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

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Question

(C) Let the moment of inertia of the bigger disc be $I = \frac{MR^{2}}{2}$.The moment of inertia of the smaller disc is $I_{2} = \frac{M(R/2)^{2}}{2}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Let the moment of inertia of the bigger disc be $I = \frac{MR^{2}}{2}$.The moment of inertia of the smaller disc is $I_{2} = \frac{M(R/2)^{2}}{2} = \frac{MR^{2}}{8} = \frac{I}{4}$.By the principle of conservation of angular momentum,the initial angular momentum equals the final angular momentum:$L_{i} = L_{f}$$I\omega_{1} + I_{2}(0) = (I + I_{2})\omega_{2}$$I\omega_{1} = (I + I/4)\omega_{2}$$I\omega_{1} = \frac{5I}{4}\omega_{2} \Rightarrow \omega_{2} = \frac{4\omega_{1}}{5}$.The initial kinetic energy is $K_{1} = \frac{1}{2}I\omega_{1}^{2}$.The final kinetic energy is $K_{2} = \frac{1}{2}(I + I_{2})\omega_{2}^{2} = \frac{1}{2}(I + I/4)(\frac{4\omega_{1}}{5})^{2} = \frac{1}{2}(\frac{5I}{4})(\frac{16\omega_{1}^{2}}{25}) = \frac{1}{2}I\omega_{1}^{2}(\frac{4}{5})$.The percentage of energy lost is $p\% = \frac{K_{1} - K_{2}}{K_{1}} 	imes 100\%$.$p\% = \frac{K_{1} - \frac{4}{5}K_{1}}{K_{1}} 	imes 100\% = (1 - 0.8) 	imes 100\% = 20\%$.Thus,the value of $p$ is $20$.

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