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

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(B) The angular momentum of the system is conserved because no external torque acts on the system.Initial angular momentum $L_i = I_1 \omega_1$,where

Vedclass · Accepted Answer

$\frac{3}{4} \omega$

Vedclass · Answer

(B) The angular momentum of the system is conserved because no external torque acts on the system.Initial angular momentum $L_i = I_1 \omega_1$,where $I_1 = \frac{1}{2} M R^2$ and $\omega_1 = \omega$.Final moment of inertia $I_2 = I_{	ext{disc1}} + I_{	ext{disc2}} = \frac{1}{2} M R^2 + \frac{1}{2} (\frac{M}{3}) R^2 = \frac{1}{2} R^2 (M + \frac{M}{3}) = \frac{1}{2} R^2 (\frac{4M}{3}) = \frac{2}{3} M R^2$.By the law of conservation of angular momentum,$I_1 \omega_1 = I_2 \omega_2$.Substituting the values: $(\frac{1}{2} M R^2) \omega = (\frac{2}{3} M R^2) \omega_2$.Solving for $\omega_2$: $\omega_2 = \frac{1/2}{2/3} \omega = \frac{3}{4} \omega$.

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