The moment of inertia of a uniform rod of mas | vedclass

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(C) Initial angular momentum of the system about the pivot point is due to the bullet: $L_i = Mv(L/2)$.After the collision,the bullet is embedded in t

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$3v/2L$

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(C) Initial angular momentum of the system about the pivot point is due to the bullet: $L_i = Mv(L/2)$.After the collision,the bullet is embedded in the rod,so the system rotates with angular velocity $\omega$. The final moment of inertia of the system is $I_f = I_{	ext{rod}} + I_{	ext{bullet}} = \frac{ML^2}{12} + M(L/2)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.By the principle of conservation of angular momentum,$L_i = L_f$,so $Mv(L/2) = I_f \omega$.Substituting the values: $\frac{MvL}{2} = \left(\frac{ML^2}{3}ight) \omega$.Solving for $\omega$: $\omega = \frac{MvL}{2} \cdot \frac{3}{ML^2} = \frac{3v}{2L}$.

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