Q87 / 90 · JEE Main 2013 (25 Apr Online) · Physics
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$\frac{m(M+2 m)}{M} \omega^2 R^2$
- B$\frac{M m}{(M+m)} \omega^2 R^2$
- Correct answerC$\frac{M m}{(M+2 m)} \omega^2 R^2$
- D$\frac{(M+m) M}{(M+2 m)} \omega^2 R^2$
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 $\mathrm{K.E.}$
$=\frac{\mathrm{Mm}}{(\mathrm{M}+2 \mathrm{~m})} \omega^2 \mathrm{R}^2$
Hence, the correct answer is option: 'C'.