A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity `omega`. Two objects each of mass m are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity of

Options

A. `(omega(M-2m))/((M + 2m))`
B. `(omega M)/(( M + 2m))`
C. `(omega M)/(( M + m))`
D. `(omega(M + 2m))/(M)`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum of the system remains constant.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial angular momentum \( L_i \) of the system consists of the angular momentum of the ring alone, as the two masses are not yet attached.
- The moment of inertia \( I_1 \) of the ring is given by:
\[ I_1 = M r^2 \]
- The initial angular velocity is \( \omega_1 = \omega \).
- Therefore, the initial angular momentum is:
\[ L_i = I_1 \omega_1 = (M r^2) \omega \]

2. **Identify Final Conditions**:
- After attaching the two masses \( m \) at opposite ends of the diameter, we need to calculate the new moment of inertia \( I_2 \).
- The moment of inertia of each mass \( m \) about the axis of rotation (the center of the ring) is given by:
\[ I_{mass} = m r^2 \]
- Since there are two masses, the total moment of inertia contributed by the masses is:
\[ I_{masses} = 2 \times (m r^2) = 2m r^2 \]
- Therefore, the total final moment of inertia \( I_2 \) is:
\[ I_2 = I_1 + I_{masses} = M r^2 + 2m r^2 = (M + 2m) r^2 \]

3. **Apply Conservation of Angular Momentum**:
- According to the conservation of angular momentum:
\[ L_i = L_f \]
- The final angular momentum \( L_f \) is given by:
\[ L_f = I_2 \omega_2 = (M + 2m) r^2 \omega_2 \]
- Setting the initial and final angular momentum equal gives:
\[ (M r^2) \omega = ((M + 2m) r^2) \omega_2 \]

4. **Solve for Final Angular Velocity \( \omega_2 \)**:
- Cancel \( r^2 \) from both sides (assuming \( r \neq 0 \)):
\[ M \omega = (M + 2m) \omega_2 \]
- Rearranging gives:
\[ \omega_2 = \frac{M \omega}{M + 2m} \]

### Final Answer:
The final angular velocity of the ring after the two masses are attached is:
\[ \omega_2 = \frac{M \omega}{M + 2m} \]