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:
\[
\text{[Use }\sin(\Delta\theta)=\Delta\theta\text{ and ignore gravity.]}
\]
Show Hint
For rolling without slipping:
\[
v=r\omega
\]
Relative angular speed for opposite motions:
\[
\Omega_{\mathrm{relative}}
=
\Omega_1+\Omega_2
\]
Step 1: Understanding the Concept:
When a small disk rolls without slipping on a stationary large disk, its center moves in an orbit. The speed of the center \( v_c \) is related to the spin \( \omega_s \) by \( v_c = \omega_s r \). Also, \( v_c = \omega_{orbit} (R+r) \). Step 3: Detailed Explanation:
1) Angular velocity of orbits:
For disk 1 (spin \( \omega \)): \( \omega_{o1} = \frac{\omega r}{R+r} = \frac{\omega(R/50)}{51R/50} = \frac{\omega}{51} \).
For disk 2 (spin \( 2\omega \)): \( \omega_{o2} = \frac{2\omega r}{R+r} = \frac{2\omega}{51} \).
Since they move in opposite directions, relative angular velocity \( \omega_{rel} = \frac{\omega}{51} + \frac{2\omega}{51} = \frac{3\omega}{51} \).
2) Angular separation:
Initial separation angle between centers \( \Delta\theta \).
In contact, the centers are separated by an arc length \( 2r \).
\( \Delta\theta \approx \frac{2r}{R+r} = \frac{2(R/50)}{51R/50} = \frac{2}{51} \).
The angular distance to cover to meet again is \( 2\pi - \Delta\theta = 2\pi - 2/51 \).
3) Time calculation:
\( \tau = \frac{\text{Angle}}{\omega_{rel}} = \frac{2\pi - 2/51}{3\omega/51} = 51 \times \frac{2\pi - 2/51}{3\omega} \). Step 4: Final Answer:
Matches (B).
