During a short collision, a fixed pivot or hinge can exert a large external impulse on the system, so the system's linear momentum need not be conserved. Angular momentum can nevertheless be conserved about a carefully chosen point.

For angular momentum about a point $P$,

$$\Delta\mathbf L_P=\int_{t_i}^{t_f}\boldsymbol\tau_{P,\rm ext},dt.$$

If a rigid body is attached to an ideal fixed pivot at $P$, the pivot force may be large during an impact, but its torque about $P$ is zero because its line of action passes through the pivot. If all other external forces produce negligible angular impulse during the short collision interval, then

$$\boxed{\mathbf L_{P,i}=\mathbf L_{P,f}}.$$

This does not imply conservation of linear momentum: the pivot can change total linear momentum while leaving angular momentum about itself unchanged.

Angular momentum of an incoming particle

A particle of mass $m$ moving with velocity $\mathbf v$ has

$$\mathbf L_P=\mathbf r\times m\mathbf v.$$

Its magnitude can be written

$$\boxed{L_P=mvb},$$

where $b$ is the perpendicular distance from the pivot to the incoming line of motion. A particle aimed directly toward the pivot has $b=0$ and therefore zero angular momentum about that pivot even if its linear momentum is large.

Particle sticks to a pivoted rigid body

Let a rigid body initially be at rest with moment of inertia $I_P$ about its fixed pivot. A particle of mass $m$ and speed $v$ strikes at distance $r$ from the pivot and sticks. Let $b$ be the perpendicular distance from the pivot to the incoming velocity line.

Initially,

$$L_{P,i}=mvb.$$

After impact, the particle and body rotate together. Their total moment of inertia is

$$I_{\rm total}=I_P+mr^2,$$

so

$$L_{P,f}=(I_P+mr^2)\omega_f.$$

Conservation about the pivot gives

$$mvb=(I_P+mr^2)\omega_f,$$

and therefore

$$\boxed{\omega_f=\frac{mvb}{I_P+mr^2}}.$$

For a tangential impact, the incoming velocity is perpendicular to the radius and $b=r$.

Worked example

A particle with

$$m=0.50,\mathrm{kg},\qquad v=6.0,\mathrm{m/s}$$

strikes tangentially at

$$r=0.40,\mathrm m$$

and sticks to a pivoted body with

$$I_P=0.24,\mathrm{kg,m^2}.$$

The initial angular momentum is

$$L_{P,i}=(0.50)(6.0)(0.40)=1.20,\mathrm{kg,m^2/s}.$$

The particle adds

$$mr^2=(0.50)(0.40)^2=0.080,\mathrm{kg,m^2},$$

so

$$I_{\rm total}=0.32,\mathrm{kg,m^2}.$$

Thus

$$\boxed{\omega_f=\frac{1.20}{0.32}=3.75,\mathrm{rad/s}}.$$

Kinetic energy is a separate condition

Because the particle sticks, the collision is perfectly inelastic. Angular momentum about the pivot is conserved under the stated impulse assumptions, but kinetic energy is not.

Initially,

$$K_i=\frac12mv^2=9.0,\mathrm J,$$

whereas finally

$$K_f=\frac12I_{\rm total}\omega_f^2 =\frac12(0.32)(3.75)^2 \approx2.25,\mathrm J.$$

The lost kinetic energy becomes deformation, heat, sound, and other internal energy.

Choosing the origin strategically

The reusable technique is to choose an angular-momentum origin about which unknown impulsive forces have zero lever arm. For a fixed pivot, the pivot itself is usually the natural choice.

Before writing a conservation equation, identify:

1. the system;
2. the external impulses acting during the collision;
3. a point about which their net angular impulse is negligible;
4. whether kinetic energy is also conserved.

Angular momentum conservation during an impact is therefore a statement about external angular impulse about the chosen point, not a blanket property of every collision.