The ball a turntable will not throw off

Turning is an acceleration found that anything moving in a circle needs a force toward the centre, and the forces that are not there described the same motion from inside a rotating frame, where the missing force appears as a centrifugal push outward and the Coriolis force turns every moving object sideways. The deflection that closes on itself followed a body with no forces on it at all, seen from the rotating Earth, and found it moving in an inertial circle at twice the rotation rate. The ratio that decides whether the planet is turning and the wave that can only travel west took the Coriolis force into the atmosphere and ocean.

Those intuitions all point the same way for a small object on a turntable: the table turns, the object is dragged along, the centrifugal force pushes it out, and it flies off the edge. That is what happens to a coin that slides. It is not what happens to a ball that rolls. A rolling ball on a spinning turntable behaves so differently that watching it is disconcerting — it circles, it returns, and on a tilted table it refuses to roll downhill — and every one of those behaviours follows from a single equation that turns out to be the equation of a charged particle in a magnetic field.

What rolling adds #

A ball rolls without slipping when the point of it touching the surface moves with the surface. On a still floor that means the contact point is at rest, and the ball’s forward speed is tied to its spin. On a turntable it means the contact point moves with the table beneath it — at the table’s own speed at that radius, $\Omega r$, round the axis. That constraint has a consequence no sliding object has: any change in the ball’s velocity must be accompanied by a matching change in its spin, and both are produced by the same single force.

That force is friction at the contact. It is the only horizontal force on the ball, and it does two jobs at once. Acting on the ball’s centre of mass it changes the ball’s velocity, at the rate $F/m$. Acting at the bottom of the ball, a radius $a$ below the centre, it exerts a torque that changes the ball’s spin, at the rate $aF/I$, where $I$ is the moment of inertia. The push that comes out sideways found a torque turning a spinning wheel’s axis rather than tipping it; here the torque turns the ball’s spin to keep its bottom in step with the table. The mass, and where it sits met this division of labour in a race down a slope, where a solid ball beats a hollow one because less of the pull goes into spinning it.

There is a way to see what that sharing does before writing any equation. Suppose the ball moves a small distance across the table. The patch of table now beneath it is moving at a slightly different velocity from the patch it left — the table’s velocity changes from place to place, by $\Omega$ times the step, and always at right angles to the step, because the table turns. The ball’s bottom must match the new patch, so the ball must acquire a sideways mismatch in the combination of its velocity and its spin. Friction supplies it, and splits it between the two in proportion to how easily each changes. For a solid ball, spinning is the easier: five-sevenths of the mismatch is taken up by a change of spin and only two-sevenths by a change of velocity. So every step the ball takes nudges its velocity sideways by two-sevenths of $\Omega$ times the step. A velocity that is always nudged at right angles to itself, by an amount proportional to how far it has gone, turns at a steady rate.

Put the two together with the rolling constraint and differentiate. The result is short:

$$\frac{d\mathbf{v}}{dt} = \frac{\Omega}{1 + ma^2/I}\;\hat{\mathbf{z}} \times \mathbf{v}.$$

For a solid ball $I = \tfrac{2}{5}ma^2$, and the factor in front is $\tfrac{2}{7}\Omega$. The ball’s velocity, seen from the room, does not grow or shrink. It turns — uniformly, in the same sense as the table, at two-sevenths of the table’s rate. A velocity that turns uniformly at a constant speed traces a circle. Every path a rolling ball can take on a flat turntable is therefore a circle, seen from the room, traversed at $2\Omega/7$.

Every path is a circle #

The figure computes four cases on a record-player turntable at 33⅓ revolutions a minute. A ball set down at rest relative to the room stays exactly where it is, forever: its velocity is zero, and a zero velocity turned is still zero. It spins on the spot, its surface moving with the table beneath it. A ball given a push of a few centimetres a second moves off in a circle whose radius is its speed divided by $2\Omega/7$, turns once every three and a half revolutions of the table — every 6.3 seconds here — and comes back to where it was pushed. The radius depends on the push; the period does not.

The fourth ball is the surprising one. It is set down moving with the table, 2.5 centimetres from the axis, so that at the moment it lands it is at rest relative to the surface beneath it. Common sense says it should simply ride round. It does not. Riding round would need a steady inward force of $m\Omega^2 r$ from friction, and a steady friction force at the contact would also steadily change the ball’s spin — but the ball’s spin is pinned to the table’s motion by the rolling condition, and cannot be changing in the way that force would require. The only motion consistent with rolling is a circle at $2\Omega/7$, and a ball starting with the table’s speed $\Omega r$ must move on a circle of radius $\Omega r/(2\Omega/7) = 3.5r$. It sweeps out to the far side of the axis, 15 centimetres away, and comes back.

This is not a small effect hiding in a large one. A ball on a turntable is simply not the object the centrifugal-force picture describes. That picture is correct for a body free to slide, whose only horizontal force is whatever friction is needed to drag it along; for a rolling body the friction is not free to be whatever is needed, because it is also turning the ball.

The view from the table #

Seen from the turntable the same motions look complicated. The ball that is at rest in the room goes round the axis backwards once per turn, as everything fixed in the room does. The others trace loops, because a circle at $2\Omega/7$ seen from a frame turning at $\Omega$ is the combination of the two rotations. An experimenter who filmed the marbles from a camera mounted on the turntable, and tried to explain the loops with centrifugal and Coriolis forces, would find those forces present and correct, and would find in addition a friction force whose size and direction change continually in exactly the way needed to cancel most of them. The simple description is the one from the room, which is a reminder that a rotating frame is a choice made for convenience and is not always the convenient one.

A magnetic field made of a turntable #

The equation $d\mathbf{v}/dt = \tfrac{2}{7}\Omega\,\hat{\mathbf{z}} \times \mathbf{v}$ has a famous twin. A charge $q$ of mass $m$ in a uniform magnetic field $B$ obeys $d\mathbf{v}/dt = (qB/m)\,\mathbf{v} \times \hat{\mathbf{z}}$: its velocity turns uniformly at the cyclotron frequency $qB/m$, and it moves on a circle whatever its speed, returning to its start once per period. The rolling ball on a turntable is a charged particle in a magnetic field, with $2\Omega/7$ in the place of the cyclotron frequency. Every result about charges in uniform magnetic fields applies to it without change.

The analogy is not decorative. The drift that does not care what the charge is found that a charge in a magnetic field acted on by a steady force in the plane does not accelerate along the force: it drifts at right angles to it, at a steady speed equal to the force divided by the charge times the field. A rolling ball on a turntable should do the same when a steady force acts on it. There is an easy way to supply one: tilt the table.

Across the slope, not down it #

On a tilted turntable, gravity pulls the ball down the slope with a force $mg\sin\alpha$, of which a rolling solid ball converts five-sevenths into acceleration — the same five-sevenths that makes it roll down a still slope more slowly than a sliding block.

The figure releases a ball from rest on a large turntable tilted by a fifth of a degree. On a still table it would roll straight downhill, ten centimetres in under three seconds. On the turning table it starts downhill, is turned across the slope, comes back up to its original height, and repeats: a chain of loops marching sideways at an average of 2.5 centimetres a second. It never gets more than 4.9 centimetres below its starting height, however long it is left. The drift speed is $\tfrac{5}{2}g\sin\alpha/\Omega$ — the push divided by the “field” — and its direction is across the slope, perpendicular to gravity, exactly as the charge’s drift is perpendicular to the electric field.

A marble on a tilted spinning table does not roll downhill. That is the most counterintuitive of the three behaviours and the one that most clearly shows what the rolling constraint is doing. The hilltop that holds the Trojans met the same trick in celestial mechanics: a force that depends on velocity and always acts at right angles to it can hold a body near the top of a hill, because it turns every attempt to slide off into a circle. The Coriolis force does it for the Trojan asteroids; the rolling constraint does it for the marble; the magnetic force does it for a charge in a magnetic trap.

A slope that sorts balls by what is inside them #

The drift across a tilted table depends on the ball, and the dependence is simple. The slope’s pull accelerates a rolling ball at $g\sin\alpha/(1 + \kappa)$, where $\kappa = I/ma^2$, and the table turns its velocity at $\Omega/(1 + 1/\kappa)$; the drift is the ratio, $g\sin\alpha/\kappa\Omega$. A solid ball, with $\kappa = 2/5$, drifts at $2.5\,g\sin\alpha/\Omega$; a hollow ball of the same size, with $\kappa = 2/3$, at $1.5\,g\sin\alpha/\Omega$. Two balls that look identical and weigh the same, one solid and one a thick shell of denser material, released together on a tilted turntable, separate across the slope at speeds in the ratio of five to three. On a still slope they would also separate, the solid one arriving first, but only along the slope and only by about nine per cent in time; on the turning table the difference becomes a difference of direction as well as of speed.

The drift also does not depend on the ball’s mass or size, only on how its mass is arranged, exactly as a charge’s drift in crossed fields does not depend on its mass or charge. It is one of the few experiments in mechanics where the answer depends on a body’s inside and on nothing else about it.

Where the mass is kept #

The two-sevenths belongs to a solid ball. The general rate is $\Omega/(1 + 1/\kappa)$, where $\kappa = I/ma^2$ measures how the ball’s mass is distributed.

A thin hollow ball, with $\kappa = 2/3$, circles at two-fifths of the table’s rate; a ball with a heavy core, less; and a frictionless puck, which has no spin to change, is not turned at all and slides in a straight line, as the centrifugal picture expects once the table’s friction is removed. No rolling ball reaches the table’s own rate, which would need an infinite moment of inertia. The rate is a direct measure of how much of the friction’s work goes into turning the ball rather than moving it: the more of the ball’s mass is far from its centre, the more the friction’s torque is resisted and the more nearly the ball moves as the table does.

That makes the turntable a way of measuring a ball’s moment of inertia without taking it apart: time its circles, divide by the table’s period, and the ratio gives $\kappa$. A solid steel ball and a hollow ping-pong ball set on the same table circle at visibly different rates.

How long rolling lasts #

The whole analysis assumes the ball rolls without slipping, and that needs enough friction.

The friction force is what turns the ball’s velocity, so it must be $m \cdot \tfrac{2}{7}\Omega v$, which grows with the ball’s speed. At 33⅓ revolutions a minute and a coefficient of friction of 0.3 the ball rolls cleanly up to three metres a second, far faster than any marble pushed by hand; a slippery surface with a coefficient of 0.1 still allows a metre a second. Above those speeds the ball skids, the constraint no longer holds, and the path is no longer a circle. Within them the circles are clean, which is why the demonstration is so easy to do.

Where the model stops #

Real marbles on real turntables do not circle forever. Rolling is never perfectly free of loss: the ball and the surface deform slightly at the contact, and a small resistance to rolling appears that has no counterpart in the ideal equation. Its effect, measured carefully in experiments with steel balls on rotating plates, is that the circles are not quite closed — the ball’s orbit drifts slowly outward, spiralling away from its initial circle over tens of revolutions, and the circling rate comes out slightly below two-sevenths. The drift is the analogue of a charge in a magnetic field subject to a weak drag, which also spirals across field lines. A turntable that is not quite flat, or not quite level, adds a small tilt and hence a steady drift. And the table’s own speed must be constant; a record player that speeds up and slows down gives the ball a varying “field” and a path that no longer closes.

The analysis also treats the ball as touching the table at one point. A soft ball on a soft surface touches over a patch, and friction then produces a torque about the vertical axis as well, which couples the ball’s spin about the vertical to its motion. For hard balls on hard surfaces that coupling is small.

What the pictures cannot show #

The figures show the ball’s centre and not its spin, which is where the interesting bookkeeping happens. The ball’s speed, seen from the room, never changes on a flat table, but its horizontal spin does, continuously, as it circles, and the energy for that comes from the table: friction does work on the ball at every instant, because the contact point moves with the table. Over each complete circle the spin returns to its starting value and the work sums to zero, which is why nothing accumulates. And they cannot show the thing that makes the effect so surprising to see: a marble that has visibly been pushed straight outward toward the edge of a spinning record, curving round and coming back to the hand that pushed it.

Still open: how a rolling ball loses its orbit #

The ideal circles depend on a contact that is both perfectly rigid and perfectly rolling, and neither is quite true. How rolling resistance, the small slip that accompanies any real rolling contact, and spin about the vertical combine to make a ball’s orbit drift outward, and at what rate, is not fully settled: experiments find drifts and rate shifts larger in some conditions than the simplest models of rolling resistance predict, and the mechanics of a ball rolling on a moving surface is sensitive to details of the contact that are hard to measure directly. The same questions arise in engineering — ball bearings, whose balls roll between races that turn at different rates, and the conveying and sorting of spherical particles on rotating plates — where the orbit of a rolling ball is not a curiosity but the thing being designed.

The habit worth carrying away is to ask what a constraint does to the forces before reaching for the fictitious ones. A ball that must roll on a turning surface has only friction to move it, and friction that also has to keep the spin matched to the surface can only turn the ball’s velocity, never grow it — so every path is a circle at two-sevenths of the table’s rate, and a tilt produces a drift across the slope, exactly as for a charge in a magnetic field. The centrifugal force is there in the table’s frame, and so is the friction that cancels most of it.