Motion of a Slipping Uniform Rigid Bar on Frictionless Floor

To determine the correct options, let us analyze the motion of the uniform rigid bar $AB$ of mass $M$ and length $L$.

1. Motion of the Midpoint (Center of Mass)

Let the horizontal floor lie along the $x$-axis, and let the vertical line passing through the initial position of the bar be the $y$-axis. The origin $(0,0)$ is chosen at the base of the initial vertical line.

• The only external forces acting on the bar are:
  1. Gravity $M\mathbf{g}$ acting vertically downward at its center of mass (the midpoint $C$).
  2. Normal reaction force $N$ from the smooth horizontal floor acting vertically upward at the contact point $B$.

Since there is no force in the horizontal direction, $F_{\text{ext}, x} = 0$. Since the bar starts from rest, the acceleration of the center of mass in the horizontal direction is $a_{C, x} = 0$. Consequently, the $x$-coordinate of the midpoint $C$ remains constant at $x_C = 0$.

Therefore, the midpoint of the bar falls purely vertically downward. Thus, Option [A] is correct.

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2. Displacement of the Midpoint

Initially, when the bar is completely vertical ($\theta = 0$), the position of the midpoint $C$ is: $y_i = \frac{L}{2}$

When the bar makes an angle $\theta$ with the vertical axis, the vertical coordinate of the midpoint becomes: $y_C = \frac{L}{2}\cos\theta$

The magnitude of the displacement of the midpoint from its initial position is: $\Delta y = y_i - y_C = \frac{L}{2} - \frac{L}{2}\cos\theta = \frac{L}{2}(1 - \cos\theta)$

Thus, the displacement of the midpoint is directly proportional to $(1 - \cos\theta)$. Therefore, Option [D] is correct.

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3. Trajectory of Point $A$

Since $x_C = 0$, the coordinates of the top end $A$ at any instant when the bar subtends an angle $\theta$ with the vertical are given by: $x_A = -\frac{L}{2}\sin\theta$ $y_A = y_C + \frac{L}{2}\cos\theta = \frac{L}{2}\cos\theta + \frac{L}{2}\cos\theta = L\cos\theta$

Eliminating $\theta$ using $\sin^2\theta + \cos^2\theta = 1$: $\left(\frac{x_A}{-L/2}\right)^2 + \left(\frac{y_A}{L}\right)^2 = 1 \implies \frac{x_A^2}{(L/2)^2} + \frac{y_A^2}{L^2} = 1$

This is the equation of an ellipse, not a parabola. Therefore, Option [B] is incorrect.

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4. Instantaneous Torque About the Contact Point

Let $B$ be the point of contact with the floor. The position of $B$ relative to $C$ along the horizontal direction is at a distance of $\frac{L}{2}\sin\theta$.

Taking torque about the contact point $B$:

• The normal force $N$ passes through point $B$, so its torque about $B$ is zero.
• The torque is due solely to the gravitational force acting at the center of mass $C$: $\tau_B = Mg \times (\text{horizontal distance between } C \text{ and } B)$ $\tau_B = Mg \left(\frac{L}{2}\sin\theta\right) = \frac{MgL}{2}\sin\theta$

Thus, the instantaneous magnitude of torque about the contact point on the floor is proportional to $\sin\theta$. Therefore, Option [C] is correct.

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Conclusion

The correct statements are A, C, and D.