The rod is pivoted at its centre and can rotate freely in a vertical plane.

During the collision, external torques about the pivot are absent (the hinge force at the pivot passes through the axis of rotation).

Hence, angular momentum about the pivot is conserved during the impact.

Before the collision, only the bullet is moving.

Its linear momentum $ = Mv $

We convert this to angular momentum about the pivot.

The bullet strikes one end of the rod, which is at a perpendicular distance of $ \frac{L}{2} $ from the pivot.

Therefore, initial angular momentum

$ L_i = Mv \left( \frac{L}{2} \right) $

Just after the collision, the bullet gets embedded in the rod end, and both rotate together with common angular velocity $ \omega $.

Moment of inertia of the combined system about the pivot:

$ I = I_{\text{rod}} + I_{\text{bullet}} $

Moment of inertia of the rod about its centre is:

$ I_{\text{rod}} = \frac{1}{12} M L^2 $

Moment of inertia of the bullet (as a point mass at a distance $ L/2 $) is:

$ I_{\text{bullet}} = M \left( \frac{L}{2} \right)^2 = \frac{1}{4} M L^2 $

Hence,

$ I = M L^2 \left( \frac{1}{12} + \frac{1}{4} \right) = M L^2 \left( \frac{1 + 3}{12} \right) = \frac{1}{3} M L^2 $

$ L_i = L_f $

$ M v \left( \frac{L}{2} \right) = \left( \frac{1}{3} M L^2 \right) \omega $

Simplify:

$ \omega = \frac{ M v \frac{L}{2} }{ \frac{1}{3} M L^2 } $

$ \omega = \frac{3v}{2L} $

$ \boxed{ \omega = \frac{3v}{2L} } $

Thus, the correct option is:

Option C: $ \displaystyle \frac{3v}{2L} $

Rotational Motion

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