Q. A thin circular ring of mass $M$ and radius $R$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $\omega$. Four small spheres each of mass $m$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the. The new angular velocity of the ring will be
A) 4$\omega$
B) $\frac{M}{4m}\omega$
C) $\left( \frac{M + 4m}{M} \right)\omega$
D) $\left( \frac{M}{M + 4m} \right)\omega$
Explanation
(d) According to conservation of angular momentum,
$I\omega =$constant
$ie$, we can write
$I_{1}\omega_{1} = I_{2}\omega_{2}$
or $MR^{2}\omega = (M + 4m)R^{2}\omega_{2}$
or $\omega_{2} = \left( \frac{M}{M + 4m} \right)\omega\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.