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

Answer: (C)

$\frac{2}{3}\omega$

Answer 2

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:
$\omega_2 = \frac{2}{3} \omega.$