Animated Solution for Physics - Rotational Motion: Consider a uniform rod of mass $M=4m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $v$ making angle $\theta =\frac{\pi }{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

A Rod and a Projectile

Imagine a uniform rod of mass $M=4m$ and length $l$, resting peacefully on a frictionless horizontal surface. It is pivoted exactly at its center, meaning it can spin freely but cannot translate. Suddenly, a small particle of mass $m$ comes hurtling towards one of its ends with a velocity $v$. It isn't a head-on collision; the particle approaches at an angle $\theta =\frac{\pi }{4}$ relative to the rod's long axis.

When the particle strikes the rod, it sticks to it. This is a classic perfectly inelastic collision, but with a rotational twist. Our goal is to find the angular speed $\omega$ of the newly formed rod-mass system immediately after the impact.

Why Angular Momentum?

In many collision problems, our first instinct is to conserve linear momentum. However, because the rod is pivoted, the pivot exerts an external impulsive force during the collision to keep the center of the rod stationary. This external force means linear momentum is not conserved.

But there is a loophole! If we calculate the torque about the pivot point itself, the torque produced by the pivot force is zero (since the lever arm is zero). With no net external torque acting about the pivot, the angular momentum of the system about the pivot is strictly conserved.

Before the collision, the rod is stationary, so all the angular momentum comes from the moving particle. The angular momentum of a point particle about an origin is given by $L=mv{r}_{\perp }$, where $r_{\perp }$ is the perpendicular distance from the pivot to the particle's line of action.

The particle strikes the end of the rod, which is at a distance of $\frac{l}{2}$ from the pivot. Using simple trigonometry, the perpendicular distance is:

Therefore, the initial angular momentum of the system is:

A Unified Rotating System

After the collision, the particle and the rod become a single rigid body rotating with an angular velocity $\omega$. The final angular momentum is given by $L_f=I_{\text{system}}\omega$.

To find $I_{\text{system}}$, we must add the moment of inertia of the rod and the moment of inertia of the stuck particle.

First, the rod. Its mass is $M=4m$, and it rotates about its center. The standard formula is $\frac{M{l}^2}{12}$:

Next, the particle. It is now a point mass located at a distance of $\frac{l}{2}$ from the pivot:

Adding these together gives the total moment of inertia:

Bringing It All Together

Now, we invoke our conservation principle, equating the initial and final angular momenta ($L_i=L_f$):

All that's left is to isolate $\omega$. Notice how the mass $m$ and one power of length $l$ beautifully cancel out from both sides:

To match the standard options, we rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

And there we have it! The elegant mechanics of rotational collisions yield a perfectly precise angular velocity.