A thin circular ring of mass `M` and radius `R` is rotating in a horizontal plane about an axis vertical to its plane with a constant angular velocity `omega`. If two objects each of mass `m` be attached gently to the opposite ends of a diameter of the ring, the ring will then rotate with an angular velocity

Options

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

Answer

To solve the problem, we need to find the new angular velocity of a thin circular ring after two masses are attached to it. We will use the principle of conservation of angular momentum.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The ring has a mass \( M \) and radius \( R \).
- It is rotating with an initial angular velocity \( \omega \).

2. **Calculate Initial Angular Momentum**:
- The moment of inertia \( I \) of a thin circular ring about an axis perpendicular to its plane is given by:
\[ I = MR^2 \]
- The initial angular momentum \( L_i \) is given by:
\[ L_i = I \cdot \omega = MR^2 \cdot \omega \]

3. **Identify Final Conditions**:
- Two masses \( m \) are attached to the ring at opposite ends of a diameter.
- The new angular velocity after the masses are attached is \( \omega' \).

4. **Calculate Final Moment of Inertia**:
- The moment of inertia of the two masses \( m \) at a distance \( R \) from the axis of rotation is:
\[ I' = I + 2 \cdot (m \cdot R^2) = MR^2 + 2mR^2 = (M + 2m)R^2 \]

5. **Calculate Final Angular Momentum**:
- The final angular momentum \( L_f \) is given by:
\[ L_f = I' \cdot \omega' = (M + 2m)R^2 \cdot \omega' \]

6. **Apply Conservation of Angular Momentum**:
- Since there are no external torques acting on the system, angular momentum is conserved:
\[ L_i = L_f \]
- Substituting the expressions for \( L_i \) and \( L_f \):
\[ MR^2 \cdot \omega = (M + 2m)R^2 \cdot \omega' \]

7. **Simplify the Equation**:
- Dividing both sides by \( R^2 \) (assuming \( R \neq 0 \)):
\[ M \cdot \omega = (M + 2m) \cdot \omega' \]

8. **Solve for the New Angular Velocity \( \omega' \)**:
- Rearranging the equation gives:
\[ \omega' = \frac{M \cdot \omega}{M + 2m} \]

### Final Answer:
The new angular velocity of the ring after attaching the two masses is:
\[ \omega' = \frac{M \cdot \omega}{M + 2m} \]