Physics: Worked Example Walkthroughs — Rolling Without Slipping

Abstract

Rolling without slipping is a fundamental constraint in classical mechanics that couples translational and rotational motion. This article develops the constraint condition from first principles, connects it to the decomposition of rigid-body motion, and works through concrete examples to build intuition for analyzing rolling systems.

Background

Understanding how objects move requires distinguishing between translation and rotation. ^{[center-of-mass-motion]} establishes that any rigid body's motion can be decomposed into two independent components: translation of the center of mass and rotation about the center of mass. When we observe a wheel rolling down a slope or a ball spinning across a floor, we are witnessing both modes simultaneously.

The constraint of rolling without slipping arises when the contact point between a rolling object and a surface has zero velocity relative to the surface. This is distinct from sliding, where the contact point moves along the surface. Rolling without slipping is the condition that couples the translational velocity of the center of mass to the angular velocity of the body.

Key Results

The Rolling Constraint

^{[rolling-without-slipping]} establishes the fundamental relationship for rolling without slipping:

$v = r\omega$

where $v$ is the linear velocity of the center of mass, $\omega$ is the angular velocity about the center of mass, and $r$ is the radius of the rolling object.

Derivation sketch: Consider a point on the rim of a rolling wheel. In the reference frame of the ground, this point's velocity is the vector sum of the center-of-mass velocity and the velocity due to rotation about the center of mass. At the contact point (bottom of the wheel), these two contributions must cancel for the point to remain stationary relative to the ground. If the center moves forward with speed $v$ and the wheel rotates with angular speed $\omega$, the rim point at the contact moves backward (relative to the center) with speed $r\omega$. Setting these equal ensures zero net velocity at the contact.

Energy Efficiency

Rolling without slipping is energetically favorable compared to sliding. When an object slides, kinetic energy is dissipated through friction at the contact point. In contrast, rolling without slipping involves static friction at the contact point, which does no work because the contact point has zero velocity. This means mechanical energy is conserved (in the absence of other dissipative forces), making rolling a more efficient mode of motion for wheels, balls, and similar objects.

Worked Examples

Example 1: A Uniform Cylinder Rolling Down an Incline

Problem: A solid cylinder of mass $m$ and radius $r$ starts from rest at the top of an incline of height $h$. Assuming it rolls without slipping, find the velocity of its center of mass at the bottom.

Solution:

We apply conservation of energy. The initial potential energy converts into translational kinetic energy of the center of mass and rotational kinetic energy about the center of mass.

Initial energy: $E_i = mgh$

Final energy: $E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a solid cylinder, the moment of inertia about the central axis is $I = \frac{1}{2}mr^2$.

Using the rolling constraint ^{[rolling-without-slipping]}, $\omega = v/r$:

$E_f = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2$

$E_f = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$

By conservation of energy: $mgh = \frac{3}{4}mv^2$

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

Physical insight: The factor $\frac{4}{3}$ (rather than $2$ for a purely sliding object) reflects the fact that some of the gravitational potential energy goes into rotational motion. A rolling object reaches the bottom slower than a sliding object because energy is partitioned between translation and rotation.

Example 2: Determining the Constraint Force

Problem: For the cylinder in Example 1, find the friction force acting on the cylinder as it rolls down the incline (angle $\theta$).

Solution:

We use Newton's second law for both translation and rotation. Let $f$ be the friction force (up the incline) and $a$ be the acceleration of the center of mass.

Translational equation (along the incline): $mg\sin\theta - f = ma$

Rotational equation (about the center of mass): $\tau = I\alpha$

The torque due to friction is $\tau = fr$, and the moment of inertia is $I = \frac{1}{2}mr^2$:

$fr = \frac{1}{2}mr^2 \cdot \alpha$

The rolling constraint ^{[rolling-without-slipping]} gives $a = r\alpha$, so $\alpha = a/r$:

$fr = \frac{1}{2}mr^2 \cdot \frac{a}{r}$

$f = \frac{1}{2}ma$

Substituting back into the translational equation: $mg\sin\theta - \frac{1}{2}ma = ma$

$mg\sin\theta = \frac{3}{2}ma$

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

Therefore: $f = \frac{1}{2}m \cdot \frac{2g\sin\theta}{3} = \frac{mg\sin\theta}{3}$

Physical insight: The friction force is not zero—it is necessary to generate the torque that accelerates the rotation. However, this friction does no work at the contact point because that point is stationary. The friction is static friction, which can be sustained as long as $f \leq \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N = mg\cos\theta$ is the normal force.

References

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

AI Disclosure

This article was drafted with the assistance of an AI language model based on personal class notes. The mathematical derivations, worked examples, and physical interpretations were generated and structured by the AI. All factual claims and equations are grounded in the cited notes and standard classical mechanics. The author is responsible for verification and accuracy of the final content.