A circular disc of mass M has an initial angu | vedclass

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(C) Since there is no external torque acting on the system,the angular momentum is conserved: $L_1 = L_2$.The initial moment of inertia of the disc is

Vedclass · Accepted Answer

$\left( \frac{M}{M + 4m} \right) \omega_1$

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(C) Since there is no external torque acting on the system,the angular momentum is conserved: $L_1 = L_2$.The initial moment of inertia of the disc is $I_1 = \frac{1}{2}MR^2$.When two spheres of mass $m$ are placed at opposite ends of a diameter (at distance $R$ from the center),the final moment of inertia becomes $I_2 = I_1 + mR^2 + mR^2 = \frac{1}{2}MR^2 + 2mR^2$.Using the conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$.$\left( \frac{1}{2}MR^2 \right) \omega_1 = \left( \frac{1}{2}MR^2 + 2mR^2 \right) \omega_2$.Dividing both sides by $R^2$ and multiplying by $2$: $M \omega_1 = (M + 4m) \omega_2$.Therefore,the final angular velocity is $\omega_2 = \left( \frac{M}{M + 4m} \right) \omega_1$.

$A$ circular disc of mass $M$ has an initial angular velocity $\omega_1$. Two small spheres,each of mass $m$,are attached to the disc at opposite points on its diameter. What will be the final angular velocity of the disc?

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