After how much time the same configuration will repeat?

Question

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)\ $ $\frac{1}{R}+\frac{1}{r}$

How do I solve this aptitude question? Thanks for your time.

Source $:$ CSIR NET JUNE $2019.$

Answer 1

As rotation and revolution are both in the same direction (say, counterclockwise), the point will come in the line connecting centers after one complete rotation. Say that happens after part revolution of angle $\theta$ against one full revolution ($2\pi$). In that time the rotation has completed by $(2\pi + \theta)$ (more than one rotation of ($2\pi$).

One rotation takes $\frac{1}{r}$ second and one revolution takes $\frac{1}{R}$ second. Equating the time taken -

$\displaystyle \frac{2\pi + \theta}{2\pi r} = \frac{\theta}{2\pi R}$

Now please find $\theta$ and then plug in either Right Hand side or in Left Hand side to find the time.

Answer 2

Given: Frequency of rotation = 'r' times/sec Frequency of revolution = 'R' times/sec.
(the rotation and the revolution are in the same sense), hence Relative frequency of rotation = (r-R) times/sec.
As per this relative frequency of rotation, the centre of the ball and the point of its surface come in line with the external fixed point O, at each full rotation.

Time taken for one full rotation = 1/f = 1/(r-R) in sec