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A rod of mass $m$ and length $l$ is sliding against a smooth vertical wall as shown. The floor is assumed to be frictionless. The speed of the bottom end of the rod at the instant shown in $v_{0}$ . The magnitude of angular momentum of the rod about $I C R$ (instantaneous axis of rotation) at the instant when angle $θ = 30 °$ is :

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None of these

$\frac{2}{3} m v_{0} l$

$\frac{1}{12} m v_{0} l$

$\frac{1}{6} m v_{0} l$

answer is A.

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Detailed Solution

$C A = \frac{l}{2} ∴ I_{A} = \frac{1}{12} m l^{2} + \frac{1}{4} m l^{2} = \frac{1}{3} m l^{2} ∴ L_{A} = I_{A} ω = \frac{1}{3} m l^{2} \frac{2 v_{0}}{l} (∵ ω = \frac{v_{0}}{\frac{l}{2}}) \Rightarrow L_{A} = \frac{2}{3} m v_{0} l$

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