1. Principle of Conservation of Angular Momentum

The system consists of the large disc, the motor (negligible mass), and the small disc. There are no external torques acting on the system about the vertical axis of rotation of the large disc. Therefore, the total angular momentum of the system about this axis is conserved.

2. Initial and Final States

• Initial State: The entire system is at rest. The initial total angular momentum, $L_i$, is zero. $L_i = 0$
• Final State: The large disc rotates with an angular speed $\Omega$. The small disc's center of mass orbits with the large disc, and the small disc also spins about its own center. The total final angular momentum, $L_f$, must also be zero due to conservation. $L_f = 0$

3. Calculating Final Angular Momentum ($L_f$)

The total final angular momentum is the sum of the angular momentum of the large disc ($L_L$) and the small disc ($L_S$). $L_f = L_L + L_S = 0$

3.1. Angular Momentum of the Large Disc ($L_L$)

The large disc has mass $M$ and radius $R$. Its moment of inertia about its axis is $I_L = \frac{1}{2}MR^2$. If its angular speed is $\Omega$, its angular momentum is: $L_L = I_L \Omega = \frac{1}{2}MR^2 \Omega$ Let's assume the large disc rotates counter-clockwise (CCW), so its angular momentum vector points upwards (positive z-direction).

3.2. Angular Momentum of the Small Disc ($L_S$)

The angular momentum of the small disc about the central axis has two components: its orbital angular momentum due to the motion of its center of mass, and its spin angular momentum about its own center of mass. $L_S = L_{S, orbital} + L_{S, spin}$

• Orbital Angular Momentum ($L_{S, orbital}$): The small disc (mass $M$) is at a distance $R$ from the central axis. Its center of mass moves in a circle with the same angular speed as the large disc, $\Omega$. Its linear speed is $v = \Omega R$. The orbital angular momentum is: $L_{S, orbital} = M v R = M (\Omega R) R = MR^2 \Omega$ This momentum is also in the same direction as the large disc's rotation (upwards).

• Spin Angular Momentum ($L_{S, spin}$): The small disc has mass $M$ and radius $R/2$. Its moment of inertia about its own center is $I_S = \frac{1}{2}M(R/2)^2 = \frac{1}{8}MR^2$. The problem states that the small disc rotates at a uniform angular speed $\omega$. This is interpreted as its absolute angular speed (speed with respect to the ground/inertial frame). The total angular momentum from the large disc and the orbital motion of the small disc is positive (upwards). For the total momentum $L_f$ to be zero, the spin angular momentum of the small disc must be negative (downwards, i.e., clockwise rotation). So, the spin angular momentum is: $L_{S, spin} = -I_S \omega = -\frac{1}{8}MR^2 \omega$

4. Applying the Conservation Law

Now, we set the total final angular momentum to zero: $L_f = L_L + L_{S, orbital} + L_{S, spin} = 0$ $\frac{1}{2}MR^2 \Omega + MR^2 \Omega - \frac{1}{8}MR^2 \omega = 0$

5. Solving for $\Omega$

Combine the terms containing $\Omega$: $\left(\frac{1}{2}MR^2 + MR^2\right) \Omega = \frac{1}{8}MR^2 \omega$ $\frac{3}{2}MR^2 \Omega = \frac{1}{8}MR^2 \omega$ Cancel the common term $MR^2$ from both sides: $\frac{3}{2} \Omega = \frac{1}{8} \omega$ Solve for $\Omega$: $\Omega = \frac{2}{3} \cdot \frac{1}{8} \omega = \frac{2}{24} \omega = \frac{1}{12} \omega$

6. Finding the Value of n

The problem states that the angular speed of the large disc is $\Omega = \omega/n$. Comparing this with our result: $\frac{\omega}{n} = \frac{1}{12} \omega$ $n = 12$