Test a rolling constraint before using its acceleration
A no-slip rolling relation is a constraint that friction must be able to sustain. Consider a rigid circular body released on a stationary incline with I = κmR². Translation down the slope gives mg sinθ − f = ma, while center torque and no-slip acceleration give fR = I a/R.
Learning outcome: Derive translation and torque equations, evaluate the friction inequality, and retain one accepted and one rejected branch.
Starting recap
Integrate mass moments about named axes, derive torque and angular-momentum constraints, and test orbital energy, angular and numerical consistency against independent evidence.
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At κ = 1 and θ = 20 degrees, required coefficient is tan(20°)/2 ≈ 0.181985, still feasible. The accepted acceleration is 5 sin(20°) ≈ 1.71010 m/s².
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If required static friction exceeds its available bound, the no-slip result should be:
Experiment
Derive translation and torque equations, evaluate the friction inequality, and retain one accepted and one rejected branch.
Check static friction before imposing rolling
Worked teaching model · synthetic values · Rigid circular body released from rest on a stationary straight incline, g=10 and static coefficient .2. Proposed no-slip rolling has I=kappa mR². Reject that constraint if required friction exceeds its bound; no kinetic slipping continuation is supplied.
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Required static coefficient versus slope
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: kappa tan(theta)/(1+kappa) (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Required static coefficient versus slope
kappa tan(theta)/(1+kappa)
| incline angle (degrees) | required static coefficient |
|---|---|
| 0 | 0 |
| 0.5 | 0.002908956 |
| 1 | 0.005818355 |
| 1.5 | 0.008728641 |
| 2 | 0.01164026 |
| 2.5 | 0.01455365 |
| 3 | 0.01746926 |
| 3.5 | 0.02038754 |
| 4 | 0.02330894 |
| 4.5 | 0.0262339 |
| 5 | 0.02916289 |
| 5.5 | 0.03209635 |
| 6 | 0.03503475 |
| 6.5 | 0.03797854 |
| 7 | 0.04092819 |
| 7.5 | 0.04388417 |
| 8 | 0.04684694 |
| 8.5 | 0.049817 |
| 9 | 0.05279481 |
| 9.5 | 0.05578087 |
| 10 | 0.05877566 |
| 10.5 | 0.06177968 |
| 11 | 0.06479344 |
| 11.5 | 0.06781743 |
| 12 | 0.07085219 |
| 12.5 | 0.07389822 |
| 13 | 0.07695606 |
| 13.5 | 0.08002625 |
| 14 | 0.08310933 |
| 14.5 | 0.08620586 |
| 15 | 0.0893164 |
| 15.5 | 0.09244151 |
| 16 | 0.0955818 |
| 16.5 | 0.09873783 |
| 17 | 0.1019102 |
| 17.5 | 0.1050996 |
| 18 | 0.1083066 |
| 18.5 | 0.1115318 |
| 19 | 0.1147759 |
| 19.5 | 0.1180395 |
| 20 | 0.1213234 |
| 20.5 | 0.1246282 |
| 21 | 0.1279547 |
| 21.5 | 0.1313035 |
| 22 | 0.1346754 |
| 22.5 | 0.1380712 |
| 23 | 0.1414916 |
| 23.5 | 0.1449375 |
| 24 | 0.1484096 |
| 24.5 | 0.1519088 |
| 25 | 0.1554359 |
| 25.5 | 0.1589918 |
| 26 | 0.1625775 |
| 26.5 | 0.1661939 |
| 27 | 0.1698418 |
| 27.5 | 0.1735224 |
| 28 | 0.1772365 |
| 28.5 | 0.1809852 |
| 29 | 0.1847697 |
| 29.5 | 0.1885909 |
| 30 | 0.1924501 |
| 30.5 | 0.1963483 |
| 31 | 0.2002869 |
| 31.5 | 0.2042669 |
| 32 | 0.2082898 |
| 32.5 | 0.2123568 |
| 33 | 0.2164692 |
| 33.5 | 0.2206285 |
| 34 | 0.2248362 |
| 34.5 | 0.2290937 |
| 35 | 0.2334025 |
| 35.5 | 0.2377644 |
| 36 | 0.2421808 |
| 36.5 | 0.2466537 |
| 37 | 0.2511847 |
| 37.5 | 0.2557757 |
| 38 | 0.2604285 |
| 38.5 | 0.2651453 |
| 39 | 0.269928 |
| 39.5 | 0.2747788 |
| 40 | 0.2796999 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Required static coefficient | 0.1213234 | ratio |
| Available static coefficient | 0.2 | ratio |
| Proposed rolling constraint feasible | 1 | flag |
| Required upslope friction per mass | 1.140067 | N/kg |
| Normal force per mass | 9.396926 | N/kg |
| Downslope center acceleration | 2.280134 | m/s² |
| Rotational fraction of kinetic energy | 0.3333333 | ratio |
The supplied static bound can support the proposed no-slip constraint.
Translation gives mg sinθ−f=ma; rotation and no-slip give fR=I a/R. Solve a=g sinθ/(1+kappa), f/m=kappa a, and require f≤mu_s mg cosθ. Static friction at stationary no-slip contact does zero total contact work in the ground frame, while its center-translation and center-torque power terms cancel. Rejection needs a new friction/rotation model.
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The supplied static bound can support the proposed no-slip constraint.
Translation gives mg sinθ−f=ma; rotation and no-slip give fR=I a/R. Solve a=g sinθ/(1+kappa), f/m=kappa a, and require f≤mu_s mg cosθ. Static friction at stationary no-slip contact does zero total contact work in the ground frame, while its center-translation and center-torque power terms cancel. Rejection needs a new friction/rotation model.
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How can a torque transfer energy into rotation while contact does no net ground-frame work?
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