Rolling Motion

Key idea: Learn rolling without slipping (no-slip condition), split kinetic energy into translation + rotation, and check friction conditions for no slip.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • show an understanding of and use the terms angular displacement, angular velocity, and angular acceleration of a rigid body with respect to a fixed axis
  • solve problems using the equations of motion for uniform angular acceleration that are analogous to the equations of motion for uniform linear acceleration
  • show an understanding of and use the terms angular momentum and moment of inertia of a rotating rigid body
  • calculate the moment of inertia about an axis for simple bodies by using calculus, the parallel-axis theorem or otherwise (knowledge of the perpendicular-axis theorem and mathematical derivation of the moment of inertia for spheres are not required)
  • show an understanding of torque produced by a force relative to a reference point, and apply the principle that torque is related to the rate of change of angular momentum to solve problems, such as those involving point masses, rigid bodies, or bodies with a variable moment of inertia e.g. an ice-skater
  • Apply Eₖ,rot = ½Iω² to the rotational kinetic energy of a rigid body.
  • Derive Eₖ,rot = ½Iω² for a rigid body from the equations of motion.
  • recall and apply the result that the motion of a rigid body can be regarded as translational motion of its centre of mass with rotational motion about an axis through the centre of mass to solve problems, including the use of F ⩽ µN for solid surfaces in no-slip contact (no distinction is made between the coefficient of static and kinetic friction)

Rolling motion combines translation of the centre of mass with rotation about the centre of mass. For rolling without slipping, the key constraint is: v_cm = ω R When a contact-force limit is relevant, the 9814 syllabus uses: F ≤ μ N

1. Definitions (Must Know)

  • Pure rolling (no slip): the point of contact with the ground has zero velocity relative to the ground at the instant of contact.
  • Rolling with slipping: the contact point slides; friction does kinetic-energy dissipation and v_cm ≠ ω R in general.
  • No-slip condition (pure rolling): in magnitude, v_cm = ω R and, for the tangential acceleration components, a_cm = α R. Signed equations depend on the chosen angular convention.
  • Kinetic energy split (about CM): Eₖ = 1/2 Mv_cm² + 1/2 I_cmω²
  • Syllabus contact-force test: F ≤ μ N. The official 2027 H3 Physics syllabus does not distinguish static and kinetic coefficients in this model.
  • Symbols used in this lesson: v_cm (m s⁻¹), a_cm (m s⁻²), ω (rad s⁻¹), α (rad s⁻²), R (m), M (kg), I_cm (kg m²), F friction (N), N normal reaction (N), μ coefficient of friction (dimensionless).

2. Key Ideas (What Earns Marks)

  • Model rolling as: translation of CM + rotation about CM (H3 syllabus statement).
  • For pure rolling:
    • Use the constraint v_cm = ω R early to reduce unknowns.
    • Use a_cm = α R for dynamics problems with acceleration.
  • A contact force may be needed to enforce no slip, but it can be zero for uniform rolling on a horizontal surface. When the problem supplies μ, check whether the required force exceeds μ N.

Quick comparison:

CaseKey conditionContact stateEnergy note
Rolling without slippingv_cm = ω Rno relative slidingno energy loss from sliding
Rolling with slippingv_cm ≠ ω Rsurfaces slide at contactmechanical energy is dissipated by friction
For a wheel rolling right without slipping, the contact point is instantaneously at rest, the centre moves at v centre of mass, and the top moves at twice that speed.
Velocity composition in the ground frame: rotation cancels translation at the contact point and adds to it at the top.

3. Detailed Explanations

A. Why “translation + rotation about CM” is the right picture

Any rigid body motion can be decomposed into:

  1. motion of the centre of mass, plus
  2. rotation about an axis through the centre of mass.

For rolling problems, this decomposition helps you:

  • write linear equations for the CM (e.g. ∑ F = Ma_cm), and
  • write rotational equations about the CM (e.g. ∑ τ_cm = I_cmα).

B. Deriving the no-slip constraint

For a wheel of radius R rolling without slipping:

  • the contact point’s speed relative to the ground is zero.

The contact point’s velocity is the vector sum of:

  • translational CM velocity v_cm (to the right, say), and
  • rotational velocity about the CM of magnitude ω R (to the left at the bottom point).

So they cancel: v_cm - ω R = 0 ⇒ v_cm = ω R

Differentiating gives: a_cm = α R

C. Contact force and the condition F ≤ μ N

In a more detailed contact model, no relative sliding corresponds to static friction. For this H3 syllabus, however, no distinction is made between static and kinetic coefficients: calculate the required contact-force magnitude F and test F ≤ μ N

If the required force exceeds μ N, the assumed no-slip motion is not possible. If the required force is zero, pure rolling can continue without a frictional force.

4. Common Mistakes

  • Assuming friction must be non-zero or must remove energy in every rolling problem.
  • Applying v_cm = ω R even when the question implies slipping.
  • Using I about the wrong axis (use I_cm when taking moments about the centre of mass).
  • Forgetting to check the no-slip condition with F ≤ μ N when μ is given.

5. Exam Tips

  • If the phrase “rolls without slipping” appears, write v_cm = ω R immediately.
  • If μ is given (or “rough surface” is emphasised), expect a slip check:
    1. solve assuming no slip,
    2. compute required F,
    3. compare with μ N.
  • Use separate equations for translation and rotation:
    • ∑ F = Ma_cm
    • ∑ τ_cm = I_cmα plus a_cm = α R for no slip.

6. Worked Examples

Modelled example 1

Pure rolling kinematics (no slip)

Core

Problem

A wheel of radius 0.20 m rolls without slipping at ω = 15 rad s⁻¹. Find v_cm.
Study the worked solution
  1. Use no slip

    Method

    v_cm = ω R.

    Reason

    The contact point is instantaneously at rest relative to the ground.

    Working

    v_cm = ω R
  2. Evaluate

    Method

    v_cm = 3.0 m s⁻¹.

    Reason

    Multiply angular speed by radius.

    Working

    (15)(0.20) = 3.0 m s⁻¹

Guided practice 2

Energy split for rolling (given v_cm)

About 6 min

Problem

A solid disc rolls without slipping at centre speed v_cm. Express total kinetic energy using only M and v_cm.

Try this before viewing the solution

Hints

Hint 1: replace angular speed
Use I_cm = (1/2)MR² and ω = v_cm/R.
View solution step by step
  1. Write both contributions

    Method

    Eₖ = (1/2)Mv_cm² + (1/2)I_cmω².

    Reason

    Rolling combines translation of the centre and rotation about it.

    Working

    E = Eₜᵣₐₙₛ + Eᵣₒₜ
  2. Substitute no slip and inertia

    Method

    The rotational term is (1/4)Mv_cm².

    Reason

    R² cancels after ω = v_cm/R.

    Working

    (1/2)((1/2)MR²)(v_cm/R)²
  3. Add

    Method

    Eₖ = (3/4)Mv_cm².

    Reason

    1/2 + 1/4 = 3/4.

    Working

    Eₖ = (3/4)Mv_cm²

Common misconception 3

No-slip check (friction limit)

Find and correct the mistake

Learner claim

A wheel requires 12 N friction for no slip, with N = 40 N and μ = 0.20. A learner says static friction automatically supplies any required value. Test the claim.

Try this before viewing the solution

Rolling outcome

View solution step by step
  1. Find the limit

    Method

    Fₗᵢₘᵢₜ = 8 N.

    Reason

    The syllabus model gives μ N.

    Working

    (0.20)(40) = 8 N
  2. Compare with requirement

    Method

    12 N exceeds 8 N.

    Reason

    The contact cannot supply enough static friction.

    Working

    F_required > Fₗᵢₘᵢₜ
  3. Conclude

    Method

    Pure rolling is impossible; the wheel slips.

    Reason

    The no-slip constraint cannot be maintained.

    Working

    v_cm ≠ ω R during slip

Examiner practice 4

Speed of the top point (ground frame)

4 marks

Examination question

A wheel rolls without slipping at v_cm = 2.5 m s⁻¹ to the right. Find the top point’s ground-frame speed. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Use no slip

    1 mark

    Method

    ω R = v_cm.

    Reason

    The bottom point is instantaneously at rest.

    Working

    ω R = 2.5 m s⁻¹
  2. Translation contribution

    1 mark

    Method

    The centre motion contributes 2.5 m s⁻¹ rightward.

    Reason

    Every point shares the centre’s translational velocity.

    Working

    + v_cm
  3. Rotation contribution

    1 mark

    Method

    Rotation contributes another 2.5 m s⁻¹ rightward at the top.

    Reason

    The top point’s rotational velocity is rightward.

    Working

    + ω R
  4. Add

    1 mark

    Method

    vₜₒₚ = 5.0 m s⁻¹.

    Reason

    The two vectors align.

    Working

    2v_cm = 5.0 m s⁻¹

Challenge 5

Rolling down an incline (solid cylinder) + minimum μ

Minimal support

Independent transfer

A solid cylinder with I_cm = (1/2)MR² rolls without slipping down a 30° incline. Find a_cm and minimum μ.

Try this before viewing the solution

Hints

Hint 1: use the rolling acceleration form
Start with a = g sin θ/[1 + I/(MR²)], then find the friction needed for the torque.
View solution step by step
  1. Find acceleration

    Method

    a_cm = (2/3)g sin θ = 3.27 m s⁻².

    Reason

    I/(MR²) = 1/2 for a solid cylinder.

    Working

    a = 9.81(0.5)/(1.5) = 3.27
  2. Find required friction

    Method

    f = (1/2)Ma_cm upslope.

    Reason

    fR = Iα and α = a_cm/R.

    Working

    f = (I/R²)a = (1/2)Ma
  3. Apply limiting condition

    Method

    μₘᵢₙ = a_cm/(2g cos θ).

    Reason

    No slip requires f ≤ μ Mg cos θ.

    Working

    μₘᵢₙ = 1/3 tan θ
  4. Evaluate

    Method

    μₘᵢₙ = 0.192.

    Reason

    tan 30° = 0.577.

    Working

    (1/3)(0.577) = 0.192

7. Mind Stretchers

Mind stretcher 1: Where is the instantaneous “pivot” point?Extension

For pure rolling without slipping, which point on the wheel has zero instantaneous velocity, and why does that make some torque/energy arguments easier?

Answer

The contact point has zero instantaneous velocity relative to the ground. You can treat the wheel as rotating instantaneously about that point, which can simplify certain kinematics or energy comparisons (when used carefully).

Mind stretcher 2: Does the contact force do work in pure rolling?Extension

In pure rolling without slipping on a stationary rigid surface, must the frictional contact force transfer energy to the wheel? Explain briefly.

Answer

No. In the ideal rigid-body model, the instantaneous contact point is at rest in the ground frame, so the instantaneous power vector F · vector v_contact is zero. The contact force can still supply a torque and change how translational and rotational energy are partitioned. This statement is scoped to a stationary, non-deforming surface; a moving or deforming contact requires separate analysis.

8. Optional/Enrichment: Rolling as Instantaneous Rotation

In pure rolling, the wheel’s motion can be seen as instantaneous rotation about the contact point. This viewpoint can be helpful, but the CM-decomposition method (translation + rotation about CM) is usually the most robust for exam problems.

Next step

Return to the Rotational Motion hub, or continue to Electric and Magnetic Fields, the next maintained H3 topic.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027