Question 763: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

We discuss the options as follows:
$$\bullet$$ For option (A) : When the bar makes an angle $$	heta$$ with the floor, the height of the midpoint is $${L \over 2}\cos 	heta $$. this is height of centre of mass.
A force mg acts vertically downward. Thus, the midpoint of bar falls vertically downward. Hence, option (A) is correct.
$$\bullet$$ For option (B) : Here, we have $$x = {L \over 2}\sin 	heta $$ and $$y = L\cos 	heta $$; therefore,
$$\sin 	heta  = {x \over {L/2}}$$ and $$\cos 	heta  = {y \over L}$$
Using the condition $${\sin ^2}	heta  + {\cos ^2}	heta  = 1$$, we get
$${{{x^2}} \over {{{(L/2)}^2}}} + {{{y^2}} \over {{L^2}}} = 1$$ (equation of ellipse)
Thus, trajectory of point A is not parabola. Hence, option (B) is incorrect.
$$\bullet$$ For option (C) : The torque acting about the point of contact with floor, that is, at point B is
$$	au  = \overrightarrow F  	imes \overrightarrow r  = mg{L \over 2}\sin 	heta $$
Thus, the torque is proportional to sin$$	heta$$. Hence, option (C) is correct.
$$\bullet$$ For option (D) : The displacement of midpoint is
$${L \over 2} - {L \over 2}\cos 	heta  = {L \over 2}(1 - \cos 	heta )$$
That is,
Displacement $$\propto$$ (1 $$-$$ cos$$	heta$$)
Hence, option (D) is correct.

Answer

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

Answer

The trajectory of the point A is parabola

Answer

The mid-point of the bar will fall vertically downward

Answer

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

Centre of Mass and Rotational Motion

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?

Choose the correct answer: