Physics: Worked Example Walkthroughs — Rolling Without Slipping

Abstract

Rolling without slipping is a foundational constraint in classical mechanics that couples translational and rotational motion. This article develops the constraint condition from first principles, explains its physical meaning, and works through representative examples to build intuition for analyzing rolling systems in engineering and physics.

Background

The motion of rigid bodies in space can be decomposed into two independent components: translation of the center of mass and rotation about the center of mass ^{[center-of-mass-motion]}. When an object rolls on a surface, these two motions are not independent—they are linked by a kinematic constraint.

The rolling-without-slipping condition arises when the contact point between a rolling object and a surface has zero velocity relative to the surface ^{[rolling-without-slipping]}. This is distinct from sliding, where the contact point moves along the surface. In rolling without slipping, no kinetic energy is dissipated at the contact point, making the motion more efficient than sliding friction would allow.

Key Results

The Rolling Constraint

For a rigid body 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 satisfy ^{[rolling-without-slipping]}:

$v = r\omega$

This relationship holds instantaneously at every moment during rolling motion. It is a kinematic constraint—a restriction on allowable velocities—not a dynamical equation derived from forces.

Derivation sketch: Consider a point on the rim of the rolling object. In the reference frame of the center of mass, this point moves in a circle of radius $r$ with speed $r\omega$. In the lab frame, the center of mass moves with velocity $v$. The contact point (instantaneously at the bottom) must have zero velocity in the lab frame. By vector addition, the velocity of the contact point in the lab frame is the sum of the center-of-mass velocity and the velocity of that point relative to the center of mass. Setting this sum to zero yields $v = r\omega$.

Physical Interpretation

Rolling without slipping is more efficient than sliding because no energy is lost to friction at the contact point ^{[rolling-without-slipping]}. The friction force that enforces the constraint does no work (since the contact point is instantaneously at rest), so mechanical energy is conserved. This makes rolling the preferred mode of motion in wheels, bearings, and other mechanical systems where efficiency matters.

Worked Examples

Example 1: A Disk Rolling Down an Incline

Setup: A uniform disk of mass $m$ and radius $r$ starts from rest at the top of a frictionless incline of height $h$. It rolls without slipping down the incline. Find the speed of the center of mass at the bottom.

Solution:

By energy conservation, the initial potential energy converts to translational and rotational kinetic energy:

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

For a uniform disk, $I = \frac{1}{2}mr^2$. Applying the rolling constraint $v = r\omega$ ^{[rolling-without-slipping]}:

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

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

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

Key insight: The center-of-mass speed is less than $\sqrt{2gh}$ (the speed of a frictionless sliding object) because kinetic energy is partitioned between translation and rotation. The rolling constraint couples these motions, reducing the translational speed.

Example 2: Comparing Rolling and Sliding

Setup: Two identical spheres start from rest at the top of the same incline of height $h$. One rolls without slipping; the other slides without friction. Compare their speeds at the bottom.

Solution:

For the sliding sphere, all potential energy becomes translational kinetic energy: $v_{\text{slide}} = \sqrt{2gh}$

For the rolling sphere with $I = \frac{2}{5}mr^2$: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

$v_{\text{roll}} = \sqrt{\frac{10gh}{7}} \approx 0.845\sqrt{2gh}$

The rolling sphere is slower because energy is shared with rotational motion. The ratio $v_{\text{roll}}/v_{\text{slide}} = \sqrt{5/7} \approx 0.845$ is independent of mass and radius—it depends only on the moment of inertia distribution.

Example 3: Acceleration of a Rolling Cylinder

Setup: A cylinder of mass $m$ and radius $r$ rolls without slipping on a horizontal surface under an applied horizontal force $F$ at the center of mass. Find the acceleration of the center of mass.

Solution:

The equations of motion are:

• Translational: $F = ma$
• Rotational about the center of mass: $\tau = I\alpha$

The torque is provided by friction at the contact point. Let $f$ be the friction force (pointing opposite to the applied force). Then: $\tau = fr = I\alpha$

The rolling constraint $a = r\alpha$ ^{[rolling-without-slipping]} gives $\alpha = a/r$: $fr = I \cdot \frac{a}{r}$ $f = \frac{Ia}{r^2}$

Substituting into the translational equation: $F - f = ma$ $F - \frac{Ia}{r^2} = ma$ $a = \frac{F}{m + I/r^2}$

For a solid cylinder, $I = \frac{1}{2}mr^2$: $a = \frac{F}{m + m/2} = \frac{2F}{3m}$

Key insight: The effective inertia is increased by the rotational inertia. The friction force is not an external dissipative force; it is a constraint force that couples translation and rotation.

References

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

AI Disclosure

This article was drafted with AI assistance. The structure, worked examples, and explanations were generated from class notes using large language models. All mathematical claims and physical principles are cited to source notes and reflect standard classical mechanics. The worked examples follow conventional problem-solving approaches found in undergraduate physics texts. A human instructor reviewed the technical content for accuracy.