# Angular Momentum

> AP Physics 1 · Unit 6: Rotational Motion
> Source: https://www.owlsprep.com/study/ap-physics-1-u6-angular-momentum/

This module covers core angular momentum concepts for AP Physics 1: definition for point masses and rigid bodies, the conservation law, angular impulse-momentum theorem, and problem-solving for collisions and changing inertia systems.

**Prerequisites:** [Rotational kinematics and moment of inertia calculations](https://www.owlsprep.com/study/ap-physics-1-u6-rotational-kinematics/); [Linear momentum and its conservation](https://www.owlsprep.com/study/ap-physics-1-u5-conservation-of-linear-momentum/); [Torque and Newton's second law for rotation](https://www.owlsprep.com/study/ap-physics-1-u6-torque-rotational-equilibrium/)

## Learning objectives

- Define angular momentum for point masses and extended rigid bodies
- Apply the law of conservation of angular momentum to solve problems
- Use the angular impulse-momentum theorem for systems with external torque
- Solve AP-style problems involving rotational collisions and changing moment of inertia

## What Is Angular Momentum?

Angular momentum is the rotational analog of linear momentum, a vector quantity that quantifies the total amount of rotational motion possessed by an object or system of objects. For the AP Physics 1 exam, it accounts for roughly 4-6% of your total score, appearing in both multiple-choice and free-response questions.

**Angular Momentum** — A vector rotational quantity analogous to linear momentum, conserved for any system with zero net external torque. It is used to analyze rotational motion, collisions, and changing systems.

*Example:* A spinning figure skater has non-zero angular momentum that changes when they pull in their arms.

> **info**
>
> Unlike energy (which can be lost to non-conservative work) or linear momentum (conserved only when net external force is zero), angular momentum is conserved when net external torque is zero, making it useful for problems that cannot be solved with other methods.

## Calculating Angular Momentum

Angular momentum is always defined relative to a specific axis of rotation; its value depends entirely on the axis you choose. For a point mass, angular momentum is defined via the cross product of the position vector $\vec{r}$ (from the axis to the mass) and linear momentum $m\vec{v}$:

$$vec{L} = vec{r} times mvec{v}$$

The magnitude simplifies to:

$$L = rmv\sin\theta$$

where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$. The term $r\sin\theta$ is the perpendicular distance from the axis to the line of the mass's velocity. For a point mass moving in a circular orbit, $\theta = 90^\circ$ so $\sin\theta = 1$, leading to $L = mvr = I\omega$. For an extended rigid body rotating about a fixed axis, the formula simplifies to:

$$L = I\omega$$

Direction is found via the right-hand rule: curl your fingers in the direction of rotation, and your thumb points in the direction of the angular momentum vector. For AP Physics 1, you only need to track sign (positive for counterclockwise, negative for clockwise) rather than full 3D components.

**Worked example:** A 2.0 kg point mass moves at 3.0 m/s in a straight line that passes 1.5 m from a fixed axis at point O. What is the magnitude of the angular momentum of the mass about O?

1. Start with the point mass angular momentum formula, where $r\sin\theta$ is the perpendicular distance from the axis to the mass's path:
2. $$L = (r\sin\theta)mv$$
3. The problem gives the perpendicular distance directly as 1.5 m, so $r\sin\theta = 1.5$ m regardless of where the mass is along its path.
4. Substitute values to solve for $L$:
5. $$L = (1.5\ \text{m})(2.0\ \text{kg})(3.0\ \text{m/s}) = 9.0\ \text{kg·m}^2/\text{s}$$
6. Confirm that the perpendicular distance gives $\sin\theta = 1$, so no correction is needed.

> **Exam tip:** Always define your axis of rotation before calculating angular momentum. If you change the axis, you change the value of $L$, so use the axis specified in the problem to avoid mistakes.

## Conservation of Angular Momentum

The law of conservation of angular momentum states that the total angular momentum of a system remains constant if and only if the net external torque acting on the system is zero ($\tau_{net,ext} = 0$). Mathematically, this is written as:

$$L_{initial} = L_{final} \\ I_i\omega_i = I_f\omega_f$$

Internal torques (torques between objects inside the system) cancel out due to Newton's third law, so they do not change the total angular momentum of the system. This makes conservation of angular momentum ideal for solving problems where the moment of inertia of a rotating system changes, or for inelastic collisions between rotating objects where kinetic energy is not conserved.

Note that angular momentum can be conserved even when linear momentum is not: if an axis exerts an external force that produces no torque (because it acts at $r=0$), linear momentum is not conserved but angular momentum is.

**Worked example:** A 100 kg merry-go-round is a solid disk with radius 2.0 m, rotating initially at 0.5 rad/s counterclockwise. A 50 kg person steps onto the edge of the merry-go-round from rest, with zero initial angular momentum about the merry-go-round's axis. What is the final angular velocity of the merry-go-round after the person steps on?

1. Check the conservation condition: the only external force at the axle acts at $r=0$, so net external torque is zero, and conservation of angular momentum applies.
2. Calculate initial angular momentum, which comes only from the merry-go-round. Moment of inertia of a solid disk is $I_{disk} = \frac{1}{2}MR^2$:
3. $$I_i = 0.5(100\ \text{kg})(2.0\ \text{m})^2 = 200\ \text{kg·m}^2 \\ L_i = I_i\omega_i = 200(0.5) = 100\ \text{kg·m}^2/\text{s}$$
4. Calculate final moment of inertia, adding the person treated as a point mass at the edge:
5. $$I_f = I_{disk} + mR^2 = 200 + (50)(2.0)^2 = 400\ \text{kg·m}^2$$
6. Set $L_i = L_f$ and solve for $\omega_f$:
7. $$\omega_f = \frac{100}{400} = 0.25\ \text{rad/s counterclockwise$$

> **Exam tip:** When calculating final moment of inertia after adding a mass to a rotating object, always include the original object's moment of inertia in your final total. Students often only add the new mass and get an incorrect result.

## Angular Impulse-Momentum Theorem

When a net external torque acts on a system, angular momentum is not conserved, and we use the angular impulse-momentum theorem to relate the change in angular momentum to the applied torque. Angular impulse is the rotational analog of linear impulse, defined for constant net torque as:

$$J_\theta = \tau_{net} \Delta t$$

The theorem states that the change in angular momentum equals the net angular impulse applied to the system:

$$\Delta L = L_f - L_i = \tau_{net} \Delta t$$

This theorem is used to solve problems involving stopping a rotating object with friction, speeding up a rotating object with a constant torque, or finding the torque required to produce a given change in rotation over time.

**Worked example:** A rotating wheel has initial angular momentum 40 kg·m²/s. A constant frictional torque of 8 N·m is applied at the axle, bringing the wheel to rest. How long does it take for the wheel to stop?

1. State the angular impulse-momentum theorem:
2. $$\Delta L = \tau_{net} \Delta t$$
3. Define initial rotation direction as positive, so frictional torque (which opposes rotation) is negative. Final angular momentum is 0, so:
4. $$\Delta L = L_f - L_i = 0 - 40 = -40\ \text{kg·m}^2/\text{s}$$
5. Rearrange to solve for $\Delta t$:
6. $$\Delta t = \frac{\Delta L}{\tau_{net}} = \frac{-40}{-8} = 5\ \text{s}$$
7. Confirm units: N·m = kg·m²/s², so (kg·m²/s) / (kg·m²/s²) = s, which matches the expected unit for time.

> **Exam tip:** If you get a negative time when solving for $\Delta t$, you almost certainly mixed up the sign of your torque. Always assign the opposite sign to torque that opposes rotation.

## AP-Style Concept Check

**Check your understanding**

Test your understanding with these original AP-style problems:

1. A figure skater is spinning at 2 rad/s with her arms extended, with a total moment of inertia of 4 kg·m². She pulls her arms in, reducing her moment of inertia to 1 kg·m². By what factor does her rotational kinetic energy change?

   - It decreases by a factor of 4
   - It stays the same
   - It increases by a factor of 2
   - It increases by a factor of 4

   *Answer:* It increases by a factor of 4

   *Why:* No net external torque acts on the skater, so angular momentum is conserved: $I_i\omega_i = I_f\omega_f$ gives $\omega_f = 8$ rad/s. Initial $K_i = 8$ J, final $K_f = 32$ J, so kinetic energy increases by a factor of 4. The extra energy comes from work the skater does pulling her arms inward.

## Common pitfalls

- **Wrong:** Calculating angular momentum of a point mass as $L = mvr$ where $r$ is the straight-line distance from the axis to the mass, regardless of velocity direction.
  - Why it fails: Students memorize the circular orbit special case and forget the $\sin\theta$ term is required for all other cases.
  - Correct: Always use $L = (r\sin\theta)mv$, where $r\sin\theta$ is the perpendicular distance from the axis to the mass's velocity vector.
- **Wrong:** Applying conservation of angular momentum to a system with a net external torque.
  - Why it fails: Students confuse angular momentum conservation with general conservation laws and automatically use it without checking the condition.
  - Correct: Before writing $L_i = L_f$, always check that net external torque on the system is zero. If not, use the angular impulse-momentum theorem instead.
- **Wrong:** Treating a small mass at the edge of a rotating object as a solid disk when calculating moment of inertia.
  - Why it fails: Students mix up moment of inertia formulas for different shapes.
  - Correct: Use $I = mr^2$ for any small mass located a fixed distance from the axis, regardless of the shape of the mass itself.
- **Wrong:** Using conservation of kinetic energy to solve perfectly inelastic rotational collisions where objects stick together.
  - Why it fails: Students associate collision problems with elastic collisions and automatically assume energy is conserved.
  - Correct: Kinetic energy is only conserved in elastic rotational collisions. For inelastic collisions, use only conservation of angular momentum (and linear momentum if applicable).
- **Wrong:** Forgetting that the direction of angular momentum matters when adding angular momentum for a system of multiple objects.
  - Why it fails: Students treat angular momentum as a scalar and add magnitudes regardless of rotation direction.
  - Correct: Assign opposite signs to clockwise and counterclockwise rotation, then add the signed values to get total angular momentum.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Angular momentum (point mass) | $L = rmv\sin\theta$ | $\theta$ = angle between position vector $\vec{r}$ (from axis) and velocity $\vec{v}$ |
| Angular momentum (rigid body) | $L = I\omega$ | For rigid bodies rotating about a fixed axis |
| Angular Impulse | $J_\theta = \tau_{net} \Delta t$ | For constant net torque |
| Angular Impulse-Momentum Theorem | $\Delta L = \tau_{net} \Delta t$ | Use when net external torque is non-zero |
| Conservation of Angular Momentum | $I_i \omega_i = I_f \omega_f$ | Only applies when net external torque = 0 |
| Rotational Kinetic Energy | $K = \frac{1}{2}I\omega^2$ | Not conserved in inelastic collisions, even if $L$ is |
| Moment of Inertia (point mass) | $I = mr^2$ | For small masses at distance $r$ from axis |
| Moment of Inertia (solid disk) | $I = \frac{1}{2}MR^2$ | For solid disks/cylinders rotating about central axis |
| Moment of Inertia (rod about center) | $I = \frac{1}{12}ML^2$ | For uniform rod rotating about its center of mass |

## What's next

Mastering angular momentum is the final core concept in rotational motion for AP Physics 1, and it prepares you to solve mixed system problems that combine rotational and linear motion, which are common high-point-value questions on the AP exam. Next, you will apply angular momentum and conservation rules to problems involving rolling motion without slipping, which combines rotational motion of the wheel and linear translation of its center of mass. Without understanding how angular momentum changes with torque or is conserved for changing systems, you will not be able to analyze mixed rotational-translational motion or solve complex collision problems involving rotating objects. Angular momentum also reinforces your understanding of conservation laws, the unifying theme across all of AP Physics 1.

- [Rotational Kinematics](https://www.owlsprep.com/study/ap-physics-1-u6-rotational-kinematics/)
- [Conservation of Angular Momentum](https://www.owlsprep.com/study/ap-physics-1-u6-conservation-of-angular-momentum/)
- [Simple Harmonic Motion and Waves Overview](https://www.owlsprep.com/study/ap-physics-1-u7-overview/)

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