A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity omega. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be

Options

A. `((M+4m) omega)/(M)`
B. `((M-4m)omega)/(M+4m)`
C. `(M omega)/(4m)`
D. `(M omega)/(M+4m)`

Answer

To solve the problem step by step, we will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial angular momentum \( L_i \) of the system is due to the ring alone, which is rotating with an angular velocity \( \omega \).
- The moment of inertia \( I \) of the ring is given by:
\[ I = MR^2 \]
- Therefore, the initial angular momentum is:
\[ L_i = I \cdot \omega = MR^2 \cdot \omega \]

2. **Identify Final Conditions**:
- When the four objects, each of mass \( m \), are placed at the ends of two perpendicular diameters of the ring, the total moment of inertia of the system changes.
- The new moment of inertia \( I_f \) of the system (ring + masses) is:
\[ I_f = I + 4 \cdot m \cdot R^2 = MR^2 + 4mR^2 \]

3. **Apply Conservation of Angular Momentum**:
- Since no external torque is acting on the system, angular momentum is conserved:
\[ L_i = L_f \]
- Therefore,
\[ MR^2 \cdot \omega = (MR^2 + 4mR^2) \cdot \omega_f \]

4. **Rearrange the Equation**:
- We can rearrange the equation to solve for the final angular velocity \( \omega_f \):
\[ MR^2 \cdot \omega = (MR^2 + 4mR^2) \cdot \omega_f \]
- Dividing both sides by \( R^2 \):
\[ M \cdot \omega = (M + 4m) \cdot \omega_f \]

5. **Solve for \( \omega_f \)**:
- Now, isolate \( \omega_f \):
\[ \omega_f = \frac{M \cdot \omega}{M + 4m} \]

### Final Answer:
The angular velocity of the ring after placing the four masses is:
\[ \omega_f = \frac{M \cdot \omega}{M + 4m} \]