Question

A thin circular ring of mass $M$ and radius $R$ is rotating about its axis with a constant velocity $\omega$. Two objects, each of mass $m$ are attached gently to the opposite ends of the diameter of the ring. The wheel now rotates with an angular velocity equal to:

• A. $\omega M/(M+m)$
• B. $(M-2M)(M+2m)\omega$
• C. $M/(M+2m)\omega$
• D. $(M+2m)/M\omega$

Answer 1

The correct option is C $M/(M+2m)\omega$

Initial Angular Momentum $L_0=M{R}^2\cdot \omega$

Final Angular Momentum $L=(M{R}^2+2m{R}^2){\omega }^{{}^{\prime }}$

Conservation of Angular Momentum gives ($\therefore$ Net Torque is 0)

$M{R}^2\omega =(M{R}^2+2m{R}^2){\omega }^{{}^{\prime }}$

$\Rightarrow {\omega }^{{}^{\prime }}=\omega \left(\frac{M}{M+2m}\right)$