Question

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is \(I\). It is rotating with angular velocity \(\omega\). Another identical ring is gently placed on it so that their centres coincide. If both the rings rotate about the same axis, then loss in kinetic energy is

• A. \( \dfrac{I\omega^2}{3} \)
• B. \( \dfrac{I\omega^2}{2} \)
• C. \( \dfrac{I\omega^2}{4} \)
• D. \( I\omega^2 \)

Hint:

In rotational collisions without external torque, angular momentum is conserved but kinetic energy is not.

Answer 1

The Correct Option is C

Step 1: Initial angular momentum.
Initially, only one ring is rotating.
\[ L_i = I\omega \]
Step 2: Final angular velocity using conservation of angular momentum.
After placing the second identical ring, total moment of inertia becomes \(2I\).
\[ I\omega = 2I\omega_f \Rightarrow \omega_f = \frac{\omega}{2} \]
Step 3: Initial kinetic energy.
\[ K_i = \frac{1}{2}I\omega^2 \]
Step 4: Final kinetic energy.
\[ K_f = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2 = \frac{I\omega^2}{4} \]
Step 5: Loss in kinetic energy.
\[ \Delta K = K_i - K_f = \frac{1}{2}I\omega^2 - \frac{I\omega^2}{4} = \frac{I\omega^2}{4} \]
Step 6: Conclusion.
The loss in kinetic energy is \( \dfrac{I\omega^2}{4} \).