Rolling without slipping: connect the two motions
You will be able to: Relate center-of-mass motion to rotation in ideal rolling on a stationary surface.
Why is the bottom of a rolling wheel instantaneously at rest?
A wheel rolls right along the floor. Its center moves forward, while rotation moves the bottom point backward relative to the center. Without slipping, these two velocities exactly cancel at the floor.
A useful starting point: Two wheels lock together: conserve L, track the energy →
Words and symbols before equations
- No-slip rolling
- The contact point has zero velocity relative to the surface at that instant.
- R
- Wheel radius, in m.
- v_CM and a_CM
- Center-of-mass velocity and acceleration along the surface.
- ω and α
- Angular velocity and angular acceleration; below, clockwise is positive to match rightward rolling.
What this picture assumes
Rigid wheel rolling without slipping on a stationary floor. Clockwise angular speed pairs with rightward translation. Velocity arrows share a scale; wheel size is schematic and physical R is labeled.
Connect the picture to the physics
For rightward rolling with clockwise taken positive, v_CM=Rω. Over a rolling interval, Δx_CM=RΔθ, and a_CM=Rα for constant R. If using counterclockwise positive instead, the corresponding signed relation has a minus sign. State the convention before substituting.
Relative to the center, a rim point has tangential speed R|ω|. Add its vector velocity to the center’s velocity. At the bottom the two cancel; at the top they add to 2v_CM. The center itself moves at v_CM. The material point touching the ground changes as the wheel turns.
Zero instantaneous contact velocity does not mean the whole wheel is at rest or that the contact point has zero acceleration. In the ideal rigid-body model, static friction at a stationary contact does not dissipate energy. Friction may be zero during steady rolling on a level surface with no other horizontal forces.
A worked example, step by step
A wheel of radius 0.5 m rolls right without slipping at v_CM=2 m/s. Find ω and the ground-frame velocities of its top and bottom points.
- Choose right and clockwise as positive for the paired linear and angular quantities.
- ω=v_CM/R=2/0.5=4 rad/s clockwise.
- Bottom: 2−Rω=2−2=0 m/s.
- Top: 2+Rω=4 m/s right. Both combine translation with local rotation.
The contact point is instantaneously at rest relative to the floor; the wheel’s center is not.
For the same center speed, does a larger wheel turn faster?
Compare with an explanation
No. ω=v_CM/R, so increasing radius reduces angular speed.
Predict. Change one thing. Explain.
Change center speed and radius. Predict which changes ω. Compare the arrows for center, top and bottom velocities in the ground frame; they share a common velocity scale.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
v_CM=2 m/s, R=0.5 m; clockwise angular speed=4 rad/s. Top=4 m/s right; contact=0 relative to the floor.
Rigid wheel rolling without slipping on a stationary floor. Clockwise angular speed pairs with rightward translation. Velocity arrows share a scale; wheel size is schematic and physical R is labeled.
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 wheel of radius 0.2 m rolls right without slipping at 3 m/s. Its center accelerates right at 1 m/s². (a) Find angular speed. (b) Find clockwise angular acceleration. (c) Find top speed. (d) Explain the contact-point velocity.
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Compare with the answer and four-point rubric
- 1 point: ω=3/0.2=15 rad/s.
- 1 point: α=1/0.2=5 rad/s² clockwise.
- 1 point: 6 m/s right.
- 1 point: Zero relative to the floor because 3 m/s translational velocity cancels the 3 m/s backward rotational contribution.
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 1Condition for v=Rω?
Rolling without slipping, with consistent signs and a stationary surface.
RECALL 2Ground-frame speed of the top point?
2v_CM during pure rolling.
RECALL 3Does no-slip rolling always require a nonzero friction force?
No; steady ideal rolling on a level surface can have zero friction.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Rolling without slipping: connect the two motions
- No slip on stationary surface: |v_CM|=R|ω|.
- Matching signs: Δx=RΔθ and a_CM=Rα.
- Top speed=2v_CM and contact speed=0 for rightward pure rolling.
Remember: The contact point is instantaneously at rest relative to the floor; the wheel’s center is not.
Conditions: Rigid wheel rolling without slipping on a stationary floor. Clockwise angular speed pairs with rightward translation. Velocity arrows share a scale; wheel size is schematic and physical R is labeled.
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.
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