Rolling

AP Physics C: Mechanics· 20 min read

1. Pure Rolling Without Slipping Fundamentals★★☆☆☆⏱ 5 min

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.

📘 Definition

Pure Rolling Condition

$v_{c m} = R ω$

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.

🔬 Derivation

Goal:

Derive the pure rolling velocity condition

Starting from:

A wheel of radius R moving right with center of mass speed $v_{c m}$ , and rotating clockwise with angular speed $ω$

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

Result:

The pure rolling condition is $v_{c m} = R ω$

📐 Worked Example

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

1
Identify given values: $R = 0.4 m$ , $v_{c m} = 12 m/s$
2
Rearrange pure rolling condition to solve for $ω$
$\omega = \frac{v_{cm}}{R}$
3
Substitute values: $ω = 12/0.4 = 30 rad/s$

✓ Quick check

What is the velocity of the top point of a pure rolling wheel relative to the ground?
Reveal answer
$2v_{cm}$ —
The top point adds the full center of mass velocity plus the tangential rotation velocity $R ω = v_{c m}$ in the same direction, for a total of $2 v_{c m}$ .

2. Total Kinetic Energy of Rolling Objects★★★☆☆⏱ 6 min

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 $v_{c m} = R ω$ condition to express total KE entirely in terms of $v_{c m}$ or entirely in terms of $ω$ .

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

📐 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.

3. Friction Behavior in Rolling Motion★★★★☆⏱ 5 min

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.

Methods compared

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: $m g sin θ - f_{s} = m a_{c m}$
2
Write torque equation around center of mass: $f_{s} R = I α = 0.5 m R^{2} (a_{c m} / R)$
3
Simplify torque equation: $f_{s} = 0.5 m a_{c m}$
4
Substitute into translation equation: $m g sin 30 = 1.5 m a_{c m}$ , so $a_{c m} = g /3$
5
Solve for friction: $f_{s} = 0.5 * 3 * (9.8/3) = 4.9 N$

4. Rolling Acceleration on Inclined Planes★★★★☆⏱ 4 min

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 $I / (m R^{2})$ ) 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	$I / (m R^{2})$ Value	Acceleration Down Incline
Solid Sphere	2/5	$(5/7) g sin θ$
Solid Cylinder / Disk	1/2	$(2/3) g sin θ$
Thin Hoop	1	$(1/2) g sin θ$
Sliding Block (no rotation)	0	$g sin θ$

📐 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: $a = (2/3) g sin 2 0^{\circ} \approx 4.47 m/s^{2}$
2
Apply translational kinematics: $v_{c m}^{2} = u^{2} + 2 a s$
3
Substitute values: $v_{c m} = 2 * 4.47 * 3 \approx 5.18 m/s$

5. Common Pitfalls

Wrong move:

Forgetting to add rotational KE to translational KE when calculating total energy of a rolling object

Why:

Students treat rolling as pure sliding, losing 50% of points on standard AP energy conservation FRQ questions

Correct move:

Always split total KE into translational (center of mass) and rotational (around center of mass) components for all rolling rigid bodies

Wrong move:

Using kinetic friction for pure rolling problems

Why:

No relative motion exists at the contact point, so only static friction acts, and it does no net work

Correct move:

Reserve kinetic friction only for rolling with slipping scenarios where the contact point slides relative to the surface

Wrong move:

Applying $v_{c m} = R ω$ to rolling with slipping scenarios

Why:

The pure rolling condition only holds when the contact point has zero velocity relative to the surface

Correct move:

Only use $v_{c m} = R ω$ for confirmed pure rolling, otherwise solve linear and angular motion equations separately

Wrong move:

Assuming all rolling objects have the same acceleration down an incline

Why:

Acceleration depends on the dimensionless moment of inertia factor $I / (m R^{2})$ , not mass or radius individually

Correct move:

Use the derived $a_{c m} = g sin θ / (1 + I / (m R^{2}))$ formula to compare accelerations for different shapes

Wrong move:

Calculating the top point of a pure rolling wheel as moving at $v_{c m}$ relative to the ground

Why:

The top point's tangential rotation velocity adds to the center of mass velocity, giving double the speed

Correct move:

Use the instantaneous axis of rotation at the contact point to quickly find velocities for any point on the rolling object

6. Quick Reference Cheatsheet

Quantity	Pure Rolling Formula	Exam Notes
Center of mass velocity	$v_{c m} = R ω$	No slipping required to apply
Total kinetic energy	$K_{t o t a l} = \frac{1}{2} m v_{c m}^{2} + \frac{1}{2} I_{c m} ω^{2}$	Sum of translational + rotational KE
Acceleration down incline	$a_{c m} = \frac{g s i n θ}{1 + I / ( m R ^{2} )}$	No slipping, no air resistance
Top point velocity	$v_{t o p} = 2 v_{c m}$	Relative to fixed ground frame
Contact point velocity	$v_{co n t a c t} = 0$	Pure rolling, no relative motion

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Physics C FRQ 1
Rolling object down rough incline
2021 · AP Physics C MCQ 2
Rolling kinetic energy comparison
2019 · AP Physics C FRQ 3
Rolling with slipping friction analysis

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.