Animated Solution for Physics - Rotational Motion: A rod of length $L$ and mass $M$ is hinged at point $O$. A small bullet of mass $m$ hits the rod as shown in the figure. The bullet gets embedded in the rod. Find angular velocity of the system just after impact.

A Collision Waiting to Happen

Imagine you are standing in a lab, observing a heavy rod of mass $M$ and length $L$ hanging perfectly still from a hinge at point $O$. Suddenly, a small bullet of mass $m$ comes flying in horizontally with a velocity $v$. It strikes the very bottom of the rod and embeds itself deep within the material.

The immediate aftermath of this collision is a violent swing. The rod, now carrying the extra mass of the bullet, begins to rotate upwards. Our goal is to find out exactly how fast this combined system starts rotating—its initial angular velocity, $\omega$.

Why Angular Momentum?

There is a catch here. Your first instinct might be to use the conservation of linear momentum. After all, it's a collision! But wait—the rod is attached to a hinge. When the bullet hits, the hinge exerts a sudden, unknown impulsive force to keep the rod attached to the wall. Because of this external force, the total linear momentum of the system is not conserved.

So, what do we do? We look for a loophole. If we choose our axis of rotation exactly at the hinge $O$, the perpendicular distance of this unknown hinge force from our axis is zero. This means the torque exerted by the hinge is zero! Since there are no other external torques acting on the system during the split-second of the collision, the angular momentum about point $O$ is perfectly conserved.

The Bullet's Momentum

Let's calculate the angular momentum just before the collision. The rod is completely at rest, so it contributes nothing. The bullet, however, is moving with a linear momentum of $mv$.

To find its angular momentum about the hinge $O$, we multiply its linear momentum by its perpendicular distance from the hinge. Since the bullet hits the very bottom of the rod, this distance is exactly $L$.

Therefore, our initial angular momentum is:

A Combined System

The moment the bullet embeds itself, the physics changes. We no longer have a separate bullet and rod; we have a single, combined rigid body rotating with an angular velocity $\omega$.

To find the final angular momentum, we need the total moment of inertia of this new system about the hinge $O$. 1. The moment of inertia of a uniform rod rotated about its end is $\frac{M{L}^2}{3}$. 2. The bullet is now effectively a point mass located at a distance $L$ from the hinge, so its moment of inertia is $m{L}^2$.

Adding these together, the total moment of inertia is:

This gives us our final angular momentum:

Bringing It All Together

Now, we simply equate the initial and final angular momenta:

Don't get intimidated by the algebra; it's a straightforward cleanup from here. Notice that we can factor out an $L^2$ from the right side:

We can cancel one $L$ from both sides. Then, let's take a common denominator inside the bracket:

Finally, we isolate $\omega$ by moving everything else to the other side:

And there we have it! The exact angular velocity of the system the moment after the bullet strikes. This elegant result shows how the mass of the bullet and the rod interact to resist the rotational motion.