USAPhO 2021, Полуфинал — Задача 4. B1 · Disk Jockey

• a. $\phi_1 + \phi_2 \approx -\frac{4m}{M}\frac{h\theta}{R}$

  The sign is convention-dependent.

• b. $\phi = \frac{4m}{M}\frac{\pi r^2}{R^2}$

a.

During the first leg of the trip, the disk has angular velocity

by conservation of angular momentum. Thus, the disk rotates through an angle

to leading order in $m/M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m/M$. The exact answer is a more complicated function of $m/M$. By going to “leading order”, we mean we are expanding that exact answer in a series in $m/M$, such as with the binomial theorem, and keeping only the first nonzero term.)

When the ant is moving radially, $\omega = 0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity

which means the disk rotates through an angle

The net rotation is

The sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)

Incidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can’t move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc’s motion coming from angular momentum conservation is not holonomic.

b.

There are many ways to do this problem, so we’ll give a selection, starting with a straightforward solution and then considering some increasingly elegant solutions.

First solution: The overall rotation angle of the disk is

where we again work to leading order in $m/M$, and $\mathbf{r}$ and $\mathbf{v}$ are the position and velocity of the ant. The coordinates of a point on the circle are given by

If the speed of the ant is $v$, the velocity is given by

To evaluate the angular momentum, note that

Thus, we have

Plugging this into the time integral above,

Changing this to an integral over $\theta$ using $d\theta = v\,dt/r$,

Of course, the problem could also be solved by parameterizing the ant’s path in a different way, such as by using polar coordinates with the origin at the center of the disk. The way we set it up here is the simplest, since it makes the integral easy. (For most students, the hardest part was finding a compact expression for $L$. A common mistake was assuming $L = mv_xr$ or a variant thereof.)

Second solution: The net effect on the disk of one ant going in the circular path is the same as two ants going along the path, each with half the mass, and thus the same as four ants each with a quarter the mass, and so on. By repeating this logic, we can thus replace the ant with a ring of radius $r$ and mass $m$ of uniform density, which rotates around once. Therefore, the rotation angle is

This is very simple, though it’s a trick that only works for a circular trajectory.

Third solution: Starting from the first line of the first solution, we notice that

where $A$ is the area of the ant’s trajectory. Thus, we have

This makes it clear why the answer had to be simple in general: the angle can only depend on a geometric property of the ant’s trajectory, namely its area. This kind of phenomenon occurs in many fields of physics, and is generally known as a geometric phase.

Fourth solution: We can decompose the circle into a stack of thin rectangles. The effect of a single ant going around the circle is the same as the effect of one ant going around each rectangle. But by slightly generalizing your result in part (a), you can show that the net rotation due to each rectangle is $(4m/M)(dA/R^2)$ where $dA$ is the area of that rectangle. Summing the areas gives the answer. Like the third solution, this works for any ant trajectory, and it makes it clear why it was the area of the trajectory that mattered.