Rotational Motion

JEE Main level, single correct, medium

280 of 280

Question 280 of 280

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

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Answer: (C) Mm(M+2m)ω2R2\frac{M m}{(M+2 m)} \omega^2 R^2

Kinetic energy (rotational) KR=12Iω2_{\text {(rotational) }} \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{I} \omega^2
Kinetic energy (translational) KT=12Mv2_{\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.\mathrm{K.E.}
=Mm(M+2 m)ω2R2=\frac{\mathrm{Mm}}{(\mathrm{M}+2 \mathrm{~m})} \omega^2 \mathrm{R}^2

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