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 mathematical framework from first principles and works through concrete examples to illustrate how the no-slip condition emerges from physical constraints and how it simplifies energy and dynamics calculations.

Background

Understanding rolling motion requires integrating two concepts: the translational motion of an object's center of mass and its rotation about that center ^{[center-of-mass-motion]}. When an object rolls on a surface, these two motions are not independent. The no-slip condition imposes a kinematic constraint that relates them.

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

Key Results

The No-Slip Constraint

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

$v = r\omega$

where $r$ 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, these must cancel for the no-slip condition to hold. If the center moves forward at speed $v$ and the wheel rotates with angular velocity $\omega$, the contact point moves backward (relative to the center) at speed $r\omega$. Setting these equal gives $v = r\omega$.

Kinetic Energy of a Rolling Object

The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy:

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

where $m$ is the mass, $I$ is the moment of inertia about the center of mass, and $v$ and $\omega$ are related by the no-slip condition.

Substituting $\omega = v/r$:

$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2 = \frac{1}{2}v^2\left(m + \frac{I}{r^2}\right)$

This form shows that rolling motion can be treated as an effective translational motion with an effective mass $m_{eff} = m + I/r^2$.

Worked Examples

Example 1: A Solid Cylinder Rolling Down an Incline

Setup: A 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 to the bottom. Find the speed of the center of mass at the bottom.

Solution:

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

Using energy conservation, the initial gravitational potential energy converts to kinetic energy:

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

Substitute $I = \frac{1}{2}mr^2$ and the no-slip condition $\omega = v/r$ ^{[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 speed is less than that of a frictionless sliding object ($v_{slide} = \sqrt{2gh}$) because some energy goes into rotational motion. The ratio $v/v_{slide} = \sqrt{2/3} \approx 0.816$.

Example 2: Comparing Rolling Objects

Setup: A solid sphere, a solid cylinder, and a thin spherical shell, all with the same mass and radius, roll without slipping down the same incline from rest. Which reaches the bottom first?

Solution:

The moments of inertia are:

• Solid sphere: $I_s = \frac{2}{5}mr^2$
• Solid cylinder: $I_c = \frac{1}{2}mr^2$
• Thin spherical shell: $I_{shell} = \frac{2}{3}mr^2$

Using the energy equation $mgh = \frac{1}{2}v^2\left(m + \frac{I}{r^2}\right)$:

$v = \sqrt{\frac{2gh}{1 + I/(mr^2)}}$

For each object:

• Sphere: $v_s = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{10gh}{7}}$
• Cylinder: $v_c = \sqrt{\frac{2gh}{1 + 1/2}} = \sqrt{\frac{4gh}{3}}$
• Shell: $v_{shell} = \sqrt{\frac{2gh}{1 + 2/3}} = \sqrt{\frac{6gh}{5}}$

Numerically: $v_s > v_c > v_{shell}$.

Key insight: Objects with more mass concentrated near the center (lower moment of inertia) roll faster. The sphere, having the lowest $I/(mr^2)$ ratio, reaches the bottom first, despite all three starting from the same height.

References

^{[rolling-without-slipping]}

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

AI Disclosure

This article was drafted with AI assistance. The mathematical derivations, worked examples, and explanations were generated based on the provided course notes and standard physics pedagogy. All factual claims are cited to the source notes. The article has been reviewed for technical accuracy and clarity.