Rolling Without Slipping: Constraint, Energy, and Rigid-Body Motion

Abstract

Rolling without slipping is a kinematic constraint that couples translational and rotational motion in rigid bodies. This article derives the fundamental relationship between linear and angular velocity, explains its physical significance, and demonstrates how it emerges from the no-slip condition at the contact point. We examine energy conservation and the role of this constraint in rigid-body dynamics.

Background

When a rigid body moves in contact with a surface, its motion can be decomposed into two components: translation of the center of mass and rotation about the center of mass ^{[center-of-mass-motion]}. For a freely rolling object—one in contact with a surface but not constrained to follow a predetermined path—these two motions are independent unless a constraint is imposed.

The rolling-without-slipping condition arises when friction is sufficient to prevent relative motion at the contact point. Unlike kinetic friction, which dissipates energy, the static friction that enforces this constraint does no work because the contact point has zero velocity. This distinction is crucial: rolling without slipping is an ideal condition that maximizes mechanical efficiency.

Key Results

The No-Slip Constraint

The fundamental constraint of rolling without slipping relates the velocity of the center of mass $v$ to the angular velocity $\omega$ ^{[rolling-without-slipping]}:

$v = r\omega$

where $r$ is the radius of the rolling object.

Derivation. Consider a rigid body rolling on a surface. The velocity of any point on the body is the sum of the center-of-mass velocity and the velocity due to rotation about the center of mass. At the contact point (instantaneously in contact with the surface), the total velocity must be zero:

$\vec{v}_{\text{contact}} = \vec{v}_{cm} + \vec{\omega} \times \vec{r}_{\text{contact}} = 0$

For a cylinder or sphere rolling in one dimension, with the center of mass moving at speed $v$ and the body rotating with angular velocity $\omega$, the contact point is at distance $r$ from the axis of rotation. The rotational contribution to velocity at the contact point is $r\omega$ in the direction opposite to $v_{cm}$. Setting the net velocity to zero:

$v = r\omega$

This is a holonomic constraint—it relates velocities and can be integrated to relate displacements. If the object rolls through an angle $\theta$, the center of mass advances by $s = r\theta$.

Energy Implications

The no-slip constraint has profound consequences for energy. A rolling object possesses both translational and rotational kinetic energy. For a rigid body with moment of inertia $I$ about its center of mass:

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

Substituting $\omega = v/r$:

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

The term $I/r^2$ acts as an effective added mass due to rotation. For a solid sphere, $I = \frac{2}{5}mr^2$, so:

$KE_{\text{total}} = \frac{1}{2}v^2\left(m + \frac{2m}{5}\right) = \frac{7}{10}mv^2$

This is less than the kinetic energy of a sliding object of the same mass and speed ($\frac{1}{2}mv^2$), reflecting the fact that some kinetic energy is "locked" in rotation. Crucially, because the contact point has zero velocity, static friction does no work, and mechanical energy is conserved during rolling on a frictionless incline.

Role in Rigid-Body Dynamics

The rolling constraint is essential for analyzing rigid-body motion ^{[center-of-mass-motion]}. Without it, a rolling object would require independent specification of both $v_{cm}$ and $\omega$. The constraint reduces the degrees of freedom: for a cylinder rolling on a plane, instead of two independent variables, only one is needed (e.g., the position of the center of mass).

Worked Examples

Example 1: Sphere Rolling Down an Incline

A uniform sphere of mass $m$ and radius $r$ starts from rest at the top of a frictionless incline of height $h$. What is its speed at the bottom?

Using energy conservation with the no-slip constraint:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a uniform sphere, $I = \frac{2}{5}mr^2$. Substituting $\omega = v/r$:

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{2}{5}mr^2\cdot\frac{v^2}{r^2} = \frac{7}{10}mv^2$

$v = \sqrt{\frac{10gh}{7}}$

Note that this is less than the speed of a frictionless sliding object ($v = \sqrt{2gh}$) because rotational kinetic energy is present.

Example 2: Cylinder vs. Sphere

Two objects—a solid cylinder ($I = \frac{1}{2}mr^2$) and a solid sphere ($I = \frac{2}{5}mr^2$)—roll down the same incline from rest. Which reaches the bottom first?

The cylinder has a smaller moment of inertia, so less energy goes into rotation. Its final speed is:

$v_{\text{cyl}} = \sqrt{\frac{4gh}{3}}$

The sphere's speed is $v_{\text{sph}} = \sqrt{\frac{10gh}{7}} \approx 1.195\sqrt{gh}$, while the cylinder's is $v_{\text{cyl}} \approx 1.155\sqrt{gh}$. The sphere is slightly faster, but both are slower than a sliding object.

References

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

AI Disclosure

This article was drafted with AI assistance from class notes. All mathematical derivations and physical reasoning were verified against the source materials. The worked examples are original applications of the principles cited.