A thin circular disc of mass M and radius R i | vedclass

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Question

(C) According to the law of conservation of angular momentum,the external torque on the system is zero,so the angular momentum remains constant.$L_i =

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$\frac{4}{5}\,\omega$

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(C) According to the law of conservation of angular momentum,the external torque on the system is zero,so the angular momentum remains constant.$L_i = L_f$$I_1 \omega_1 = I_2 \omega_2$Here,$I_1 = \frac{1}{2} M R^2$ and $\omega_1 = \omega$.When the second disc is placed,the new moment of inertia $I_2$ is the sum of the moments of inertia of both discs:$I_2 = I_1 + I_{	ext{disc2}} = \frac{1}{2} M R^2 + \frac{1}{2} (\frac{M}{4}) R^2 = \frac{1}{2} M R^2 (1 + \frac{1}{4}) = \frac{1}{2} M R^2 (\frac{5}{4}) = \frac{5}{8} M R^2$.Now,applying the conservation law:$(\frac{1}{2} M R^2) \omega = (\frac{5}{8} M R^2) \omega_2$$\omega_2 = \frac{1/2}{5/8} \omega = \frac{1}{2} 	imes \frac{8}{5} \omega = \frac{4}{5} \omega$.

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