Animated Solution for Physics - Rotational Motion: A uniform rod of length $l$ is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is $T(x)$ at a distance $x$ from the axis, then which of the following graphs depicts it most closely?

Unraveling the Mystery of Internal Tension

Imagine a uniform rod of length $l$ and mass $m$ resting on a smooth horizontal table, pinned at one end, and spinning with a constant angular velocity $\omega$. It looks peaceful from the outside, but internally, it's a microscopic tug-of-war. Every single atom in the rod wants to fly off in a straight line due to inertia. What keeps them moving in a circle? The internal tension of the rod.

Let's dive into the mathematics to find out exactly how this tension varies along the length of the rod.

Building from the Ground Up

To find the tension $T(x)$ at an arbitrary distance $x$ from the pivot, we must realize a fundamental physical truth: The tension at any cross-section $x$ is solely responsible for providing the centripetal force for all the mass *outside* of that point (from $x$ to the free end $l$).

Let's consider a tiny element of the rod of length $dx$ located at a distance $r$ from the pivot. Since the rod is uniform, its linear mass density is $\lambda =\frac{m}{l}$. The mass of our tiny element is $dm=\lambda dr=\frac{m}{l}dr$.

This tiny element is moving in a circle of radius $r$, so it requires a tiny centripetal force $d{F}_c$:

To find the total tension $T(x)$ at our point $x$, we must sum up (integrate) all these tiny centripetal forces required by every element from $x$ all the way to the end of the rod $l$:

Pulling the constants out of the integral, we get:

Integrating $r$ gives $\frac{r^2}{2}$. Applying the limits from $x$ to $l$:

The Center of Mass Method

If you want to save time in competitive exams like JEE, there is a brilliant shortcut. Instead of integrating, we can treat the entire outer portion of the rod (from $x$ to $l$) as a single point mass located at its center of mass.

1. Mass of the outer portion ($m^{\prime }$):

2. Center of Mass of the outer portion ($r_{cm}$): The outer portion spans from $x$ to $l$. Its center is exactly in the middle:

3. Total Centripetal Force ($T(x)$):

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$, we instantly arrive at:

Look closely at our final equation: $T(x)=\frac{m{\omega }^2}{2l}(l^2-x^2)$. Because the highest power of $x$ is $2$ and it has a negative coefficient, the graph of $T(x)$ versus $x$ is a parabola opening downwards.

Let's check the boundary conditions to ensure it makes physical sense: - At the pivot ($x=0$): $T(0)=\frac{m{\omega }^{2}l}{2}$. The tension is maximum here because this cross-section has to hold the entire rod together. - At the free end ($x=l$): $T(l)=0$. This makes perfect sense! There is no mass beyond the free end that needs to be pulled inwards, so the tension drops to zero.

Therefore, the correct graphical representation is a downward-facing parabola that starts at a maximum value and smoothly curves down to zero at $x=l$.