Time for Rolling Disks on Circular Path to Recontact

To find the time $\tau$ at which the two smaller disks come into contact again, we analyze the geometric initial conditions and the kinematics of rolling without slipping.

1. Initial Angular Separation ($\Delta\theta$)

Let $O$ be the center of the large disk of radius $R$. The radius of each small disk is $r = \frac{R}{50}$.

The centers $C_1$ and $C_2$ of the two small disks are at a distance of $R + r$ from $O$: $OC_1 = OC_2 = R + r = R + \frac{R}{50} = \frac{51}{50}R = 51r$

Since the two smaller disks are initially in contact, the distance between their centers is $C_1 C_2 = 2r$. In the isosceles triangle $\Delta OC_1C_2$, the angle subtended at $O$ is $\Delta\theta$: $\sin\left(\frac{\Delta\theta}{2}\right) = \frac{C_1 C_2 / 2}{OC_1} = \frac{r}{R+r} = \frac{r}{51r} = \frac{1}{51}$

Using the given small-angle approximation $\sin(\Delta\theta) \approx \Delta\theta$, we get: $\frac{\Delta\theta}{2} \approx \frac{1}{51} \implies \Delta\theta = \frac{2}{51}\text{ rad}$

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2. Relation Between Spin and Orbital Angular Velocities

Let $\omega_1 = \omega$ and $\omega_2 = 2\omega$ be the spin angular velocities of the two disks about their respective centers.

For a disk rolling without slipping on a fixed circular path of radius $R$, the speed $v$ of its center of mass is: $v = r \omega_{\text{spin}}$

The orbital angular velocity $\Omega$ of the center of mass about the center $O$ of the stationary large disk is related to $v$ by: $v = (R + r)\Omega$

Equating the two expressions for $v$: $(R + r)\Omega = r \omega_{\text{spin}} \implies \Omega = \frac{r}{R + r} \omega_{\text{spin}} = \frac{r}{51r} \omega_{\text{spin}} = \frac{\omega_{\text{spin}}}{51}$

Thus, the orbital angular velocities of the centers of the two disks are: $\Omega_1 = \frac{\omega}{51}, \quad \Omega_2 = \frac{2\omega}{51}$

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3. Condition for Recontact

Let the initial angular positions of the centers of disk 1 and disk 2 be $\theta_1(0) = \frac{\Delta\theta}{2}$ and $\theta_2(0) = -\frac{\Delta\theta}{2}$, respectively.

Since they roll in opposite directions around the large disk, their angular positions at time $t$ are: $\theta_1(t) = \frac{\Delta\theta}{2} + \Omega_1 t$ $\theta_2(t) = -\frac{\Delta\theta}{2} - \Omega_2 t$

The disks come into contact again at time $\tau$ when the angular separation between their centers becomes $\Delta\theta$ on the opposite side of the circle: $\theta_1(\tau) - \theta_2(\tau) = 2\pi - \Delta\theta$

Substituting $\theta_1(\tau)$ and $\theta_2(\tau)$: $\left(\frac{\Delta\theta}{2} + \Omega_1 \tau\right) - \left(-\frac{\Delta\theta}{2} - \Omega_2 \tau\right) = 2\pi - \Delta\theta$ $\Delta\theta + (\Omega_1 + \Omega_2)\tau = 2\pi - \Delta\theta$ $(\Omega_1 + \Omega_2)\tau = 2\pi - 2\Delta\theta$

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4. Calculation of Time $\tau$

Substituting $2\Delta\theta = 2 \times \frac{2}{51} = \frac{4}{51}$ and $\Omega_1 + \Omega_2 = \frac{\omega}{51} + \frac{2\omega}{51} = \frac{3\omega}{51}$:

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

$\tau = 51 \times \left(2\pi - \frac{4}{51}\right) / 3\omega$

Thus, the correct option is C.