Physics: Worked Example Walkthroughs—Rolling Motion and Rigid Body Decomposition

Abstract

This article presents a structured approach to understanding rolling motion and rigid body kinematics through worked examples. We examine how the decomposition of rigid body motion into translation and rotation ^{[center-of-mass-motion]} simplifies analysis, and how the rolling-without-slipping constraint ^{[rolling-without-slipping]} connects linear and angular velocities. Two detailed walkthroughs illustrate these principles in practice.

Background

Rigid body mechanics presents a conceptual challenge: objects in the real world rarely move in only one way. A wheel rolling down a hill translates and rotates. A thrown baseball follows a parabolic path while spinning. Analyzing such motion naively—tracking every point on the object—quickly becomes intractable.

The key insight is decomposition. Any rigid body motion can be understood as two independent, simultaneous processes: translation of the center of mass and rotation about the center of mass ^{[center-of-mass-motion]}. This separation allows us to apply familiar kinematics and dynamics to each component separately, then combine results.

Rolling motion introduces an additional constraint. When an object rolls without slipping on a surface, the contact point has zero velocity relative to that surface. This kinematic condition ^{[rolling-without-slipping]} links the linear velocity of the center of mass to the angular velocity:

$v = r\omega$

where $v$ is the center-of-mass velocity, $\omega$ is the angular velocity, and $r$ is the radius. This relationship is not a law of nature but a constraint—a consequence of sufficient friction preventing sliding. It reduces the degrees of freedom in the problem, making analysis tractable.

Key Results

Decomposition of Rigid Body Motion

A rigid body of mass $m$ undergoing general motion can be analyzed as ^{[center-of-mass-motion]}:

• Translation with velocity $\vec{V}_{cm}$
• Rotation about the center of mass with angular velocity $\vec{\omega}_{cm}$

These occur independently and simultaneously. The translational motion is governed by Newton's second law applied to the center of mass; the rotational motion is governed by the rotational equation of motion about the center of mass.

Rolling Without Slipping Constraint

When rolling without slipping occurs, the kinematic relationship ^{[rolling-without-slipping]} is:

$v = r\omega$

This constraint reduces the system's degrees of freedom. For a rolling object, specifying either $v$ or $\omega$ automatically determines the other. Physically, this occurs when friction is sufficient to prevent sliding; the contact point remains stationary relative to the surface.

Worked Examples

Example 1: A Solid Cylinder Rolling Down an Incline

Problem: A solid cylinder of mass $m$, radius $r$, and moment of inertia $I = \frac{1}{2}mr^2$ rolls without slipping down a frictionless incline at angle $\theta$. Find the acceleration of the center of mass.

Solution:

We decompose the motion into translation and rotation ^{[center-of-mass-motion]}. Let $a$ be the acceleration of the center of mass down the incline.

Translational equation: The forces acting on the cylinder are gravity (component $mg\sin\theta$ down the incline) and friction $f$ (up the incline, necessary to prevent sliding):

$mg\sin\theta - f = ma$

Rotational equation: Friction creates a torque about the center of mass:

$\tau = fr = I\alpha$

where $\alpha$ is the angular acceleration. Substituting $I = \frac{1}{2}mr^2$:

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

Constraint: Rolling without slipping ^{[rolling-without-slipping]} gives $v = r\omega$. Differentiating:

$a = r\alpha$

From the rotational equation: $f = \frac{I\alpha}{r} = \frac{\frac{1}{2}mr^2 \alpha}{r} = \frac{1}{2}mr\alpha = \frac{1}{2}ma$

Substituting into the translational equation:

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

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

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

Key insight: The acceleration is less than $g\sin\theta$ (which would apply to a sliding block) because rotational inertia "absorbs" some of the gravitational potential energy. The rolling constraint couples the two motions, reducing the net acceleration.

Example 2: A Ball Rolling Horizontally, Then Up an Incline

Problem: A uniform sphere (moment of inertia $I = \frac{2}{5}mr^2$) rolls without slipping on a horizontal surface with speed $v_0$. It then encounters a frictionless incline. How high does it rise before stopping?

Solution:

On the horizontal surface, the sphere rolls without slipping ^{[rolling-without-slipping]}, so $v_0 = r\omega_0$, giving $\omega_0 = \frac{v_0}{r}$.

The total kinetic energy is the sum of translational and rotational contributions:

$KE = \frac{1}{2}mv_0^2 + \frac{1}{2}I\omega_0^2 = \frac{1}{2}mv_0^2 + \frac{1}{2} \cdot \frac{2}{5}mr^2 \cdot \frac{v_0^2}{r^2}$

$KE = \frac{1}{2}mv_0^2 + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$

On the frictionless incline, the sphere cannot maintain the rolling constraint—there is no friction to enforce it. The sphere will slip. However, angular momentum about the contact point is conserved during the transition (the impulse from the incline acts at the contact point). For a frictionless incline, the sphere's rotational motion is unaffected by the surface, so $\omega$ remains $\omega_0$.

As the sphere rises, translational kinetic energy converts to gravitational potential energy, while rotational kinetic energy remains constant (no torque acts on the sphere on a frictionless surface):

$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega_0^2 = \frac{7}{10}mv_0^2$

At maximum height, $v = 0$:

$mgh + \frac{1}{5}mv_0^2 = \frac{7}{10}mv_0^2$

$mgh = \frac{1}{2}mv_0^2$

$h = \frac{v_0^2}{2g}$

Key insight: The sphere rises to a height determined by its translational kinetic energy alone; the rotational energy persists but does not contribute to height gain on a frictionless surface. This illustrates how decomposition ^{[center-of-mass-motion]} allows us to track each component of motion independently.

References

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

AI Disclosure

This article was drafted with the assistance of an AI language model. The structure, worked examples, and explanatory text were generated based on class notes provided as input. All mathematical derivations and physical reasoning have been reviewed for technical accuracy against the source notes. The article does not represent original research but rather a pedagogical synthesis of existing course material.