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Rolling without slipping: velocities

Key idea: For rolling without slipping on a stationary surface, the contact point is instantaneously at rest and vcm = ωR. Add translation and rotation as velocity vectors for other points.

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  • GCE A-Level H3 Physics 2027
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A wheel can translate and rotate at the same time. To find the velocity of any point on it, add two vectors: the centre-of-mass velocity and that point’s rotational velocity relative to the centre.

We model a rigid circular wheel of radius R on a stationary, non-deforming surface. Rolling without slipping means that the wheel and surface have equal velocity at their contact: the wheel’s contact point is instantaneously at rest in the ground frame. This does not make it a permanently fixed pivot.

Predict velocities around a rolling wheel

Check your understanding 1 Predict first

A bicycle wheel rolls forwards along a road without slipping. Relative to the road, which point of the wheel is moving fastest?

Fastest point
Show answer

The top. There, the forward motion of the whole wheel and the motion around the hub point the same way, so the top moves at twice the hub’s speed. At the bottom they cancel, and the contact point is momentarily at rest.

Combine translation and rotation

Take rightward motion as positive. For a wheel rolling right, rotation is clockwise. Every point shares the centre’s rightward translational velocity v_cm; its rotational velocity is tangent to its circle about the centre.

At the bottom, the rotational velocity is leftward with magnitude ω R. Hence

v_contact = v_cm-ω R = 0 ⇒ v_cm = ω R.

Here v_cm is in m s⁻¹, angular speed ω is in rad s⁻¹ and R is in metres. With a fixed radius and a consistent clockwise-positive convention, differentiating the no-slip relation gives a_cm = α R for the tangential acceleration magnitudes during speeding up.

Scroll across the figure to read all labels.

Ground-frame velocities are vector sums. A zero contact-point velocity at one instant does not mean that point has zero acceleration.

Worked example 1

Centre speed from angular speed

Problem

A wheel of radius 0.20 m rolls without slipping at ω = 15 rad s⁻¹. Find v_cm.
Show full solution

  1. Use no slip

    Method

    v_cm = ω R.

    Reason

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

    Working

    Answer

    v_cm = ω R

  2. Evaluate

    Method

    v_cm = 3.0 m s⁻¹.

    Reason

    Multiply angular speed by radius.

    Working

    Answer

    (15)(0.20) = 3.0 m s⁻¹

Exam-style question 1

Speed of the top point

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

Unit: m s⁻¹
Show solution step by step

  1. Use no slip

    1 mark

    Method

    ω R = v_cm.

    Reason

    The bottom point is instantaneously at rest.

    Working

    Answer

    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

    Answer

    + 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

    Answer

    + ω R

  4. Add

    1 mark

    Method

    vₜₒₚ = 5.0 m s⁻¹.

    Reason

    The two vectors align.

    Working

    Answer

    2v_cm = 5.0 m s⁻¹

Check your understanding 2 A car tyre

A car travels at 20 m s⁻¹ on tyres of radius 0.30 m that roll without slipping. Find the tyres’ angular speed and the ground-frame speed of each tyre’s contact point.

Unit: rad s⁻¹
Unit: m s⁻¹
Show answer

ω = v_cm/R = 20/0.30 = 67 rad s⁻¹. The contact point is instantaneously at rest: v_cm-ω R = 20-20 = 0.

Check the direction before adding magnitudes

At the top, both contributions point right, so vₜₒₚ = 2v_cm. At either side, the contributions are perpendicular: the ground-frame speed is square root of 2 v_cm, not 2v_cm.

A rim point’s speed relative to the centre is always ω R. Its speed relative to the ground varies with position. Stating the frame prevents these two statements from appearing contradictory.

Try it yourself 2

A side point, then a different frame

Minimal support

Problem

A wheel rolls right without slipping with centre speed 2.0 m s⁻¹. Find the ground-frame velocity of its rightmost rim point, then its velocity relative to an observer moving right with the centre.

Try this before viewing the solution

Unit: m s⁻¹
Unit: °
Velocity relative to an observer moving with the centre

Hints

Hint 1

draw each contribution

Draw the tangent at the rightmost point for clockwise rotation. Which shared motion disappears in the centre’s frame?
Show solution step by step

  1. Resolve in the ground frame

    Method

    The components are 2.0 m s⁻¹ rightward and 2.0 m s⁻¹ downward.

    Reason

    No slip gives ω R = v_cm, and the rotational tangent at the right side points down.

    Working

    Answer

    vₓ = 2.0 m s⁻¹, vy = -2.0 m s⁻¹, with upwards positive.

  2. Find the resultant

    Reason

    The components are perpendicular, so use Pythagoras rather than adding their magnitudes.

    Working

    Answer

    |vector v| = square root of (2.0² + 2.0²) = 2.8 m s⁻¹, at 45° below the rightward horizontal.

  3. Move with the centre

    Method

    The point moves vertically downward at 2.0 m s⁻¹ in the translating observer’s frame.

    Reason

    Subtracting the observer’s rightward velocity removes the shared translation, leaving only rotation.

    Working

    Answer

    v'ₓ = 2.0-2.0 = 0, v'y = -2.0 m s⁻¹

No slip is a constraint, not a force law

The surface may need to exert a contact force to maintain no slip. Friction can be zero for an ideal wheel already rolling uniformly on a horizontal surface. A spinning wheel that slides does not generally satisfy v_cm = ω R.

For a surface moving horizontally at speed u, no slip instead requires v_cm-ω R = u with the conventions above. The zero-contact-velocity form applies to the stationary surface used in this lesson.

Check your understanding 3 A wheel on a moving belt

A wheel of radius 0.25 m turns clockwise at 12 rad s⁻¹ on a horizontal belt that moves right at 1.0 m s⁻¹. The wheel does not slip on the belt. Taking right as positive, find the velocity of the wheel’s centre.

Show the steps Optional working
Unit: m s⁻¹
Unit: m s⁻¹
Show answer

ω R = 12 × 0.25 = 3.0 m s⁻¹. No slip requires v_cm-ω R = u, so v_cm = 1.0 + 3.0 = 4.0 m s⁻¹ to the right. On a stationary road the same wheel would move at 3.0 m s⁻¹.

An instantaneous pivot helps with energy

The contact point is instantaneously at rest, so pure rolling can be described kinematically as rotation about that point at that instant. Using the parallel-axis theorem, I_contact = I_cm + MR², gives

(1/2)I_contactω² = (1/2)I_cmω² + (1/2)Mv_cm².

This is an energy identity. Applying a fixed-pivot torque equation about a moving contact requires additional justification; taking moments about the centre avoids that problem.

Check your understanding 4 A rolling ring’s energy

A thin ring of mass 2.0 kg rolls without slipping at v_cm = 3.0 m s⁻¹. Using I_contact, find its total kinetic energy.

Unit: J
Show answer

I_contact = MR² + MR² = 2MR². Then E = (1/2)(2MR²)(v/R)² = Mv² = 2.0 × 3.0² = 18 J. As a check, translation gives (1/2)Mv² = 9.0 J and rotation about the centre another 9.0 J.

Common mistakes

  • Adding speeds at a side point instead of adding velocity vectors.
  • Treating the contact point as a fixed pivot. It is at rest only at that instant.
  • Using v_cm = ω R for a wheel that slips.
  • Mixing a point’s speed relative to the centre, ω R, with its speed relative to the ground.

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