Dynamics of Rolling Without Slipping

An object rolls without slipping when the point in contact with the surface is momentarily at rest, so vcm = ωR and acm = αR; its kinetic energy is the sum of translational and rotational parts, Ek = ½Mv2 + ½Iω2 = ½Mv2(1 + k), where k = I ÷ MR2. In brief: k = 1/2 for a solid cylinder, 2/5 for a solid sphere and 1 for a hoop. On an incline of angle θ, the acceleration is a = g sin θ ÷ (1 + k), so objects with a smaller k roll down faster: sphere before solid cylinder before hoop, regardless of their masses and radii. Static friction makes the rolling possible but does no work.

Wheels, balls, cans and cylinders roll rather than slide, and the rotational energy changes how they accelerate. This guide explains the no-slip condition, the energy and force analysis of rolling down a slope, compares different shapes in a table, works through a race of four objects and corrects common mistakes.

What is the condition for rolling without slipping?

For a wheel of radius R rolling on a flat surface, in one revolution the center advances by 2πR, the circumference. So vcm = ωR, and acm = αR. The point of the wheel touching the ground has velocity vcm − ωR = 0 relative to the ground, so it is instantaneously at rest (static friction applies, not kinetic friction). The top of the wheel moves at 2vcm. The motion can be seen as a pure rotation about the contact point, or as a translation of the center of mass plus a rotation about the center. If the surface is too smooth, or the acceleration too large, the wheel slips (wheel spin when a car accelerates on ice, or skid under braking), and the condition no longer holds.

What is the kinetic energy of a rolling body?

The total kinetic energy is the sum of the translational energy of the center of mass and the rotational energy about it: Ek = ½Mvcm2 + ½Icmω2. With ω = v ÷ R and I = kMR2, Ek = ½Mv2(1 + k). For a solid cylinder, rotation accounts for one third of the total; for a hoop, for half. Since static friction acts at the contact point, which is at rest, it does no work, and mechanical energy is conserved for rolling without slipping (ignoring rolling resistance). A sliding block on a frictionless slope has k = 0, so all its energy is translational.

How does a body roll down an incline?

Using energy: a body starting from rest and descending a height h gains kinetic energy Mgh = ½Mv2(1 + k), so v = √(2gh ÷ (1 + k)). Using forces: along the incline, Mg sin θ − f = Ma, and for the rotation about the center, fR = Iα = kMR2(a ÷ R), so f = kMa. Then a = g sin θ ÷ (1 + k). The friction force is f = [k ÷ (1 + k)] Mg sin θ, and no slipping requires f ≤ μsN = μsMg cos θ, so μs ≥ [k ÷ (1 + k)] tan θ. A steeper slope needs a higher coefficient of friction. All bodies with the same k reach the bottom together, whatever their mass or size. The body with the smaller k wins: more of its energy is in translation.

Bar chart of speeds at the bottom of a 1.0 meter drop for a sliding block, sphere, cylinder and hoop 0 1 2 3 4 5 Speed at the bottom (m/s) Sliding block 4.43 Solid sphere 3.74 Solid cylinder 3.61 Hoop 3.13
Speeds at the bottom of a 1.0 m drop, from the worked example. A frictionless block puts all its energy into motion and wins. The hoop keeps the largest share of its energy in rotation, so it is the slowest. Mass and size make no difference, only the shape.

Rolling shapes compared

Rolling down a 30° incline from a height of 1.0 m
Body k = I ÷ MR2 Acceleration (m/s2) Speed at the bottom (m/s)
Sliding block (no friction)04.904.43
Solid sphere0.403.503.74
Solid cylinder0.503.273.61
Hoop or thin pipe1.002.453.13

Worked example: a race down a ramp

  1. Solid cylinder. From energy: Mgh = ½Mv2 + ½(½MR2)(v ÷ R)2 = ¾Mv2. So v = √(4gh ÷ 3) = √(4 × 9.8 × 1.0 ÷ 3) = √13.07 = 3.61 m/s.
  2. Other shapes (h = 1.0 m): sphere v = √(2 × 9.8 ÷ 1.4) = 3.74 m/s; hoop v = √(2 × 9.8 ÷ 2) = 3.13 m/s; frictionless sliding block v = √(2gh) = 4.43 m/s. The order of finishing: block, sphere, cylinder, hoop.
    4.43 m/s 3.74 m/s 3.61 m/s 3.13 m/s
  3. Acceleration on a 30° slope for the solid cylinder: a = g sin 30° ÷ 1.5 = 4.9 ÷ 1.5 = 3.27 m/s2. The length of the slope for h = 1.0 m is 2.0 m, so the time = √(2L ÷ a) = √(4.0 ÷ 3.27) = 1.11 s. For the hoop: a = 2.45 m/s2, t = √(4.0 ÷ 2.45) = 1.28 s.
  4. Friction required. For the solid cylinder on the 30° slope: μs ≥ [0.5 ÷ 1.5] × tan 30° = 0.333 × 0.577 = 0.19. For a hoop: 0.5 × 0.577 = 0.29. If the coefficient of friction is lower than this, the body slips as it rolls.
  5. Fraction of energy in rotation. For the solid sphere, the rotational part is k ÷ (1 + k) = 0.4 ÷ 1.4 = 29 percent of the total kinetic energy.
  6. Car wheel. A wheel of radius 0.30 m and mass 20 kg (taken as a solid disc, k = 0.5) rolls at 10 m/s. The kinetic energy is ½ × 20 × 100 × 1.5 = 1,500 J, of which 1,000 J is translational and 500 J is rotational.

How was rolling studied?

Galileo used balls rolling down inclined planes to study accelerated motion in the early 1600s; he did not take the rotation of the balls into account, which affects the absolute value of the acceleration, though the proportionality of distance to time squared was correct. The correct treatment with the moment of inertia was developed after Euler’s work on rigid bodies in the 1750s. Today, the analysis of rolling is applied to wheels, ball bearings, rolling robots and vehicle stability.

Which mistakes are common?

Mistake 1: Saying that friction dissipates energy in pure rolling

Static friction does no work at the contact point, which is at rest. Mechanical energy is conserved.

Mistake 2: Omitting the rotational kinetic energy

A rolling body has both translational and rotational energy. Using only ½Mv2 gives a speed that is too high.

Mistake 3: Thinking that heavier or bigger objects roll faster

For a given shape, mass and radius cancel. Only the shape (k) matters.

Mistake 4: Using kinetic friction for a rolling wheel

When there is no slipping, the friction is static, f ≤ μsN.