Q. A thin circular ring of mass $m$ and radius $R$ is rotating about its axis with a constant angular velocity $\omega$. Two objects each of mass $M$ are attached gently to the opposite ends of a diameter of the ring the ring. The ring now rotates with an angular velocity $\omega'$ is equal to
A) $\frac{\omega(m + 2M)}{m}$
B) $\frac{\omega(m - 2M)}{(m + 2M)}$
C) $\frac{\omega m}{(m + M)}$
D) $\frac{\omega m}{(m + 2M)}$
Explanation
(d) As no external torque is acting about the axis, angular momentum of system remains conserved.
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$\therefore$ $I_{1}\omega = I_{2}\omega''$
$\Rightarrow$ $mR^{2}\omega = (mR^{2} + 2MR^{2})\omega''$
$\Rightarrow$ $\omega^{''} = \left( \frac{m}{m + 2M)} \right)\omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.