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]

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 $θ$ . Which of the following statements about its motion is/are correct? [2017]

Smart Achievers · Accepted Answer

(1, 3, 4)

Force acting on COM $F_{x} = 0,$ $∴$ $a_{x} = 0 .$ Therefore the force acting in vertical direction will move the mid point or COM of the bar fall vertically downwards.

When the bar makes an angle $θ$ the height of its COM $= \frac{L}{2} \cos θ$

$∴$ Displacement of its mid point from the initial position $= \frac{L}{2} - \frac{L}{2} \cos θ = \frac{L}{2} (1 - \cos θ)$

Instantaneous torque about the point of contact P

$τ = m g \times \frac{L}{2} \sin θ$

$Now x = \frac{L}{2} \sin θ, y = L \sin (90 ° - θ) = L \cos θ$

$∴$ ${(\frac{2 x}{L})}^{2} + {(\frac{y}{L})}^{2} = 1 or \frac{4 x^{2}}{L^{2}} + \frac{y^{2}}{L^{2}} = 1$

Thus path of A is an ellipse not parabola.

Solution

Q.
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]

1 The midpoint of the bar will fall vertically downward

2 The trajectory of the point A is a parabola

3 Instantaneous torque about the point in contact with the floor is proportional to $\sin θ$

4 When the bar makes an angle $θ$ with the vertical, the displacement of its midpoint from the initial position is proportional to $(1 - \cos θ)$

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