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

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(B) Since there are no external torques acting on the system,the angular momentum $L_{ang}$ is conserved.$L_{initial} = L_{final}$$I_1 \omega_0 = I_2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Since there are no external torques acting on the system,the angular momentum $L_{ang}$ is conserved.$L_{initial} = L_{final}$$I_1 \omega_0 = I_2 \omega_2$Initially,the beads are at the center,so their distance from the axis of rotation is $0$. The moment of inertia of the system is just that of the rod:$I_1 = \frac{ML^2}{12}$When the beads reach the ends of the rod,their distance from the axis of rotation is $L/2$. The moment of inertia of the system becomes:$I_2 = \frac{ML^2}{12} + 2 	imes m\left(\frac{L}{2}ight)^2 = \frac{ML^2}{12} + \frac{2mL^2}{4} = \frac{ML^2}{12} + \frac{mL^2}{2} = \frac{ML^2 + 6mL^2}{12} = \frac{(M + 6m)L^2}{12}$Using conservation of angular momentum:$\frac{ML^2}{12} \omega_0 = \frac{(M + 6m)L^2}{12} \omega_2$$\omega_2 = \frac{M\omega_0}{M + 6m}$

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