Discussion Forum



1)

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?

71202093_tri.PNG


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 midpoint of the bar will fall vertically downward

D) When the bar makes an angle $\theta$ with the vertical, the displacement of its mid point from intial position is proportional to $(1-\cos\theta)$

Answer:

Option A,C,D

Explanation:

when the bar makes an angle of the height of its COM (midpoint)  $\frac{L}{2}\cos \theta$

$\therefore$    Displacement

                           =$L-\frac{L}{2}\sin\theta=\frac{L}{2}(1-\cos\theta)$

Since, force on COM is only along the vertical direction, Hence COM is falling vertically downwards instantaneous  torque about point of contact is

  $\tau =mg \times \frac{L}{2}\sin\theta$   or $\tau\propto sin\theta$

712020306_8888.PNG

Now,

                     $x=\frac{L}{2}\sin\theta$

               $y=L\cos\theta$

$\frac{X^{2}}{(\frac{L}{2})^{2}}+\frac{Y^{2}}{L^{2}}=1$

  Path of A is an ellipse