A uniform rod AB of mass m and length 2a is falling freely without rotation under gravity with AB horizontal. Suddenly the end A is fixed when the speed of rod is V. The angular speed with which the rod begins to rotate is

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$\frac{V}{3 a}$

$\frac{4 V}{3 a}$

$\frac{3 V}{4 a}$

$\frac{V}{2 a}$

answer is D.

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angular momentum about A will be conserved ie,

$\begin{array}{l} L_{i} = L_{f} \\ m V a = I ω \\ m V a = \frac{m {(2 a)}^{2}}{3} ω \\ ω = \frac{3 v}{4 a} \end{array}$

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