Angular Momentum — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of angular momentum for point masses and extended rigid bodies, conservation of angular momentum, the angular impulse-momentum theorem, right-hand rule for direction, and problem-solving for rotational collisions and changing moment of inertia systems.

You should already know: Rotational kinematics and moment of inertia calculations, linear momentum and its conservation, torque and Newton's second law for rotation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. 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 CED, angular momentum is a core topic within Unit 6 Rotational Motion, accounting for roughly 4-6% of the total AP exam score, with questions appearing on both multiple-choice (MCQ) and free-response (FRQ) sections. It is frequently tested as a stand-alone concept or combined with linear momentum, energy, and torque to solve problems involving changing rotational systems or collisions between moving and rotating objects. Unlike energy, which can be lost to non-conservative work, or linear momentum which is only conserved if net external force is zero, angular momentum is conserved when net external torque on a system is zero. This makes it an extremely powerful tool for solving problems that would be impossible to solve with force or energy analysis alone.

2. 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 $m\vec{v}$: $\vec{L}=\vec{r}\times m\vec{v}$. 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 around the axis, $\theta =90^{\circ }$ so $\sin \theta =1$, giving $L=mvr=m{r}^2\omega =I\omega$. For an extended rigid body rotating about a fixed axis, the total angular momentum is the sum of angular momentum of all individual point masses in the body, leading to the simple formula: $$L=I\omega$$ 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 vector 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: $L=(r\sin \theta )mv$, where $r\sin \theta$ is the perpendicular distance from the axis to the mass’s path.
2. 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.
3. Substitute values: $L=(1.5\text{m})(2.0\text{kg})(3.0\text{m/s})=9.0{\text{kgcdotpm}}^2/\text{s}$.
4. Confirm that the velocity is perpendicular to the perpendicular position vector, so $\sin \theta =1$ and 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

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 (e.g., a figure skater pulling in their arms, a person walking on a rotating merry-go-round) or for inelastic collisions between rotating objects, where kinetic energy is not conserved but angular momentum is. 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: only the merry-go-round contributes. The moment of inertia of a solid disk is $I_{disk}=\frac{1}{2}M{R}^2=0.5(100\text{kg})(2.0\text{m}{)}^2=200{\text{kgcdotpm}}^2$. Initial angular momentum $L_i=I_i{\omega }_i=200(0.5)=100{\text{kgcdotpm}}^2/\text{s}$.
3. Calculate final moment of inertia: add the person, treated as a point mass at the edge: $I_f=I_{disk}+m{R}^2=200+(50)(2.0{)}^2=400{\text{kgcdotpm}}^2$.
4. Set $L_i=L_f=I_f{\omega }_f$, so ${\omega }_f=\frac{100}{400}=0.25$ 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 a result that is twice the correct value.

4. 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. It is the rotational equivalent of the impulse-momentum theorem for linear motion, and follows the same logic.

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: $\Delta L={\tau }_{net}\Delta t$.
2. Define initial rotation direction as positive, so frictional torque (which opposes rotation) is negative. The final angular momentum is 0, so $\Delta L=L_f-L_i=0-40=-40$ kg·m²/s.
3. Rearrange to solve for $\Delta t$: $\Delta t=\frac{\Delta L}{{\tau }_{net}}=\frac{-40}{-8}=5$ s.
4. 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.

Common Pitfalls (and how to avoid them)

• Wrong move: 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: Students memorize the circular orbit special case and forget the $\sin \theta$ term is required for all other cases. Correct move: 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 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.

Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

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? A) It decreases by a factor of 4 B) It stays the same C) It increases by a factor of 2 D) It increases by a factor of 4

Worked Solution: There is no net external torque on the skater, so angular momentum is conserved. Applying conservation: $I_i{\omega }_i=I_f{\omega }_f$, so $(4)(2)=(1){\omega }_f$, giving ${\omega }_f=8$ rad/s. Rotational kinetic energy is $K=\frac{1}{2}I{\omega }^2$. Initial $K_i=0.5(4)(2{)}^2=8$ J, final $K_f=0.5(1)(8{)}^2=32$ J. The ratio of final to initial kinetic energy is 4, so kinetic energy increases by a factor of 4 (the extra energy comes from work the skater does pulling her arms inward). The correct answer is D.

Question 2 (Free Response)

A child builds a toy seesaw that pivots without friction at its center. The seesaw is a uniform rod of mass 2.0 kg and length 3.0 m, initially at rest. A 1.0 kg ball of clay is thrown horizontally at 6.0 m/s perpendicular to the end of the seesaw, where it sticks and rotates with the seesaw after collision. (a) Explain why angular momentum is conserved about the pivot axis during the collision, but linear momentum is not. (b) Calculate the angular velocity of the seesaw immediately after the collision. (c) If the clay had hit the seesaw 1.0 m from the center instead of at the end, would the final angular velocity be larger, smaller, or the same? Justify your answer.

Worked Solution: (a) Angular momentum is conserved because the only external torque during the collision comes from the force at the pivot, which acts at $r=0$ from the axis, so its torque is zero, meaning net external torque is zero. Linear momentum is not conserved because the pivot exerts an external force on the seesaw during the collision, so net external force is non-zero. (b) Initial angular momentum comes only from the clay: $L_i=rmv=(1.5\text{m})(1.0\text{kg})(6.0\text{m/s})=9.0{\text{kgcdotpm}}^2/\text{s}$. Final moment of inertia: $I=I_{rod}+I_{clay}=\frac{1}{12}M{L}^2+m{r}^2=\frac{1}{12}(2.0)(3.0{)}^2+(1.0)(1.5{)}^2=1.5+2.25=3.75{\text{kgcdotpm}}^2$. By conservation of angular momentum: ${\omega }_f=\frac{L_i}{I_f}=\frac{9.0}{3.75}=2.4$ rad/s. (c) The final angular velocity would be smaller. Initial angular momentum is proportional to $r$, the distance from the pivot where the clay hits, so decreasing $r$ from 1.5 m to 1.0 m decreases $L_i$. The total moment of inertia decreases by a smaller factor than $L_i$, so ${\omega }_f=L_i/I_f$ decreases overall.

Question 3 (Application / Real-World Style)

A utility-scale wind turbine has a total moment of inertia of $3.2\times 10^6$ kg·m² about its rotation axis. When the wind stops blowing, the turbine is rotating at 0.4 rad/s, and a constant braking torque of 20,000 N·m is applied to bring it to a complete stop for maintenance. How long does it take for the turbine to stop, and what is the total angular impulse applied by the brake?

Worked Solution: Use the angular impulse-momentum theorem: $\Delta L={\tau }_{net}\Delta t$. Initial angular momentum $L_i=I{\omega }_i=(3.2\times 10^6)(0.4)=1.28\times 10^6$ kg·m²/s. Final angular momentum $L_f=0$, so $\Delta L=-1.28\times 10^6$ kg·m²/s. Braking torque is ${\tau }_{net}=-20,000$ N·m. Solving for $\Delta t$: $\Delta t=\frac{-1.28\times 10^6}{-2\times 10^4}=64$ s. The magnitude of the total angular impulse equals the change in angular momentum, so $J_{\theta }=1.28\times 10^6$ N·m·s. In context, this means the brake takes just over one minute to bring the large turbine to a stop, which matches typical real-world maintenance timelines for utility-scale turbines.

Quick Reference Cheatsheet

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 Torque and Rotational Equilibrium Rolling Motion Without Slipping Conservation of Linear Momentum