# Rolling

> AP Physics C: Mechanics · AP Physics C Mechanics
> Source: https://www.owlsprep.com/study/ap-physics-c-mech-u5-rolling/

This module covers pure rolling conditions, rolling with slipping, total kinetic energy of rolling rigid bodies, friction in rolling, and standard problem frameworks for AP exam questions.

**Prerequisites:** [Master translational kinematics and Newton's laws for point masses](https://www.owlsprep.com/study/ap-physics-c-mech-u2-translational-kinematics/); [Understand rotational kinematics and moment of inertia definitions](https://www.owlsprep.com/study/ap-physics-c-mech-u5-rotational-kinematics/)

## Learning objectives

- Distinguish between pure rolling, rolling with slipping, and sliding motion
- Derive the linear-angular velocity relationship for pure rolling without slipping
- Calculate total kinetic energy of a rolling object as sum of translational and rotational components
- Solve standard rolling problems for inclined planes and horizontal applied forces

## Pure Rolling Without Slipping Fundamentals

Pure rolling is the most commonly tested rolling scenario on the AP exam, where no relative motion exists between the bottom of the rolling object and the surface it moves on. This condition eliminates sliding at the contact point, and creates a fixed relationship between the linear speed of the center of mass and the angular speed of rotation.

**Pure Rolling Condition** — For a round rigid object of radius R, the center of mass speed equals the product of radius and angular speed, so the contact point has zero net velocity relative to the surface.

*Notation:* $v_{cm} = R\omega$

**Derivation:** Derive the pure rolling velocity condition

*Starting from:* A wheel of radius R moving right with center of mass speed $v_{cm}$, and rotating clockwise with angular speed $\omega$

1. Velocity of contact point from translation alone: $+v_{cm}$ (right)
2. Velocity of contact point from rotation alone: $-R\omega$ (left, tangential direction)
3. Set net contact point velocity to 0 for no slipping: $v_{cm} - R\omega = 0$

*Conclusion:* The pure rolling condition is $v_{cm} = R\omega$

**Worked example:** A 0.4 m radius car wheel rolls purely at 12 m/s. Find its angular rotation rate.

1. Identify given values: $R = 0.4 \text{ m}$, $v_{cm} = 12 \text{ m/s}$
2. Rearrange pure rolling condition to solve for $\omega$

   $$\omega = \frac{v_{cm}}{R}$$
3. Substitute values: $\omega = 12 / 0.4 = 30 \text{ rad/s}$

**Check your understanding**

1. What is the velocity of the top point of a pure rolling wheel relative to the ground?

   *Why:* The top point adds the full center of mass velocity plus the tangential rotation velocity $R\omega = v_{cm}$ in the same direction, for a total of $2v_{cm}$.

## Total Kinetic Energy of Rolling Objects

A rolling object has two independent components of kinetic energy: translational kinetic energy associated with the motion of its entire center of mass, and rotational kinetic energy associated with its rotation around its own center of mass. For pure rolling, you can substitute the $v_{cm} = R\omega$ condition to express total KE entirely in terms of $v_{cm}$ or entirely in terms of $\omega$.

$$K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$

> **Energy Shortcut**
>
> For pure rolling, substitute $\omega = v_{cm}/R$ to rewrite total KE as $K_{total} = \frac{1}{2}mv_{cm}^2 \left(1 + \frac{I_{cm}}{mR^2}\right)$ to simplify energy conservation calculations.

**Worked example:** A 2 kg solid sphere (I = 2/5 mR²) and a 2 kg thin hoop (I = mR²) both roll purely at 5 m/s. Compare their total kinetic energies.

1. Calculate total KE for solid sphere first

   $$K_{sphere} = 0.5*2*(5)^2 + 0.5*(0.4*2*R^2)*(5/R)^2 = 25 + 10 = 35 \text{ J}$$
2. Calculate total KE for thin hoop next

   $$K_{hoop} = 0.5*2*(5)^2 + 0.5*(2*R^2)*(5/R)^2 = 25 + 25 = 50 \text{ J}$$
3. Result: The hoop has 15 J more total kinetic energy, as a larger fraction of its energy is stored in rotation.

**Exam command terms**

AP exam questions use specific cues to test rolling KE understanding:

- **Total kinetic energy** — You must include BOTH translational and rotational components, not just one *("Calculate the total kinetic energy of the rolling disk at the bottom of the incline")*

- **Translational kinetic energy** — Only the 1/2 m v_cm² term, no rotational contribution

## Friction Behavior in Rolling Motion

Friction is required to initiate pure rolling, but it does zero net work on a pure rolling object because there is no relative displacement at the contact point. For pure rolling, only static friction acts, not kinetic friction. Kinetic friction only appears when the object is slipping relative to the surface.

> **Common Misconception**
>
> Static friction for pure rolling does NOT dissipate energy as heat, unlike sliding friction. This is why a pure rolling object can travel a long distance without slowing down significantly.

**Comparing methods**

Compare friction behavior for different rolling scenarios:

- **Pure Rolling Without Slipping** — Static friction acts at the contact point, no relative motion, zero work done, no energy loss
  - Pros: Simple to model with energy conservation
  - Cons: Requires sufficient static friction coefficient to prevent slipping

- **Rolling With Slipping** — Kinetic friction acts at the contact point, relative motion exists, work done dissipates energy as heat
  - Pros: Easy to solve linear and angular motion equations separately
  - Cons: Cannot use $v_{cm} = R\omega$ directly

**Worked example:** Find the static friction force acting on a 3 kg solid cylinder (I = 0.5 mR²) rolling purely down a 30° incline.

1. Write Newton's second law for translation: $mg\sin\theta - f_s = ma_{cm}$
2. Write torque equation around center of mass: $f_s R = I \alpha = 0.5 m R^2 (a_{cm}/R)$
3. Simplify torque equation: $f_s = 0.5 m a_{cm}$
4. Substitute into translation equation: $mg\sin30 = 1.5 m a_{cm}$, so $a_{cm} = g/3$
5. Solve for friction: $f_s = 0.5 * 3 * (9.8/3) = 4.9 \text{ N}$

## Rolling Acceleration on Inclined Planes

For any pure rolling object down a fixed incline, the center of mass acceleration depends only on the shape of the object (via its moment of inertia factor $I/(mR^2)$) and the incline angle, not on the object's total mass or radius. This means a small solid sphere and a large solid sphere will roll down the incline at exactly the same rate.

| Object Shape | $I/(mR^2)$ Value | Acceleration Down Incline |
| --- | --- | --- |
| Solid Sphere | 2/5 | $(5/7)g\sin\theta$ |
| Solid Cylinder / Disk | 1/2 | $(2/3)g\sin\theta$ |
| Thin Hoop | 1 | $(1/2)g\sin\theta$ |
| Sliding Block (no rotation) | 0 | $g\sin\theta$ |

**Worked example:** A solid disk rolls purely down a 20° incline from rest. Find its speed after traveling 3 m along the surface.

1. Use acceleration formula for solid disk: $a = (2/3) g \sin 20^\circ \approx 4.47 \text{ m/s}^2$
2. Apply translational kinematics: $v_{cm}^2 = u^2 + 2 a s$
3. Substitute values: $v_{cm} = \sqrt{2 * 4.47 * 3} \approx 5.18 \text{ m/s}$

## Common pitfalls

- **Wrong:** Forgetting to add rotational KE to translational KE when calculating total energy of a rolling object
  - Why it fails: Students treat rolling as pure sliding, losing 50% of points on standard AP energy conservation FRQ questions
  - Correct: Always split total KE into translational (center of mass) and rotational (around center of mass) components for all rolling rigid bodies
- **Wrong:** Using kinetic friction for pure rolling problems
  - Why it fails: No relative motion exists at the contact point, so only static friction acts, and it does no net work
  - Correct: Reserve kinetic friction only for rolling with slipping scenarios where the contact point slides relative to the surface
- **Wrong:** Applying $v_{cm} = R\omega$ to rolling with slipping scenarios
  - Why it fails: The pure rolling condition only holds when the contact point has zero velocity relative to the surface
  - Correct: Only use $v_{cm} = R\omega$ for confirmed pure rolling, otherwise solve linear and angular motion equations separately
- **Wrong:** Assuming all rolling objects have the same acceleration down an incline
  - Why it fails: Acceleration depends on the dimensionless moment of inertia factor $I/(mR^2)$, not mass or radius individually
  - Correct: Use the derived $a_{cm} = g\sin\theta / (1 + I/(mR^2))$ formula to compare accelerations for different shapes
- **Wrong:** Calculating the top point of a pure rolling wheel as moving at $v_{cm}$ relative to the ground
  - Why it fails: The top point's tangential rotation velocity adds to the center of mass velocity, giving double the speed
  - Correct: Use the instantaneous axis of rotation at the contact point to quickly find velocities for any point on the rolling object

## Cheatsheet

| Quantity | Pure Rolling Formula | Exam Notes |
| --- | --- | --- |
| Center of mass velocity | $v_{cm} = R\omega$ | No slipping required to apply |
| Total kinetic energy | $K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$ | Sum of translational + rotational KE |
| Acceleration down incline | $a_{cm} = \frac{g\sin\theta}{1 + I/(mR^2)}$ | No slipping, no air resistance |
| Top point velocity | $v_{top} = 2v_{cm}$ | Relative to fixed ground frame |
| Contact point velocity | $v_{contact} = 0$ | Pure rolling, no relative motion |

## What's next

Mastering rolling motion is a critical bridge between translational and rotational dynamics, and it is a frequent anchor for AP Physics C FRQ questions that combine energy conservation, Newton's laws, and rotational kinematics. After completing this module, you will be ready to tackle more advanced rigid body problems including rotational collisions, angular momentum conservation for extended systems, and precessional motion. The concepts you learned here will also be reused heavily when you study simple harmonic motion for physical pendulums later in the unit, and they will appear in nearly all cumulative AP exam practice tests.

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