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

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(D) Let $\omega$ be the angular velocity of the rod about the point of contact with the surface.The velocity of the center of mass $(COM)$ is given by

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$v_0 	an 	heta$

Vedclass · Answer

(D) Let $\omega$ be the angular velocity of the rod about the point of contact with the surface.The velocity of the center of mass $(COM)$ is given by $v_0 = \omega \cdot r_{COM}$,where $r_{COM}$ is the distance from the point of contact to the $COM$. Since the $COM$ is at the center of the rod,$r_{COM} = \frac{L}{2}$.However,the velocity $v_0$ given is the vertical component of the velocity of the $COM$. The distance of the $COM$ from the point of contact along the rod is $\frac{L}{2}$. The vertical component of the velocity of the $COM$ is $v_{0,y} = \omega \cdot (\frac{L}{2} \cos 	heta) = v_0$.The velocity of the end of the rod in contact with the surface is horizontal. Let this velocity be $v$. The horizontal component of the velocity of the $COM$ is $v_{COM,x} = v - \omega \cdot (\frac{L}{2} \sin 	heta)$.Since the rod is rotating about the instantaneous center of rotation at the point of contact,the velocity of the $COM$ is perpendicular to the rod. The vertical component of the velocity of the $COM$ is $v_0 = \omega \frac{L}{2} \cos 	heta$.The horizontal velocity of the end of the rod is $v = \omega \frac{L}{2} \sin 	heta$.Dividing the two equations: $\frac{v}{v_0} = \frac{\omega \frac{L}{2} \sin 	heta}{\omega \frac{L}{2} \cos 	heta} = 	an 	heta$.Therefore,$v = v_0 	an 	heta$.

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