# 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?

• You should specify the frame of reference. You seem to use the reference frame as the table the stationary table is on, at a distance from the quarters. An observer in small scale sitting at the center of the stationary quarter would say the moving quarter makes one rotation. This is similar to asking how long it takes for Earth's moon to rotate. The answer is 27 1/3 days (relative to the stars), 29 1/2 days (relative to the sun, which also means according to the moon's phases), and forever (relative to the earth). – Rory Daulton Nov 22 '15 at 18:23
• I think the video does a good job explaining all the points of view, so what is the question? – Morgan Rodgers Nov 22 '15 at 18:26
• Which do you think is correct? – Ahmed S. Attaalla Nov 22 '15 at 18:28
• @RoryDaulton: The OP specifically says, "We pick one [to] be stationary, and we call it $B$." This constitutes specifying a frame of reference, surely? – TonyK Nov 22 '15 at 23:03
• @RoryDaulton: We are not talking about suns and moons here, we are talking about coins on a tabletop. The situation is clear, in my opinion. You seem to be making difficulties just for the sake of it! – TonyK Nov 23 '15 at 0:39

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.