Kinetic energy (rotational) KR=21Iω2 Kinetic energy (translational) KT=21Mv2 $\begin{aligned}
& (\mathrm{v}=\mathrm{R} \omega) \
& \text { M.I. }{\text {(initial) }} \mathrm{I}{\text {ring }}=\mathrm{MR}^2 ; \omega_{\text {initial }}=\omega \
& \text { M.I. } {\text { (new) }}{\mathrm{I}^{\prime}}{ }{\text {(system) }}=\mathrm{MR}^2+2 \mathrm{mR}^2 \
& \omega_{(\text {system) }}^{\prime}=\frac{M \omega}{M+2 m} \
&
\end{aligned}Solvingwegetlossin\mathrm{K.E.}=\frac{\mathrm{Mm}}{(\mathrm{M}+2 \mathrm{~m})} \omega^2 \mathrm{R}^2$
A ring of mass M and radius R is rotating about its axis with angular velocity ω. Two identical bodies each of mass m are now gently attached at the…
JEE Main 2013 — Physics Mechanics
2013mcqmedium
A ring of mass M and radius R is rotating about its axis with angular velocity ω. 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 :
Official previous-year question
Held on 25 Apr 2013 · Verified 6 Jul 2026.
Options
- A
Mm(M+2m)ω2R2
- B
(M+m)Mmω2R2
- C
(M+2m)Mmω2R2
- D
(M+2m)(M+m)Mω2R2
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