A rigid uniform bar AB of length ? is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is `theta.` Which of the following statements about its motion is/are correct?

Options

A. Instantaneous torque about the point in contact with the floor is proportional to `sin theta`
B. The trajectory of the point A is parabola
C. The mid-point of the bar will fall vertically downward
D. When the bar makes an angle with the vertical, the displacement of its mid-point from the initial position is proportional to `(1- cos theta)`

Answer

A, C, D

Explanation

(i) `x=L/2 sin theta, y = L cos theta`
Hence, `x^(2)/(L/2)^(2) + y^(2)/L^(2) =1`
Hence NOT parabola.

(ii) `("Torque about")_(B) = (mg L/2 sin theta )` Hence correct.
(iii) As there is no external horizontal force and zero initial velocity. Hence, it will move downwards.
(iv) Displacement of mid point `=(L/2 -L/2 cos theta)`