---
title: "Unit 6.2 | Angular Momentum and Angular Impulse"
description: "Learn angular momentum in AP Physics 1: L = Iω, L = mvr⊥ for a moving object, angular impulse τΔt = ΔL, and torque and angular momentum graphs."
featured_image: "https://nerd-notes.com/wp-content/uploads/2026/10/133457.jpg"
url: "https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/"
date_modified: "2026-10-31T09:35:31+00:00"
---

# Unit 6.2 | Angular Momentum and Angular Impulse

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Momentum gets a rotational twin. This lesson covers angular momentum for spinning and moving objects, angular impulse, and the two graphs the AP exam loves.

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### Unit 6 Breakdown

You are on Lesson 2 of 4

1. [Unit 6.1 | Rotational Kinetic Energy and Work](https://nerd-notes.com/unit-6-1-rotational-kinetic-energy-and-work/)
2. [Unit 6.2 | Angular Momentum and Angular Impulse](https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/)
3. [Unit 6.3 | Conservation of Angular Momentum](https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/)
4. [Unit 6.4 | Rolling and Orbiting Satellites](https://nerd-notes.com/unit-6-4-rolling-and-orbiting-satellites/)

#### In this lesson: 

- Angular momentum of a spinning object, [katex] L = I\omega [/katex], and of an object moving in a straight line, [katex] L = mvr_\perp [/katex]
- Angular impulse, [katex] \tau\Delta t = \Delta L [/katex], the rotational twin of the impulse-momentum theorem
- Reading the two graphs: the slope of [katex] L [/katex] vs. [katex] t [/katex] and the area under [katex] \tau [/katex] vs. [katex] t [/katex]
- The 4 problem types: a moving point, angular impulse, torque vs. time graphs, and angular momentum vs. time graphs

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### Introduction

This lesson covers **angular momentum** and **angular impulse**: how much rotational motion an object has, and what it takes to change it.

In [Unit 4.1](https://nerd-notes.com/unit-4-1-momentum-and-impulse/), momentum [katex] p = mv [/katex] measured how hard an object is to stop, and an impulse [katex] F\Delta t [/katex] changed it. [Unit 6.1](https://nerd-notes.com/unit-6-1-rotational-kinetic-energy-and-work/) gave kinetic energy a rotational twin. This lesson does the same for momentum and impulse.

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### Angular Momentum of a Spinning Object

For a rigid object turning about a fixed axis, angular momentum is rotational inertia times angular velocity:

[katex] L = I\omega [/katex]

- Units: [katex] \text{kg·m}^2\text{/s} [/katex] (the same as [katex] \text{N·m·s} [/katex]).
- It has a direction, like [katex] \omega [/katex]: counterclockwise positive, clockwise negative.
- Two objects spinning opposite ways can have angular momenta that cancel.

A [katex] 4.0\,\text{kg} [/katex] solid disk of radius [katex] 0.50\,\text{m} [/katex] spins at [katex] 6.0\,\text{rad/s} [/katex]. [katex] I = \tfrac{1}{2}(4.0)(0.50)^2 = 0.50 \text{ kg·m}^2 [/katex], so [katex] L = (0.50)(6.0) = 3.0 \text{ kg·m}^2\text{/s} [/katex]. A hoop of the same mass, radius and spin has twice the [katex] I [/katex], so twice the [katex] L [/katex].

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| **Linear** | **Rotational** |
| --- | --- |
| Momentum [katex] p = mv [/katex] | Angular momentum [katex] L = I\omega [/katex] |
| Impulse [katex] J = F\Delta t [/katex] | Angular impulse [katex] \tau\Delta t [/katex] |
| [katex] F\Delta t = \Delta p [/katex] | [katex] \tau\Delta t = \Delta L [/katex] |
| Slope of [katex] p [/katex] vs. [katex] t [/katex] = net force | Slope of [katex] L [/katex] vs. [katex] t [/katex] = net torque |

### Angular Momentum of an Object Moving in a Line

An object does not have to spin, or even move in a circle, to have angular momentum about a point. A puck sliding in a straight line past a post has angular momentum about the post:

[katex] L = mvr_\perp [/katex]

\(v\)

\(m\)

\(v\)

\(m\)

\(\vec r\)

\(r_{\perp}\)

\(L = m v r_{\perp}\)

*\(r_\perp\) is the perpendicular distance from the point to the line the object moves along. On a straight path at constant speed, \(r_\perp\) never changes, so neither does \(L\).*

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Here [katex] r_\perp [/katex] is the closest distance between the point and the object’s path, the same lever-arm idea as torque in Unit 5.2. An object moving straight toward or away from the point has [katex] L = 0 [/katex] about it. This is the piece you need for collisions where a moving object hits something that spins (Unit 6.3).

### Angular Impulse

A net torque changes angular momentum, just as a net force changes momentum. The change equals the torque times the time it acts, the **angular impulse**:

[katex] \tau_{net}\Delta t = \Delta L = I\omega_f – I\omega_i [/katex]

If the torque changes with time, the angular impulse is the **area under the torque vs. time graph**. And since [katex] \tau_{net} = \frac{\Delta L}{\Delta t} [/katex], the **slope of an angular momentum vs. time graph** is the net torque. With no net torque, [katex] L [/katex] stays constant: that is the next lesson.

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A force can change an object’s momentum without changing its angular momentum: a push straight through the center of mass moves the object but does not spin it. A push off-center does both.

Every angular momentum question on the AP exam is one of the four types below.

| **Problem type** | **The clue** | **The setup** |
| --- | --- | --- |
| 1. Angular momentum of a moving point | an object moving in a line “relative to point P” | [katex] L = mvr_\perp [/katex], with the perpendicular distance to the path |
| 2. Angular impulse | a torque acts “for [katex] 3.0\,\text{s} [/katex]”, a brake stops a wheel | [katex] \tau\Delta t = I\omega_f – I\omega_i [/katex] |
| 3. Torque vs. time graphs | a graph of [katex] \tau [/katex] against [katex] t [/katex] | Area = [katex] \Delta L [/katex] |
| 4. Angular momentum vs. time graphs | a graph of [katex] L [/katex] against [katex] t [/katex] | Slope = net torque |

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### Problem Type 1: Angular momentum of a moving point

**How to spot it:** An object moves in a straight line (or passes by), and the question asks for its angular momentum about a point that is not on its path, or ranks it for several points.

**Setup:** Find [katex] r_\perp [/katex], the perpendicular distance from the point to the line of motion, not the straight-line distance to the object. Then [katex] L = mvr_\perp [/katex]. Give it a sign: + if the object goes counterclockwise around the point.

A [katex] 0.20\,\text{kg} [/katex] puck slides at [katex] 3.0\,\text{m/s} [/katex] in a straight line. Its path passes [katex] 0.50\,\text{m} [/katex] from a post at its closest. Find its angular momentum about the post when it is closest, and again when it is [katex] 2.0\,\text{m} [/katex] farther along.

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[katex] L = (0.20)(3.0)(0.50) = 0.30 \text{ kg·m}^2\text{/s} [/katex] at the closest point. Farther along, the puck is farther from the post, but [katex] r_\perp [/katex] is still [katex] 0.50\,\text{m} [/katex], so [katex] L [/katex] is still [katex] 0.30\,\text{kg·m}^2\text{/s} [/katex]. No torque acts on the puck about the post, so its angular momentum cannot change.

**AP trap:** Using the distance from the point to the object instead of [katex] r_\perp [/katex]. Only the part of the position perpendicular to the velocity counts, exactly like the lever arm of a force.

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Question 1

An object of mass m moves in the xy -plane with a constant velocity of magnitude v in the positive x -direction along the line y = 0. Three fixed…

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Question 2

A drone of mass m is flying in a straight horizontal line at a constant speed v and a constant altitude H above the ground. A stationary observer is…

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Question 3

Car A of mass m_A is moving to the east along a straight road. Car B of mass m_B is moving to the north along another straight road. At the instant t…

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### Problem Type 2: Angular impulse

**How to spot it:** A torque acts for a known time (“for [katex] 4.0\,\text{s} [/katex]”), a brake or friction stops a spinning object, or a question asks how long a spin-up takes.

**Setup:** Find the net torque (a brake or friction torque is negative). Then [katex] \tau_{net}\Delta t = I\omega_f – I\omega_i [/katex]. Use it whenever time is given or asked for; use energy (Unit 6.1) when an angle is.

A grinding wheel with [katex] I = 0.40 \text{ kg·m}^2 [/katex] spins at [katex] 30\,\text{rad/s} [/katex]. When the motor is switched off, a friction torque of [katex] 2.0\,\text{N·m} [/katex] acts at the axle. How long does the wheel take to stop?

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[katex] \Delta L = 0 – (0.40)(30) = -12 \text{ kg·m}^2\text{/s} [/katex]. [katex] -(2.0)\Delta t = -12 [/katex], so [katex] \Delta t = 6.0 \text{ s} [/katex]. Same answer as [katex] \alpha = \frac{\tau}{I} [/katex] plus kinematics, in one line.

**AP trap:** Mixing up linear and angular impulse. A force through the center of mass gives a linear impulse but no angular impulse about the center. The same force off-center gives the same linear impulse and an angular impulse as well.

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Question 4

An object's angular momentum changes by 10 kgcdotm²/s in 2.0 s. What magnitude average torque acted on this object?

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Question 5

A uniform rod of length L is at rest on a frictionless horizontal surface. A student conducts two trials where a horizontal force of constant…

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Question 6

An object is experiencing a nonzero net force. Which of the following statements is most accurate?

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Question 7

A disk of known radius and rotational inertia can rotate without friction in a horizontal plane around its fixed central axis. The disk has a cord of…

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### Problem Type 3: Torque vs. time graphs

**How to spot it:** A graph of net torque against time, followed by “angular momentum at t = …”, “angular speed at …” or “when is L greatest”.

**Setup:** The area under the [katex] \tau [/katex] vs. [katex] t [/katex] graph is [katex] \Delta L [/katex]. Area below the axis counts as negative. [katex] L [/katex] keeps growing as long as the torque is positive, even while the torque shrinks.

Net torque on a wheel

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*Area = \(\Delta L\): a triangle \(\tfrac{1}{2}(2)(4) = 4\) plus a rectangle \((3)(4) = 12\), so \(\Delta L = 16\,\text{kg·m}^2\text{/s}\).*

A wheel with [katex] I = 2.0 \text{ kg·m}^2 [/katex] starts from rest. The net torque on it is shown above. How fast is it spinning at [katex] 5\,\text{s} [/katex]?

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Area: [katex] \tfrac{1}{2}(2)(4) + (3)(4) = 16 \text{ kg·m}^2\text{/s} [/katex]. Starting from rest, [katex] L = 16 [/katex], so [katex] \omega = \frac{L}{I} = \frac{16}{2.0} = 8.0 \text{ rad/s} [/katex].

**AP trap:** Picking the time of the largest torque as the time of the largest angular momentum. [katex] L [/katex] peaks where the torque crosses zero, the moment it stops adding and starts taking away.

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Question 8

A wheel with rotational inertia I = 2.0 kg*m^2 is mounted on a fixed axle with negligible friction and is initially at rest. A net torque tau is…

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Question 9

A space probe with a fixed rotational inertia about its central axis is rotating in deep space. Thrusters exert a variable net torque τₙₑₜ on the…

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Question 10

A net torque is applied to the edge of a spinning object as it rotates about its internal axis. The table shows the net torque exerted on the object…

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### Problem Type 4: Angular momentum vs. time graphs

**How to spot it:** A graph of [katex] L [/katex] (or [katex] \omega [/katex]) against time, and the question asks about the torque: when it is largest, when it is zero, or how two torques compare.

**Setup:** The slope is the net torque: [katex] \tau_{net} = \frac{\Delta L}{\Delta t} [/katex]. Steep means a large torque, flat means zero torque, and a straight line means a constant torque.

A flywheel’s angular momentum drops in a straight line from [katex] 24\,\text{kg·m}^2\text{/s} [/katex] to zero in [katex] 8.0\,\text{s} [/katex]. What does the graph tell you about the brake?

Slope: [katex] \frac{0 – 24}{8.0} = -3.0 \text{ N·m} [/katex]. The brake applies a constant torque of [katex] 3.0\,\text{N·m} [/katex] against the rotation. A straight line means the torque never changed.

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**AP trap:** Reading the height of the graph instead of its slope. Where [katex] L [/katex] is largest and flat, the net torque is zero. Where [katex] L [/katex] crosses zero, the torque can be large.

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Question 11

A rigid disk rotates about a fixed, frictionless axle perpendicular to the disk through its center. External forces exert torques on the disk along…

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Question 12

Two disks, A and B, each experience a net external torque that varies over an interval of 5 s. Disk B has a rotational inertia that is twice that of…

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Question 13

During the experiment, students collect data about the angular momentum of a rigid, uniform spinning wheel about an axle as a function of time, which…

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Question 14

A flywheel with rotational inertia I rotates about a fixed axle with an initial angular momentum L₀. At time t = 0, a brake applies a constant…

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### Lesson 6.3 Preview

In the [next lesson](https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/) the net torque is zero, so angular momentum is conserved. That one idea explains spinning skaters, collapsing stars and clay that sticks to a spinning disk. ## Attribution Guidelines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarizing, or training on this content, please credit Nerd Notes and link to https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/
