Quarter rolled around a quarter. This question is inspired by this YouTube video which asks a question about how many times a circle revolves when rolled around another circle and comes up with a counterintuitive answer. I have a related question:

Suppose we have two (perfectly equal in size) quarters. We pick one to be stationary and call it $B$. We pick the other $A$ to be in motion. We roll $A$ around $B$, how many revolutions does $A$ revolve?

I approached this problem by noticing that the center of $A$ will stay a constant distance around the center of $B$. If $r$ represents the radius of each of the coins, that distance is $2r$. The center of $A$ will travel in a circular path fixed about radius $2r$.
The distance travelled by the center is $4\pi r$. In order the revolve once, the center of circle $A$ must travel a distance equal to its own circumference, which is $2\pi r$. Taking the ratio of the two we get that the moving coin should have made $2$ revolutions.
Is this a valid approach to the problem?
 A: Your reasoning is correct and elegant, but it's difficult to see that it's correct without further arguments because it's not clear that the length of the curve traversed by the center is so directly related to how many times it spun. I propose the following argument to that end:

For a coin to be rolling on another surface without slipping, it is necessary that the velocity of the coin at the point of contact is $0$. To make use of this, let the speed of the coin at its center be $v$, the radius to the point of contact be $r$, and its angular velocity be $\omega$. We find that the speed at the contact point will be $v-r\omega$ since the velocity at the contact point due to rotation about the center directly opposes the coin's velocity. Since this speed is zero, we get $v-r\omega=0$ or $\frac{v}r=\omega$.

This relates the coin's speed to its angular velocity. Integrating over these quantities gives that the angle which the moving coin rotates is $\frac{\ell}r$ where $\ell$ is the distance the center travels. Noting that "rotating by $2\pi$" is the same as "revolving once", we get that the number of revolutions is $\frac{\ell}{2\pi r}$, which is exactly the result you claim.
