A thin smooth rod of length L and mass M is r | vedclass

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(D) Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.The initial moment of inertia of the rod is $I_{rod} =

Vedclass · Accepted Answer

$\frac{M\omega_0}{M + 6m}$

Vedclass · Answer

(D) Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.The initial moment of inertia of the rod is $I_{rod} = \frac{ML^2}{12}$. The beads are at the centre,so their initial distance from the axis is $0$. Thus,$I_i = \frac{ML^2}{12}$.The initial angular momentum is $L_i = I_i \omega_0 = \left( \frac{ML^2}{12} \right) \omega_0$.When the beads reach the ends,their distance from the axis is $L/2$. The final moment of inertia is $I_f = \frac{ML^2}{12} + 2 \left( m \left( \frac{L}{2} \right)^2 \right) = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.Using $L_i = L_f$: $\left( \frac{ML^2}{12} \right) \omega_0 = \left( \frac{L^2(M + 6m)}{12} \right) \omega_f$.Solving for $\omega_f$: $\omega_f = \frac{M\omega_0}{M + 6m}$.

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