Angular momentum of a rigid system
You will be able to: Calculate signed angular momentum and distinguish it from kinetic energy.
Why can two wheels spinning at the same rate have different angular momentum?
A light wheel and a heavy flywheel can both turn at 3 rad/s. If the flywheel has more rotational inertia, it has more angular momentum about that axle. Their common turning rate alone does not tell the whole story.
A useful starting point: Read work from a torque–angle graph →
Words and symbols before equations
- Angular momentum L
- A rotational state quantity; for fixed-axis rigid rotation L=Iω, in kg·m²/s.
- Signed convention
- Here counterclockwise is positive and clockwise negative from a fixed viewing side.
- Rigid system
- Its mass distribution and shape stay fixed during the motion.
What this picture assumes
Separate coaxial wheels: I_A=2 kg·m², ω_A=+3 rad/s; I_B=1 kg·m². CCW positive from the same side. These are state comparisons, not a coupling simulation.
Connect the picture to the physics
Rotational inertia describes the mass distribution about the chosen axis; ω describes the turning rate. Their product L=Iω combines both. At fixed I, doubling ω doubles L, whereas it quadruples kinetic energy.
For multiple parts rotating about the same axis, add their signed angular momenta. Opposite rotations can cancel in total L even though both parts have positive kinetic energy. Energy and angular momentum answer different questions.
In this course we use a chosen positive turning sense for calculations; full three-dimensional angular-momentum directions are beyond the AP Physics 1 scope. Keep the same axis and viewpoint throughout a comparison. A torque changes L; L is not itself a torque.
A worked example, step by step
Wheel A has I=2 kg·m² and ω=+3 rad/s. Coaxial wheel B has I=1 kg·m² and ω=−4 rad/s. Find total L and total K.
- Use the same axis and counterclockwise-positive convention for both.
- L_A=2(3)=+6 and L_B=1(−4)=−4 kg·m²/s.
- L_total=+2 kg·m²/s.
- K_total=½(2)(3²)+½(1)(4²)=9+8=17 J. The energies add without cancellation.
Zero total L does not mean every part is at rest or that total kinetic energy is zero.
Two identical wheels counterrotate at equal speed. Is total K zero?
Compare with an explanation
No. Their L values cancel, but their positive kinetic energies add.
Predict. Change one thing. Explain.
Keep wheel A at I=2 kg·m² and ω=+3 rad/s. Adjust the second wheel’s angular velocity. Find zero total L, then inspect the positive total energy.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
L_A=6, L_B=-4, total L=2 kg·m²/s. Total kinetic energy=17 J.
Separate coaxial wheels: I_A=2 kg·m², ω_A=+3 rad/s; I_B=1 kg·m². CCW positive from the same side. These are state comparisons, not a coupling simulation.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant physical relationship or contact condition to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionTwo coaxial wheels each have I=1 kg·m². Their ω values are +4 and −4 rad/s. (a) Find each L. (b) Find total L. (c) Find total K. (d) Explain why zero total L does not establish rest.
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Compare with the answer and four-point rubric
- 1 point: +4 and −4 kg·m²/s.
- 1 point: 0 kg·m²/s.
- 1 point: 8+8=16 J.
- 1 point: The signed momenta cancel while both wheels still rotate.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Rigid-body angular momentum?
L=Iω.
RECALL 2What must match when adding angular momenta?
The reference axis and sign convention.
RECALL 3How do K and L depend on ω at fixed I?
K depends on ω²; L depends on ω.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Angular momentum of a rigid system
- L=Iω; add signed values about the same axis.
- K=½Iω²: scalar, unlike signed L in fixed-axis problems.
Remember: Zero total L does not mean every part is at rest or that total kinetic energy is zero.
Conditions: Separate coaxial wheels: I_A=2 kg·m², ω_A=+3 rad/s; I_B=1 kg·m². CCW positive from the same side. These are state comparisons, not a coupling simulation.
Refresh Kid · Unit 6 · Objectives 6.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.3, objectives 6.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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