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

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(C) Since there is no external torque acting on the system,the angular momentum is conserved.$L_i = L_f$$I_1 \omega_1 = I_2 \omega_2$For the first dis

Vedclass · Accepted Answer

$4\omega / 5$

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(C) Since there is no external torque acting on the system,the angular momentum is conserved.$L_i = L_f$$I_1 \omega_1 = I_2 \omega_2$For the first disc,the moment of inertia is $I_1 = \frac{1}{2} M R^2$ and angular velocity is $\omega_1 = \omega$.When the second disc of mass $M/4$ is placed coaxially,the new moment of inertia of the system is $I_2 = I_1 + I_{disc2} = \frac{1}{2} M R^2 + \frac{1}{2} (M/4) R^2$.$I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{4+1}{8} M R^2 = \frac{5}{8} M R^2$.Using the conservation of angular momentum:$\frac{1}{2} M R^2 \cdot \omega = \frac{5}{8} M R^2 \cdot \omega_2$$\omega_2 = \frac{1/2}{5/8} \omega = \frac{1}{2} \cdot \frac{8}{5} \omega = \frac{4}{5} \omega$.

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