JEE Advanced 2017 · Paper 2 · Physics Q13

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PhysicsMultiple correct+4 / -2Question 13 of 54

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 $\theta$ . Which of the following statements about its motion is/are correct?

A bar of length L standing vertically on a frictionless floor at O, and a dashed copy tilted at angle θ to the vertical with its upper end A and lower end B on the floor to the right of O.

The midpoint of the bar will fall vertically downward

The trajectory of the point

A

is a parabola

Instantaneous torque about the point in contact with the floor is proportional to

\sin\theta

When the bar makes an angle

\theta

with the vertical, the displacement of its midpoint from the initial position is proportional to

(1 - \cos\theta)

Answer and worked solution

Official answer: A, C, D

(A) The floor is frictionless, so there is no horizontal force. The centre of mass falls straight down. True.
(B) With the midpoint at $(0,\tfrac L2\cos\theta)$ , end $A$ is at $\left(-\tfrac L2\sin\theta,\ L\cos\theta\right)$ . That traces an ellipse, not a parabola. False.
(C) About the contact point only gravity has a moment, since the normal force passes through it: $\tau=mg\tfrac L2\sin\theta\propto\sin\theta$ . True.
(D) The midpoint drops by $\tfrac L2(1-\cos\theta)$ . True.

(A), (C), (D)

Topics:Torque and rotational dynamics Centre of mass

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JEE Advanced 2017 · Paper 2, Physics question 13