Question

A thin circular ring of mass \( M \) and radius \( R \) is rotating about a transverse axis passing through its centre with constant angular velocity \( \omega \). Two objects each of mass \( m \) are attached gently to the opposite ends of a diameter of the ring. What is the new angular velocity?

• A. \( \frac{M \omega}{M + 2m} \)
• B. \( \frac{M \omega}{M + m} \)
• C. \( (M + 2m) \omega \)
• D. \( \frac{(M - 2m) \omega}{M + 2m} \)

Hint:

For systems with rotating objects, use the conservation of angular momentum. When mass is added to a rotating system, the angular velocity decreases because the moment of inertia increases.

Answer 1

The Correct Option is A

Step 1: Using the principle of conservation of angular momentum.
The angular momentum of the system must be conserved because there is no external torque. Initially, the angular momentum of the rotating ring is given by: \[ L_{\text{initial}} = I_{\text{ring}} \omega = M R^2 \omega \] where \( I_{\text{ring}} = M R^2 \) is the moment of inertia of the ring. After the masses are added, the new system has an additional moment of inertia due to the two masses at the ends of the diameter. The moment of inertia of each mass is \( m R^2 \), so the total moment of inertia becomes: \[ I_{\text{total}} = M R^2 + 2 m R^2 = (M + 2m) R^2 \] The new angular velocity is found by using the conservation of angular momentum: \[ L_{\text{initial}} = L_{\text{final}} \quad \Rightarrow \quad M R^2 \omega = (M + 2m) R^2 \omega' \] Solving for \( \omega' \), we get: \[ \omega' = \frac{M \omega}{M + 2m} \] Step 2: Conclusion.
Thus, the correct answer is (A) \( \frac{M \omega}{M + 2m} \).