Two men each of mass m stand on the rim of a horizontal circular disc, diametrically opposite to each other. The disc has a mass M and is free to rotate about a vertical axis passing through its centre of mass. Each mass start simultaneously along the rim clockwise and reaches their original starting positions on the disc. The angle turned through by disc with respect to the ground (in radian) is

Options

A. `(8mpi)/(4m+M)`
B. `(2mpi)/(4m+M)`
C. `(mpi)/(M+m)`
D. `(4mpi)/(2M+m)`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. Let's go through the solution step by step.

### Step 1: Understand the System
We have a disc of mass \( M \) with two men, each of mass \( m \), standing on its rim, diametrically opposite to each other. When the men start moving clockwise, the disc will rotate in the opposite direction due to conservation of angular momentum.

### Step 2: Define the Angular Displacements
Let:
- The angle turned by the disc be \( \theta \).
- The total angle covered by each man when they return to their starting position is \( 2\pi \).

Since the men move clockwise, the disc will rotate counterclockwise. Therefore, the angle covered by the men relative to the ground is \( 2\pi - \theta \).

### Step 3: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
The initial angular momentum \( L_{\text{initial}} \) is zero because the system is initially at rest.

The final angular momentum can be expressed as:
\[ L_{\text{final}} = L_{\text{men}} + L_{\text{disc}} \]
Where:
- \( L_{\text{men}} = I_{\text{men}} \cdot \omega_{\text{men}} \)
- \( L_{\text{disc}} = I_{\text{disc}} \cdot \omega_{\text{disc}} \)

### Step 4: Calculate the Angular Momentum of the Men
The moment of inertia of each man about the center of the disc is:
\[ I_{\text{men}} = 2 \cdot m \cdot R^2 = 2mR^2 \]
The angular velocity of the men is:
\[ \omega_{\text{men}} = \frac{2\pi - \theta}{t} \]
Thus, the angular momentum of the men is:
\[ L_{\text{men}} = 2mR^2 \cdot \frac{2\pi - \theta}{t} \]

### Step 5: Calculate the Angular Momentum of the Disc
The moment of inertia of the disc about its center is:
\[ I_{\text{disc}} = \frac{1}{2} M R^2 \]
The angular velocity of the disc is:
\[ \omega_{\text{disc}} = \frac{\theta}{t} \]
Thus, the angular momentum of the disc is:
\[ L_{\text{disc}} = \frac{1}{2} M R^2 \cdot \frac{\theta}{t} \]

### Step 6: Set Up the Equation
Setting the total angular momentum of the men equal to that of the disc:
\[ 2mR^2 \cdot \frac{2\pi - \theta}{t} = \frac{1}{2} M R^2 \cdot \frac{\theta}{t} \]
Cancel \( R^2 \) and \( t \) from both sides:
\[ 2m(2\pi - \theta) = \frac{1}{2} M \theta \]

### Step 7: Solve for \( \theta \)
Expanding and rearranging:
\[ 4m\pi - 2m\theta = \frac{1}{2} M \theta \]
\[ 4m\pi = 2m\theta + \frac{1}{2} M \theta \]
\[ 4m\pi = \theta \left(2m + \frac{1}{2} M\right) \]
Thus,
\[ \theta = \frac{4m\pi}{2m + \frac{1}{2} M} \]
Multiplying through by 2 to simplify:
\[ \theta = \frac{8m\pi}{M + 4m} \]

### Final Answer
The angle turned through by the disc with respect to the ground is:
\[ \theta = \frac{8m\pi}{M + 4m} \]