Rolling Without Slipping: Unifying Translation and Rotation

Abstract

Rolling without slipping is a fundamental kinematic constraint that couples translational and rotational motion in rigid bodies. This article examines the constraint condition, derives its relationship to center-of-mass motion, and demonstrates how it simplifies the analysis of rolling systems. We show that the constraint $v = r\omega$ emerges naturally from the requirement that the contact point has zero velocity relative to the surface.

Background

When a rigid body moves through space, its motion can be decomposed into two independent components: translation of the center of mass and rotation about the center of mass ^{[center-of-mass-motion]}. For many practical systems—wheels, balls, cylinders—these two motions are not independent. Instead, they are coupled by a constraint imposed by the contact surface.

Rolling without slipping occurs when the point of contact between a rolling object and a surface does not slide relative to that surface. This is distinct from sliding friction, which dissipates energy; in rolling without slipping, the contact point instantaneously has zero velocity, and energy is conserved more efficiently ^{[rolling-without-slipping]}.

Key Results

The Rolling Constraint

For an object of radius $r$ rolling without slipping on a flat surface, the linear velocity $v$ of the center of mass and the angular velocity $\omega$ about the center of mass are related by ^{[rolling-without-slipping]}:

$v = r\omega$

This relationship is a kinematic constraint, not a dynamical law. It holds regardless of the forces acting on the object, provided the no-slip condition is maintained.

Derivation from Contact Velocity

The constraint emerges from requiring that the velocity of the contact point equals the velocity of the surface (zero, for a stationary surface). The velocity of any point on a rigid body is the sum of the center-of-mass velocity and the velocity due to rotation about the center of mass:

$\vec{v}_{\text{contact}} = \vec{v}_{cm} + \vec{\omega} \times \vec{r}_{\text{contact}}$

For a point at the bottom of a rolling object, $\vec{r}_{\text{contact}}$ points downward (perpendicular to the direction of motion) with magnitude $r$. The rotational contribution has magnitude $r\omega$ and points backward (opposite to the direction of motion). For the contact point to have zero velocity:

$v_{cm} = r\omega$

This confirms the rolling constraint.

Decomposition of Kinetic Energy

The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy ^{[rolling-without-slipping]}:

$KE_{\text{total}} = \frac{1}{2}m v^2 + \frac{1}{2}I\omega^2$

where $I$ is the moment of inertia about the center of mass. Substituting the rolling constraint $\omega = v/r$:

$KE_{\text{total}} = \frac{1}{2}m v^2 + \frac{1}{2}I\frac{v^2}{r^2} = \frac{1}{2}\left(m + \frac{I}{r^2}\right)v^2$

The effective mass is increased by the rotational inertia, which is why rolling objects accelerate more slowly down an incline than sliding objects.

Worked Examples

Example 1: Cylinder Rolling Down an Incline

A uniform cylinder of mass $m$ and radius $r$ rolls without slipping down an incline of angle $\theta$. Find the acceleration of the center of mass.

Solution:

The moment of inertia of a uniform cylinder about its center is $I = \frac{1}{2}mr^2$. Using the rolling constraint $\omega = v/r$, the total kinetic energy is:

$KE = \frac{1}{2}m v^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\frac{v^2}{r^2} = \frac{3}{4}m v^2$

By energy conservation, as the cylinder descends a height $h$:

$mgh = \frac{3}{4}m v^2$

$v = \sqrt{\frac{4gh}{3}}$

Differentiating with respect to time:

$a = \frac{dv}{dt} = \frac{2}{3}g\sin\theta$

Note that this is less than $g\sin\theta$, the acceleration of a frictionless sliding object, because some gravitational potential energy goes into rotational kinetic energy.

Example 2: Wheel Rolling at Constant Velocity

A wheel of radius $0.5$ m rolls without slipping at a linear velocity of $10$ m/s. What is its angular velocity?

Solution:

Using the rolling constraint ^{[rolling-without-slipping]}:

$\omega = \frac{v}{r} = \frac{10 \text{ m/s}}{0.5 \text{ m}} = 20 \text{ rad/s}$

References

^{[rolling-without-slipping] [center-of-mass-motion]}

AI Disclosure

This article was drafted with AI assistance from class notes. All factual claims and mathematical derivations are grounded in cited source material. The worked examples were generated to illustrate the concepts but follow standard textbook treatments of rolling motion.