Consider a large disk of radius $R$ and two smaller disks, each of radius $r=\frac{R}{50}$ lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation $\Delta \theta$ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities $\omega$ and $2\omega$ while the large disk is held stationary. The time at which the smaller disks are again in contact is:

[Use $\sin \left(\Delta \theta \right)=\Delta \theta$ and ignore gravity.]

The correct option is:
$\tau =51\times \frac{2\pi -\frac{2}{51}}{3\omega }$

Step 1: Determine the initial angular separation $\Delta \theta$
As shown in the figure, two smaller disks of radius $r$ are placed externally on the circumference of a large disk of radius $R$, in contact with each other.

The distance from the center of the large disk to the center of each smaller disk is $R+r$.
When the two small disks touch, the distance between their centers is $2r$.
Using geometry, for the angle $\Delta \theta$ subtended at the center of the large disk:
$\sin \left(\frac{\Delta \theta }{2}\right)=\frac{r}{R+r}$

Given that $r=\frac{R}{50}$, we have:
$R+r=R+\frac{R}{50}=\frac{51R}{50}=51r$

Substituting this ratio into the sine expression:
$\sin \left(\frac{\Delta \theta }{2}\right)=\frac{r}{51r}=\frac{1}{51}$
Using the small-angle approximation $\sin \left(x\right)\approx x$:
$\frac{\Delta \theta }{2}=\frac{1}{51}\Rightarrow \Delta \theta =\frac{2}{51}$

Step 2: Relate linear motion (rolling without slipping) to angular motion along the circumference
The smaller disks roll without slipping on the stationary large disk.
For a small disk rolling with angular speed ${\omega }_{\mathrm{disk}}$, the speed of its contact point relative to its center is $v=r{\omega }_{\mathrm{disk}}$.
Since the large disk is held stationary, the velocity of the center of the small disk along the circular path is:
$v_c=r{\omega }_{\mathrm{disk}}$

The rate of change of the position angle $\Theta$ of the center of the small disk about the center of the large disk is:
$\Omega =\frac{v_c}{R+r}=\frac{r{\omega }_{\mathrm{disk}}}{51r}=\frac{{\omega }_{\mathrm{disk}}}{51}$

Step 3: Calculate the relative angular velocity
The two small disks move in opposite directions along the circumference of the large disk with angular spin speeds $\omega$ and $2\omega$.
Their respective orbital angular speeds around the center of the large disk are:
${\Omega }_1=\frac{\omega }{51}$
${\Omega }_2=\frac{2\omega }{51}$

Since they move in opposite directions, the total relative angular speed at which the angle between their centers increases is:
${\Omega }_{\mathrm{rel}}={\Omega }_1+{\Omega }_2=\frac{\omega }{51}+\frac{2\omega }{51}=\frac{3\omega }{51}$

Step 4: Calculate the time required to meet again
Initially, the disks are at an angular separation of $\Delta \theta$.
To come into contact again after traveling around the large disk, the total relative angular distance covered by their centers around the large disk must be $2\pi -\Delta \theta$.
$\theta =2\pi -\frac{2}{51}$

Therefore, the required time $\tau$ is given by:
$\tau =\frac{\theta }{{\Omega }_{\mathrm{rel}}}=\frac{2\pi -\frac{2}{51}}{\frac{3\omega }{51}}$
$\tau =51\times \frac{2\pi -\frac{2}{51}}{3\omega }$