Conservation of Angular Momentum kept $L$ constant for a spinning skater. Here a particle with Angular Momentum of a Particle hits an object that can only turn about a pivot, whose moment of inertia comes from The Parallel-Axis Theorem.
Why it works
During the brief impact, the pivot pushes on the rod; otherwise the rod's top would swing sideways along with its bottom. That push is an outside force, so linear momentum is not conserved. But the pivot's force acts at the pivot, where its lever arm is zero, so it exerts no torque about the pivot. Gravity's torque matters only over time, and the impact is very short. So angular momentum about the pivot is the conserved quantity.
Key idea
For a ball (mass $m$, speed $v$) hitting a pivoted object at perpendicular distance $d$ from the pivot and sticking: $$mvd = (I_{\text{object}} + md^2)\,\omega.$$ Kinetic energy is not conserved in a sticking collision. After the impact, use energy conservation for any swing.
Worked example: Clay hits a hanging rod
A uniform $2\ \text{kg}$, $1.0\ \text{m}$ rod hangs from a pivot at its top. A $0.20\ \text{kg}$ lump of clay moving horizontally at $10\ \text{m/s}$ hits its bottom end and sticks. Find $\omega$ just after.
- Before: $L = mvd = 0.2(10)(1.0) = 2\ \text{kg·m}^2\text{/s}$ about the pivot.
- After: $I = \tfrac{1}{3}(2)(1)^2 + 0.2(1)^2 = 0.867\ \text{kg·m}^2$.
- $\omega = 2/0.867 = 2.31\ \text{rad/s}$.
- Kinetic energy: $10\ \text{J}$ before, $\tfrac{1}{2}(0.867)(2.31)^2 = 2.31\ \text{J}$ after; 77% is lost to deforming the clay.
Answer: $\omega = 2.31\ \text{rad/s}$.
Worked example: How high does it swing?
How far up does the rod (with the clay) swing after the impact above?
- Now energy is conserved: $\tfrac{1}{2}I\omega^2 = \left(M\tfrac{L}{2} + md\right)g(1 - \cos\theta)$.
- $$2.31 = (1 + 0.2)(9.8)(1 - \cos\theta) = 11.76(1 - \cos\theta)$$ so $1 - \cos\theta = 0.196$.
- $\theta = 36.5^\circ$ from vertical.
Answer: About $36.5^\circ$.
There is one special spot. For a uniform rod pivoted at its end, a hit at $d = \tfrac{2}{3}L$ from the pivot (the center of percussion) leaves the pivot with no sideways jolt at all; hit anywhere else and the pivot must push. It's the reason a bat or a hammer has a "sweet spot" that doesn't sting your hands.
Common mistake
Conserving linear momentum through the impact. If you write $mv = (m + M)v_{\text{cm}}$, you ignore the pivot's sideways push, which can be large. Check: is any object in the system attached to something that can push on it during the impact? If so, find a point where that push has no torque (here, the pivot) and conserve angular momentum about that point instead.