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

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(A) $1$. The impulse $P$ applied at end $B$ provides a linear momentum $P = mv_{cm}$,where $v_{cm}$ is the velocity of the center of mass. Thus,$v_{cm

Vedclass · Accepted Answer

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

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(A) $1$. The impulse $P$ applied at end $B$ provides a linear momentum $P = mv_{cm}$,where $v_{cm}$ is the velocity of the center of mass. Thus,$v_{cm} = \frac{P}{m}$.$2$. The impulse also provides an angular impulse about the center of mass: $J_{	heta} = P \cdot \frac{\ell}{2}$.$3$. Since $J_{	heta} = I\omega$,where $I = \frac{m\ell^2}{12}$ is the moment of inertia about the center of mass,we have $\frac{P\ell}{2} = \frac{m\ell^2}{12} \omega$.$4$. Solving for angular velocity $\omega$,we get $\omega = \frac{6P}{m\ell}$.$5$. The time $t$ taken to rotate through an angle $	heta = \frac{\pi}{2}$ is given by $t = \frac{	heta}{\omega} = \frac{\pi/2}{6P/m\ell} = \frac{\pi m \ell}{12P}$.

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

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