Animated Solution for Physics - Rotational Motion: A rod of mass $m$ and length $L$, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed ${\omega }_M$ is achieved for $x=x_M$. Then

The beauty of rotational dynamics lies in its ability to describe complex, real-world collisions using elegant conservation laws. In this classic JEE Advanced problem, we are presented with a vertical rod pivoted at its top end and a bullet that strikes it horizontally. Let's embark on a thrilling journey to decode the physics behind this collision and find the optimal striking point for maximum rotation.

Imagine a uniform rod of mass $m$ and length $L$ hanging peacefully from a pivot. Suddenly, a bullet, also of mass $m$, comes flying in with a horizontal velocity $v$. It strikes the rod at a distance $x$ from the pivot and gets embedded. This is a classic perfectly inelastic collision.

Your first instinct might be to conserve linear momentum. However, there is a catch! The rod is attached to a pivot. During the violent impact of the collision, the pivot exerts a massive, unknown impulsive force on the rod to keep it attached. Because there is a net external force acting on our bullet-rod system, linear momentum is strictly not conserved.

Conservation of Angular Momentum

If linear momentum fails us, what tool do we have left? We look at the torques. The only external impulsive force acts exactly at the pivot point. The torque of a force about its own point of application is zero (${\tau }_{\text{pivot}}=0$). Therefore, the angular momentum of the system about the pivot is perfectly conserved.

Let's set up the equation. Before the collision, only the bullet is moving. Its angular momentum about the pivot is the product of its linear momentum and the perpendicular distance to the pivot:

After the collision, the bullet is embedded in the rod, and the entire system rotates together with an angular velocity $\omega$. The final angular momentum is given by $I_{\text{system}}\omega$.

We must calculate the total moment of inertia of the combined system about the pivot. The rod, pivoted at its end, has a standard moment of inertia of $\frac{m{L}^2}{3}$. The embedded bullet acts as a point mass at a distance $x$, contributing $m{x}^2$. Thus, the total moment of inertia is:

Equating the initial and final angular momenta, we get:

Solving for $\omega$, we can cancel out the common mass $m$ and multiply the numerator and denominator by 3 to clean up the fraction:

This elegant expression perfectly matches option (A)!

We now have a function $\omega (x)$ that tells us how fast the rod will swing based on where the bullet hits. The problem asks us to find the maximum possible angular speed, ${\omega }_M$, and the specific striking distance, $x_M$, that achieves it.

To find the maximum of a function, we turn to calculus. We must differentiate $\omega$ with respect to $x$ and set the derivative to zero:

Applying the quotient rule to the function $\frac{x}{L^2+3x^2}$, we get:

For the derivative to be zero, the numerator must be zero:

Taking the positive square root (since distance must be positive), we find the optimal striking point:

This confirms that option (C) is absolutely correct.

To complete our analysis, we must find the actual maximum angular speed, ${\omega }_M$. We simply substitute our optimal distance $x_M$ back into the original $\omega$ equation:

Let's simplify the denominator first. Squaring $\frac{L}{\sqrt{3}}$ gives $\frac{L^2}{3}$. Multiplying by 3 gives $L^2$. So the denominator is simply $L^2+L^2=2L^2$.

The numerator is $\frac{3vL}{\sqrt{3}}=\sqrt{3}vL$. Putting it all together:

This matches option (D) perfectly.

Through a beautiful synthesis of conservation laws and differential calculus, we have completely unraveled the dynamics of this system. The correct options are indeed (A), (C), and (D).