A) \[\frac{\operatorname{m}(\operatorname{M}+2m)}{\operatorname{M}}{{\omega }^{2}}{{R}^{2}}\]
B) \[\frac{\operatorname{Mm}}{(\operatorname{M}+m)}{{\omega }^{2}}{{R}^{2}}\]
C) \[\frac{\operatorname{Mm}}{(\operatorname{M}+2m)}{{\omega }^{2}}{{R}^{2}}\]
D) (d)\[\frac{\left( \operatorname{M}+m \right)\operatorname{M}}{(\operatorname{M}+2m)}{{\omega }^{2}}{{R}^{2}}\]
Correct Answer: C
Solution :
Kinetic energy \[_{\text{(rotational)}}{{K}_{R}}=\frac{1}{2}I{{\omega }^{2}}\] Kinetic energy\[_{\text{(translational)}}{{K}_{T}}=\frac{1}{2}M{{v}^{2}}\]\[(v=R\omega )\] \[M.l{{.}_{(initial)}}{{I}_{ring}}=M{{R}^{2}};{{\omega }_{initial}}=\omega \] \[M.I{{.}_{(new)}}I{{'}_{(system)}}=M{{R}^{2}}+2m{{R}^{2}}\] \[\omega {{'}_{(system)}}=\frac{M\omega }{M+2m}\] Solving we get loss in K.E. \[=\frac{Mm}{(M+2m)}{{\omega }^{2}}{{R}^{2}}\]
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