Rolling Kinematics

Sit on the axle and the wheel simply spins: every rim point moves at Rω along its tangent, the top forward and the bottom backward. Stand on the road and that whole pattern is carried forward at v_cm. So for any point P, v⃗_P = v⃗_cm + ω⃗ × r⃗′, where r⃗′ runs from the axle to P, and the two terms are added as vectors, not as speeds. Take R = 0.30 m, v_cm = 6.0 m s⁻¹ and ω = 20 rad s⁻¹, so Rω = 6.0 m s⁻¹. At the top the spin velocity points forward: 6.0 + 6.0 = 12 m s⁻¹. At the bottom it points backward: 6.0 − 6.0 = 0. At the front-most rim point it points straight down, at right angles to the axle's velocity, so that point moves at √(6.0² + 6.0²) = 8.5 m s⁻¹, aimed 45° below the horizontal. Same wheel, same instant, three different speeds. Rω is the rim's speed relative to the axle; only the vector sum is its speed over the road.

Nothing in that sum ties ω to v_cm. The tie comes from the contact. If the tyre is not sliding, the rubber on the road has the road's velocity, which is zero, and at the bottom the sum reads v_cm − Rω. Hence v_cm = Rω, with v_cm positive forward and ω positive in the sense that rolls the wheel forward. The condition holds at every instant of rolling, so calculus applies to it. Differentiate: a_cm = Rα. Integrate: Δx_cm = RΔθ, the distance rolled equals the arc laid down on the road. A trolley wheel of radius 0.050 m that turns through 240 rad has carried its axle 12 m, and if the trolley gains speed at 0.80 m s⁻² its wheels must gain spin at 0.80/0.050 = 16 rad s⁻². If the surface itself moves, the contact point matches it instead: v_cm − Rω = v_surface. None of this used a force or an energy. The constraint is geometry, and it holds whether the wheel is coasting, driven or braked, for as long as it does not slide.

A rigid body with one point at rest can, at that instant, only be turning about that point, and its rate of turning is the same ω. So while a wheel rolls without slipping, every point moves at |v_P| = ωd, where d is its distance from the contact point, in the direction perpendicular to the line joining them. Check the three cases: the axle has d = R, giving Rω = v_cm; the top has d = 2R, giving 2v_cm; the front-most point has d = √2 R, giving 8.5 m s⁻¹ at 45°, as before. A rim point an angle φ round from the top sits d = 2R|cos(φ/2)| from the contact, so its speed is 2v_cm|cos(φ/2)|: at φ = 60° on the 6.0 m s⁻¹ wheel, 12 × cos 30° = 10.4 m s⁻¹. On the vertical diameter the speed is simply ω times the height above the road. Every point above the axle has a larger forward velocity than the vehicle and every point below it a smaller one, and the speeds near the road are the smallest of all, which is why a photograph of a rolling wheel shows crisp spokes near the road and smeared ones at the top.

The pivot picture is about velocities only. A real hinge has zero velocity and zero acceleration; the contact point has only the first. Roll the 0.30 m wheel steadily at 6.0 m s⁻¹. The axle is unaccelerated, so each rim point's acceleration is just its centripetal acceleration about the axle, ω²R = 20² × 0.30 = 120 m s⁻², directed at the axle. For the contact point that is straight up: it touches down, stops for an instant, and is already accelerating off the road at about 12g. That is how a different piece of rubber can be the contact point a moment later. When the vehicle itself accelerates, add a_cm to every point and a tangential Rα about the axle; at the contact these two are horizontal and cancel exactly when a_cm = Rα, leaving ω²R upward. So use the contact point to find velocities, and go back to translation plus rotation for accelerations.

Start the clock when a rim point P touches the road and let the wheel turn through θ = ωt. The axle has advanced Rθ, and P sits R sin θ behind it and R cos θ below it, so x = R(θ − sin θ) and y = R(1 − cos θ), a cycloid. Differentiate: dx/dt = Rω(1 − cos θ) and dy/dt = Rω sin θ, giving a speed √((dx/dt)² + (dy/dt)²) = 2Rω|sin(θ/2)|, the pivot result again, zero at θ = 0 and 2v_cm at θ = π. Two things follow. dx/dt is never negative, so no point of the tyre ever moves backward over the road. And the distance P travels in one turn is the integral of its speed, ∫ 2R sin(θ/2) dθ from 0 to 2π, which is 8R. For R = 0.30 m that is 2.40 m of path while the axle advances 2πR = 1.88 m: the rim point's average speed is 4/π = 1.27 times the vehicle's, though it spends part of each turn almost stationary.

Define the contact-point velocity v_c = v_cm − Rω and the constraint becomes a test. v_c = 0 is rolling. v_c > 0 means the tyre is sliding forward over the road, as under hard braking; a locked wheel has ω = 0 and v_c = v_cm. v_c < 0 is wheel-spin, the rubber sliding backward. A car doing 20 m s⁻¹ whose 0.30 m wheels turn at 50 rad s⁻¹ has Rω = 15 m s⁻¹, so its tyres are sliding forward at 5.0 m s⁻¹. The point of zero velocity still exists, a distance v_cm/ω = 0.40 m below the axle, but that is now 0.10 m under the road; with wheel-spin it rises into the wheel instead. Anti-lock brakes and traction control work this way: compare a wheel-speed sensor with the vehicle's speed and act on the difference. Integrate the difference and you get the length of the slide, ∫ v_c dt = Δx_cm − RΔθ, the road covered minus the arc rolled out. Which way friction then acts, and what it does to the energy, is dynamics. Kinematics only says what is sliding past what, and how fast.