A rod of mass $^{'} M^{'}$ and length $^{'} L^{'}$ lies on horizontal table and hinged at its one end. A particle of mass $^{'} m^{'}$ is moving with velocity $^{'} v^{'}$ and it hits perpendicularly to other end of the rod. If particles sticks to the rod after collision, then find the angular velocity of the rod.

\frac{2 m v}{(M + 3 m) L}

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\frac{3 m v}{(M + 3 m) L}

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\frac{2 m v}{(M + 2 m) L}

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\frac{3 m v}{(M + 2 m) L}

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Solution

The correct option is B $\frac{3 m v}{(M + 3 m) L}$
When the particle strikes the rod, an external force (reaction of hinge) acts on the rod, so the linear momentum of the system will not remains conserved.
Using conservation of angular momentum,
$L_{i} = L_{f}$
$m v L = I ω$
$m v L = (\frac{M L^{2}}{3} + m L^{2}) ω$
$ω L^{2} (M + 3 m) = 3 m v L$
$ω = \frac{3 m v}{(M + 3 m) L}$
.

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