Down a ramp: share energy between translation and rotation
You will be able to: Use energy conservation to compare ideal rolling objects with different mass distributions.
Why can a solid sphere reach the bottom before a hoop?
Release a hoop and a solid sphere from rest at the same height on an incline. If both roll without slipping, the sphere reaches a greater center speed at the bottom. More of the hoop’s kinetic energy is tied up in rotation relative to its translational energy.
A useful starting point: Rolling without slipping: connect the two motions →
Words and symbols before equations
- Height drop h
- Vertical loss in center-of-mass height, in m.
- Shape factor β (beta)
- Defined by I_CM=βMR²; dimensionless.
- Gravitational potential energy Mgh
- Near-Earth energy of the object–Earth system, with chosen zero height.
- Ideal rolling assumptions
- Rigid objects, enough static friction for no slip, no air drag or rolling resistance.
What this picture assumes
M=2 kg, g=10 m/s², I_CM=βMR². Ideal rigid no-slip rolling with sufficient static friction, no drag or rolling resistance. Intermediate β values represent model mass distributions. Each setting is a separate release.
Connect the picture to the physics
Choose object plus Earth so gravity is internal. On a stationary incline with ideal no-slip contact, Mgh=½Mv²+½I_CMω². Insert ω=v/R and I_CM=βMR² to obtain Mgh=½Mv²(1+β), hence v²=2gh/(1+β). Mass and radius cancel under these assumptions.
A hoop has β=1, a uniform solid disk β=1/2 and a uniform solid sphere β=2/5. Smaller β gives larger final v at the same h. Their equal lost potential energies are divided differently: K_rot/K_trans=β. These formulas assume the stated shapes and axes.
For an object released freely down an incline, static friction points uphill and supplies the torque about the center that spins it up. It does not dissipate energy at an ideal stationary contact. If friction is insufficient, slipping occurs and this no-slip energy result no longer applies.
A worked example, step by step
A uniform solid disk rolls from rest through a vertical drop of 1.5 m. Use g=10 m/s² and β=1/2. Find bottom speed and the translational fraction of its kinetic energy.
- Write Mgh=½Mv²(1+β).
- v²=2(10)(1.5)/(1.5)=20 m²/s².
- v=√20≈4.47 m/s.
- K_trans/K_total=1/(1+β)=2/3; rotation holds the remaining 1/3.
Static friction is not automatically an energy loss. Also, mass cancellation is conditional on the same shape factor and ideal rolling.
Two uniform solid disks differ in mass and radius. Same h, ideal no-slip release. Same final center speed?
Compare with an explanation
Yes. Both have β=1/2; mass and radius cancel in this ideal model.
Predict. Change one thing. Explain.
At a fixed height, compare β=0.4, 0.5 and 1. Predict the speed ranking and energy split. Then double the height and check the speed factor; the model uses g=10 m/s².
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At h=1.5 m and β=0.5, bottom speed=4.47 m/s. 30 J = 20 J translation + 10 J rotation. K_rot/K_trans=0.5 when h>0.
M=2 kg, g=10 m/s², I_CM=βMR². Ideal rigid no-slip rolling with sufficient static friction, no drag or rolling resistance. Intermediate β values represent model mass distributions. Each setting is a separate release.
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 hoop of mass 2 kg rolls from rest through h=1 m. Use g=10 m/s². (a) Find initial gravitational energy relative to the bottom. (b) Find bottom speed. (c) Find each kinetic-energy contribution. (d) Explain whether ideal static friction turned energy into heat.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Mgh=20 J.
- 1 point: β=1 gives v²=10, so v≈3.16 m/s.
- 1 point: K_trans=10 J and K_rot=10 J.
- 1 point: No. In ideal rigid no-slip rolling on a stationary surface, contact does not slip and friction does not dissipate energy.
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 does β describe?
Mass distribution: I_CM/(MR²).
RECALL 2Why does a hoop roll more slowly than a solid sphere from the same height?
Its larger β allocates a larger fraction of kinetic energy to rotation.
RECALL 3When is the ideal no-slip energy formula invalid?
When slipping or other dissipative effects violate the assumptions.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Down a ramp: share energy between translation and rotation
- From rest, ideal rolling: v=√[2gh/(1+β)], I_CM=βMR².
- K_rot/K_trans=β. Hoop β=1; solid disk β=1/2; solid sphere β=2/5.
Remember: Static friction is not automatically an energy loss. Also, mass cancellation is conditional on the same shape factor and ideal rolling.
Conditions: M=2 kg, g=10 m/s², I_CM=βMR². Ideal rigid no-slip rolling with sufficient static friction, no drag or rolling resistance. Intermediate β values represent model mass distributions. Each setting is a separate release.
Refresh Kid · Unit 6 · Objectives 6.5.A, 6.5.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 6.5, objectives 6.5.A, 6.5.B. 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.
Want to work through this with a tutor?
Bring your question about Down a ramp: share energy between translation and rotation. Your explanation and answers remain free to access.