A rigid uniform bar AB of length L is slippin | vedclass

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(C) $1$. Since the floor is frictionless,there is no horizontal force acting on the rod. Therefore,the center of mass $(C.M.)$ of the rod will move on

Vedclass · Accepted Answer

$A, B, C$

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(C) $1$. Since the floor is frictionless,there is no horizontal force acting on the rod. Therefore,the center of mass $(C.M.)$ of the rod will move only in the vertical direction. Thus,the midpoint of the bar falls vertically downward. Statement $[A]$ is correct.$2$. The trajectory of the top end $A$ is not a parabola; it follows an elliptical path. Statement $[B]$ is incorrect.$3$. The instantaneous torque about the point of contact with the floor is $	au = mg 	imes (\frac{L}{2} \sin 	heta)$,which is proportional to $\sin 	heta$. Statement $[C]$ is correct.$4$. The initial height of the midpoint is $h_i = \frac{L}{2}$. When the bar makes an angle $	heta$ with the vertical,the height of the midpoint is $h_f = \frac{L}{2} \cos 	heta$. The vertical displacement is $\Delta h = h_i - h_f = \frac{L}{2}(1 - \cos 	heta)$. Thus,the displacement is proportional to $(1 - \cos 	heta)$. Statement $[D]$ is correct.Therefore,statements $[A], [C],$ and $[D]$ are correct.

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