Rotation 5 Question 12

22. 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?

(2017 Adv.)

(a) Instantaneous torque about the point in contact with the floor is proportional to sinθ

(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 θ with the vertical, the displacement of its mid-point from the initial position is proportional to (1cosθ)

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Answer:

Correct Answer: 22. (a,c,d)

Solution:

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

Displacement =LL2cosθ=L2(1cosθ)

Since, force on COM is only along the vertical direction, hence COM is falling vertically downward. Instantaneous torque about point of contact is

or

τ=mg×L2sinθ

τsinθ

Now,

x=L2sinθy=Lcosθx2(L/2)2+y2L2=1

Path of A is an ellipse.



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