Rolling Without Slipping

Rolling Without Slipping: Kinematics and Energy

When a rigid body rolls down an incline without slipping, its translational motion and rotational motion are strongly coupled. The static friction acting at the contact point provides the torque necessary to spin the object, causing its mechanical energy to be split between translation and rotation. Because static friction acts at an instantaneously stationary contact point, it does no macroscopic work, and mechanical energy is conserved when rolling resistance is negligible.

The non-slip condition enforces that the linear speed of the center of mass, v, and the angular speed, ω, are related by v = ωR, where R is the radius. Writing the moment of inertia as I = κmR² (with κ = 1 for a hoop, 1/2 for a solid cylinder, and 2/5 for a solid sphere), energy conservation over a vertical drop h gives mgh = (1/2)mv² + (1/2)Iω², which simplifies to v = √(2gh/(1+κ)).

Worked Example: Solid Cylinder Dropping 0.5 m

Consider a 1 kg solid cylinder (κ = 1/2) of radius R = 0.5 m rolling down a 30° incline over a vertical drop of h = 0.5 m from rest, with g = 10 m/s²:

1. Determine total available potential energy: E = mgh = (1)(10)(0.5) = 5 J.
2. Calculate center speed: v = √(2gh/(1+κ)) = √((2 × 10 × 0.5)/(1+0.5)) = √(20/3) ≈ 2.58198889747 m/s.
3. Calculate angular speed: ω = v/R = √(20/3) / 0.5 = 2√(20/3) ≈ 5.16397779494 rad/s.
4. Check translational energy: K_trans = (1/2)mv² = (1/2)(1)(20/3) = 10/3 ≈ 3.33333333333 J.
5. Check rotational energy: K_rot = (1/2)Iω² = (1/4)mv² = 5/3 ≈ 1.66666666667 J, summing to 10/3 + 5/3 = 5 J.
6. Find down-slope acceleration: a = g sin(θ)/(1+κ) = (10 × 0.5)/1.5 = 10/3 ≈ 3.33333333333 m/s².
7. Find required static friction and minimum static coefficient: f_s = κma = 5/3 ≈ 1.66666666667 N and μ_s,min = (κ tan θ)/(1+κ) = tan(30°)/3 ≈ 0.19245008973.

Experiment with the Module

Open the Rolling Without Slipping interactive tool to compare shapes, radii, drops, and incline angles:

1. Set Shape to "Solid cylinder", Radius R to "1/2", Vertical drop h to "1/2", and Incline angle degrees to "30". Verify that translational energy is 3.33333333333 J and rotational energy is 1.66666666667 J.
2. Change Shape to "Hoop" with Radius R set to "1" and Vertical drop h set to "1". Because κ = 1, translational and rotational kinetic energy split equally at 5 J each.
3. Change Radius R between "1/2", "1", and "2" for a fixed shape and drop. Observe that center speed and energies stay constant while angular speed scales inversely with R.
4. Set Vertical drop h to "0". The center speed and kinetic energies are zero at the release instant, whereas down-slope acceleration and static friction remain nonzero.

Limitations

The model assumes that the static friction coefficient is always large enough to prevent slipping; it does not simulate kinetic friction or mixed slipping-and-rolling regimes. Rolling resistance from surface deformation and aerodynamic drag are neglected, the rolling body is treated as perfectly rigid with fixed mass m = 1 kg and g = 10 m/s², and the incline schematic uses scaled force vectors for visual clarity.

References

OpenStax University Physics Volume 1, Section 11.1: Rolling Motion — Coupled translational and rotational rolling kinematics, dynamics, and mechanical energy conservation down an incline.