Question

A thin circular disc of mass \(M\) and radius \(R\) is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity \(\omega\). If another disc of same dimensions but of mass \(\frac{M}{2}\) 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. \(\frac{3}{2}\omega\)

Answer 1

The Correct Option is C

Using the law of conservation of angular momentum:
\[I_1\omega = I_2\omega_2.\]
The moment of inertia of the first disc:
\[I_1 = \frac{MR^2}{2}.\]
The combined moment of inertia of both discs:
\[I_2 = \frac{MR^2}{2} + \frac{1}{2} \left(\frac{MR^2}{2}\right) = \frac{3MR^2}{4}.\]
Applying conservation of angular momentum:
\[\frac{MR^2}{2} \times \omega = \frac{3MR^2}{4} \times \omega_2.\]
Simplifying:
\[\omega_2 = \frac{\frac{MR^2}{2}}{\frac{3MR^2}{4}} \times \omega = \frac{2}{3} \times \omega.\]
Thus, the new angular velocity of the system is:
\[\omega_2 = \frac{2}{3} \omega.\]