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# Conservation of Angular Momentum

## Angular Momentum

“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.

A tornado touches down in Spring Hill, Kansas, May 20, 1957.

These observations have the hallmarks of a conservation law, but 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. For example, the twirling figure skater in figure a can pull her arms in to increase her r.p.m.'s.

Figure a: A figure skater pulls in her arms so that she can execute a spin more rapidly.

The example of the figure skater suggests that this conserved quantity depends on distance from the axis of rotation. We'll notate this distance as r, since, for an object moving in a circle around an axis of rotation, its distance from the axis equals the radius of the circle.

Once we realize that r is a variable that matters, it becomes clear that the examples we've been considering were all examples that would be fairly complicated mathematically, because different parts of these objects' masses have different values of r. For example, the figure skater's front teeth are farther from the axis than her back teeth. That suggests that instead of objects with complicated shapes, we should consider the simplest possible example, which is a single particle, of mass m, traveling in a circle of radius r at speed v. Experiments show that the conserved quantity in this situation is

± mvr .

We call this quantity angular momentum. The symbol ± indicates that angular momentum has a positive or negative sign to represent the direction of rotation; for example, in a given problem, we could choose to represent clockwise angular momenta as positive numbers, and counterclockwise ones as negative. In this equation, the only velocity that matters is velocity that is perpendicular to the radius line; motion parallel to the radius line, i.e., directly in our out, is neither clockwise nor counterclockwise.

#### Example 1: A figure skater pulls her arms in

When the skater in figure a 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 would be closer to the axis of rotation and therefore have a smaller r (as well as a smaller v because they would be completing a smaller circle in the same time). This is impossible because it would violate conservation of angular momentum. If her total angular momentum is to remain constant, the decrease in angular momentum 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.

#### Example 2: A longjump

In figure b, 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.

b / Example 2: An early photograph of an old-fashioned long-jump.

#### Example 3: Changing the axis

An object's angular momentum can be different depending on the axis about which it rotates, because r is defined relative to the axis. Figure c shows shows two double-exposure photographs a viola player tipping the bow in order to cross from one string to another. Much more angular momentum is required when playing near the bow's handle, called the frog, as in the panel on the right; not only are most of the atoms in the bow are at greater distances, r, from the axis of rotation, but the ones in the tip also have more velocity, v. It is difficult for the player to quickly transfer a large angular momentum into the bow, and then transfer it back out just as quickly. This is one of the reasons that string players tend to stay near the middle of the bow as much as possible.

Figure c: Example 3.

#### Example 4: Kepler's equal-area law

The hypothetical planet in figure d has an orbit in which its closest approach to the sun is at half the distance compared to the point at which it recedes the farthest. Since angular momentum, mvr, is conserved, and the planet's mass is constant, the quantity vr must be the same at both ends of the orbit. Doubling r therefore requires cutting v in half. If the time interval from A to B is the same as that from C to D, then the distance from C to D must be half as much. But this is exactly what Kepler's equal area law requires, since the triangular pie wedge on top needs to have half the width to compensate for its doubled height. In other words, the equal area law is a direct consequence of conservation of angular momentum.

Figure d: Example 4.

##### Discussion Questions
• 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.”
• 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?

Figure: Discussion question B.

## 3.3 Noether's Theorem

Suppose a sunless planet is sitting all by itself in interstellar space, not rotating. Then, one day, it decides to start spinning. This doesn't necessarily violate conservation of energy, because it could have energy stored up, e.g., the heat in a molten core, which could be converted into kinetic energy. It does violate conservation of angular momentum, but even if we didn't already know about that law of physics, the story would seem odd. How would it decide which axis to spin around? If it was to spontaneously start spinning about some axis, then that axis would have to be a special, preferred direction in space. That is, space itself would have to have some asymmetry to it.

In reality, experiments show to a very high degree of precision that the laws of physics are completely symmetric with respect to different directions. The story of the planet that abruptly starts spinning is an example of Noether's theorem, applied to angular momentum. We now have three such examples:

## Homework Problems

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)

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)

3. 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.]

4. (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.

h / Problem 4.

5. 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.]

6. (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.

7. 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. (The tension is a measure of how tight the calf muscle has been pulled; it has units of newtons, and equals the amount of force applied by the muscle where it attaches to the heel.) For simplicity, assume that all the forces are at 90-degree angles to the foot. Suggestion: Write down an equation that says the total force on the foot is zero, and another equation saying that the total torque on the foot is zero; solve the two equations for the two unknowns.

i / Problem 7.