Physics · Mechanics

Rolling Motion

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: Angular MomentumCarry the same rotational bookkeeping into a conserved quantity.

Key takeaway

Rolling without slipping keeps center speed and angular speed locked by $v_{cm} = r ω$ .
Shape changes the race because the inertia factor changes how the same downhill pull is split between translation and rotation.
Static friction supplies the rolling torque even while the object still accelerates down the incline.

Common misconception

A larger or heavier-looking rolling object must always lose because size alone controls the race.

In this bounded no-slip model, the key race variable is not total mass by itself. It is how the mass is distributed about the axis, summarized by the inertia factor $k$ .

Carry rolling ideas into later mechanics

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Angular Momentum

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Review the rotation languageMechanics

Uniform Circular Motion

Track a particle moving at constant speed around a circle and connect radius, angular speed, tangential speed, centripetal acceleration, and the inward-force requirement to the same live state.

Linear analogueMechanics

Momentum and Impulse

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Reference and support

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Equation snapshotRolling without slipping · Rolling acceleration on the inclineShow 3 equations

Keep the no-slip link in view first, then use the acceleration and energy split to explain why different shapes race differently.

Rolling without slipping
$v_{cm} = r ω, a_{cm} = r α$
The contact point does not slide, so the linear and angular motions must stay linked.
Rolling acceleration on the incline
$a_{cm} = \frac{g sin θ}{1 + k}$
On the same incline, a larger inertia factor means more of the gravitational pull is tied up in spinning the object, so the center accelerates less.
Energy split in rolling
$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
The drop in gravitational potential energy is shared between translational and rotational kinetic energy.

Why it behaves this way

Explanation

Rolling without slipping links the motion of the center of mass to the spin of the object. As the roller speeds up down the incline, it must also spin so that the point touching the ramp does not slide relative to the surface.

On this bench, every setup feels the same downhill component of gravity on the same incline. What changes is how the object's mass is distributed about its axis. That distribution changes how the motion is shared between translation and rotation, which is why different shapes reach the bottom at different times.

Key ideas

01No-slip rolling means

v_{cm} = r ω

and

a_{cm} = r α

, so the linear and angular motions are locked together.

02The inertia factor

k = I / (m r^{2})

tells you how hard the shape is to spin for its mass and radius. Larger

k

means more of the same downhill pull must go into building rotation.

03On a fixed incline,

a_{cm} = g sin θ / (1 + k)

, so smaller inertia factor gives larger center-of-mass acceleration and a faster run.

04Changing radius for the same shape mostly changes the spin rate through

ω = v / r

. In this bounded model, it does not by itself change the center-of-mass acceleration.

Worked examples

Solve the live rolling state

Open examples when you want to see the same idea walked through step by step.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough

Use the live slope, shape, and inspected time as evidence. First predict the center-of-mass acceleration and full-ramp travel time from the current slope and inertia factor. Then read the linked linear and angular speeds at one instant and check how the kinetic energy is split.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current setup, what center-of-mass acceleration should this solid cylinder have, and how long should it take to travel the full $2.4 m$ ramp?

θ

Slope angle

12 °

k

Inertia factor

0.5

1. Write the rolling-acceleration rule

For this bounded incline use

a_{cm} = \frac{g sin θ}{1 + k}

, with

g = 9.81 m/ s^{2}

and

k = 0.5

2. Insert the live slope and inertia factor

With

θ = 1 2^{\circ}

, the center-of-mass acceleration becomes

a_{cm} = 1.36 m/ s^{2}

for the current shape or custom mass distribution.

3. Use the acceleration to find the ramp time

The ramp length is

L = 2.4 m

, so from rest use

L = \frac{1}{2} a t^{2}

. That gives a travel time of

1.88 s

Center-of-mass acceleration and ramp time

a_{cm} = 1.36 m/ s^{2}, t_{bottom} = 1.88 s

The solid cylinder sits between the sphere and hoop cases, so its acceleration and travel time stay in the middle as well.

Quick test

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Accessibility

Open the text-first descriptions when you need the simulation and graph translated into words.

The simulation shows one roller descending an incline from left to right. The interior graphic changes to represent a solid sphere, solid cylinder, hoop, or custom mass distribution, while the center marker and spoke rotate with the motion.

Optional overlays can show the no-slip link between center speed and angular speed, a mass-layout cue for the inertia factor, an energy-split bar, and an uphill static-friction arrow with its torque cue. The same live state drives the stage, readouts, and graphs.

Graph summary

The distance graph curves upward because the center of mass accelerates down the incline. The speed-link graph plots center speed and $r ω$ , which should overlap whenever rolling without slipping holds.

The energy graph shows translational, rotational, and total kinetic energy together. The acceleration map falls as inertia factor rises, which is why a hoop rolls more slowly than a sphere on the same incline.

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Starter track

Step 4 of 5

Rotational Mechanics

Rolling Motion appears later in this track, so it is cleaner to start from the beginning first.

Previous step: Rotational Inertia / Moment of Inertia

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