Unit 6 · Topic 6.3

6.3 Angular Momentum and Angular Impulse

Angular momentum is rotation's version of momentum: $L = I\omega$ for a rigid object and $\vec{L} = \vec{r} \times \vec{p}$ for a particle. A net torque changes it: $\tau_{\text{net}} = \dfrac{dL}{dt}$ , and the angular impulse $\int \tau\,dt$ equals ΔL.

Key terms

angular momentum
angular impulse
angular impulse–momentum theorem
point particle
torque–time graph

What changed in the 2024 update

Before the Fall 2024 course update, this topic was part of the Rotation unit (old Unit 5). It's now in Unit 6, Energy and Momentum of Rotating Systems, so older videos may file it under rotation.

Angular momentum of a rigid object

For a rigid object spinning about a fixed axis, $L = I\omega$ The unit is kg·m²/s. A heavy, fast-spinning flywheel has a lot of angular momentum, and it takes a large torque, or a long time, to stop it.

Like ω, its direction on this exam is just clockwise or counterclockwise about the axis.

Angular momentum of a particle

A single moving particle has angular momentum about any point you choose: $\vec{L} = \vec{r} \times \vec{p}$ , where $\vec{r}$ runs from the point to the particle and $\vec{p} = m\vec{v}$ . Its magnitude is $L = mvr\sin\theta = mv\,r_\perp$ where $r_\perp$ is the perpendicular distance from the point to the particle's line of motion.

So a particle moving in a straight line has angular momentum about a point off that line. As it moves, r and θ change, but r sin θ stays the same. If no force acts, L about that point is constant.

The value depends on the reference point. About a point on the particle's path, r⊥ = 0 and L = 0. Always say which point you're using.

The two formulas agree. A small ball moving in a circle of radius r has $v = r\omega$ , so $mvr = mr^2\omega = I\omega$ . A rigid object's L = Iω is just the total of mvr⊥ for all its pieces.

Torque changes angular momentum

Newton's second law for rotation can be written $\tau_{\text{net}} = \frac{dL}{dt}$ For a rigid object with constant I, this is the same as $\tau_{\text{net}} = I\alpha$ . On a graph of L against time, the slope is the net torque.

Integrate over time and you get the angular impulse–momentum theorem: $\Delta L = \int \tau_{\text{net}}\,dt$ The angular impulse $\int \tau\,dt$ is the area under a torque–time graph. It points in the same direction as the torque, and its unit is N·m·s, which equals kg·m²/s.

Linear	Rotational
p = mv	L = Iω
F = dp/dt	τ = dL/dt
impulse J = ∫F dt	angular impulse ∫τ dt
J = Δp	∫τ dt = ΔL

Comparing initial and final

When an object reverses its spin, the change in angular momentum is bigger than either value alone. Going from +5 to −3 kg·m²/s is a change of −8, not −2. Keep track of signs, just as you did for linear momentum.

The same angular impulse can come from a big torque for a short time or a small torque for a long time. That's why a long, gentle push on a merry-go-round can spin it up as much as a quick hard shove.

Worked examples

Try each one yourself first, then open the solution.

Example 1
A particle moving in a straight line
A 0.20 kg puck slides at 5.0 m/s in the +x direction along the line y = 3.0 m. Find its angular momentum about the origin. Does it change as the puck slides?
Show the solutionHide the solution
1. Step 1: The perpendicular distance from the origin to the puck's line of motion is $r_\perp = 3.0$ m.
2. Step 2: $L = mv\,r_\perp = (0.20)(5.0)(3.0) = 3.0$ kg·m²/s.
3. Step 3: Direction: the puck is above the origin and moving to the right, so it sweeps clockwise around the origin (viewed with +y up and +x to the right).
4. Step 4: No force acts, so v and r⊥ stay the same, and L stays 3.0 kg·m²/s clockwise as it slides.
Answer: 3.0 kg·m²/s, clockwise about the origin, and constant
Example 2Calculator allowed
Angular impulse from a torque–time graph
A wheel with I = 0.80 kg·m² starts at rest. A torque acts on it that rises steadily from 0 to 6.0 N·m between t = 0 and t = 2.0 s, then stays at 6.0 N·m until t = 5.0 s. Find the wheel's angular velocity at t = 5.0 s.
Show the solutionHide the solution
1. Step 1: Angular impulse is the area under the τ–t graph. Triangle from 0 to 2.0 s: ½(2.0)(6.0) = 6.0 N·m·s. Rectangle from 2.0 to 5.0 s: (3.0)(6.0) = 18 N·m·s.
2. Step 2: Total: ΔL = 24 kg·m²/s. The wheel started at rest, so L = 24 kg·m²/s.
3. Step 3: $\omega = \dfrac{L}{I} = \dfrac{24}{0.80} = 30$ rad/s.
Answer: 30 rad/s
Example 3
Torque from an angular momentum function
A spinning object's angular momentum is $L(t) = 2.0t^2 + 3.0$ , in kg·m²/s with t in seconds. Find the net torque at t = 1.5 s and the angular impulse from t = 0 to t = 2.0 s.
Show the solutionHide the solution
1. Step 1: Net torque is the slope: $\tau = \dfrac{dL}{dt} = 4.0t$ . At t = 1.5 s, τ = 6.0 N·m.
2. Step 2: Angular impulse equals ΔL: L(2.0) − L(0) = 11 − 3.0 = 8.0 kg·m²/s.
3. Step 3: You don't need to integrate the torque. The change in L already gives the impulse.
Answer: τ = 6.0 N·m at 1.5 s; angular impulse = 8.0 N·m·s

Common mistakes

Saying an object moving in a straight line has no angular momentum. About a point off its path, it does.
Using the full distance r instead of the perpendicular distance r⊥ (or r sin θ) for a particle.
Giving a particle's angular momentum without naming the reference point.
Subtracting magnitudes when the spin reverses. Use signs for clockwise and counterclockwise.

On the exam

Graph questions are common: slope of L–t is torque, and area under τ–t is change in L. Say which one you're using.
In collisions between a particle and a rod, you'll need the particle's mvr⊥ about the pivot. Draw the perpendicular distance on the diagram.

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Check yourself

4 questions on 6.3 Angular Momentum and Angular Impulse. Pick an answer to see if you got it, and why.

A 2.0 kg particle moves at a constant 3.0 m/s in the +x direction along the line y = 4.0 m. Point O is the origin.

Described situation

Question 1 of 4Calculator allowed

What is the magnitude of the particle's angular momentum about O?

Question 2 of 4Calculator allowed

As the particle moves from x = −5 m to x = +5 m, what happens to its angular momentum about O?

A net torque acts on a wheel with rotational inertia 3.0 kg·m². The wheel starts at rest. The torque grows steadily from 0 at t = 0 to 6.0 N·m at t = 2.0 s. It then stays at 6.0 N·m until t = 5.0 s and then stops.

Described situation

Question 3 of 4Calculator allowed

What angular impulse does the torque deliver from t = 0 to t = 5.0 s?

Question 4 of 4Calculator allowed

What is the wheel's angular speed at t = 5.0 s?

0 of 4 answered