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, connects it to the broader framework of rigid-body kinematics, and works through concrete examples to build intuition for how this principle governs real systems.

Background

Understanding how objects move requires separating the motion into simpler components. ^{[center-of-mass-motion]} establishes that any rigid body's motion can be decomposed into translation of its center of mass and rotation about that center. When a wheel rolls down a slope or a ball rolls across a floor, both modes of motion occur simultaneously—the center of mass translates forward while the body spins.

The rolling-without-slipping condition is a kinematic constraint that relates these two modes. ^{[rolling-without-slipping]} defines this constraint mathematically and explains why it matters: when an object rolls without slipping, mechanical energy is conserved more efficiently than when sliding occurs, because no kinetic energy is dissipated at the contact point.

Key Results

The Rolling Constraint

For an object of radius $r$ rolling without slipping on a surface, the linear velocity of the center of mass $v$ and the angular velocity $\omega$ satisfy ^{[rolling-without-slipping]}:

$v = r\omega$

This equation encodes a geometric fact: if the object rotates through angle $\theta$, a point on its rim travels arc length $s = r\theta$. For rolling without slipping, the center of mass must advance by exactly this distance. Over a time interval $dt$:

$v \, dt = r \, \omega \, dt$

which yields the constraint directly.

Why This Matters

The no-slip condition eliminates one degree of freedom from the system. Instead of specifying both $v$ and $\omega$ independently, specifying one determines the other. This simplification is crucial for solving dynamics problems: it reduces the number of unknowns and often allows energy methods to be applied more directly.

Worked Examples

Example 1: Solid Cylinder Rolling Down an Incline

Setup: A uniform solid cylinder 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 slope. Find the speed of its center of mass at the bottom.

Solution:

Since the cylinder rolls without slipping, we use energy conservation. The gravitational potential energy converts into translational kinetic energy of the center of mass and rotational kinetic energy about the center of mass.

Initial energy: $E_i = mgh$ (taking the bottom as reference).

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

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

Apply the rolling constraint ^{[rolling-without-slipping]}: $v = r\omega$, so $\omega = \frac{v}{r}$.

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

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

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

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

Key insight: The final speed is less than $\sqrt{2gh}$ (the speed of a frictionless sliding object) because some energy goes into rotation. The rolling constraint couples the two motions, so rotational inertia effectively increases the system's resistance to acceleration.

Example 2: Identifying the No-Slip Condition

Setup: A wheel of radius $r = 0.5$ m rotates at $\omega = 10$ rad/s. Does it roll without slipping on a surface if its center moves at $v = 4$ m/s?

Solution:

Check whether $v = r\omega$ ^{[rolling-without-slipping]}:

$r\omega = 0.5 \times 10 = 5 \text{ m/s}$

Since $v = 4$ m/s $\neq 5$ m/s, the wheel does not roll without slipping. The center of mass is moving slower than the rolling constraint requires, which means the wheel is spinning faster than its forward motion justifies—the wheel is slipping backward relative to the surface.

Key insight: The constraint is an equality, not an inequality. Deviations in either direction indicate slipping, which introduces friction losses and complicates the energy analysis.

Example 3: Two Objects, Different Moments of Inertia

Setup: A solid sphere and a thin spherical shell, both of mass $m$ and radius $r$, roll without slipping down the same incline from height $h$. Which reaches the bottom first?

Solution:

Use energy conservation for each. The moments of inertia are:

• Solid sphere: $I_s = \frac{2}{5}mr^2$
• Thin shell: $I_{sh} = \frac{2}{3}mr^2$

For the solid sphere: $mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v_s}{r}\right)^2 = \frac{1}{2}mv_s^2 + \frac{1}{5}mv_s^2 = \frac{7}{10}mv_s^2$

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

For the thin shell: $mgh = \frac{1}{2}mv_{sh}^2 + \frac{1}{2}\left(\frac{2}{3}mr^2\right)\left(\frac{v_{sh}}{r}\right)^2 = \frac{1}{2}mv_{sh}^2 + \frac{1}{3}mv_{sh}^2 = \frac{5}{6}mv_{sh}^2$

$v_{sh} = \sqrt{\frac{6gh}{5}}$

Since $\frac{10}{7} \approx 1.43 > \frac{6}{5} = 1.2$, we have $v_s > v_{sh}$. The solid sphere reaches the bottom faster.

Key insight: Objects with smaller moments of inertia (more mass concentrated near the center) roll faster down an incline. The rolling constraint ^{[rolling-without-slipping]} couples translational and rotational motion, so the distribution of mass matters—it is not just about total mass.

References

^{[rolling-without-slipping]}

^{[center-of-mass-motion]}

AI Disclosure

This article was drafted with AI assistance. The structure, worked examples, and pedagogical framing were generated based on the source notes. All mathematical derivations and numerical results were verified against the source material. The article represents an original synthesis intended for publication on a personal scholarly site.