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What is the condition for rolling without slipping, and how do energy methods predict which object reaches the bottom of a ramp first?

Topic 6.5 Rolling: analyze objects that roll without slipping using the v = R omega condition and the partition of energy between translation and rotation.

A focused answer to AP Physics 1 Topic 6.5, covering rolling without slipping, the constraint v_cm = R omega, the total kinetic energy of a rolling object, why mass distribution decides the race down a ramp, and the role of static friction, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Rolling without slipping means the contact point is instantaneously at rest, which forces the constraint $v_{cm} = R\omega$ (and $a_{cm} = R\alpha$ ) linking the center-of-mass motion to the rotation. A rolling object has total kinetic energy $K = \tfrac{1}{2}mv_{cm}^2 + \tfrac{1}{2}I\omega^2$ , split between translation and rotation. The split depends on the shape factor $c$ in $I = cmR^2$ : a hoop ( $c = 1$ ) stores more energy in rotation than a solid sphere ( $c = \tfrac{2}{5}$ ). Down a ramp, energy conservation gives $v_{cm} = \sqrt{\dfrac{2gh}{1 + c}}$ , so the object with the smaller shape factor reaches the bottom first, independent of mass and radius. Rolling without slipping is made possible by static friction, which does no work.

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What this topic is asking
The rolling constraint
The total kinetic energy splits two ways
The race down a ramp
Try this

What this topic is asking

The College Board (Topic 6.5) wants you to analyze objects that roll without slipping, using the constraint $v_{cm} = R\omega$ to link translation and rotation, and the total kinetic energy $\tfrac{1}{2}mv_{cm}^2 + \tfrac{1}{2}I\omega^2$ to solve problems with energy methods. The headline result is that mass distribution (through $I$ ) decides how quickly an object rolls down a ramp.

The rolling constraint

This constraint is the heart of the topic and ties Unit 6 back to the linear-rotational link of Topic 5.2. Because the contact point is momentarily stationary, the center moves forward exactly as fast as the rim turns, giving $v_{cm} = R\omega$ . If the object skids, this relation fails and the motion is no longer pure rolling. The constraint lets you reduce a problem with two unknowns ( $v_{cm}$ and $\omega$ ) to one.

The total kinetic energy splits two ways

This partition is what makes rolling problems distinctive. For a given amount of energy, an object with a large shape factor moves its center of mass more slowly, because more of the energy is "spent" on spinning. This single idea, the $(1 + c)$ factor, drives the ramp-race result and the most common exam questions in the topic.

The race down a ramp

Releasing an object from rest at height $h$ and applying energy conservation (with static friction doing no work) gives:

mgh = \tfrac{1}{2}mv_{cm}^2(1 + c) \implies v_{cm} = \sqrt{\frac{2gh}{1 + c}}.

The mass and radius cancel, so the only thing that distinguishes the objects is the shape factor $c$ . A solid sphere ( $c = \tfrac{2}{5}$ ) beats a solid disc ( $c = \tfrac{1}{2}$ ), which beats a hoop ( $c = 1$ ), every time, regardless of how heavy or large they are. This counterintuitive result, that a hollow hoop loses to a solid sphere purely because of where its mass sits, is the signature insight of the topic. The reason static friction does no work is that it acts at the contact point, which is instantaneously at rest, so there is no displacement at the point of application; this is why mechanical energy is conserved even though friction is present. Understanding rolling as an energy partition governed by mass distribution connects Topic 5.4's rotational inertia, Topic 6.1's rotational kinetic energy, and Unit 3's energy conservation into a single, satisfying picture.

A disc rolling down a ramp

A solid disc ( $I = \tfrac{1}{2}mR^2$ , so $c = \tfrac{1}{2}$ ) is released from rest and rolls without slipping down a ramp of vertical height $1.2$ m. Find the speed of its center of mass at the bottom. Take $g = 9.8$ m/s squared.

step 1 Write energy conservation

All the gravitational potential energy becomes kinetic energy (friction does no work): $mgh = \tfrac{1}{2}mv_{cm}^2(1 + c)$ .

step 2 Insert the shape factor

With $c = \tfrac{1}{2}$ , the factor is $1 + \tfrac{1}{2} = \tfrac{3}{2}$ . So $mgh = \tfrac{1}{2}mv_{cm}^2 \cdot \tfrac{3}{2} = \tfrac{3}{4}mv_{cm}^2$ .

step 3 Cancel the mass and solve

$gh = \tfrac{3}{4}v_{cm}^2$ , so $v_{cm}^2 = \tfrac{4}{3}gh = \tfrac{4}{3}(9.8)(1.2) = 15.7$ . Then $v_{cm} = 3.96$ m/s.

step 4 Compare with a non-rolling block

A frictionless sliding block would reach $v = \sqrt{2gh} = \sqrt{2(9.8)(1.2)} = 4.85$ m/s. The disc is slower because some energy went into rotation; the mass cancelled out entirely.

Try this

Q1. A hoop of radius $0.25$ m rolls without slipping with center-of-mass speed $2.0$ m/s. Calculate its angular velocity. [2 points]

Cue. $\omega = v_{cm}/R = 2.0/0.25 = 8.0$ rad/s.

Q2. A solid sphere ( $c = \tfrac{2}{5}$ ) and a hoop ( $c = 1$ ) roll from the same height. State which reaches the bottom first. [1 point]

Cue. The sphere, because its smaller shape factor puts more energy into translation.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)7 marksSection II (long FRQ). A solid sphere (

I = \tfrac{2}{5}mR^2

) and a hoop (

I = mR^2

) of the same mass and radius are released from rest at the top of the same ramp and roll without slipping. (a) Write the total kinetic energy of a rolling object in terms of

v_{cm}

and the shape factor. (b) Using energy conservation, derive an expression for the speed of each at the bottom in terms of

g

and the drop height

h

. (c) State which object reaches the bottom first and justify your answer physically.

Show worked answer →

A 7-point FRQ on rolling energy and the race down a ramp.

(a) Total kinetic energy (2 points): $K = \tfrac{1}{2}mv_{cm}^2 + \tfrac{1}{2}I\omega^2$ . With $\omega = v_{cm}/R$ and $I = cmR^2$ (shape factor $c$ ), $K = \tfrac{1}{2}mv_{cm}^2(1 + c)$ .
(b) Speeds (3 points): energy conservation gives $mgh = \tfrac{1}{2}mv_{cm}^2(1 + c)$ , so $v_{cm} = \sqrt{\dfrac{2gh}{1 + c}}$ . Sphere ( $c = \tfrac{2}{5}$ ): $v = \sqrt{\dfrac{2gh}{1.4}} = \sqrt{1.43\,gh}$ . Hoop ( $c = 1$ ): $v = \sqrt{\dfrac{2gh}{2}} = \sqrt{gh}$ .
(c) Winner (2 points): the sphere reaches the bottom first. It has the smaller shape factor, so less of its energy is locked into rotation and more goes into translation, giving a higher center-of-mass speed for the same drop.

Markers reward the $(1 + c)$ energy split, the derived speed expression, and identifying the sphere as faster because of its smaller rotational share.

AP 2024 (style)1 marksSection I (multiple choice). A wheel of radius

R

rolls without slipping with center-of-mass speed

v

. What is its angular velocity? (A)

v

(B)

vR

(C)

v/R

(D)

R/v

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the rolling constraint. The answer is (C).

Rolling without slipping requires $v_{cm} = R\omega$ , so $\omega = v/R$ . The trap is multiplying instead of dividing; the angular velocity is the center-of-mass speed divided by the radius.

Related dot points

Sources & how we know this

AP Physics 1: Algebra-Based Course and Exam Description — College Board (2024)