Question 1

Given $r = R/50$, the distance from the center of the large disk to the center of each smaller disk is $D = R + r = R + R/50 = \frac{51R}{50} = 51r$.

Initially, the disks are in contact, so the distance between their centers is $2r$. Using the approximation $\sin(\Delta \theta) \approx \Delta \theta$, the initial angular separation $\Delta \theta$ between the centers (subtended at the center of the large disk) is: $\Delta \theta = \frac{2r}{D} = \frac{2r}{51r} = \frac{2}{51} \text{ rad}$

When a disk of radius $r$ rolls without slipping on a fixed surface, the velocity of its center is $v = \omega' r$. The angular velocity of the center of disk 1 (with angular velocity $\omega$) about the center of the large disk is: $\Omega_1 = \frac{v_1}{D} = \frac{\omega r}{51r} = \frac{\omega}{51}$

Similarly, for disk 2 (with angular velocity $2\omega$): $\Omega_2 = \frac{2\omega r}{51r} = \frac{2\omega}{51}$

Since they move in opposite directions, the relative angular velocity of their centers is: $\Omega_{rel} = \Omega_1 + \Omega_2 = \frac{\omega}{51} + \frac{2\omega}{51} = \frac{3\omega}{51}$

For the disks to be in contact again on the opposite side, the relative angular distance to be covered by the centers is $2\pi$ minus the initial and final angular gaps required for contact: $\theta_{total} = 2\pi - \Delta \theta - \Delta \theta = 2\pi - 2\Delta \theta = 2\pi - \frac{4}{51}$

The time $\tau$ is: $\tau = \frac{\theta_{total}}{\Omega_{rel}} = \frac{2\pi - 4/51}{3\omega / 51} = 51 \times \frac{(2\pi - 4/51)}{3\omega}$