Rolling Without Slipping: Deriving the Constraint from First Principles

Abstract

Rolling without slipping is a fundamental kinematic constraint in mechanics that couples linear and rotational motion. This article derives the constraint relationship $v = r\omega$ from the condition that the contact point between a rolling object and a surface has zero velocity, then explores its physical implications and worked examples. The derivation clarifies why this constraint emerges naturally under sufficient friction and why it simplifies energy analysis in rolling systems.

Background

The motion of any rigid body can be decomposed into two independent components: translation of its center of mass and rotation about that center ^{[center-of-mass-motion]}. When a wheel rolls down a road or a ball rolls across a table, both motions occur simultaneously. The center of mass translates with velocity $\vec{V}_{cm}$, while the body rotates about that center with angular velocity $\vec{\omega}_{cm}$.

Rolling without slipping is a kinematic constraint that relates these two motions ^{[rolling-without-slipping]}. Unlike a sliding object (where the contact point moves relative to the surface) or a spinning object (where the center of mass is stationary), a rolling object satisfies a special condition: the point of contact between the rolling object and the surface has zero velocity relative to the surface. This constraint is not a law of nature but rather a consequence of sufficient friction preventing relative motion at the contact point.

Key Results

Derivation of the Rolling Constraint

Consider a rigid body of radius $r$ rolling on a flat surface. At any instant, the body has:

• Linear velocity of the center of mass: $v$ (in the direction of motion)
• Angular velocity about the center of mass: $\omega$ (perpendicular to the plane of motion)

The velocity of any point on the body is the sum of the translational velocity of the center of mass and the rotational velocity about the center of mass. For a point at position $\vec{r}$ relative to the center of mass:

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

At the contact point (the point touching the surface), the position vector relative to the center of mass points downward with magnitude $r$. The rotational contribution to velocity at this point is $\omega r$ in the direction opposite to the center-of-mass motion. For the contact point to have zero velocity (the no-slip condition):

$v - \omega r = 0$

Therefore:

$v = r\omega$

This is the rolling-without-slipping constraint ^{[rolling-without-slipping]}.

Physical Interpretation

The constraint $v = r\omega$ means that once the center-of-mass velocity is specified, the angular velocity is completely determined, and vice versa. This reduces the degrees of freedom in the problem. Rather than treating translation and rotation as independent, they become coupled by a single relationship.

The no-slip condition also has important energy implications. In sliding friction, kinetic energy is dissipated at the contact point because relative motion occurs there. In rolling without slipping, the contact point has zero velocity, so no energy is dissipated by friction at that point ^{[rolling-without-slipping]}. This makes rolling far more efficient than sliding, which is why wheels are ubiquitous in transportation and machinery.

Worked Examples

Example 1: A Wheel Rolling Down an Incline

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

Solution:

Using the constraint $v = r\omega$, we can differentiate with respect to time:

$a = r\alpha$

where $a$ is the linear acceleration of the center of mass and $\alpha$ is the angular acceleration.

Applying Newton's second law in the direction along the incline:

$mg\sin\theta - f = ma$

where $f$ is the friction force. For rotation about the center of mass:

$\tau = I\alpha$

The torque due to friction is $\tau = fr$, and for a solid disk, $I = \frac{1}{2}mr^2$:

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

Substituting $\alpha = a/r$:

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

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

Substituting back into the force equation:

$mg\sin\theta - \frac{1}{2}ma = ma$

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

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

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

Example 2: Comparing Rolling and Sliding

A sphere of mass $m$ and radius $r$ is released from rest at the top of an incline of height $h$. Compare the final velocity of the center of mass for rolling without slipping versus frictionless sliding.

Solution:

For frictionless sliding, energy conservation gives:

$mgh = \frac{1}{2}mv_{\text{slide}}^2$

$v_{\text{slide}} = \sqrt{2gh}$

For rolling without slipping, the total kinetic energy includes both translational and rotational parts:

$mgh = \frac{1}{2}mv_{\text{roll}}^2 + \frac{1}{2}I\omega^2$

For a solid sphere, $I = \frac{2}{5}mr^2$. Using the constraint $v = r\omega$:

$mgh = \frac{1}{2}mv_{\text{roll}}^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \frac{v_{\text{roll}}^2}{r^2}$

$mgh = \frac{1}{2}mv_{\text{roll}}^2 + \frac{1}{5}mv_{\text{roll}}^2$

$mgh = \frac{7}{10}mv_{\text{roll}}^2$

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

The rolling sphere reaches the bottom with a lower velocity than the sliding block because kinetic energy is partitioned between translation and rotation. The ratio is:

$\frac{v_{\text{roll}}}{v_{\text{slide}}} = \sqrt{\frac{10}{14}} \approx 0.845$

References

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

AI Disclosure

This article was drafted with AI assistance. The structure, mathematical derivations, and worked examples were generated based on class notes and verified against the source material. All factual claims are cited to the original notes. The article has been reviewed for technical accuracy and clarity.