Question

A thin circular ring of mass 0.2 kg is rotating about its axis with an angular speed of 51 rad/s. Two particles having mass 2 g each are now attached at diametrically opposite points on the ring. Then the angular speed of the system is:

• A. 100 rad/s
• B. 50 rad/s
• C. 51 rad/s
• D. 102 rad/s

Hint:

In a system with rotating bodies, angular momentum is conserved unless acted upon by external forces.

Answer 1

The Correct Option is B

- Angular momentum is conserved since no external torque acts on the system.
- Initial angular momentum is only due to the ring: \[ L_{\text{initial}} = I_{\text{ring}} \, \omega_{\text{initial}} \]
- Moment of inertia of the ring: \[ I_{\text{ring}} = m_{\text{ring}} r^2 \]
- After two particles stick to the ring, total moment of inertia becomes: \[ I_{\text{final}} = m_{\text{ring}} r^2 + 2 m_{\text{particle}} r^2 \]
- Using conservation of angular momentum: \[ I_{\text{ring}} \, \omega_{\text{initial}} = I_{\text{final}} \, \omega_{\text{final}} \]
- Solving for final angular speed: \[ \omega_{\text{final}} = \frac{I_{\text{ring}}}{I_{\text{final}}} \, \omega_{\text{initial}} \]
- Substituting values gives: \[ \omega_{\text{final}} = 50 \,\text{rad/s} \]