A rod of length L is held vertically on a smo | vedclass

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(D) Let the angular velocity of the rod about the point of contact be $\omega$. The center of mass $(COM)$ of the rod is at a distance $L/2$ from the

Vedclass · Accepted Answer

$v_0 	an 	heta$

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(D) Let the angular velocity of the rod about the point of contact be $\omega$. The center of mass $(COM)$ of the rod is at a distance $L/2$ from the end in contact with the surface. The vertical component of the velocity of the $COM$ is given as $v_0$. Since the $COM$ is rotating about the contact point,its velocity is $v_{COM} = \omega (L/2)$. The vertical component of this velocity is $v_{COM, y} = \omega (L/2) \cos 	heta = v_0$. Therefore,$\omega = \frac{2v_0}{L \cos 	heta}$. The horizontal velocity of the end of the rod in contact with the surface is $v = \omega (L/2) \sin 	heta$. Substituting the value of $\omega$: $v = \left( \frac{2v_0}{L \cos 	heta} ight) \left( \frac{L}{2} ight) \sin 	heta = v_0 	an 	heta$.

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