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

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(D) From the equation of linear momentum,$m v_{cm} = J$,so $v_{cm} = \frac{J}{m}$.From the equation of angular momentum about the center of mass $(CM)

Vedclass · Accepted Answer

$\sqrt{2} \frac{J}{m}$

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(D) From the equation of linear momentum,$m v_{cm} = J$,so $v_{cm} = \frac{J}{m}$.From the equation of angular momentum about the center of mass $(CM)$,$J \cdot \frac{l}{2} = I \omega = \frac{m l^2}{12} \cdot \omega$,which gives $\omega = \frac{6 J}{m l}$.At time $t = \frac{\pi m l}{12 J}$,the angle rotated by the rod is $	heta = \omega t = \left( \frac{6 J}{m l} ight) \left( \frac{\pi m l}{12 J} ight) = \frac{\pi}{2}$.The tangential velocity of particle $P$ due to rotation about the $CM$ is $v_t = r \omega = \frac{l}{6} \cdot \frac{6 J}{m l} = \frac{J}{m}$.Since the rod has rotated by $\frac{\pi}{2}$,the velocity vector due to rotation is now perpendicular to the initial velocity vector $v_{cm}$.The resultant velocity of particle $P$ is $v_p = \sqrt{v_{cm}^2 + v_t^2} = \sqrt{\left( \frac{J}{m} ight)^2 + \left( \frac{J}{m} ight)^2} = \sqrt{2} \frac{J}{m}$.

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