This was something in a course of mine I'm a bit too thick to see. If one takes a circle of radius $3$ and a circle of radius $1$, and rolls the smaller circle smoothly inside the larger one until the point $P$ which starts out in contact with the larger circle is in the same place, how can one prove that the smaller circle goes through exactly two rotations?
The curve generated by the path of the vector $\mathbf{P}$ (the vector from the center of the larger circle to $P$) is called a roulette – more specifically, a deltoid.
If you trust the picture, then you can see that the smaller circle does go through two rotations: letting $\mathbf{C}$ be the center of the smaller circle, the little vector pointing from $\mathbf{C}$ to $\mathbf{P}$ rotates around to pointing due right exactly twice, so the circle rotates twice also. But I don't know how to prove this.
This question came up because we wanted to find the parametric equation of $\mathbf{P}$. It'll be the sum of the vectors $\mathbf{C}$ and $\mathbf{P- C}$; the first one has equation $(2\cos(t), 2\sin(t))$, and if the smaller circle rotates around (clockwise) exactly two times, it'll have the equation $(\cos(2t), -\sin(2t))$. This would give the parametrization of the curve immediately. But how to prove that?
(More generally, it seems like a circle of radius $1$ rotating in a circle of radius $n$ will go around $n-1$ times.)
Best Answer
From the following figure it can immediately been read off that when the point of contact $Q$ moves counterclockwise by the angle $\phi$ then the inner circle (as a rigid structure) rotates by the amount $2\phi$ clockwise. Therefore after a full turn of $Q$ the point $P$ has made $2$ turns.