A disc of mass 𝑀 and radius 𝑅 is free to rotate about its vertical axis as shown in the figure. A battery operated motor of negligible mass is fixed to this disc at a point on its circumference. Another disc of the same mass 𝑀 and radius 𝑅/2 is fixed to the motor’s thin shaft. Initially, both the discs are at rest. The motor is switched on so that the smaller disc rotates at a uniform angular speed 𝜔. If the angular speed at which the large disc rotates is 𝜔/𝑛, then the value of 𝑛 is _____.

The correct answer is 12.

Step-by-step derivation:

1. Identify the components of the system:
- A large disc of mass M and radius R, which rotates about its own center (let this axis be O).
- A motor of negligible mass fixed at the circumference of the large disc, i.e., at a distance R from the center O.
- A smaller disc of mass M and radius R/2, mounted on the motor's shaft. Its axis of rotation is vertical and passes through its center of mass, which is situated at a distance of R from the center O.

2. Moment of inertia calculations:
The moment of inertia of the large disc about its central vertical axis is:

$I_1=\frac{1}{2}M{R}^2$

The moment of inertia of the smaller disc about its own central axis is:

$I_2=\frac{1}{2}M{\left(\frac{R}{2}\right)}^2=\frac{1}{8}M{R}^2$

3. Conservation of angular momentum:
Since the system is initially at rest and no external torque acts on the system about the vertical axis passing through O, the total angular momentum L of the system about this axis must remain conserved and equal to zero.

$L_{\text{total}}=0$

4. Formulate the angular momentum terms:
Let the large disc rotate with an angular speed ${\omega }_L=\frac{\omega }{n}$ in one direction (say, counter-clockwise).
The angular momentum of the large disc about the axis O is:

$L_1=I_1{\omega }_L=\left(\frac{1}{2}M{R}^2\right){\omega }_L$

For the smaller disc, its total angular momentum about the axis O consists of two parts:
- The orbital angular momentum due to the circular motion of its center of mass (which is at a distance R and rotates with the large disc at angular speed ${\omega }_L$):

$L_{\text{orbit}}=M{R}^2{\omega }_L$

- The spin angular momentum due to its rotation about its own center of mass with angular speed $\omega$ in the opposite direction:

$L_{\text{spin}}=-I_2\omega =-\left(\frac{1}{8}M{R}^2\right)\omega$

5. Solve for n:
Set the sum of these angular momenta to zero:

$L_1+L_{\text{orbit}}+L_{\text{spin}}=0$

$\left(\frac{1}{2}M{R}^2\right){\omega }_L+M{R}^2{\omega }_L-\left(\frac{1}{8}M{R}^2\right)\omega =0$

Divide the entire equation by $M{R}^2$:

$\frac{3}{2}{\omega }_L=\frac{1}{8}\omega$

${\omega }_L=\frac{2}{3\times 8}$

Comparing this with the given expression ${\omega }_L=\frac{\omega }{n}$, we find:

$n=12$