Test the no-slip condition before partitioning rolling energy

Choose the axis, reference or material model

For a rigid body rolling without slipping on a stationary surface, center speed and angular speed obey v=Rω, with consistent signs. That relation is a constraint supported by contact forces, not a universal description of every moving, spinning object. While slipping, translation and rotation must be analyzed separately.

Build the algebraic relationship

Write I=βMR² using a supplied shape factor β. On a slope, downhill force balance is Mg sinθ−f=Ma and the uphill friction torque about the center gives fR=Iα. Combining with a=Rα yields a=g sinθ/(1+β), f=βMg sinθ/(1+β). Required static coefficient is βtanθ/(1+β), because normal force is Mg cosθ.

Check the original numerical packet

The workbench allows coefficient 0.30. With β=0.5,θ=30°, required coefficient is 1/(3√3)≈0.19245, so the no-slip solution is supported. At β=1,θ=40°, requirement≈0.41955 exceeds the available value; the model omits a final rolling-speed prediction rather than passing a conditional result off as actual motion.

Test the boundary and independent evidence

For a feasible release through vertical drop0.6 m under g=10, gravitational energy 12 J becomes translation plus rotation: Mv²(1+β)/2. Thus v²=12/(1+β). At β=0.5, translational K is 8 J and rotational K4 J. Ideal static friction does no work at the instantaneously stationary contact point, although it affects both motions. Kinetic slip or real rolling resistance would require another energy account.

Worked original case

For β=0.5 andθ=30°, acceleration is 10/3 m/s² and friction 10/3 N. The slope distance for a 0.6 m vertical drop is 1.2 m, so v²=2as=8, agreeing with the energy result.

Bounded investigation

Compare β=0.5 and 1 at 30° and 40°. Derive required friction before evaluating the energy split. For one feasible state compare energy and constant-acceleration geometry, and explain why a failed state needs independent slip dynamics.

Changed-case numerical practice

Try the changed case before feedback

A body with suppliedβ=1 rolls without slipping from rest through height 0.4 m under g=10. Find final speed squared in m²/s².

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. All records are supplied or generated teaching evidence. No rotating equipment, applied loads, material tests or physical assembly is requested. An admissible virtual model state does not certify real equipment safety, supervised practical skill or independent mastery.