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

Explanation

To solve the problem step by step, we will use the principle of conservation of angular momentum. Here's how we can derive the angular velocity of the ring after attaching the two masses.

### Step 1: Understand the initial conditions
We have a thin circular ring of mass \( M \) and radius \( R \) rotating with an initial angular velocity \( \omega \). The moment of inertia \( I \) of the ring about its axis is given by:
\[ I_{\text{initial}} = M R^2 \]

### Step 2: Calculate the initial angular momentum
The initial angular momentum \( L_{\text{initial}} \) of the system is given by:
\[ L_{\text{initial}} = I_{\text{initial}} \cdot \omega = M R^2 \cdot \omega \]

### Step 3: Consider the final conditions after attaching the masses
When two objects, each of mass \( m \), are attached to the opposite ends of a diameter of the ring, the total mass of the system becomes \( M + 2m \). The moment of inertia of the two masses about the axis of rotation is:
\[ I_{\text{masses}} = 2m \cdot R^2 \]
Thus, the total moment of inertia of the system after the masses are added is:
\[ I_{\text{final}} = I_{\text{initial}} + I_{\text{masses}} = M R^2 + 2m R^2 = (M + 2m) R^2 \]

### Step 4: Calculate the final angular momentum
Let the new angular velocity after the masses are attached be \( \omega' \). The final angular momentum \( L_{\text{final}} \) is given by:
\[ L_{\text{final}} = I_{\text{final}} \cdot \omega' = (M + 2m) R^2 \cdot \omega' \]

### Step 5: Apply conservation of angular momentum
According to the conservation of angular momentum, the initial angular momentum must equal the final angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
Substituting the expressions we derived:
\[ M R^2 \cdot \omega = (M + 2m) R^2 \cdot \omega' \]

### Step 6: Simplify the equation
We can cancel \( R^2 \) from both sides (assuming \( R \neq 0 \)):
\[ M \omega = (M + 2m) \omega' \]

### Step 7: Solve for the new angular velocity \( \omega' \)
Rearranging the equation to solve for \( \omega' \):
\[ \omega' = \frac{M \omega}{M + 2m} \]

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