Angular Momentum

AP Physics 1· AP Physics 1 CED — Rotational Motion· 14 min read

1. What Is Angular Momentum?★★☆☆☆⏱ 3 min

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.

📘 Definition

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.

2. Calculating Angular Momentum★★★☆☆⏱ 4 min

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 $r$ (from the axis to the mass) and linear momentum $m v$ :

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

The magnitude simplifies to:

L = rmv\sin\theta

where $θ$ is the angle between $r$ and $v$ . The term $r sin θ$ is the perpendicular distance from the axis to the line of the mass's velocity. For a point mass moving in a circular orbit, $θ = 9 0^{\circ}$ so $sin θ = 1$ , leading to $L = m v r = I ω$ . 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 $r sin θ$ 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 $r sin θ = 1.5$ 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 $sin θ = 1$ , 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.

3. Conservation of Angular Momentum★★★☆☆⏱ 4 min

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 ( $τ_{n e t, e x t} = 0$ ). 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 $I_{d i s k} = \frac{1}{2} M R^{2}$ :
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 $ω_{f}$ :
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.

4. Angular Impulse-Momentum Theorem★★★☆☆⏱ 3 min

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 $Δ t$ :
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 $Δ t$ , you almost certainly mixed up the sign of your torque. Always assign the opposite sign to torque that opposes rotation.

5. AP-Style Concept Check★★★★☆⏱ 4 min

✓ Quick check

Test your understanding with these original AP-style problems:

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
Reveal answer
3 —
No net external torque acts on the skater, so angular momentum is conserved: $I_{i} ω_{i} = I_{f} ω_{f}$ gives $ω_{f} = 8$ 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.

6. Common Pitfalls

Wrong move:

Calculating angular momentum of a point mass as $L = m v r$ where $r$ is the straight-line distance from the axis to the mass, regardless of velocity direction.

Why:

Students memorize the circular orbit special case and forget the $sin θ$ term is required for all other cases.

Correct move:

Always use $L = (r sin θ) m v$ , where $r sin θ$ is the perpendicular distance from the axis to the mass's velocity vector.

Wrong move:

Applying conservation of angular momentum to a system with a net external torque.

Why:

Students confuse angular momentum conservation with general conservation laws and automatically use it without checking the condition.

Correct move:

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 move:

Treating a small mass at the edge of a rotating object as a solid disk when calculating moment of inertia.

Why:

Students mix up moment of inertia formulas for different shapes.

Correct move:

Use $I = m r^{2}$ for any small mass located a fixed distance from the axis, regardless of the shape of the mass itself.

Wrong move:

Using conservation of kinetic energy to solve perfectly inelastic rotational collisions where objects stick together.

Why:

Students associate collision problems with elastic collisions and automatically assume energy is conserved.

Correct move:

Kinetic energy is only conserved in elastic rotational collisions. For inelastic collisions, use only conservation of angular momentum (and linear momentum if applicable).

Wrong move:

Forgetting that the direction of angular momentum matters when adding angular momentum for a system of multiple objects.

Why:

Students treat angular momentum as a scalar and add magnitudes regardless of rotation direction.

Correct move:

Assign opposite signs to clockwise and counterclockwise rotation, then add the signed values to get total angular momentum.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Angular momentum (point mass)	$L = r m v sin θ$	$θ$ = angle between position vector $r$ (from axis) and velocity $v$
Angular momentum (rigid body)	$L = I ω$	For rigid bodies rotating about a fixed axis
Angular Impulse	$J_{θ} = τ_{n e t} Δ t$	For constant net torque
Angular Impulse-Momentum Theorem	$Δ L = τ_{n e t} Δ t$	Use when net external torque is non-zero
Conservation of Angular Momentum	$I_{i} ω_{i} = I_{f} ω_{f}$	Only applies when net external torque = 0
Rotational Kinetic Energy	$K = \frac{1}{2} I ω^{2}$	Not conserved in inelastic collisions, even if $L$ is
Moment of Inertia (point mass)	$I = m r^{2}$	For small masses at distance $r$ from axis
Moment of Inertia (solid disk)	$I = \frac{1}{2} M R^{2}$	For solid disks/cylinders rotating about central axis
Moment of Inertia (rod about center)	$I = \frac{1}{12} M L^{2}$	For uniform rod rotating about its center of mass

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Conservation on rotating merry-go-round
2022 · 1
Angular impulse stopping turbine

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.