Momentum $p = mv$ measures how hard straight-line motion is to stop. Angular momentum does the same for turning, and the rotational second law gives it in the same way that $F\Delta t = \Delta p$ came from $F = ma$.
Why it works
Start from $\Sigma\tau = I\alpha = I\dfrac{\Delta\omega}{\Delta t}$ and multiply by $\Delta t$: $\Sigma\tau\,\Delta t = \Delta(I\omega)$. The quantity $L = I\omega$ is the angular momentum, and torque times time, the angular impulse, changes it. A single small mass going around a circle of radius $r$ has $I = mr^2$ and $\omega = v/r$, so $L = mr^2\cdot\dfrac{v}{r} = mvr$. That last form works even for a mass moving in a straight line, measured about any point: use $L = mvr_\perp$, where $r_\perp$ is the perpendicular distance from the point to the line of motion.
Key idea
Angular momentum: $L = I\omega$ for a spinning object, and $L = mvr_\perp$ for a point mass about a chosen point ($r_\perp$ = perpendicular distance from the point to its line of motion). Unit: $\text{kg·m}^2\text{/s}$. Angular impulse changes it: $\tau\,\Delta t = \Delta L$.
Worked examples
Worked example: A spinning wheel
A wheel with $I = 0.40\ \text{kg·m}^2$ spins at $15\ \text{rad/s}$. Find its angular momentum.
- $L = I\omega = (0.40)(15) = 6.0\ \text{kg·m}^2\text{/s}$.
Answer: $6.0\ \text{kg·m}^2\text{/s}$
Worked example: A skater passing a post
A $60\ \text{kg}$ skater glides in a straight line at $4.0\ \text{m/s}$, passing $1.5\ \text{m}$ from a post at the closest point. Find her angular momentum about the post.
- The perpendicular distance from the post to her path is $r_\perp = 1.5\ \text{m}$.
- $$L = mvr_\perp = (60)(4.0)(1.5) = 360\ \text{kg·m}^2\text{/s},$$ the same at every point along her path.
- If she grabbed a rope tied to the post, she would swing around it with exactly this angular momentum.
Answer: $360\ \text{kg·m}^2\text{/s}$
Worked example: Angular impulse
A steady torque of $2.0\ \text{N·m}$ acts for $3.0\ \text{s}$ on a wheel ($I = 0.25\ \text{kg·m}^2$) that starts at rest. Find its final $\omega$.
- $$\Delta L = \tau\Delta t = (2.0)(3.0) = 6.0\ \text{kg·m}^2\text{/s}.$$
- $\omega = \dfrac{L}{I} = \dfrac{6.0}{0.25} = 24\ \text{rad/s}$.
Answer: $24\ \text{rad/s}$
Common mistake
Thinking an object moving in a straight line can't have angular momentum. Angular momentum is measured about a point. About any point that isn't on its path, a moving object has $L = mvr_\perp$, even though nothing spins. It is zero only about points on the line itself. Check: a puck aimed straight at the post has $r_\perp = 0$, and one passing to the side does not.