A rigid uniform bar A B of length L is slipping from its vertical position on a frictionless floor (as shown…

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 correct?

  1. When the bar makes an angle θ with the vertical, the displacement of its midpoint from the initial position is proportional to 1-cosθ
  2. The midpoint of the bar will fall vertically downward
  3. Instantaneous torque about the point in contact with the floor is proportional to sinθ
  4. The trajectory of the point A is a parabola

Solution

When the bar makes an angle θ, the height of its COM (mid point) is  L2cosθ.

∴Displacement =L-L2cosθ=L21-cosθ

Since, force on COM is only along the vertical direction, hence, COM is falling vertically downward.

Now, x=L2sinθ

y=Lcosθ

x2L22+y2L2=1

Put of A is an ellipse.

Torques about the point of contact, τ→=r→×F→⇒τ=mgL2sinθ

Asked in: JEE Advanced 2017 (Paper 2)

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