A uniform thin rod of mass M and length L is standing vertically along the Y-axis on a smooth horizo

Q: Wait, why do we assume the Centre of Mass stays exactly on the Y-axis?

Because the floor is smooth, there is zero friction. The only forces are gravity (down) and the normal force (up). With zero horizontal force, the horizontal acceleration of the CM is zero. Since it started from rest, its horizontal velocity remains zero forever. Thus, it can only fall straight down!

Q: Why is the distance $OB$ equal to $\frac{L}{2} \cos\theta$?

Look at the right-angled triangle formed by the origin $O$, the bottom of the rod $B$, and the Centre of Mass $C$. The hypotenuse is the segment $CB$, which is half the rod's length ($\frac{L}{2}$). The angle at $B$ is $\theta$. Using basic trigonometry, the adjacent side $OB$ is hypotenuse $\times \cos\theta$.

Q: What happens if the point $P$ is exactly at the Centre of Mass?

If $P$ is the Centre of Mass, then its distance from the bottom is $r = \frac{L}{2}$. If you plug this into our trajectory equation, the denominator for the $x^2$ term becomes zero, which implies $x$ must be exactly $0$. This perfectly matches our earlier conclusion: the CM falls in a straight vertical line along the Y-axis!

Q: Why do we use $\sin^2\theta + \cos^2\theta = 1$ to find the trajectory?

A trajectory is a path in the $x-y$ plane, meaning we need an equation that relates $x$ and $y$ directly, without any time or angle variables. Since our coordinates depend on $\theta$, we use this fundamental trigonometric identity as a mathematical trick to completely eliminate $\theta$ from the system.

Q: Would the path still be an ellipse if the floor had friction?

No! If the floor had friction, there would be a horizontal force acting on the rod. This would cause the Centre of Mass to accelerate horizontally, meaning it would no longer fall in a straight vertical line. The geometry would become much more complex, and the path of point $P$ would not be a simple ellipse.

Question

A uniform thin rod of mass $M$ and length $L$ is standing vertically along the $Y$-axis on a smooth horizontal surface, with its lower end at the origin $(0, 0)$. A slight disturbance at $t = 0$ causes the lower end to slip on the smooth surface along the positive $X$-axis, and the rod starts falling. (1993)
(a) What is the path followed by the centre of mass of the rod during its fall?
(b) Find the equation of the trajectory of a point on the rod located at a distance $r$ from the lower end. What is the shape of the path of this point ?

Sigma Percentile · Accepted Answer

### The Physics of a Falling Rod

Imagine a uniform rod standing perfectly vertical on a frictionless floor. A slight nudge causes it to slip. To understand its motion, we must first analyze the forces acting on it.

The only external forces are gravity pulling it downwards (its weight, $Mg$) and the normal force from the floor pushing it upwards ($N$). Because the floor is perfectly smooth, there is absolutely zero friction. This means there are **no horizontal forces** acting on the rod.

According to Newton's Second Law, the acceleration of the Centre of Mass (CM) is directly proportional to the net external force. Since the net horizontal force is zero, the horizontal acceleration of the CM is zero.

Furthermore, the rod starts from rest, meaning its initial horizontal velocity is zero. With no horizontal acceleration and no initial horizontal velocity, the CM cannot move horizontally. It is constrained to fall vertically along the Y-axis. This elegant deduction tells us that the path of the Centre of Mass is a **straight vertical line**.

### Tracking a Random Point

Now, let's find the trajectory of an arbitrary point $P$ on the rod, located at a distance $r$ from the lower end $B$.

Let's set up our coordinate system. The CM is at point $C(0, y_c)$ on the Y-axis, and the lower end is at point $B(x_b, 0)$ on the X-axis. The distance between the CM and the lower end is exactly half the length of the rod, $\frac{L}{2}$. Let the rod make an angle $	heta$ with the horizontal X-axis.

To find the coordinates $(x, y)$ of point $P$, we drop a perpendicular from $P$ to the X-axis at point $M$. The x-coordinate is the distance $OM$, which can be written as the total distance $OB$ minus the segment $MB$.

Using basic trigonometry in the right-angled triangle $OBC$, the base $OB$ is:

$$OB = \frac{L}{2} \cos	heta$$

Similarly, in the smaller right-angled triangle $PMB$, the base $MB$ is:

$$MB = r \cos	heta$$

Subtracting these gives us the x-coordinate of point $P$:

$$x = \frac{L}{2} \cos	heta - r \cos	heta$$

Factoring out $\cos	heta$, we get:

$$x = \left(\frac{L}{2} - right) \cos	heta$$

Rearranging this to isolate $\cos	heta$, we obtain our first parametric equation:

$$\cos	heta = \frac{x}{\frac{L}{2} - r}$$

### The Vertical Coordinate

Finding the y-coordinate is much simpler. It is just the height of point $P$ above the ground, which corresponds to the perpendicular $PM$ in our smaller triangle.

Using trigonometry again:

$$y = r \sin	heta$$

Isolating $\sin	heta$, we get our second parametric equation:

$$\sin	heta = \frac{y}{r}$$

### The Grand Synthesis

We now have expressions for both $\sin	heta$ and $\cos	heta$. To find the trajectory, we must eliminate the variable $	heta$. We can do this by invoking the fundamental trigonometric identity:

$$\sin^2	heta + \cos^2	heta = 1$$

Substituting our derived expressions into this identity yields:

$$\left(\frac{y}{r}ight)^2 + \left(\frac{x}{\frac{L}{2} - r}ight)^2 = 1$$

Rearranging this into the standard mathematical form, we get the final equation of the trajectory:

$$\frac{x^2}{\left(\frac{L}{2} - right)^2} + \frac{y^2}{r^2} = 1$$

This is the unmistakable standard equation of an **ellipse**. Therefore, as the rod slips and falls, any given point on it (except the CM and the ends) traces out a beautiful elliptical path!

Visualized Solution

Initial Setup and Forces

Motion of the Centre of Mass

Defining the Coordinates

Coordinates of a General Point $P$

Calculating Horizontal Distances

Isolating $cos θ$

Calculating the Vertical Coordinate

Eliminating $θ$ to Find the Trajectory

The Sigma Insight: Motion of Centre of Mass

The Physics of a Falling Rod

Tracking a Random Point

The Vertical Coordinate

The Grand Synthesis

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