A circular disc of mass M and radius R is rot | vedclass

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Question

Let moment of inertia of bigger disc is $I =\frac{ MR ^{2}}{2}$

$\Rightarrow$ $MOI$ of small disc $I_{2}=\frac{M\left(\frac{R}{2}\right)^{2}}{2}=\f

Vedclass · Accepted Answer

$20$

Vedclass · Answer

Let moment of inertia of bigger disc is $I =\frac{ MR ^{2}}{2}$

$\Rightarrow$ $MOI$ of small disc $I_{2}=\frac{M\left(\frac{R}{2}ight)^{2}}{2}=\frac{I}{4}$

by angular momentum conservation

$I \omega_{1}+\frac{ I }{4}( D )= I \omega_{2}+\frac{ I }{4} \omega_{2} \Rightarrow \omega_{2}=\frac{4 \omega_{1}}{5}$

initial kinetic energy $K _{1}=\frac{1}{2} I \omega_{1}^{2}$

final kinetic energy $K _{2}$

$=\frac{1}{2}\left( I +\frac{ I }{4}ight)\left(\frac{4 \omega_{1}}{5}ight)^{2}=\frac{1}{2} I \omega_{1}^{2}\left(\frac{4}{5}ight)$

$P \%=\frac{ K _{1}- K _{2}}{ K _{1}} 	imes 100 \%=\frac{1-4 / 5}{1} 	imes 100=20 \%$

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