Angular Momentum and Angular Impulse used $L = I\omega$ for spinning objects and $L = mvr_\perp$ for a point mass going by. With Torque as a Cross Product, both come from one vector definition.
Key idea
The angular momentum of a particle about a point O is $$\vec{L} = \vec{r}\times\vec{p} = m\,\vec{r}\times\vec{v},$$ with $\vec{r}$ measured from O. Its size is $mvr\sin\phi = mvd$, where $d$ is the perpendicular distance from O to the line of motion. Newton's second law gives $\dfrac{d\vec{L}}{dt} = \vec{\tau}$.
Why it works
Differentiate with the product rule: $\dfrac{d\vec{L}}{dt} = \dfrac{d\vec{r}}{dt}\times\vec{p} + \vec{r}\times\dfrac{d\vec{p}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times\vec{F}$. The first term is zero (a vector crossed with itself), and the second is the torque. So with no torque, $\vec{L}$ is constant. A free particle moving in a straight line has no force, so no torque about any point, and its angular momentum about any point never changes, even though it doesn't go around anything.
Worked example: A puck passing a post
A $2\ \text{kg}$ puck slides at $3\ \text{m/s}$ along a line that passes $1.5\ \text{m}$ from a post. Find its angular momentum about the post.
- $|\vec{L}| = mvr\sin\phi$, and $r\sin\phi = d = 1.5\ \text{m}$ wherever the puck is on its line.
- $L = 2(3)(1.5) = 9\ \text{kg·m}^2\text{/s}$, the same before, at and after the closest approach.
Answer: $9\ \text{kg·m}^2\text{/s}$.
Worked example: The full vector
A $2\ \text{kg}$ particle at $\langle 1, 2, 0 \rangle\ \text{m}$ moves with $\vec{v} = \langle 3, 0, 1 \rangle\ \text{m/s}$. Find $\vec{L}$ about the origin.
- $\vec{p} = m\vec{v} = \langle 6, 0, 2 \rangle\ \text{kg·m/s}$.
- $$\vec{L} = \vec{r}\times\vec{p} = \langle 2(2) - 0(0),\ 0(6) - 1(2),\ 1(0) - 2(6) \rangle = \langle 4, -2, -12 \rangle\ \text{kg·m}^2\text{/s}$$
Answer: $\langle 4, -2, -12 \rangle\ \text{kg·m}^2\text{/s}$.
Common mistake
Thinking a particle needs to move in a circle to have angular momentum. Any motion that doesn't head straight at (or away from) the point has some. A meteor flying past the Earth, a ball thrown past a door's hinge, a puck sliding past a post: each carries $mvd$ about that point. Check: compute $\vec{r}\times\vec{p}$. It is zero only if $\vec{v}$ points along $\vec{r}$, toward or away from the point.
For a rigid body spinning about a fixed axis, add up $\vec{r}\times m\vec{v}$ over all its pieces: each piece has $v = r\omega$ around the axis, and the axial part of the total is $\sum mr^2\omega = I\omega$. So the $L = I\omega$ from Physics 1 is this same definition, summed.
This is why a ball thrown at a door far from its hinges swings it more than one thrown near the hinge, and why Collisions with Rotation conserve angular momentum about a pivot.