Test Index

Rotational Motion

Section: Physics

Question : 22 of 54

Marks: +1, -0

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

Solution:

(A)

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

y = L \cos\theta

Hence

\frac{X^2}{\left(\frac{L}{2}\right)^2} + \frac{y^2}{L^2} = 1

Hence NOT parabola

(B)

\left(\text{Torque about}\right)_B = \left(\frac{mgL}{2} \sin\theta\right)

Hence correct

(D) Displacement of mid point

= \left(\frac{L}{2} - \frac{L}{2} \cos\theta\right)

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