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

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(A) Concept: The angular impulse applied to the rod is equal to the change in its angular momentum.Angular impulse about the center of mass $O$ is giv

Vedclass · Accepted Answer

$\frac{\pi}{12} \frac{m l}{P}$

Vedclass · Answer

(A) Concept: The angular impulse applied to the rod is equal to the change in its angular momentum.Angular impulse about the center of mass $O$ is given by $J_{	heta} = P \cdot \frac{l}{2}$.The angular momentum of the rod about its center of mass is $L = I \omega$,where $I = \frac{m l^2}{12}$ is the moment of inertia of the rod about its center.Equating angular impulse to change in angular momentum:$P \cdot \frac{l}{2} = I \omega$$P \cdot \frac{l}{2} = \left( \frac{m l^2}{12} ight) \omega$Solving for angular velocity $\omega$:$\omega = \frac{P \cdot l}{2} \cdot \frac{12}{m l^2} = \frac{6 P}{m l}$The rod rotates with a constant angular velocity $\omega$. The time $\Delta t$ taken to turn through an angle $\Delta 	heta = \frac{\pi}{2}$ is:$\Delta t = \frac{\Delta 	heta}{\omega} = \frac{\pi / 2}{6 P / (m l)} = \frac{\pi}{2} \cdot \frac{m l}{6 P} = \frac{\pi m l}{12 P}$

$A$ 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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