A uniform rod AB of length L and mass M is pi | vedclass

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(C) Since there are no external torques about the pivot point $O$ during the collision,the angular momentum of the system is conserved.Initial angular

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$\frac{12v}{7L}$

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(C) Since there are no external torques about the pivot point $O$ during the collision,the angular momentum of the system is conserved.Initial angular momentum of the particle about $O$ is $L_i = Mv 	imes r$,where $r = OC = L/4$.So,$L_i = Mv(L/4)$.After the collision,the particle sticks to the rod (assuming an inelastic collision),and the system rotates with angular velocity $\omega$.The total moment of inertia $I$ of the system about $O$ is the sum of the moment of inertia of the rod and the particle.$I = I_{rod} + I_{particle} = \frac{ML^2}{12} + M(L/4)^2 = \frac{ML^2}{12} + \frac{ML^2}{16} = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.Applying conservation of angular momentum: $L_i = I\omega$.$Mv(L/4) = (7ML^2/48) \omega$.Solving for $\omega$: $\omega = \frac{MvL}{4} 	imes \frac{48}{7ML^2} = \frac{12v}{7L}$.

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