---
title: "Unit 6.3 | Conservation of Angular Momentum"
description: "Learn conservation of angular momentum in AP Physics 1: spinning skaters, objects landing on turntables, clay hitting a pivoted rod, and collide then swing."
featured_image: "https://nerd-notes.com/wp-content/uploads/2023/06/12672.jpg"
url: "https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/"
date_modified: "2026-10-31T09:33:52+00:00"
---

# Unit 6.3 | Conservation of Angular Momentum

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Take away the torque and angular momentum cannot change. That one idea explains spinning skaters, collapsing stars and clay that sticks to a spinning disk.

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### Unit 6 Breakdown

You are on Lesson 3 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: 

- When angular momentum is conserved: zero net external torque on the system
- Why pulling mass inward speeds up a spin, [katex] I_i\omega_i = I_f\omega_f [/katex], and where the extra kinetic energy comes from
- Rotational collisions: objects that land on, stick to or hit something that rotates
- The 4 problem types: changing shape, landing on a turntable, hitting a pivoted object, and collide-then-swing

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### Introduction

This lesson covers **conservation of angular momentum**, the rotational version of the most useful idea in Unit 4.

In [Unit 4.2](https://nerd-notes.com/unit-4-2-conservation-of-momentum/), momentum was conserved when no net external force acted on a system. In [Unit 6.2](https://nerd-notes.com/unit-6-2-angular-momentum-and-angular-impulse/) you saw that a net torque changes angular momentum. Take the torque away and angular momentum cannot change. That is the whole lesson, and it explains a surprising amount.

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### Conservation of Angular Momentum

If the net external torque on a system is zero, the total angular momentum of the system stays constant:

[katex] \sum L_i = \sum L_f [/katex]

- **Pick the system** so the interaction is internal: the skater and her arms, the disk and the clay, the student and the stool.
- **Check the torques** about the axis. A frictionless axle pushes on the system but exerts no torque about itself, so angular momentum about the axle is conserved even though linear momentum may not be.
- **Add the angular momenta** with signs: spinning objects [katex] I\omega [/katex], moving objects [katex] mvr_\perp [/katex].

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Angular momentum and kinetic energy are different stories. Angular momentum is conserved whenever the external torque is zero. Kinetic energy can go up (a skater pulling in) or down (clay sticking) at the same time.

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

| **Problem type** | **The clue** | **The setup** |
| --- | --- | --- |
| 1. Changing shape | a skater pulls in her arms, a star collapses, a student pulls in weights | [katex] I_i\omega_i = I_f\omega_f [/katex] |
| 2. Landing on a turntable | something dropped on, or a person stepping onto, a rotating platform | [katex] I_1\omega_1 = (I_1 + I_2)\omega_f [/katex] |
| 3. Hitting a pivoted object | a ball or clay strikes a rod, door or disk that can rotate | [katex] mvr_\perp + I\omega_i = I_{total}\,\omega_f [/katex] |
| 4. Collide, then swing | after the hit the rod “swings up to an angle” | Angular momentum for the hit, then energy for the swing |

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### Problem Type 1: Changing shape

**How to spot it:** One object changes how its mass is spread out while spinning freely: a skater or diver pulls in, a student on a stool moves weights, a star collapses, beads slide outward along a rod.

**Setup:** No external torque, so [katex] I_i\omega_i = I_f\omega_f [/katex]. Mass moved inward lowers [katex] I [/katex] and raises [katex] \omega [/katex]. Then, if asked, compare [katex] K = \frac{L^2}{2I} [/katex]: with [katex] L [/katex] fixed, the energy goes as [katex] \frac{1}{I} [/katex].

\(\omega_i\)

\(I_i \text{ large}, \ \omega_i \text{ small}\)

\(\omega_f\)

\(I_f \text{ small}, \ \omega_f \text{ large}\)

\(L = I_i\omega_i = I_f\omega_f\)

*Same angular momentum in both pictures. Pulling the masses in lowers \(I\), so \(\omega\) must rise.*

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A skater spins at [katex] 2.0\,\text{rev/s} [/katex] with her arms out, where her rotational inertia is [katex] 3.6\,\text{kg·m}^2 [/katex]. She pulls her arms in, lowering it to [katex] 1.2\,\text{kg·m}^2 [/katex]. Find her new spin rate and how her kinetic energy changes.

[katex] (3.6)(2.0) = (1.2)\omega_f [/katex], so [katex] \omega_f = 6.0 \text{ rev/s} [/katex]. (Revolutions are fine here: both sides use the same unit.) Her kinetic energy is [katex] \frac{L^2}{2I} [/katex] with [katex] L [/katex] fixed, so it triples.

Where does the extra energy come from? Her muscles. Her hands move in a circle, so they need an inward force just to stay on it. To bring them closer she pulls inward while they move inward, which is positive work, and that work becomes kinetic energy.

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**AP trap:** Conserving kinetic energy instead of angular momentum. [katex] L [/katex] stays the same, [katex] K [/katex] does not. If [katex] I [/katex] drops to a third, [katex] \omega [/katex] triples and [katex] K [/katex] triples too.

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Question 15

An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about…

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Question 16

A figure skater is spinning on a horizontal sheet of ice with negligible friction. Initially, the skater spins with their arms extended horizontally…

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Question 17

A figure skater is spinning on a horizontal sheet of ice with negligible friction. Initially, the skater has rotational inertia I₀ and is spinning…

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Question 18

A light horizontal rod of negligible mass is mounted on a vertical, frictionless axle and rotates freely with an initial angular speed. Two identical…

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### Problem Type 2: Landing on a turntable

**How to spot it:** Something with no spin lands on, or is placed on, a rotating platform or disk and moves with it: clay dropped on a turntable, a child jumping onto a merry-go-round, a disk dropped onto a spinning disk.

**Setup:** It is a perfectly inelastic collision, rotated: [katex] I_1\omega_1 = (I_1 + I_2)\omega_f [/katex]. A small object at radius [katex] r [/katex] adds [katex] mr^2 [/katex]. Kinetic energy is lost.

A turntable with [katex] I = 0.020 \text{ kg·m}^2 [/katex] spins freely at [katex] 3.0\,\text{rad/s} [/katex]. A [katex] 0.10\,\text{kg} [/katex] lump of clay is dropped straight down onto it and sticks [katex] 0.20\,\text{m} [/katex] from the center. Find the new angular speed and the kinetic energy lost.

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The clay adds [katex] mr^2 = (0.10)(0.20)^2 = 0.0040 \text{ kg·m}^2 [/katex]. [katex] (0.020)(3.0) = (0.024)\omega_f [/katex], so [katex] \omega_f = 2.5 \text{ rad/s} [/katex].

Energy: [katex] K_i = \tfrac{1}{2}(0.020)(3.0)^2 = 0.090 \text{ J} [/katex] and [katex] K_f = \tfrac{1}{2}(0.024)(2.5)^2 = 0.075 \text{ J} [/katex]. Friction between the clay and the disk turned [katex] 0.015\,\text{J} [/katex] into thermal energy.

**AP trap:** Saying the turntable alone keeps its angular momentum. The clay exerts a friction torque on the turntable while it speeds up, so the turntable loses angular momentum to the clay. Only the total, turntable plus clay, is conserved.

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Question 19

A uniform horizontal disk of mass M and radius R is rotating with an initial angular velocity ω₀ about a fixed vertical axis through its center. The…

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Question 20

A horizontal, uniform disk of mass M and radius R rotates freely about a vertical axle through its center with an initial angular velocity ω₀…

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Question 21

A platform is initially rotating on smooth ice with negligible friction, as shown above. A stationary disk is dropped directly onto the center of the…

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Question 22

A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kgcdotm² and is rotating at 10.0 rev/min about a frictionless…

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### Problem Type 3: Hitting a pivoted object

**How to spot it:** A ball, bullet or lump of clay moving in a straight line hits a rod, door or disk that can rotate about a fixed pivot, and sticks or bounces.

**Setup:** Use angular momentum about the pivot, so the pivot force drops out. Before: [katex] mvr_\perp [/katex] for the moving object (plus [katex] I\omega [/katex] if the target already spins). After: [katex] I_{total}\,\omega_f [/katex]. Linear momentum is usually not conserved, because the pivot pushes during the hit.

\(v\)

\(m\)

\(L\)

\(L_i = m v L\)

\(\omega\)

\(L_f = \left(\tfrac{1}{3}ML^2 + mL^2\right)\omega\)

*Before the hit, the ball’s angular momentum about the pivot is \(mvL\). After, the rod and ball turn together with \(I = \tfrac{1}{3}ML^2 + mL^2\).*

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A uniform [katex] 0.60\,\text{kg} [/katex] rod, [katex] 1.2\,\text{m} [/katex] long, hangs from a frictionless pivot at its top. A [katex] 0.10\,\text{kg} [/katex] lump of clay moving horizontally at [katex] 4.0\,\text{m/s} [/katex] hits the bottom end and sticks. How fast does the rod turn just after the hit?

Before: [katex] L = mvr_\perp = (0.10)(4.0)(1.2) = 0.48 \text{ kg·m}^2\text{/s} [/katex]. After: [katex] I = \tfrac{1}{3}(0.60)(1.2)^2 + (0.10)(1.2)^2 = 0.288 + 0.144 = 0.432 \text{ kg·m}^2 [/katex]. So [katex] \omega_f = \frac{0.48}{0.432} \approx 1.1 \text{ rad/s} [/katex].

Linear momentum is not conserved here. Before the hit it is [katex] (0.10)(4.0) = 0.40 \text{ kg·m/s} [/katex]. Just after, the rod’s center of mass and the clay move at [katex] 0.60\omega [/katex] and [katex] 1.2\omega [/katex], for a total of [katex] (0.60)(0.60)(1.11) + (0.10)(1.2)(1.11) \approx 0.53 \text{ kg·m/s} [/katex]. The pivot pushed forward on the rod during the hit.

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**AP trap:** Using conservation of linear momentum, or [katex] v = r\omega [/katex] with the rod’s center. When something is pinned at a pivot, conserve angular momentum about that pivot.

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Question 23

A small ball of mass m moves horizontally with speed v toward a uniform thin rod of mass M and length L that is hanging vertically from a…

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Question 24

A small ball of mass m slides across a horizontal, frictionless surface with a velocity v toward a uniform rod of mass M and length L. The rod is…

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Question 25

A uniform thin rod of length L and mass M lies on a frictionless horizontal tabletop and is constrained to rotate about a fixed vertical axle through…

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Question 26

A small ball of mass m travels horizontally with speed v on a frictionless surface. It strikes the bottom end of a uniform rod of mass M and length L…

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### Problem Type 4: Collide, then swing

**How to spot it:** After the hit, the rod swings up to some angle or height. Or before it, the rod swings down and then hits something.

**Setup:** Split it at the collision, exactly like [Unit 4.4](https://nerd-notes.com/unit-4-4-momentum-and-energy-together/). Angular momentum through the hit. Energy before and after it, with the rotational kinetic energy [katex] \tfrac{1}{2}I\omega^2 [/katex] and the rise of each center of mass.

Continue the problem above. How high does the rod swing?

Just after the hit: [katex] K = \tfrac{1}{2}(0.432)(1.11)^2 \approx 0.267 \text{ J} [/katex]. If the rod swings to an angle [katex] \theta [/katex], its center rises [katex] 0.60(1 – \cos\theta) [/katex] and the clay rises [katex] 1.2(1 – \cos\theta) [/katex]:

[katex] 0.267 = (9.8)(1 – \cos\theta)\left[(0.60)(0.60) + (0.10)(1.2)\right] [/katex], so [katex] 1 – \cos\theta \approx 0.057 [/katex] and [katex] \theta \approx 19^\circ [/katex].

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The clay brought [katex] 0.80\,\text{J} [/katex] of kinetic energy, and only [katex] 0.267\,\text{J} [/katex] survived the sticky collision. Using energy across the hit would give a much bigger swing.

▶ **Watch:** a worked collision and angular momentum problem (starts at 14:17).

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**AP trap:** Using energy conservation across the collision. A sticky hit loses kinetic energy. Use angular momentum for the hit, then energy for the swing.

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Question 27

A small projectile of mass m is moving horizontally with speed v₀ when it strikes the bottom end of a uniform thin rod of mass M and length L. The…

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Question 28

A uniform rod of mass M and length L is pivoted at one end and held in a horizontal position, as shown in the figure. The rod is released from rest…

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Applying The Conservation of Angular Momentum

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Quiz Lab

Angular Momentum and Its Conservation

Spinning skaters and turntables. Know when angular momentum is conserved, and what happens to the energy.

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### Lesson 6.4 Preview

In the [last lesson](https://nerd-notes.com/unit-6-4-rolling-and-orbiting-satellites/) of the unit, objects move and spin at once. A ball rolling down a ramp carries both kinds of kinetic energy, and a planet orbiting a star keeps both its energy and its angular momentum. ## 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-6-3-conservation-of-angular-momentum/. **Publication:** Nerd Notes **Original URL:** https://nerd-notes.com/unit-6-3-conservation-of-angular-momentum/
