Question

A ring of mass $M$ and radius $R$ is rotating about its axis with angular velocity $\omega$. Two identical bodies each of mass $m$ are now gently attached at the two ends of a diameter of the rin. Because of this, the kinetic energy loss will be:

• A. $\frac{m(M+2m)}{M}{\omega }^2{R}^2$
• B. $\frac{Mm}{(M+m)}{\omega }^2{R}^2$
• C. $\frac{Mm}{(M+2m)}{\omega }^2{R}^2$
• D. $\frac{(M+m)}{(M+2m)}{\omega }^2{R}^2$

Answer 1

The correct option is C $\frac{Mm}{(M+2m)}{\omega }^2{R}^2$

By using angular momentum conservation
$M{R}^2\times \omega =(M{R}^2+M{R}^2+M{R}^2){\omega }^{\prime }$
${\omega }^{\prime }=\frac{M\omega }{M+2m}$
Loss in kinetic energy$=\frac{1}{2}m{R}^2\times {\omega }^2-\frac{1}{2}(M+2m)R^2\times \frac{M^2{\omega }^2}{(M+2m{)}^2}=\frac{Mn{\omega }^2{R}^2}{(M+2m)}$