A rigid uniform bar AB of length L 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 θ. Which of the following statements about its motion is/are correct?
(A)The midpoint of the bar will fall vertically downward✓ Correct
(B)The trajectory of the point A is a parabola
(C)Instantaneous torque about the point in contact with the floor is proportional to sinθ✓ Correct
(D)When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to (1−cosθ)✓ Correct
Answer: (A), (C), (D)
Solution
Since there is no horizontal force on the rod (frictionless surfaces), the center of mass falls vertically. The torque about point B (on the ground) is τ=mg2Lsinθ∝sinθ. The vertical displacement of the midpoint is 2L−2Lcosθ=2L(1−cosθ)∝(1−cosθ). The trajectory of point A is not a parabola.
JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options
