Rolling Without Slipping

You’ll be able to

State the rolling constraint v_cm = Rω and a_cm = Rα
Analyze an object rolling down an incline using both force and torque equations
Compare the accelerations of different shapes rolling down the same incline

The rolling constraint

When a wheel rolls without slipping, the contact point is instantaneously at rest — the wheel pivots about it rather than skidding. This locks the translation of the center to the rotation: the center moves at v_cm = Rω and accelerates at a_cm = Rα. Static friction at the contact point provides the torque that makes the wheel spin up, and because that point does not slide, this friction does no work.

Rolling-without-slipping constraint

v_cm = Rω a_cm = Rα

The center-of-mass motion and the rotation are locked together by the wheel radius R. If the wheel slips, this constraint no longer holds.

Racing shapes down an incline

For an object rolling down an incline, combining ΣF = ma_cm and τ = Iα with the constraint gives a_cm = g sin θ / (1 + I/MR²). The shape enters only through the dimensionless ratio I/MR²: a smaller ratio means a larger acceleration. A solid sphere (2/5) beats a solid disk (1/2), which beats a hoop (1) — and mass and radius cancel out entirely, so the winner is the same for any size.

Acceleration rolling down an incline

a_cm = g sin θ / (1 + I/MR²)

Only the shape factor I/MR² matters. Smaller I/MR² (mass nearer the axis) rolls faster; the plain block with no rotation (ratio 0) accelerates fastest of all at g sin θ.

Worked example

A uniform solid sphere (I = 2/5 MR²) rolls without slipping down an incline of angle 30°. Using g = 10 m/s², find the acceleration of its center of mass.

1.Use a_cm = g sin θ / (1 + I/MR²) with the shape factor I/MR² = 2/5 for a solid sphere.
2.Numerator: g sin 30° = 10 × 0.5 = 5 m/s².
3.Denominator: 1 + 2/5 = 7/5 = 1.4.
4.Divide: a_cm = 5 / 1.4 ≈ 3.6 m/s² — less than g sin θ because energy also goes into spinning the sphere.

Answer: a_cm = g sin θ / (1 + 2/5) = 5 / 1.4 ≈ 3.6 m/s²

Checkpoint

A wheel of radius 0.3 m rolls without slipping, and its center moves at 6 m/s. What is its angular velocity?

Checkpoint

A solid sphere, a solid disk, and a hoop — all of equal mass and radius — are released from rest and roll without slipping down the same incline. Which reaches the bottom first?

On the exam

For rolling problems, always bring in the constraint a_cm = Rα to connect the force equation to the torque equation. Remember that the shape factor I/MR² alone decides the race down an incline — mass and radius drop out.

Answer the 2 checkpoints as you read.