Subfield: Mechanics

## Disk Jockey



A disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a frictionless floor. The disk is attached to the floor by a frictionless pivot at its center, which keeps the center of the disk in place, but allows the disk to rotate freely. An ant of mass $m \ll M$ is initially standing on the edge of the disk; you may give your answers to leading order in $m / M$.
Context question:
a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$.



<img_4433>



Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?
Context answer:
\boxed{$\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right)$}


Question: b. Now suppose the ant walks with speed $v$ along a circle of radius $r$, tangent to its starting point.



<img_4453>



Through what net angle does the disk rotate?

Solution:
First solution: The overall rotation angle of the disk is



$$

\phi=\int \omega d t=\frac{2}{M R^{2}} \int L d t=\frac{2 m}{M R^{2}} \int \mathbf{r} \times \mathbf{v} d t

$$



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



$$

(r \sin \theta,(R-r)+r \cos \theta)

$$







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



$$

(v \cos \theta,-v \sin \theta)

$$



To evaluate the angular momentum, note that



$$

|\mathbf{v} \times \mathbf{r}|=v r \cos ^{2} \theta+v r \sin ^{2} \theta+v(R-r) \cos \theta=v r+v(R-r) \cos \theta

$$



Thus, we have



$$

L=m v(r+(R-r) \cos \theta)

$$



Plugging this into the time integral above,



$$

\phi=\int \frac{2 m v}{M R^{2}}(r+(R-r) \cos (\theta(t))) \mathrm{d} t

$$



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



$$

\phi=\int_{0}^{2 \pi} \frac{2 m r}{M R^{2}}(r+(R-r) \cos (\theta(t))) \mathrm{d} \theta=\frac{4 m}{M} \frac{\pi r^{2}}{R^{2}}

$$



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 $L=m v_{x} r$ or a variant thereof.)
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



$$

\phi=2 \pi \frac{I_{\mathrm{disk}}}{I_{\text {ring }}}=2 \pi \frac{m r^{2}}{M R^{2} / 2}=\frac{4 m}{M} \frac{\pi r^{2}}{R^{2}}

$$



This is very simple, though it's a trick that only works for a circular trajectory.
Starting from the first line of the first solution, we notice that



$$

\int \mathbf{r} \times \mathbf{v} d t=\int \mathbf{r} \times d \mathbf{r}=2 A

$$



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



$$

\phi=\frac{4 m}{M} \frac{A}{R^{2}}=\frac{4 m}{M} \frac{\pi r^{2}}{R^{2}}

$$



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.
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 $(4 m / M)\left(d A / R^{2}\right)$ where $d A$ 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.

Final answer: $\frac{4 m}{M} \frac{\pi r^{2}}{R^{2}}$