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

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(C) The impulse applied is $P = F 	imes \Delta t$.Angular impulse about the center of mass is given by $J_{	heta} = P 	imes \frac{l}{2}$.Since angu

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$\frac{\pi}{12} \frac{ml}{P}$

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(C) The impulse applied is $P = F 	imes \Delta t$.Angular impulse about the center of mass is given by $J_{	heta} = P 	imes \frac{l}{2}$.Since angular impulse equals the change in angular momentum,we have $P 	imes \frac{l}{2} = I \omega$,where $I = \frac{ml^2}{12}$ is the moment of inertia about the center of mass.Thus,$P 	imes \frac{l}{2} = \frac{ml^2}{12} \omega$,which simplifies to $\omega = \frac{6P}{ml}$.The rod rotates with a constant angular velocity $\omega$ after the impulse. The time taken to turn through a right angle ($\pi/2$ radians) is $t = \frac{	heta}{\omega} = \frac{\pi/2}{\omega}$.Substituting $\omega$,we get $t = \frac{\pi/2}{6P/ml} = \frac{\pi ml}{12P}$.

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $P$ is applied to the end $B$. The time taken by the rod to turn through a right angle is

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