A smooth uniform rod of length L and mass M h | vedclass

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(D) Since there are no external torques acting on the system,the angular momentum of the system remains constant.Initially,when the beads are at the c

Vedclass · Accepted Answer

$\frac{M{\omega _0}}{M + 6m}$

Vedclass · Answer

(D) Since there are no external torques acting on the system,the angular momentum of the system remains constant.Initially,when the beads are at the center of the rod,the moment of inertia of the system is $I_1 = \frac{ML^2}{12}$.The initial angular momentum is $L_1 = I_1{\omega _0} = \left( \frac{ML^2}{12} \right){\omega _0}$.When the beads reach the ends of the rod,the moment of inertia of the system becomes $I_2 = \frac{ML^2}{12} + m\left( \frac{L}{2} \right)^2 + m\left( \frac{L}{2} \right)^2 = \frac{ML^2}{12} + \frac{mL^2}{4} + \frac{mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{L^2(M + 6m)}{12}$.The final angular momentum is $L_2 = I_2\omega ' = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$.By the principle of conservation of angular momentum,$L_1 = L_2$:$\left( \frac{ML^2}{12} \right){\omega _0} = \left( \frac{L^2(M + 6m)}{12} \right)\omega '$$\omega ' = \frac{M{\omega _0}}{M + 6m}$.

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