AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-3)
Unit 6 · Topic 6.3
6.3 Angular Momentum and Angular Impulse
Angular momentum measures how much rotational motion something has: L = Iω for a spinning rigid object, and L = rmv sin θ for a small object moving past a chosen point. A torque acting for a time delivers angular impulse, τΔt, which equals the change in angular momentum.
Key terms
- angular momentum
- angular momentum of a point object
- angular impulse
- torque–time graph
- angular impulse–momentum theorem
Angular momentum of a spinning object
Linear momentum, p = mv, describes how hard it is to stop something moving in a line. Angular momentum plays the same role for rotation. For a rigid object spinning about an axis, L = Iω, measured in kg·m²/s.
Like ω, it has a sense, clockwise or counterclockwise, which you show with a sign. Angular momentum is really a vector along the axis, but that direction is beyond the course.
Angular momentum of a moving point object
A small object doesn't have to go in a circle to have angular momentum. About a chosen point, L = rmv sin θ. Here r is the distance from the point to the object and θ is the angle between the line from the point to the object and the object's velocity.
There's an easier way to see it: r sin θ is the perpendicular distance from the point to the object's line of motion, so L = mv × (that distance).
This means L depends on the point you choose. An object moving in a straight line at constant velocity has the same L about a given point the whole time, because the perpendicular distance never changes. About any point on its path, its L is zero.
For an object moving in a circle of radius r, the velocity is perpendicular to r, so L = mvr. Since v = rω, that's (mr²)ω = Iω, matching the rigid-object formula.
Angular impulse
Angular impulse is a torque multiplied by the time it acts: τΔt, in N·m·s (the same as kg·m²/s). It points the same way as the torque. If the torque changes, the angular impulse is the area under the torque–time graph.
The angular impulse–momentum theorem
The net angular impulse on an object equals its change in angular momentum: τ_netΔt = ΔL = L_f − L_i. It comes straight from τ_net = Iα when I is constant: τ_net = IΔω/Δt, so τ_netΔt = IΔω = ΔL.
Turned around, the net torque is the rate of change of angular momentum. On a graph of L against time, the slope is the net torque. A flat L–t graph means zero net torque.
| Linear | Rotational |
|---|---|
| p = mv | L = Iω or L = rmv sin θ |
| impulse J = F_avgΔt | angular impulse = τΔt |
| J = Δp | τΔt = ΔL |
| slope of p–t graph = F_net | slope of L–t graph = τ_net |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A puck that isn't spinning (classic trap)
A 0.20 kg puck slides in a straight line at 3.0 m/s. Its path passes 0.50 m from point P at closest approach. Find its angular momentum about P, and about a point Q that lies on its path.
Show the solutionHide the solution
- Step 1: About P: L = mv × (perpendicular distance from P to the line of motion) = (0.20)(3.0)(0.50) = 0.30 kg·m²/s.
- Step 2: This stays 0.30 kg·m²/s at every point along the path, because the perpendicular distance stays 0.50 m.
- Step 3: About Q: the line of motion passes through Q, so the perpendicular distance is zero and L = 0.
- Step 4: The trap is thinking something must spin to have angular momentum, or that L doesn't depend on the reference point.
Answer: 0.30 kg·m²/s about P; 0 about Q
- Example 2Calculator allowed
Angular impulse from a graph
A wheel (I = 0.30 kg·m²) starts from rest. The net torque on it rises steadily from 0 to 6.0 N·m during the first 0.40 s, then stays at 6.0 N·m until t = 1.0 s. Find the angular impulse and the wheel's final angular speed.
Show the solutionHide the solution
- Step 1: Area from 0 to 0.40 s (triangle): ½(0.40)(6.0) = 1.2 N·m·s.
- Step 2: Area from 0.40 to 1.0 s (rectangle): (0.60)(6.0) = 3.6 N·m·s.
- Step 3: Angular impulse = 1.2 + 3.6 = 4.8 N·m·s, which equals ΔL.
- Step 4: L_f = 0 + 4.8 = Iω, so ω = 4.8 ÷ 0.30 = 16 rad/s.
Answer: 4.8 N·m·s; 16 rad/s
- Example 3Calculator allowed
How long friction takes to stop a wheel
A wheel with I = 0.40 kg·m² spins at 25 rad/s. A constant 0.50 N·m friction torque acts at the axle. How long does the wheel take to stop?
Show the solutionHide the solution
- Step 1: L_i = Iω = (0.40)(25) = 10 kg·m²/s. L_f = 0.
- Step 2: Angular impulse = ΔL: −(0.50)Δt = 0 − 10, so Δt = 20 s. The friction torque is negative because it opposes the spin.
- Step 3: Check: with constant torque the average ω is 12.5 rad/s, so the wheel turns (12.5)(20) = 250 rad, the same angle the energy method gives in 6.2.
Answer: 20 s
Common mistakes
- Saying an object moving in a straight line has no angular momentum. About any point not on its path, it does.
- Forgetting to name the reference point or axis. L changes when you change the point.
- Using the area under a torque–angle graph as angular impulse. Angular impulse is the area under torque against time.
- Mixing up angular momentum (Iω) and rotational kinetic energy (½Iω²). Doubling ω doubles L but quadruples K.
On the exam
- Graph questions may give torque against time (find the area to get ΔL) or L against time (find the slope to get the net torque).
- Choosing between time-based and angle-based reasoning matters: if a question gives time, use angular impulse; if it gives an angle, use work and energy (6.2).
Connected topics
Videos
Check yourself
4 questions on 6.3 Angular Momentum and Angular Impulse. Pick an answer to see if you got it, and why.
A disk with rotational inertia 0.25 kg·m² spins at 12 rad/s. What is the magnitude of its angular momentum?
A 2.0 kg puck slides on frictionless ice at a constant 3.0 m/s along a straight line. The line passes 0.50 m from a point P at its closest. The puck approaches P, passes its closest point, and then moves away.
Described scenario
What is the magnitude of the puck's angular momentum about P at a moment when the puck is 2.0 m from P?
How does the puck's angular momentum about P change as it moves past P and away from it?
A ball moves in a straight line at constant speed. About which point is its angular momentum zero?
0 of 4 answered