Question 9

Let the angular velocity of the large disc be $\Omega$. Since no external torque acts on the system about the central vertical axis, the total angular momentum of the system remains conserved.

Initial angular momentum of the system is $L_i = 0$.

The final angular momentum $L_f$ consists of three components evaluated about the central vertical axis:

1. Spin angular momentum of the large disc: $L_{large} = I_{large} \Omega = (\frac{1}{2} M R^2) \Omega$
2. Orbital angular momentum of the small disc's center of mass (CM): The motor is fixed on the circumference, so the small disc's CM is at a distance $R$ from the central axis and moves with a linear velocity $v_{cm} = \Omega R$. $L_{orbit} = M v_{cm} R = M (\Omega R) R = M R^2 \Omega$
3. Spin angular momentum of the small disc about its own CM: Because the problem states the small disc rotates at an angular speed $\omega$ without specifying a relative reference frame (like "relative to the motor"), it is conventionally taken with respect to the ground (inertial frame). The radius of the small disc is $R/2$. $L_{spin} = I_{small} \omega = [\frac{1}{2} M (R/2)^2] \omega = \frac{1}{8} M R^2 \omega$

Equating the total final angular momentum to zero: $L_{large} + L_{orbit} + L_{spin} = 0$ $(\frac{1}{2} M R^2) \Omega + M R^2 \Omega + \frac{1}{8} M R^2 \omega = 0$

Dividing the entire equation by $M R^2$: $\frac{1}{2} \Omega + \Omega + \frac{1}{8} \omega = 0$ $\frac{3}{2} \Omega = - \frac{1}{8} \omega$ $\Omega = - \frac{\omega}{12}$

The magnitude of the angular speed of the large disc is $\omega/12$. Therefore, the value of $n$ is $12$.