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SCIENCE · INVESTIGATE AND EXPLAIN

Test the energy and friction conditions for rolling

Rolling without slipping on a stationary surface requires the point of contact to have zero instantaneous velocity relative to that surface. For a circular body, vCM=Rω in magnitude and aCM=Rα for the corresponding motion. The center moves while the contact point is instantaneously at rest; neither statement means the whole body is at rest.

Learning outcome: Compare all three supplied shape factors at h=0.6 m. Verify energy subtotals and derive the friction requirement for each. For a separate assigned surface μs=0.15, decide which no-slip predictions are feasible rather than applying every displayed speed unconditionally.

Starting recap

Review axis-specific inertia, force application locations, torque-angle versus torque-time area, energy conservation and system boundaries.

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Try

For a disk with β=½, m=2 kg, R=0.2 m, g=10 m/s² and h=0.6 m, initial gravitational energy is 12 J. Final center speed is √8≈2.828 m/s; translation has 8 J and rotation 4 J. On a 30° incline, acceleration is 10/3 m/s², uphill static friction 10/3 N, and minimum μs≈0.19245.

Open the experiment

Predict

At the same starting height, does a hoop or uniform solid sphere have the greater center speed if both roll without slipping?

Experiment

Compare all three supplied shape factors at h=0.6 m. Verify energy subtotals and derive the friction requirement for each. For a separate assigned surface μs=0.15, decide which no-slip predictions are feasible rather than applying every displayed speed unconditionally.

Check energy and friction for ideal downhill rolling

Worked teaching model · synthetic values · Rigid body mass2 kg,radius0.2 m; supplied inertia I=βmR² with β=.4,.5,1. Incline30°,g=10 m/s². Assume no slipping and sufficient static friction; report the minimum coefficient required. No rolling resistance, deformation, air drag or moving support.

Reset keeps your prediction, captures and practice history.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Inertia factor beta0.5ratio
Center speed2.82843m/s
Angular speed14.14214rad/s
Translational kinetic energy8J
Rotational kinetic energy4J
Released gravitational potential energy12J
Downhill center acceleration3.33333m/s²
Uphill static friction required3.33333N
Minimum static friction coefficient0.19245ratio
Accessible plot coordinates
Primary plotted points
coordinatevalue

The no-slip energy result is conditional on the required static friction being available.

mgh=½mv²(1+β), and a=g sinθ/(1+β). In the stationary-surface idealization, the instantaneous contact point is at rest, so static friction does not dissipate mechanical energy. If μs is below the displayed requirement, this rolling solution is not valid.

The no-slip energy result is conditional on the required static friction being available.

mgh=½mv²(1+β), and a=g sinθ/(1+β). In the stationary-surface idealization, the instantaneous contact point is at rest, so static friction does not dissipate mechanical energy. If μs is below the displayed requirement, this rolling solution is not valid.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

1. Frame the question

Rolling without slipping on a stationary surface requires the point of contact to have zero instantaneous velocity relative to that surface. For a circular body, vCM=Rω in magnitude and aCM=Rα for the corresponding motion. The center moves while the contact point is instantaneously at rest; neither statement means the whole body is at rest.

2. Build the mechanism

Total kinetic energy is ½mvCM²+½ICMω². If the supplied shape formula is ICM=βmR², substitution of the no-slip relation gives K=½mvCM²(1+β). A body released from rest through vertical height h without dissipative losses therefore has vCM=√[2gh/(1+β)]. More energy goes into spin when β is larger, leaving a smaller center speed at the same height.

For an ideal incline at angle θ, translation gives mg sinθ−f=ma and torque about the center gives fR=Iα. With a=Rα, these imply a=g sinθ/(1+β) and uphill static friction f=βmg sinθ/(1+β). The required coefficient is at least f/(mg cosθ)=β tanθ/(1+β). A neat energy answer is invalid if the available static friction cannot maintain the assumed constraint.

Static friction here does not dissipate mechanical energy: on the stationary rigid surface its instantaneous application point has zero velocity. This is a carefully stated idealization, not a claim that friction never transfers energy or that real rolling has no resistance. Moving surfaces, deformation and rolling resistance need different accounting and are outside this model.

3. Work an example

For a disk with β=½, m=2 kg, R=0.2 m, g=10 m/s² and h=0.6 m, initial gravitational energy is 12 J. Final center speed is √8≈2.828 m/s; translation has 8 J and rotation 4 J. On a 30° incline, acceleration is 10/3 m/s², uphill static friction 10/3 N, and minimum μs≈0.19245.

4. Predict and investigate

Predict first: At the same starting height, does a hoop or uniform solid sphere have the greater center speed if both roll without slipping?

Compare all three supplied shape factors at h=0.6 m. Verify energy subtotals and derive the friction requirement for each. For a separate assigned surface μs=0.15, decide which no-slip predictions are feasible rather than applying every displayed speed unconditionally. Retain settings and raw evidence in the lesson notebook.

5. Transfer and explain

Try a new practice case

A hoop with I=mR² rolls from rest through height 0.9 m with g=10 m/s² and no losses. Find center speed.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Reflect and retain evidence

Why can static friction produce a torque about the center without dissipating energy in this particular stationary-surface model?

Retain your first response, system and axis choices, separate scalar energy and signed angular-momentum ledgers, assumptions and a changed-case prediction. Public worked checks are practice, not independent mastery evidence. Synthetic captures are not a performed physical experiment.

Framework and source notes

Original teaching prose and numerical cases. College Board AP Physics 1 CED, Unit 6 topic mapping: 6.1; 6.5. Checked 30 October 2026. Planar signed analysis without required three-dimensional angular-momentum direction. Slipping evolution is qualitative, in accordance with the framework boundary; no rolling-resistance model. Framework cross-reference, not AP authorization.

Apply

A disk with I=½mR² rolls without slipping from height h. Its kinetic energy divides as:

Public practice checks a specific claim. It does not establish independent mastery.

Why can static friction produce a torque about the center without dissipating energy in this particular stationary-surface model?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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