A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (a

Q: Why doesn't the center of mass move horizontally?

Because the floor is frictionless, there is no horizontal force acting on the rod. According to Newton's second law, zero net force means zero acceleration. Since it started from rest, the horizontal velocity remains zero.

Q: Why is the torque only due to $mg$?

We are calculating the torque about the point of contact with the floor. The normal reaction force acts exactly at this point, so its perpendicular distance from the pivot is zero, resulting in zero torque.

Q: How did we get the coordinates of point A?

By setting the initial point of contact as the origin, the center of mass (which only moves vertically) has an $x$-coordinate of 0. Using the rod's length and angle, we can use basic trigonometry to find the positions of both ends relative to the center of mass.

Q: Why isn't the path of point A a parabola?

A parabola typically arises in projectile motion where there is a constant acceleration in one direction and constant velocity in another. Here, the motion is constrained by the rod's fixed length, leading to a trigonometric relationship between the $x$ and $y$ coordinates that forms an ellipse.

Q: What would happen if the floor had friction?

If the floor had friction, there would be a horizontal force acting on the rod. The center of mass would then have a horizontal acceleration, and its path would no longer be a straight vertical line. The trajectory of point A would also change significantly.

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?

Options:

(A) Instantaneous torque about the point in contact with the floor is proportional to $\sin	heta$

(B) The trajectory of the point $A$ is parabola

(C) The mid-point of the bar will fall vertically downward

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a rigid rod standing perfectly vertical on a smooth, frictionless floor. The moment it starts to slip, we need to understand the forces at play.

Because the floor is frictionless, there is absolutely no horizontal force acting on the rod. The only forces are the gravitational pull $mg$ acting downwards at the center of mass, and the normal reaction $N$ pushing upwards from the floor.

### The Motion of the Center of Mass

According to Newton's second law, the acceleration of the center of mass in any direction is directly proportional to the net force in that direction.

Since $\sum F_x = 0$, the horizontal acceleration of the center of mass is zero ($a_{cx} = 0$).

Because the rod was initially at rest, a zero horizontal acceleration means the center of mass will never acquire any horizontal velocity. It is constrained to move purely in the vertical direction, falling straight down. This immediately confirms that statement (c) is correct!

### Calculating the Displacement

Let's quantify this vertical fall. If the rod has a length $L$, its center of mass is located exactly at the midpoint, a distance of $L/2$ from either end.

When the rod is perfectly vertical, the initial height of the center of mass is $y_i = \frac{L}{2}$. As the rod slips and makes an angle $	heta$ with the vertical, we can use simple trigonometry to find its new height. The vertical component of the upper half of the rod is $y_f = \frac{L}{2} \cos	heta$.

The displacement is simply the difference between the initial and final heights:

$$\Delta y = y_i - y_f = \frac{L}{2} - \frac{L}{2} \cos	heta = \frac{L}{2}(1 - \cos	heta)$$

This clearly shows that the displacement is directly proportional to $(1 - \cos	heta)$, making statement (d) correct.

### Evaluating the Torque

Now, let's shift our focus to the rotational dynamics, specifically the instantaneous torque about the point of contact with the floor (let's call it point $B$).

Torque is defined as the force multiplied by the perpendicular distance from the pivot point to the line of action of the force. The normal reaction $N$ acts exactly at point $B$, so its perpendicular distance is zero, resulting in zero torque.

The only force generating torque is the weight $mg$. The horizontal distance from point $B$ to the vertical line passing through the center of mass is $\frac{L}{2} \sin	heta$. Therefore, the torque is:

$$	au_B = mg \left( \frac{L}{2} \sin	heta ight)$$

This reveals that the instantaneous torque is proportional to $\sin	heta$, confirming that statement (a) is correct.

### The Trajectory of the Top End

Finally, let's determine the path traced by the top end of the rod, point $A$. Many students intuitively guess this to be a parabola, but let's let the math speak.

If we set the initial point of contact as the origin $(0,0)$, the center of mass is always on the y-axis. Using the geometry of the rod, the coordinates of point $A$ at any angle $	heta$ are:

$$x_A = -\frac{L}{2} \sin	heta$$

$$y_A = L \cos	heta$$

To find the equation of the trajectory, we need to eliminate the parameter $	heta$. We can rearrange the equations to isolate the trigonometric functions:

$$\sin	heta = -\frac{2x_A}{L} \quad 	ext{and} \quad \cos	heta = \frac{y_A}{L}$$

By squaring both equations and adding them together, we utilize the fundamental trigonometric identity $\sin^2	heta + \cos^2	heta = 1$:

$$\frac{x_A^2}{(L/2)^2} + \frac{y_A^2}{L^2} = 1$$

This is the standard equation of an ellipse! The path of point $A$ is an elliptical arc, not a parabola. Thus, statement (b) is incorrect.

Select Answer:

Visualized Solution

$F_{net}$ \text{ Analysis}

\text{Motion of COM}

\text{Height of COM}

\text{Displacement of COM}

\text{Torque Setup}

\text{Evaluating Torque}

\text{Coordinates of A}

\text{Trajectory Equation}

The Sigma Insight: Kinematics of Rotational Motion

Analyzing the Setup

The Motion of the Center of Mass

Calculating the Displacement

Evaluating the Torque

The Trajectory of the Top End

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