Q. A thin uniform circular disc of mass $m$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to the plane with an angular velocity $\omega$. Another disc of same dimensions but of mass $\frac{1}{4}m$ is placed gently on the first disc co-axially. The angular velocity of the system is
A) $\sqrt{2}\omega$
B) $\frac{4}{5}\omega$
C) $\frac{3}{4}\omega$
D) $\frac{1}{3}\omega$
Explanation
(b) Using conservation of angular momentum
$\left( \frac{1}{2}mR^{2} \right)\omega = \{\frac{1}{2}mR^{2} + \frac{1}{2}\left( \frac{m}{4} \right)R^{2}\}\omega'$
$\left( \frac{1}{2}mR^{2} \right)\omega = \frac{5}{8}mR^{2}\omega'$
$\omega^{'} = \frac{4}{5}\ \omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.