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 ring. Because of this, the kinetic energy loss will be:

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

Answer 1

Answer: (C)

$Kinetic \, energy _{\left(rotational\right)} K_{R}=\frac{1}{2} I \omega^{2}$
$Kinetic \, energy _{\left(rotational\right)} K_{r}=\frac{1}{2} MV^{2}$
(v=$R\omega$)
$M. I. _{\left(initia\right)} \, I_{ring} =MR^{2}; \omega_{initial} =\omega$
$M. I. _{\left(new\right)} I'_{\left(system\right)} =MR^{2}+2mR^{2}$
$\omega'_{\left(system\right)} =\frac{M \omega}{M+2m}$
Solving we get loss in $K.E.$
$=\frac{M m}{\left(M+2m\right)} \omega^{2} R^{2}$