an arrow in the figure). This force does no work, because the elbow joint is not moving.

Because the elbow joint is motionless, it is natural to define our torques using the joint as the axis. The situation now becomes quite simple, because the upper arm bone's force exerted at the elbow neither does work nor creates a torque. We can ignore it completely. In any lever there is such a point, called the fulcrum.

If we restrict ourselves to the case in which the forearm rotates with constant angular momentum, then we know that the total torque on the forearm is zero,

\[\begin{equation*}
\tau_{muscle} + \tau_{load} = 0 .
\end{equation*}\]

If we choose to represent counterclockwise torques as positive, then the muscle's torque is positive, and the load's is negative. In terms of their absolute values,

\[\begin{equation*}
|\tau_{muscle}| = |\tau_{load}| .
\end{equation*}\]

Assuming for simplicity that both forces act at angles of 90° with respect to the lines connecting the axis to the points at which they act, the absolute values of the torques are

\[\begin{equation*}
r_{muscle} F_{muscle} = r_{load} F_{arm} ,
\end{equation*}\]

where \(r_{muscle}\), the distance from the elbow joint to the biceps' point of insertion on the forearm, is only a few cm, while \(r_{load}\) might be 30 cm or so. The force exerted by the muscle must therefore be about ten times the force exerted by the load. We thus see that this lever is a force reducer. In general, a lever may be used either to increase or to reduce a force.

Why did our arms evolve so as to reduce force? In general, your body is built for compactness and maximum speed of motion rather than maximum force. This is the main anatomical difference between us and the Neanderthals (their brains covered the same range of sizes as those of modern humans), and it seems to have worked for us.

As with all machines, the lever is incapable of changing the amount of mechanical work we can do. A lever that increases force will always reduce motion, and vice versa, leaving the amount of work unchanged.

It is worth noting how simple and yet how powerful this analysis was. It was simple because we were well prepared with the concepts of torque and mechanical work. In anatomy textbooks, whose readers are assumed not to know physics, there is usually a long and complicated discussion of the different types of levers. For example, the biceps lever, aa, would be classified as a class III lever, since it has the fulcrum and load on the ends and the muscle's force acting in the middle. The triceps, ab, is called a class I lever, because the load and muscle's force are on the ends and the fulcrum is in the middle. How tiresome! With a firm grasp of the concept of torque, we realize that all such examples can be analyzed in much the same way. Physics is at its best when it lets us understand many apparently complicated phenomena in terms of a few simple yet powerful concepts.

Kepler determined purely empirically that the planets' orbits were ellipses, without understanding the underlying reason in terms of physical law. Newton's proof of this fact based on his laws of motion and law of gravity was considered his crowning achievement both by him and by his contemporaries, because it showed that the same physical laws could be used to analyze both the heavens and the earth. Newton's proof was very lengthy, but by applying the more recent concepts of conservation of energy and angular momentum we can carry out the proof quite simply and succinctly, and without calculus.

The basic idea of the proof is that we want to describe the shape of the planet's orbit with an equation, and then show that this equation is exactly the one that represents an ellipse. Newton's original proof had to be very complicated because it was based directly on his laws of motion, which include time as a variable. To make any statement about the shape of the orbit, he had to eliminate time from his equations, leaving only space variables. But conservation laws tell us that certain things don't change over time, so they have already had time eliminated from them.

There are many ways of representing a curve by an equation, of which the most familiar is \(y=ax+b\) for a line in two dimensions. It would be perfectly possible to describe a planet's orbit using an \(x-y\) equation like this, but remember that we are applying conservation of angular momentum, and the space variables that occur in the equation for angular momentum are the distance from the axis, \(r\), and the angle between the velocity vector and the \(r\) vector, which we will call \(\phi \). The planet will have \(\phi \)=90° when it is moving perpendicular to the \(r\) vector, i.e., at the moments when it is at its smallest or greatest distances from the sun. When \(\phi \) is less than 90° the planet is approaching the sun, and when it is greater than 90° it is receding from it. Describing a curve with an \(r-\phi \) equation is like telling a driver in a parking lot a certain rule for what direction to steer based on the distance from a certain streetlight in the middle of the lot.

The proof is broken into the three parts for easier digestion. The first part is a simple and intuitively reasonable geometrical fact about ellipses, whose proof we relegate to the caption of figure ad; you will not be missing much if you merely absorb the result without reading the proof.

(1) If we use one of the two foci of an ellipse as an axis for defining the variables \(r\) and \(\phi \), then the angle between the tangent line and the line drawn to the other focus is the same as \(\phi \), i.e., the two angles labeled \(\phi \) in figure ad are in fact equal.

The other two parts form the meat of our proof. We state the results first and then prove them.

(2) A planet, moving under the influence of the sun's gravity with less than the energy required to escape, obeys an equation of the form

\[\begin{equation*}
\sin\phi = \frac{1}{\sqrt{-pr^2+qr}} ,
\end{equation*}\]

where \(p\) and \(q\) are positive constants that depend on the planet's energy and angular momentum.

(3) A curve is an ellipse if and only if its \(r-\phi \) equation is of the form

\[\begin{equation*}
\sin\phi = \frac{1}{\sqrt{-pr^2+qr}} ,
\end{equation*}\]

where \(p\) and \(q\) are constants that depend on the size and shape of the ellipse and \(p\) is greater than zero.

The component of the planet's velocity vector that is perpendicular to the \(\mathbf{r}\) vector is \(v_\perp=v \sin \phi \), so conservation of angular momentum tells us that \(L=mrv \sin\phi\) is a constant. Since the planet's mass is a constant, this is the same as the condition

\[\begin{equation*}
rv \sin \phi = \text{constant} .
\end{equation*}\]

Conservation of energy gives

\[\begin{equation*}
\frac{1}{2}mv^2-\frac{GMm}{r} = \text{constant} .
\end{equation*}\]

We solve the first equation for \(v\) and plug into the second equation to eliminate \(v\). Straightforward algebra then leads to the equation claimed above, with the constant \(p\) being positive because of our assumption that the planet's energy is insufficient to escape from the sun, i.e., its total energy is negative.

We define the quantities \(\alpha\), \(d\), and \(s\) as shown in the figure. The law of cosines gives

\[\begin{equation*}
d^2 = r^2+s^2-2rs\cos\alpha .
\end{equation*}\]

Using \(\alpha =180°-2\phi\) and the trigonometric identities \(\cos (180°-x)=-\cos x\) and \(\cos 2x= 1-2 \sin^2 x\), we can rewrite this as

\[\begin{equation*}
d^2 = r^2 + s^2 - 2rs\left(2\sin^2\phi-1\right) .
\end{equation*}\]

Straightforward algebra transforms this into

\[\begin{equation*}
\sin\:\phi = \sqrt{\frac{(r+s)^2-d^2}{4rs}} .
\end{equation*}\]

Since \(r+s\) is constant, the top of the fraction is constant, and the denominator can be rewritten as \(4rs=4r(\text{constant}-r)\), which is equivalent to the desired form.

*angular momentum* — a measure of rotational motion; a conserved
quantity for a closed system

*axis* — An arbitrarily chosen point used in the definition of
angular momentum. Any object whose direction changes
relative to the axis is considered to have angular momentum.
No matter what axis is chosen, the angular momentum of a
closed system is conserved.

*torque* — the rate of change of angular momentum; a numerical
measure of a force's ability to twist on an object

*equilibrium* — a state in which an object's momentum and
angular momentum are constant

*stable equilibrium* — one in which a force always acts to bring
the object back to a certain point

*unstable equilibrium* — one in which any deviation of the
object from its equilibrium position results in a force
pushing it even farther away

\(L\) — angular momentum

\(t\) — torque

\(T\) — the time required for a rigidly rotating body to complete one rotation

period — a name for the variable \(T\) defined above

moment of inertia, \(I\) — the proportionality constant in the equation \(L = 2\pi I / T\)

{}

Angular momentum is a measure of rotational motion which is
conserved for a closed system. This book only discusses
angular momentum for rotation of material objects in two
dimensions. Not all rotation is rigid like that of a wheel
or a spinning top. An example of nonrigid rotation is a
cyclone, in which the inner parts take less time to complete
a revolution than the outer parts. In order to define a
measure of rotational motion general enough to include
nonrigid rotation, we define the angular momentum of a
system by dividing it up into small parts, and adding up all
the angular momenta of the small parts, which we think of as
tiny particles. We arbitrarily choose some point in space,
the *axis*, and we say that anything that changes its
direction relative to that point possesses angular momentum.
The angular momentum of a single particle is

\[\begin{equation*}
L = mv_{\perp}r ,
\end{equation*}\]

where \(v_{\perp}\) is the component of its velocity perpendicular to the line joining it to the axis, and \(r\) is its distance from the axis. Positive and negative signs of angular momentum are used to indicate clockwise and counterclockwise rotation.

The *choice of axis theorem* states that any axis may
be used for defining angular momentum. If a system's angular
momentum is constant for one choice of axis, then it is also
constant for any other choice of axis.

The *spin theorem* states that an object's angular
momentum with respect to some outside axis A can be found by
adding up two parts:

(1) The first part is the object's angular momentum found by using its own center of mass as the axis, i.e., the angular momentum the object has because it is spinning.

(2) The other part equals the angular momentum that the object would have with respect to the axis A if it had all its mass concentrated at and moving with its center of mass.

Torque is the rate of change of angular momentum. The torque a force can produce is a measure of its ability to twist on an object. The relationship between force and torque is

\[\begin{equation*}
|\boldsymbol{\tau} | = r |F_{\perp} | ,
\end{equation*}\]

where \(r\) is the distance from the axis to the point where the force is applied, and \(F_{\perp}\) is the component of the force perpendicular to the line connecting the axis to the point of application. Statics problems can be solved by setting the total force and total torque on an object equal to zero and solving for the unknowns.

\begin{homeworkforcelabel}{rv}{1}{}{1}You are trying to loosen a stuck bolt on your RV using a big wrench that is 50 cm long. If you hang from the wrench, and your mass is 55 kg, what is the maximum torque you can exert on the bolt? (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{therapy}{1}{}{2}A physical therapist wants her patient to rehabilitate his injured elbow by laying his arm flat on a table, and then lifting a 2.1 kg mass by bending his elbow. In this situation, the weight is 33 cm from his elbow. He calls her back, complaining that it hurts him to grasp the weight. He asks if he can strap a bigger weight onto his arm, only 17 cm from his elbow. How much mass should she tell him to use so that he will be exerting the same torque? (He is raising his forearm itself, as well as the weight.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{equilibrium-top-of-arc}{1}{}{3}An object thrown straight up in the air is momentarily at rest when it reaches the top of its motion. Does that mean that it is in equilibrium at that point? Explain. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{const-l-no-torques}{1}{}{4}An object is observed to have constant angular momentum. Can you conclude that no torques are acting on it? Explain. [Based on a problem by Serway and Faughn.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tiptoe}{1}{}{5} A person of weight \(W\) stands on the ball of one foot. Find the tension in the calf muscle and the force exerted by the shinbones on the bones of the foot, in terms of \(W,a\), and \(b\). For simplicity, assume that all the forces are at 90-degree angles to the foot, i.e., neglect the angle between the foot and the floor. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pandl}{1}{}{6}Two objects have the same momentum vector. Assume that they are not spinning; they only have angular momentum due to their motion through space. Can you conclude that their angular momenta are the same? Explain. [Based on a problem by Serway and Faughn.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{whitedwarf}{1}{}{7}The sun turns on its axis once every 26.0 days. Its mass
is \(2.0\times10^{30}\) kg and its radius is \(7.0\times10^8\) m.
Assume it is a rigid sphere of uniform density.

(a) What is the sun's angular momentum? (answer check available at lightandmatter.com)

In a few billion years, astrophysicists predict that the sun
will use up all its sources of nuclear energy, and will
collapse into a ball of exotic, dense matter known as a
white dwarf. Assume that its radius becomes \(5.8\times10^6\) m
(similar to the size of the Earth.) Assume it does not lose
any mass between now and then. (Don't be fooled by the
photo, which makes it look like nearly all of the star was
thrown off by the explosion. The visually prominent gas
cloud is actually thinner than the best laboratory vacuum
ever produced on earth. Certainly a little bit of mass is
actually lost, but it is not at all unreasonable to make an
approximation of zero loss of mass as we are doing.)

(b) What will its angular momentum be?

(c) How long will it take to turn once on its axis? (answer check available at lightandmatter.com)

\end{homeworkforcelabel}

\begin{homeworkforcelabel}{ladderwarmup}{1}{}{8}A uniform ladder of mass \(m\) and length \(L\) leans against
a smooth wall, making an angle \(q\) with respect to the
ground. The dirt exerts a normal force and a frictional
force on the ladder, producing a force vector with magnitude
\(F_1\) at an angle \(\phi \) with respect to the ground. Since
the wall is smooth, it exerts only a normal force on the
ladder; let its magnitude be \(F_2\).

(a) Explain why \(\phi \) must be greater than \(\theta\).
No math is needed.

(b) Choose any numerical values you like for \(m\) and \(L\),
and show that the ladder can be in equilibrium (zero torque
and zero total force vector) for \(\theta=45.00°\)
and \(\phi=63.43°\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{ladder}{2}{}{9}Continuing the previous problem, find an equation for \(\phi \) in terms of \(\theta \), and show that \(m\) and \(L\) do not enter into the equation. Do not assume any numerical values for any of the variables. You will need the trig identity \(\sin (a-b)=\sin a \cos b - \sin b \cos a\). (As a numerical check on your result, you may wish to check that the angles given in part \(b\) of the previous problem satisfy your equation.) (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{wheel-over-step}{1}{}{10}(a) Find the minimum horizontal force which, applied at
the axle, will pull a wheel over a step. Invent algebra
symbols for whatever quantities you find to be relevant, and
give your answer in symbolic form. [Hints: There are four
forces on the wheel at first, but only three when it lifts
off. Normal forces are always perpendicular to the surface
of contact. Note that the corner of the step cannot be
perfectly sharp, so the surface of contact for this force
really coincides with the surface of the wheel.]

(b) Under what circumstances does your result become
infinite? Give a physical interpretation.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{yoyo}{2}{}{11} A yo-yo of total mass \(m\) consists of two solid cylinders of radius \(R\), connected by a small spindle of negligible mass and radius \(r\). The top of the string is held motionless while the string unrolls from the spindle. Show that the acceleration of the yo-yo is \(g/(1+R^2/2r^2)\). [Hint: The acceleration and the tension in the string are unknown. Use \(\tau =\Delta L/\Delta t\) and \(F=ma\) to determine these two unknowns.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{tetherball}{1}{}{12}A ball is connected by a string to a vertical post. The ball is set in horizontal motion so that it starts winding the string around the post. Assume that the motion is confined to a horizontal plane, i.e., ignore gravity. Michelle and Astrid are trying to predict the final velocity of the ball when it reaches the post. Michelle says that according to conservation of angular momentum, the ball has to speed up as it approaches the post. Astrid says that according to conservation of energy, the ball has to keep a constant speed. Who is right? [Hint: How is this different from the case where you whirl a rock in a circle on a string and gradually reel in the string?] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{orbital-bomb}{1}{}{13}In the 1950's, serious articles began appearing in
magazines like *Life* predicting that world domination
would be achieved by the nation that could put nuclear bombs
in orbiting space stations, from which they could be dropped
at will. In fact it can be quite difficult to get an
orbiting object to come down. Let the object have energy
\(E=KE+PE\) and angular momentum \(L\). Assume
that the energy is negative, i.e., the object is moving at
less than escape velocity. Show that it can never reach a radius less than

\[\begin{equation*}
r_{min} =
\frac{GMm}{2E}\left(-1+\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}\right)
.
\end{equation*}\]

[Note that both factors are negative, giving a positive result.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bridge}{1}{}{14}[Problem 14 has been deleted.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{angle-bars}{2}{}{15}[Problem 15 has been deleted.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{hinge-on-cylinder}{2}{}{16}Two bars of length \(L\) are connected with a hinge and placed on a frictionless cylinder of radius \(r\). (a) Show that the angle \(\theta \) shown in the figure is related to the unitless ratio \(r/L\) by the equation

\[\begin{equation*}
\frac{r}{L} = \frac{\cos^2\theta}{2\tan\theta} .
\end{equation*}\]

(b) Discuss the physical behavior of this equation for very
large and very small values of \(r/L\).
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{ship-in-bottle}{1}{}{17}You wish to determine the mass of a ship in a bottle without taking it out. Show that this can be done with the setup shown in the figure, with a scale supporting the bottle at one end, provided that it is possible to take readings with the ship slid to several different locations. Note that you can't determine the position of the ship's center of mass just by looking at it, and likewise for the bottle. In particular, you can't just say, “position the ship right on top of the fulcrum” or “position it right on top of the balance.” \end{homeworkforcelabel}

\begin{homeworkforcelabel}{lennardjones}{1}{1}{18}Two atoms will interact via electrical forces between their protons and electrons. One fairly good approximation to the potential energy is the Lennard-Jones formula,

\[\begin{equation*}
PE (r) = k\left[\left(\frac{a}{r}\right)^{12}-2\left(\frac{a}{r}\right)^{6}\right],
\end{equation*}\]

where \(r\) is the center-to-center distance between the atoms and \(k\) is a positive constant.
Show that (a) there is an equilibrium point at \(r=a\),

(b) the equilibrium is stable, and

(c) the energy required to bring
the atoms from their equilibrium separation to infinity is
\(k\). \hwhint{hwhint:lennardjones}
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{alternate-universe}{1}{}{19}Suppose that we lived in a universe in which Newton's law of gravity gave forces proportional to \(r^{-7}\) rather than \(r^{-2}\). Which, if any, of Kepler's laws would still be true? Which would be completely false? Which would be different, but in a way that could be calculated with straightforward algebra? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pliers}{1}{}{20} (solution in the pdf version of the book) The figure shows scale drawing of a pair of pliers being used to crack a nut, with an appropriately reduced centimeter grid. Warning: do not attempt this at home; it is bad manners. If the force required to crack the nut is 300 N, estimate the force required of the person's hand. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{rolling-ratio}{1}{}{21}Show that a sphere of radius \(R\) that is rolling without slipping has angular momentum and momentum in the ratio \(L/p=(2/5)R\). \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bowling}{1}{}{22} Suppose a bowling ball is initially thrown so that it has no angular momentum at all, i.e., it is initially just sliding down the lane. Eventually kinetic friction will get it spinning fast enough so that it is rolling without slipping. Show that the final velocity of the ball equals 5/7 of its initial velocity. [Hint: You'll need the result of problem 21.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{rod-finger-string}{1}{}{23}The rod in the figure is supported by the finger and the string.

(a) Find the tension, \(T\), in the string, and the force,
\(F\), from the finger, in terms of \(m,b,L\), and \(g\).(answer check available at lightandmatter.com)

(b) Comment on the cases \(b=L\) and \(b=L/2\).

(c) Are any values of \(b\) unphysical?
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{two-branches}{1}{}{24}Two horizontal tree branches on the same tree have equal diameters, but one branch is twice as long as the other. Give a quantitative comparison of the torques where the branches join the trunk. [Thanks to Bong Kang.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{zero-and-nonzero-p-and-l}{1}{}{25}(a) Alice says Cathy's body has zero momentum, but Bob says Cathy's momentum is nonzero.
Nobody is lying or making a mistake. How is this possible? Give a concrete example.

(b) Alice and Bob agree that Dong's body has nonzero momentum, but disagree about Dong's
angular momentum, which Alice says is zero, and Bob says is nonzero. Explain.
\end{homeworkforcelabel}

\begin{homeworkforcelabel}{surfin-tux}{1}{}{26}Penguins are playful animals. Tux the Penguin invents a new game using a natural circular depression in the ice. He waddles at top speed toward the crater, aiming off to the side, and then hops into the air and lands on his belly just inside its lip. He then belly-surfs, moving in a circle around the rim. The ice is frictionless, so his speed is constant. Is Tux's angular momentum zero, or nonzero? What about the total torque acting on him? Take the center of the crater to be the axis. Explain your answers. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{corkscrew}{1}{}{27}Make a rough estimate of the mechanical advantage of the lever shown in the figure. In other words, for a given amount of force applied on the handle, how many times greater is the resulting force on the cork? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{toppling-rod}{2}{}{28} (solution in the pdf version of the book) In example 8 on page 391, prove that if the rod is sufficiently thin, it can be toppled without scraping on the floor. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{check-mg-at-cm}{1}{}{29}A massless rod of length \(\ell\) has weights, each of mass \(m\), attached to its ends. The rod is initially put in a horizontal position, and laid on an off-center fulcrum located at a distance \(b\) from the rod's center. The rod will topple. (a) Calculate the total gravitational torque on the rod directly, by adding the two torques. (b) Verify that this gives the same result as would have been obtained by taking the entire gravitational force as acting at the center of mass. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{motorcycle-jump}{1}{}{30}A skilled motorcyclist can ride up a ramp, fly through the air, and land on another ramp. Why would it be useful for the rider to speed up or slow down the back wheel while in the air? \end{homeworkforcelabel}

\begin{handson}{}{Torque}{\onecolumn}

Equipment:

- rulers with holes in them
- spring scales (two per group)

\widefignocaptionnofloat{../../../share/mechanics/figs/ex-torque}

While one person holds the pencil which forms the axle for the ruler, the other members of the group pull on the scale and take readings. In each case, calculate the total torque on the ruler, and find out whether it equals zero to roughly within the accuracy of the experiment. Finish the calculations for each part before moving on to the next one. \end{handson}

(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.