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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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.
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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.
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| Quantity | Value | Unit |
|---|---|---|
| Inertia factor beta | 0.5 | ratio |
| Center speed | 2.82843 | m/s |
| Angular speed | 14.14214 | rad/s |
| Translational kinetic energy | 8 | J |
| Rotational kinetic energy | 4 | J |
| Released gravitational potential energy | 12 | J |
| Downhill center acceleration | 3.33333 | m/s² |
| Uphill static friction required | 3.33333 | N |
| Minimum static friction coefficient | 0.19245 | ratio |
Accessible plot coordinates
| coordinate | value |
|---|
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.
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Why can static friction produce a torque about the center without dissipating energy in this particular stationary-surface model?
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