Q. A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with a constant angular velocity $\omega$. Four objects each of mass $m$, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be
A) $\frac{M\omega}{M + 4m}$
B) $\frac{(M + 4m)\omega}{M}$
C) $\frac{(M - 4m)\omega}{M + 4m}$
D) $\frac{M\omega}{4m}$
Explanation
(a) Initial angular momentum of ring, $L = I\omega = Mr^{2}\omega$
Final angular momentum of ring and four particles system
$L = (Mr^{2} + 4mr^{2})\omega'$
As there is no torque on the system therefore angular momentum remains constant
$Mr^{2}\omega = \left( Mr^{2} + 4mr^{2} \right)\omega^{'} \Rightarrow \omega^{'} = \frac{M\omega}{M + 4m}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.