Consider a large disk of radius $R$ and two smaller disks, each of radius $r = R/50$, 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 $\tau$ at which the smaller disks are again in contact is: [Use $\sin(\Delta\theta) = \Delta\theta$ and ignore gravity.]

Step 1: Define the system parameters.

The system consists of a stationary disk of radius $R$ and two smaller disks of radius $r = \frac{R}{50}$ rolling on its outer circumference.

Step 2: Establish the rolling without slipping condition.

The relation between the orbital angular velocity of a small disk's center, $\dot{\phi}$, and its spin angular velocity, $\omega_{\text{spin}}$, is determined by the constraint of rolling without slipping.

$$(R + r)\dot{\phi} = \omega_{\text{spin}} r$$

Substitute $R = 50r$ into the relation.

$$\dot{\phi} = \frac{\omega_{\text{spin}} r}{51r} = \frac{\omega_{\text{spin}}}{51}$$

Step 3: Calculate the relative orbital angular velocity.

The two disks roll in opposite directions with spin angular velocities $\omega$ and $2\omega$.

$$\dot{\phi}_1 = \frac{\omega}{51}$$ $$\dot{\phi}_2 = -\frac{2\omega}{51}$$

The magnitude of their relative orbital angular velocity is the sum of their individual orbital angular speeds.

$$\omega_{\text{rel}} = \frac{\omega + 2\omega}{51} = \frac{3\omega}{51}$$

Step 4: Determine the initial angular separation.

When the two smaller disks are in contact, the distance between their centers is $2r$.

$$2(R + r)\sin\left(\frac{\Delta\theta}{2}\right) = 2r$$ $$\sin\left(\frac{\Delta\theta}{2}\right) = \frac{r}{51r} = \frac{1}{51}$$

Using the small-angle approximation $\sin x \approx x$:

$$\frac{\Delta\theta}{2} \approx \frac{1}{51}$$ $$\Delta\theta \approx \frac{2}{51}$$

Step 5: Calculate the time $\tau$ until the next contact.

The disks separate and travel around the circumference until they meet again, requiring a relative angular sweep of $2\pi - 2\Delta\theta$.

$$\tau = \frac{2\pi - 2\Delta\theta}{\omega_{\text{rel}}}$$ $$\tau = \frac{2\pi - 2\left(\frac{2}{51}\right)}{\frac{3\omega}{51}}$$ $$\tau = \frac{51\left(2\pi - \frac{4}{51}\right)}{3\omega}$$

Answer: $\tau = \dfrac{51\left(2\pi - \dfrac{4}{51}\right)}{3\omega}$