Derive and test a no-slip rolling constraint

State the geometric condition

For a rigid body rolling without slipping on a stationary surface, center travel equals R times angular displacement; vCM=R omega and aCM=R alpha with compatible signs. These relations are constraints, not universal definitions of rolling motion. The contact point is instantaneously at rest relative to the surface, although the center is moving.

Derive acceleration and required friction

The body has I=beta M R² and rolls down an incline at angle theta. Along the incline, Mg sin(theta)−f=Ma. About the center, uphill friction supplies fR=I alpha=I a/R. Thus a=g sin(theta)/(1+beta), f=beta Mg sin(theta)/(1+beta). Static friction points uphill in this particular gravity-driven case; that direction is not a rule for every rolling body with applied forces or torques.

Reject an unsupported branch

The normal force is Mg cos(theta), so no slip requires mu_s≥beta tan(theta)/(1+beta). The workbench compares this requirement with a stipulated available coefficient 0.30. If it fails, the calculated no-slip acceleration is conditional and not a prediction of actual motion. A slipping analysis must replace the invalid constraint.

Reconcile force and energy

K=MvCM²/2+Iomega²/2=MvCM²(1+beta)/2. Descending a vertical height h from rest gives v²=2gh/(1+beta) when no mechanical energy is dissipated. Ideal static friction does no work at its stationary contact point, yet transfers the required influence between center translation and rotation. This is not a claim that kinetic friction or real rolling resistance never dissipates energy. Rolling resistance is outside this model.

Worked original case

At beta=.5 and theta=30 degrees, acceleration is 10/3 m/s², required friction is 10/3 N for M=2 kg, and mu_s,min=1/(3sqrt(3))≈.19245. Available .30 is sufficient. At beta=1 and theta=40 degrees, mu_s,min≈.41955, so the no-slip branch fails.

Investigation

Compare beta=.5 and 1 at 30 and 40 degrees. Draw the force diagram and derive the required friction, then test the coefficient before interpreting acceleration. For a feasible case, independently obtain the speed after a specified vertical drop from energy and from the constant-acceleration geometry.

Public numerical transfer practice

Try a new practice case

A solid cylinder beta=.5 rolls without slipping from rest through a vertical drop .6 m with g=10 m/s². What is its final speed squared in m²/s²?

Numerical result in the requested units Check this practice result

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.