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, we will use the principle of conservation of angular momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a ring with an initial angular velocity \( \omega \) and two masses \( m \) placed at diametrically opposite points on the ring. The mass of the ring is denoted as \( M \).

### Step 2: Calculate the Initial Moment of Inertia
The moment of inertia \( I \) of the ring about its axis is given by:
\[ I = M R^2 \]
where \( R \) is the radius of the ring.

### Step 3: Calculate the Moment of Inertia with Added Masses
When we add two masses \( m \) at diametrically opposite points, the total moment of inertia \( I' \) of the system becomes:
\[ I' = I + 2 \cdot m \cdot R^2 \]
Substituting the value of \( I \):
\[ I' = M R^2 + 2m R^2 = (M + 2m) R^2 \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum, the initial angular momentum must equal the final angular momentum:
\[ I \omega = I' \omega' \]
Substituting the values we have:
\[ M R^2 \omega = (M + 2m) R^2 \omega' \]

### Step 5: Simplify the Equation
We can cancel \( R^2 \) from both sides (assuming \( R \neq 0 \)):
\[ M \omega = (M + 2m) \omega' \]

### Step 6: Solve for the Resultant Angular Velocity
Rearranging the equation to solve for \( \omega' \):
\[ \omega' = \frac{M \omega}{M + 2m} \]

### Step 7: Conclusion
Thus, the resultant angular velocity \( \omega' \) of the ring after placing the two masses is:
\[ \omega' = \frac{M \omega}{M + 2m} \]