A thin uniform circular disc of mass M and ra | vedclass

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(B) The initial moment of inertia of the first disc is $I_1 = \frac{1}{2} M R^2$.The second disc has mass $M$ and radius $R/2$. Its moment of inertia

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$(4 / 5) \omega$

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(B) The initial moment of inertia of the first disc is $I_1 = \frac{1}{2} M R^2$.The second disc has mass $M$ and radius $R/2$. Its moment of inertia is $I_2 = \frac{1}{2} M (R/2)^2 = \frac{1}{8} M R^2$.When the second disc is placed coaxially,the total moment of inertia of the system becomes $I_{total} = I_1 + I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{5}{8} M R^2$.According to the law of conservation of angular momentum,the initial angular momentum equals the final angular momentum: $L_i = L_f$.$I_1 \omega = I_{total} \omega_{final}$$\left(\frac{1}{2} M R^2\right) \omega = \left(\frac{5}{8} M R^2\right) \omega_{final}$$\omega_{final} = \frac{(1/2)}{(5/8)} \omega = \frac{4}{5} \omega$.

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