Find the ratio of the velocity of the centre of a sphere

Find the ratio of the velocity of the centre of a sphere of radius $R$, rolling without slipping on two parallel rails a distance $d$ apart, to the angular velocity of the sphere.

Step 1: Where the sphere touches

By symmetry the sphere rests with its centre above the midline, so each contact point lies a horizontal distance $\dfrac{d}{2}$ from the vertical through the centre. The contact point is on the sphere's surface, so the radius to it has length $R$, and the right triangle with legs $\dfrac{d}{2}$ and $\rho$ gives

\[ \rho = \sqrt{R^{2}-\frac{d^{2}}{4}}. \tag{1} \]

Step 2: The instantaneous axis

Rolling without slipping means both contact points are momentarily at rest, so the instantaneous axis of rotation is the horizontal line joining them. The centre's distance from that axis is exactly $\rho$.

Step 3: Relate the two velocities

A point at distance $\rho$ from the instantaneous axis moves with speed $\omega\rho$: $v = \omega\rho = \omega\sqrt{R^{2}-\frac{d^{2}}{4}}$.

Step 4: The two extremes

Rails brought together $\bigl(d\to0\bigr)$: the two contacts merge into one and the sphere is effectively on a plane,

\[ \frac{v}{\omega} \to R, \]

the familiar result for rolling on a flat surface.

Rails at maximum separation $\bigl(d = 2R\bigr)$: the sphere touches at the ends of a diameter,

\[ \frac{v}{\omega} = \sqrt{R^{2}-R^{2}} = 0, \]

so the centre does not advance at all — the sphere merely spins in place between the rails.

Step 5: Verification

\[R = 10,\ d = 12:\ \rho = \sqrt{100-36} = 8, \text{ so } v = 8\omega \ \text{(against } 10\omega \text{ on a plane)} \;\;\checkmark\]

\[\text{The ratio decreases monotonically in } d, \text{ from } R \text{ down to } 0 \;\;\checkmark\]

\[\text{Requires } d\le2R, \text{ otherwise the sphere falls through} \;\;\checkmark\]