Rolling

AP Physics 1· 12 min read

1. The Pure Rolling Condition★★☆☆☆⏱ 3 min

Pure rolling describes ideal motion where a round object moves across a surface without sliding at the point of contact. The bottom of the object touching the ground is instantaneously at rest relative to the surface, so no kinetic friction acts at the contact point.

📘 Definition

Pure Rolling Condition

$v_{c m} = R ω$

The center of mass translational velocity equals the product of the object's radius and its angular velocity, for no slip motion.

🔬 Derivation

Goal:

Prove the pure rolling velocity relation

Starting from:

The point of contact has zero total velocity relative to the ground

1
Total velocity of contact point = translational velocity of center of mass + tangential rotational velocity at contact point
2
For zero relative motion: $v_{c m} - R ω = 0$ (tangential velocity points opposite to translational motion at the bottom)
3
Rearranging gives the standard pure rolling relation $v_{c m} = R ω$

Result:

This relation only holds when no slipping occurs at the contact surface.

📐 Worked Example

A solid disk of radius 0.2 m rolls without slipping with a center of mass speed of 3 m/s. Calculate its angular velocity.

1
Start with the pure rolling condition: $v_{c m} = R ω$
2
Rearrange to solve for angular velocity: $ω = \frac{v _{c m}}{R}$
3
$\omega = \frac{3 \text{ m/s}}{0.2 \text{ m}} = 15 \text{ rad/s}$

Exam tip:

AP exam questions almost always explicitly state 'rolling without slipping' to signal you can use the $v_{c m} = R ω$ relation.

2. Total Kinetic Energy of Rolling Objects★★★☆☆⏱ 4 min

A rolling object has two independent forms of kinetic energy: translational kinetic energy from the motion of its entire center of mass, and rotational kinetic energy from its spinning motion around its center of mass.

K_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2

📐 Worked Example

A 2 kg solid sphere (moment of inertia $I = \frac{2}{5} M R^{2}$ ) of radius 0.1 m rolls without slipping at 5 m/s. Calculate its total kinetic energy.

1
First calculate translational kinetic energy: $K_{t r an s} = \frac{1}{2} M v_{c m}^{2} = 0.5 * 2 * 5^{2} = 25$ J
2
Use pure rolling to find $ω = 5/0.1 = 50$ rad/s, then calculate rotational KE: $K_{r o t} = 0.5 * (0.4 * 2 * 0. 1^{2}) * 5 0^{2} = 10$ J
3
Sum the two components: $K_{t o t a l} = 25 + 10 = 35$ J

✓ Quick check

Confirm you can distinguish KE components for rolling objects:

Which fraction of a solid sphere's total rolling KE is rotational?
- 1/5
- 2/7
- 1/2
- 2/5
Reveal answer
2/7 —
The ratio $I / (M R^{2}) = 2/5$ , so rotational KE is 2/7 of total KE.

3. Rolling Down an Incline★★★★☆⏱ 4 min

🚫 No Calculator

For pure rolling down an incline, static friction does no net work because the contact point has zero displacement, so mechanical energy is fully conserved. Gravitational potential energy at the top of the incline converts entirely to total kinetic energy at the bottom.

🔬 Derivation

Goal:

Find final speed of a rolling object at the bottom of an incline of height h

Starting from:

Gravitational potential energy at top = total kinetic energy at bottom

1
$Mgh = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2$
2
Substitute $ω = v_{c m} / R$ and factor out $\frac{1}{2} M v_{c m}^{2}$
3
$Mgh = \frac{1}{2}Mv_{cm}^2 \left(1 + \frac{I}{MR^2}\right)$

Result:

Mass and radius cancel out, so final speed $v = \frac{2 g h}{1 + I / ( M R ^{2} )}$ depends only on the shape of the object.

📐 Worked Example

Compare the final speeds of a solid sphere and a hollow hoop rolling down a 3 m high incline without slipping.

1
For solid sphere: $I / (M R^{2}) = 2/5$ , so $v_{s p h er e} = 2 * 9.8 * 3/ (1 + 0.4) \approx 6.48$ m/s
2
For hollow hoop: $I / (M R^{2}) = 1$ , so $v_{h oo p} = 2 * 9.8 * 3/2 \approx 5.42$ m/s
3
The solid sphere reaches the bottom faster, as less of its energy is stored as rotational KE.

4. Static Friction and Slipping★★★★☆⏱ 3 min

Static friction is required to generate the torque that spins up the rolling object as it accelerates down the incline. If the maximum available static friction force is too low, the object will slip, and the pure rolling condition no longer holds.

📘 Definition

Slipping Motion

Occurs when the applied torque from static friction is not large enough to maintain $a_{c m} = R α$ , so the contact point slides relative to the surface.

📐 Worked Example

Calculate the minimum coefficient of static friction required for a solid sphere to roll without slipping down a 30° incline.

1
Use linear Newton's second law: $M g sin θ - f_{s} = M a_{c m}$
2
Use rotational Newton's second law: $f_{s} R = I α$ , substitute $I = 2/5 M R^{2}$ and $a_{c m} = R α$
3
Solve to find $f_{s} = \frac{2}{7} M g sin θ$ , set $f_{s} \leq μ_{s} M g cos θ$ , so $μ_{s} \geq \frac{2}{7} tan 3 0^{\circ} \approx 0.165$

5. Common Pitfalls

Wrong move:

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

Why:

Students incorrectly treat rolling as pure translational motion, leading to overestimated final speeds

Correct move:

Always split kinetic energy into separate translational and rotational components for no-slip rolling.

Wrong move:

Applying the $v_{c m} = R ω$ relation to slipping objects

Why:

This relation only holds when there is zero relative motion at the contact point

Correct move:

Only use $v_{c m} = R ω$ if the problem explicitly states rolling without slipping.

Wrong move:

Including a non-zero work term for static friction during pure rolling

Why:

Static friction acts at a point with zero instantaneous displacement, so it does no net work

Correct move:

Apply conservation of mechanical energy directly, no extra work term for static friction.

Wrong move:

Assuming all objects of the same mass roll down an incline at the same speed

Why:

Final speed depends only on the moment of inertia ratio $I / (M R^{2})$ , mass and radius cancel out completely

Correct move:

Rank speeds by comparing the $I / (M R^{2})$ factor, ignore mass and radius for identical shape objects.

Wrong move:

Treating rolling friction as equivalent to kinetic sliding friction

Why:

Rolling friction is a tiny dissipative force from surface deformation, not the same as sliding friction

Correct move:

AP Physics 1 problems almost always ignore rolling friction unless it is explicitly mentioned.

6. Quick Reference Cheatsheet

Quantity	Pure Rolling Formula	Exam Note
Velocity relation	$v_{c m} = R ω$	Only valid for no-slip motion
Total kinetic energy	$K_{t o t a l} = \frac{1}{2} M v_{c m}^{2} + \frac{1}{2} I ω^{2}$	Sum of translational + rotational components
Final speed from height h	$v = \frac{2 g h}{1 + I / ( M R ^{2} )}$	Mass and radius cancel out
Min static friction for incline	$μ_{s} \geq \frac{t a n θ}{1 + M R ^{2} / I}$	Prevents slipping during acceleration

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 · FRQ 2
Rolling hoop on inclined plane
2021 · MCQ 12
Rolling sphere vs sliding block
2019 · FRQ 3
Rolling disk with applied torque

What's Next

Mastering rolling motion is a critical stepping stone to solving more complex rotational motion problems on the AP Physics 1 exam, including systems with combined translation and rotation, angular momentum conservation for rolling collisions, and rotational equilibrium scenarios. The no-slip condition you learned here will reappear frequently in FRQ and multi-part MCQ sets that combine multiple units, such as linking rolling kinematics to work and energy or gravitation. After completing this module, you should practice mixed problem sets that ask you to rank rolling object speeds, calculate total kinetic energy, and identify minimum friction values for no-slip motion to lock in your understanding before exam day.