A thin uniform rod AB of mass m and length l | vedclass

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Question

(C) According to the law of conservation of energy,the loss in potential energy $(P.E.)$ is equal to the gain in rotational kinetic energy $(K.E.)$.Wh

Vedclass · Accepted Answer

$\sqrt{\frac{3g}{l}}$

Vedclass · Answer

(C) According to the law of conservation of energy,the loss in potential energy $(P.E.)$ is equal to the gain in rotational kinetic energy $(K.E.)$.When the rod falls from a vertical position to the ground,its center of mass falls through a height of $h = \frac{l}{2}$.Loss in $P.E. = mgh = mg \left( \frac{l}{2} ight) = \frac{mgl}{2}$.The moment of inertia of the rod about the hinged end $A$ is $I = \frac{ml^2}{3}$.Gain in rotational $K.E. = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{ml^2}{3} ight) \omega^2 = \frac{ml^2 \omega^2}{6}$.Equating the loss in $P.E.$ to the gain in rotational $K.E.$:$\frac{mgl}{2} = \frac{ml^2 \omega^2}{6}$.Solving for $\omega^2$:$\omega^2 = \frac{mgl}{2} 	imes \frac{6}{ml^2} = \frac{3g}{l}$.Therefore,the angular velocity is $\omega = \sqrt{\frac{3g}{l}}$.

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