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

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(A) The moment of inertia of the first disc about the central axis is $I_1 = \frac{1}{2} MR^2$.Since no external torque acts on the system,the angular

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$\frac{2}{3} \omega$

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(A) The moment of inertia of the first disc about the central axis is $I_1 = \frac{1}{2} MR^2$.Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.Initially,$L_i = I_1 \omega = \frac{1}{2} MR^2 \omega$.When the second disc of mass $\frac{M}{2}$ is placed on the first,the total moment of inertia becomes $I_2 = I_1 + I_{	ext{disc2}} = \frac{1}{2} MR^2 + \frac{1}{2} (\frac{M}{2}) R^2 = \frac{1}{2} MR^2 + \frac{1}{4} MR^2 = \frac{3}{4} MR^2$.Using conservation of angular momentum: $I_1 \omega = I_2 \omega_2$.$\frac{1}{2} MR^2 \omega = \frac{3}{4} MR^2 \omega_2$.Solving for $\omega_2$: $\omega_2 = (\frac{1}{2} 	imes \frac{4}{3}) \omega = \frac{2}{3} \omega$.

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