Account for energy in rolling contact

For a circular body of radius R rolling straight on a stationary plane, choose positive center velocity down the ramp and positive angular speed in the corresponding rolling sense. No slip requires v = Rω. More generally, the contact-point velocity is the center-of-mass velocity plus ω×r from the center to contact; that sum must equal the surface velocity. The simple scalar relation depends on the geometry and sign convention. A sphere rolling along a changing direction has a more involved nonholonomic constraint.

Let the center-of-mass moment around the rolling axis be I = κmR². Then the total kinetic energy is one half of m(1+κ)v² after imposing no slip. The rotational part is not an extra mass sliding down the ramp; it is a compact way to express energy stored in the same body. The relation is useful only while the body maintains circular contact, remains rigid, and rolls without slip. It does not justify imposing v = Rω during an arbitrary impact or initial sliding transient.

For downhill rolling under gravity alone, write translation as ma = mg sin α − f, where f denotes the uphill static-friction magnitude. Taking torques about the center gives I(a/R) = fR with the chosen positive rolling sense. Combine these equations before making any friction assumption. The result is a = g sin α/(1+κ), with f determined by the rotational inertia. Static friction is an unknown constraint force bounded by μsN; it is not automatically equal to that upper bound.

The normal reaction for this simple straight incline is mg cos α. Requiring f ≤ μsN produces a minimum coefficient for no-slip rolling. If the available coefficient is smaller, the assumed solution is inconsistent and slipping must be analyzed using an appropriate contact law. An external torque or a force applied away from the center can reverse the needed friction direction. Thus the uphill direction found here is a conclusion for this particular free-body diagram, not a universal property of all rolling wheels.

The power delivered by a contact force is its dot product with the velocity of its point of application. For ideal rolling on a fixed plane that velocity vanishes, so static friction and the normal force do zero total power. In a center-of-mass decomposition, uphill friction does negative translational power while its torque supplies equal positive rotational power. Their sum is zero. Gravity supplies the increase in total kinetic energy. A moving belt, deforming tire, rolling resistance, or slipping contact changes this accounting and can exchange or dissipate energy.

Use the quadratic-function lab to inspect the exact speed dependence of the ideal rolling kinetic energy after choosing consistent scaled units. A unit-mass solid cylinder has κ = one half, so its energy coefficient is a = 0.75. Predict the energy at x = 2 before moving the position control, treating x as signed speed. The lab also shows a tangent and an integral; the integral of kinetic energy over speed is not work. This display is an algebraic energy plot, not a contact or friction simulator.