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?

From the figure, x = L/2 Sinθ ⇒ Sinθ = x/(L/2) [ 1 ]
y = L Cosθ ⇒ Cosθ = y/L [ 2 ]
Squaring and adding equations 1 and 2, we have
Sin²θ + Cos²θ = x²/(L/2)² + y²/L²
1 = 4x²/L² + y²/L²
4x² + y² = L²
This is the equation of an ellipse. Therefore, the trajectory of the point A is an ellipse and not a parabola.
The instantaneous torque about the point in contact with the floor is given by:
τ = mg(L/2)sinθ
Therefore, the torque is proportional to sinθ.
The center of mass of the bar is at its midpoint. The horizontal acceleration of the midpoint is non-zero, while the vertical acceleration is also non-zero, because there is a net vertical force acting on it. The midpoint will not fall vertically downwards.
Let the initial position of the midpoint be (0, L/2). When the bar makes an angle θ with the vertical, the coordinates of the midpoint are (L/2)sinθ, (L/2)cosθ.
The displacement of the midpoint from the initial position is given by:
√[((L/2)sinθ)² + ((L/2)cosθ - L/2)²]
= (L/2)√(sin²θ + cos²θ - 2cosθ + 1)
= (L/2)√(2 - 2cosθ)
= (L/2)√(4sin²(θ/2))
= Lsin(θ/2)
This is approximately proportional to θ/2 for small θ, which in turn is proportional to 1 - cosθ for small θ.
However, for large θ, the displacement is not exactly proportional to (1 - cosθ). Therefore, this statement is incorrect.