A ball rotates at a rate $r$ rotations per second and simultaneously revolves around a stationary point $O$ at a rate $R$ revolutions per second $(R<r)$.The rotation and revolution are in the same sense.A certain point on the ball is in the line of the centre of the ball and point $O$ at a certain time.This configuration repeats after a time
$(1)\frac{1}{r-R}$
$(2)\frac{1}{R}-\frac{1}{r}$
$(3)\frac{1}{r+R}$
$(4)\frac{1}{R}+\frac{1}{r}$
I will give this a try. Shift to the reference frame of the center of the ball. The point on the ball (call it A) now rotates wrt the center at an angular velocity $\omega_{1} = 2\pi r$ whereas the point O rotates around it at an angular velocity $\omega_{2} = 2\pi R$ in the opposite direction. [Since rotation and revolution of the ball around O are in the same sense, the rotation of points A and O wrt the center of the ball are in opposite sense). This gives a relative velocity of $\omega_{rel} = \omega_{1} + \omega_{2} = 2\pi (r+R)$. This means that the configuration will repeat after a time of $\frac{1}{r+R}$.
For sanity check (and utilizing the options): since the rotation and revolution are in the same sense, the configuration will get repeated before one rotation is over, i.e answer should be less than 1/r - can eliminate options 1 and 4. Also when R = 0, the answer is obviously 1/r, hence can eliminate option 2. Yay, we are left with 1/(r+R).
