Rolling Without Slipping: Deriving the Constraint Relationship

Abstract

Rolling without slipping is a fundamental kinematic constraint in mechanics that couples linear and rotational motion. This article derives the constraint equation $v = r\omega$ from first principles, explains its physical meaning, and demonstrates how it simplifies the analysis of rolling systems. The constraint emerges naturally when friction prevents relative motion at the contact point, reducing degrees of freedom and enabling efficient energy analysis.

Background

Understanding rigid body motion requires decomposing complex trajectories into simpler components. ^{[Any rigid body's motion can be separated into translation of its center of mass and rotation about that center]}. These two motions occur simultaneously and independently, allowing us to analyze rolling objects by tracking both the path of the center of mass and the body's spin rate.

Rolling is ubiquitous in engineering and nature—wheels, balls, cylinders, and countless mechanical systems rely on rolling motion. However, not all rolling is the same. An object can roll while simultaneously sliding (like a skidding tire), or it can roll without slipping, where the contact point has zero velocity relative to the surface. The latter case, rolling without slipping, is the focus here because it represents the idealized, efficient regime where friction prevents sliding and energy is conserved more effectively.

Key Results

The Rolling Without Slipping Constraint

^{[For an object rolling without slipping, the linear velocity $v$ of the center of mass and angular velocity $\omega$ are related by:]}

$v = r\omega$

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

Derivation from the Contact Point Condition

The constraint emerges from a single physical requirement: the contact point has zero velocity relative to the surface. Let us derive this rigorously.

Consider a rigid body rolling on a flat surface. The velocity of any point on the body can be expressed as the sum of two contributions:

1. The translational velocity of the center of mass, $\vec{V}_{cm}$
2. The rotational velocity about the center of mass, $\vec{\omega}_{cm} \times \vec{r}_{rel}$, where $\vec{r}_{rel}$ is the position vector from the center of mass to that point.

For a point at the contact (bottom) of the rolling object, the position vector from the center of mass points downward with magnitude $r$. If we align our coordinate system so that the center of mass moves in the positive $x$-direction with speed $v$, and the body rotates about the $z$-axis with angular velocity $\omega$, then:

• Translational contribution to velocity at contact: $v$ (in the $x$-direction)
• Rotational contribution to velocity at contact: $\omega \times r$ (in the negative $x$-direction, since the bottom of a rolling wheel moves backward relative to the center)

The total velocity of the contact point is:

$v_{contact} = v - \omega r$

For rolling without slipping, the contact point must be stationary relative to the surface:

$v_{contact} = 0$

Therefore:

$v = \omega r$

This is the rolling without slipping constraint. It is a kinematic relationship—a purely geometric condition that does not depend on forces or energy, though it has profound consequences for both.

Physical Interpretation

^{[The constraint reduces degrees of freedom in the problem: once you know how fast the center of mass moves, you automatically know the spin rate, and vice versa]}. This reduction is powerful because it simplifies analysis. Instead of treating translational and rotational motion as independent, we can express one in terms of the other.

The constraint also ensures energy efficiency. ^{[Unlike sliding friction, which dissipates energy, rolling friction is minimal because the contact point isn't sliding]}. This is why wheels are so effective in vehicles and machinery—the no-slip condition minimizes energy loss to friction.

Worked Example: Rolling Cylinder Down an Incline

Consider a solid cylinder of mass $m$ and radius $r$ released from rest at the top of a frictionless incline of height $h$. Assume it rolls without slipping.

Step 1: Apply energy conservation.

The initial potential energy converts to kinetic energy of translation and rotation:

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

For a solid cylinder, $I = \frac{1}{2}mr^2$.

Step 2: Apply the rolling constraint.

From $v = r\omega$, we have $\omega = \frac{v}{r}$.

Step 3: Substitute into the energy equation.

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

$mgh = \frac{1}{2}m v^2 + \frac{1}{4}m v^2$

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

Step 4: Solve for velocity.

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

The rolling constraint allowed us to eliminate $\omega$ and solve the problem with a single energy equation. Without the constraint, we would need additional information about the torques and forces acting on the cylinder.

References

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

AI Disclosure

This article was drafted with the assistance of an AI language model based on class notes and source materials. All mathematical derivations and physical reasoning have been reviewed for technical accuracy against the cited source notes. The worked example is original but follows standard mechanics pedagogy.