Angular velocity of a ring is `omega`. If we put two masses each of mass m at the diametrically opposite points then the resultant angular velocity (m=mass of ring)

Answer

To solve the problem of finding the resultant angular velocity of a ring with two masses added at diametrically opposite points, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**:
We have a ring with an initial angular velocity \( \omega \) and a mass \( m \). Two additional masses, each of mass \( m \), are placed at diametrically opposite points on the ring.

2. **Identify the Moment of Inertia**:
The moment of inertia \( I \) of a ring about its center is given by:
\[ I_{\text{ring}} = m R^2 \]
where \( R \) is the radius of the ring.

3. **Calculate the Moment of Inertia of the Added Masses**:
Each mass \( m \) is located at a distance \( R \) from the center of the ring. Since there are two masses, the total moment of inertia contributed by the two masses is:
\[ I_{\text{masses}} = 2 \cdot m R^2 = 2m R^2 \]

4. **Total Moment of Inertia**:
The total moment of inertia \( I_{\text{total}} \) of the system (ring + two masses) is:
\[ I_{\text{total}} = I_{\text{ring}} + I_{\text{masses}} = m R^2 + 2m R^2 = 3m R^2 \]

5. **Apply Conservation of Angular Momentum**:
According to the conservation of angular momentum, the initial angular momentum \( L_i \) must equal the final angular momentum \( L_f \):
\[ L_i = I_{\text{ring}} \cdot \omega = m R^2 \cdot \omega \]
\[ L_f = I_{\text{total}} \cdot \omega' = 3m R^2 \cdot \omega' \]
Setting these equal gives:
\[ m R^2 \cdot \omega = 3m R^2 \cdot \omega' \]

6. **Solve for the Resultant Angular Velocity \( \omega' \)**:
Dividing both sides by \( m R^2 \) (assuming \( m \neq 0 \) and \( R \neq 0 \)):
\[ \omega = 3 \omega' \]
Rearranging gives:
\[ \omega' = \frac{\omega}{3} \]

### Final Result:
The resultant angular velocity \( \omega' \) of the ring after adding the two masses is:
\[ \omega' = \frac{\omega}{3} \]