JEE Main 2013 — Rotational Motion Question with Solution

Kinetic energy ${}_{\text{(rotational) }}{\mathrm{K}}_{\mathrm{R}}=\frac{1}{2}\mathrm{I}{\omega }^2$ Kinetic energy ${}_{\text{(translational) }}{\mathrm{K}}_{\mathrm{T}}=\frac{1}{2}{\mathrm{Mv}}^2$ $\begin{aligned} & (\mathrm{v}=\mathrm{R} \omega) \\ & \text { M.I. }_{\text {(initial) }} \mathrm{I}_{\text {ring }}=\mathrm{MR}^2 ; \omega_{\text {initial }}=\omega \\ & \text { M.I. } _{\text { (new) }}{\mathrm{I}^{\prime}}{ }_{\text {(system) }}=\mathrm{MR}^2+2 \mathrm{mR}^2 \\ & \omega_{(\text {system) }}^{\prime}=\frac{M \omega}{M+2 m} \\ & \end{aligned}$ Solving we get loss in $K.E.$ $=\frac{\mathrm{Mm}}{(\mathrm{M}+2\mathrm{m})}{\omega }^2{\mathrm{R}}^2$