Question

A thin circular disc of mass \( M \) and radius \( R \) is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity \( \omega \). If another disc of the same dimensions but of mass \( M/2 \) is placed gently on the first disc co-axially, then the new angular velocity of the system is:

• A. \( \frac{4}{5} \omega \)
• B. \( \frac{5}{4} \omega \)
• C. \( \frac{2}{3} \omega \)
• D. \( \frac{3}{2} \omega \)

Hint:

For conservation of angular momentum, always ensure that \( I \omega \) remains constant for a closed system. Carefully update the moment of inertia \( I \) when new masses are added or redistributed in the system.

Answer 1

The Correct Option is C

Using the principle of conservation of angular momentum: \[ I_1 \omega_1 = I_2 \omega_2. \] The initial moment of inertia of the system is: \[ I_1 = \frac{1}{2} M R^2. \] The final moment of inertia after placing the second disc is: \[ I_2 = \frac{1}{2} M R^2 + \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{3}{4} M R^2. \] Substituting these into the conservation of angular momentum equation: \[ \frac{1}{2} M R^2 \omega = \frac{3}{4} M R^2 \omega_2. \] Simplifying: \[ \omega_2 = \frac{\frac{1}{2}}{\frac{3}{4}} \omega = \frac{2}{3} \omega. \] Final Answer: \[ \boxed{\frac{2}{3} \omega} \]