A train is moving slowly at 2 , m/s next to a | vedclass

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(B) When the man alights,his angular momentum about the point of contact on the platform is conserved.By the principle of conservation of angular mome

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(B) When the man alights,his angular momentum about the point of contact on the platform is conserved.By the principle of conservation of angular momentum:$L_{	ext{initial}} = L_{	ext{final}}$Initially,the man acts as a particle of mass $m$ moving with velocity $v$ at a height $h = l/2$ from the ground. So,$L_{	ext{initial}} = m v (l/2)$.After alighting,the man acts as a rigid rod of length $l$ rotating about one end fixed on the ground. The moment of inertia of a rod about one end is $I = \frac{m l^2}{3}$.Thus,$m v (l/2) = I \omega$$m v (l/2) = \left( \frac{m l^2}{3} ight) \omega$Solving for $\omega$:$\omega = \frac{3 v (l/2)}{l^2} = \frac{3 v}{2 l}$Given $v = 2 \, m/s$ and $l = 1.5 \, m$:$\omega = \frac{3 	imes 2}{2 	imes 1.5} = \frac{6}{3} = 2 \, rad/s$.

$A$ train is moving slowly at $2 \, m/s$ next to a railway platform. $A$ man,$1.5 \, m$ tall,alights from the train such that his feet are fixed on the ground. Taking him to be a rigid body,the instantaneous angular velocity (in $rad/s$) is

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