Pull inward: faster spin, unchanged angular momentum
You will be able to: Relate changes in inertia, angular speed and energy when external torque is negligible.
Where does the extra kinetic energy come from when a skater pulls inward?
A skater pulls their arms closer to the rotation axis and spins faster. The surrounding ice supplies very little torque about the vertical axis. Angular momentum stays nearly constant, but the skater’s muscles supply energy as the mass distribution changes.
A useful starting point: Choose a system before conserving angular momentum →
Words and symbols before equations
- Nonrigid system
- A system whose mass distribution can change.
- I_i and I_f
- Initial and final rotational inertias about the same axis.
- Internal energy conversion
- Energy such as chemical energy becoming mechanical energy within a system.
What this picture assumes
Initial I=4 kg·m², ω=2 rad/s, L=8 kg·m²/s. External torque negligible. Compare endpoints after radial motion stops. This model omits intermediate radial kinetic energy.
Connect the picture to the physics
With negligible net external torque, I_iω_i=I_fω_f. Solve for ω_f=(I_i/I_f)ω_i. Moving mass inward reduces I, so the angular speed must increase to keep the product constant. This is not a claim that torque is required to maintain motion.
Substitute ω=L/I into K=½Iω² to obtain K=L²/(2I). At fixed L, smaller inertia means larger rotational kinetic energy. This differs from comparing separate objects at fixed ω, where K increases with I. Always name what you hold fixed.
The increased rotational energy comes from internal work and energy conversion as the masses move inward. Total energy is conserved for an isolated system, but rotational kinetic energy alone need not be constant. Compare initial and final states after radial motion has stopped; intermediate radial kinetic energy is outside this simple two-state model.
| Quantity | No net external torque | Pulling masses inward |
|---|---|---|
| Angular momentum | Constant | Same before and after |
| Rotational inertia | Can change | Decreases |
| Angular speed | Can change | Increases |
| Rotational kinetic energy | Need not stay constant | Increases through internal energy conversion |
A worked example, step by step
A rotating system starts with I_i=4 kg·m² and ω_i=2 rad/s. It contracts to I_f=2 kg·m² with negligible external torque. Find final ω and the change in rotational K.
- L_i=4(2)=8 kg·m²/s, which stays constant.
- ω_f=8/2=4 rad/s.
- K_i=½(4)(2²)=8 J; K_f=½(2)(4²)=16 J.
- Rotational K increases by 8 J through internal energy conversion. Constant L did not require constant K.
Conserving angular momentum does not automatically conserve rotational kinetic energy.
If final inertia is one-third of initial inertia, how does ω change at constant L?
Compare with an explanation
Its magnitude triples in the same turning sense.
Predict. Change one thing. Explain.
Change final inertia while the model keeps L=8 kg·m²/s. Compare final ω and K to the initial state. Predict whether expanding outward increases or decreases each quantity.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Final I=2 kg·m², ω=4 rad/s. L stays 8 kg·m²/s. K changes from 8 to 16 J (ΔK=8 J). Internal energy conversion accounts for the difference.
Initial I=4 kg·m², ω=2 rad/s, L=8 kg·m²/s. External torque negligible. Compare endpoints after radial motion stops. This model omits intermediate radial kinetic energy.
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 questionA system changes from I=6 to 2 kg·m² with negligible external torque, starting at ω=1 rad/s. (a) Find L. (b) Find final ω. (c) Find initial and final rotational K. (d) Account for the difference.
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Compare with the answer and four-point rubric
- 1 point: L=6 kg·m²/s.
- 1 point: ω_f=3 rad/s.
- 1 point: K_i=3 J; K_f=9 J.
- 1 point: 6 J is converted into rotational kinetic energy by internal work; L conservation does not fix K.
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 1What stays constant when a skater contracts without external torque?
Total angular momentum about the axis.
RECALL 2At fixed L, what happens to ω if I falls?
Angular speed increases.
RECALL 3Where can extra rotational K come from?
Internal energy converted by work as the shape changes.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Pull inward: faster spin, unchanged angular momentum
- No external torque: I_iω_i=I_fω_f.
- At fixed L: K_rot=L²/(2I).
Remember: Conserving angular momentum does not automatically conserve rotational kinetic energy.
Conditions: Initial I=4 kg·m², ω=2 rad/s, L=8 kg·m²/s. External torque negligible. Compare endpoints after radial motion stops. This model omits intermediate radial kinetic energy.
Refresh Kid · Unit 6 · Objectives 6.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.4, objectives 6.4.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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