The push that comes out sideways

A bicycle wheel is spun up, and one end of its axle is rested on a stand. It ought to fall. It does not: it swings slowly round the stand, horizontally, staying level, and it will keep doing so until friction takes the spin out of it. The demonstration is a century and a half old and it is still the most reliable way to make a room of people believe that they do not understand rotation. Two pushes in the wrong order leave a turn behind for the same reason: rotations do not commute.

The common explanation — that the spin somehow holds it up — is false, and the way to see that it is false is to notice that nothing about the forces has changed. The weight is the same weight. The pivot’s reaction is the same reaction. The torque about the pivot is the same torque, spinning or not. What has changed is what the body does with a torque.

A torque changes angular momentum, and that is all it does #

The rotational equation of motion is one line:

$$\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t} = \boldsymbol{\tau}.$$

Nothing in it says that a torque makes something fall. It says that a torque changes the angular momentum vector at a rate equal to itself, in its own direction. Whether that shows up as falling depends entirely on what angular momentum was there already.

For a top the torque is the weight acting at the centre of mass, at the distance from the pivot to that point, and its direction is the part worth being careful about. It is the cross product of a vector pointing up along the axis with a vector pointing straight down, so it is horizontal: at right angles to both, and at right angles in particular to the angular momentum it is about to change.

For a body at rest, $\mathbf{L}$ is zero, so the torque creates angular momentum where there was none, in its own direction, and the body starts rotating about the horizontal axis the torque points along — which is falling over. For a body already spinning, $\mathbf{L}$ is a large vector along the spin axis, and the torque adds a small vector at right angles to it. Adding a perpendicular vector to a long one barely changes its length and turns its direction.

That is the same structure as a velocity turned by a perpendicular acceleration. There the speed never changes and the direction goes round a circle; here the length of $\mathbf{L}$ never changes and its direction goes round a cone. One picture serves both, because both are $\mathrm{d}\mathbf{v}/ \mathrm{d}t = \mathbf{a}$ with $\mathbf{a}$ perpendicular to $\mathbf{v}$, and the only difference is which pair of letters is written on the arrows.

So the rate at which the axis swings round follows from a triangle. In a time $\mathrm{d}t$ the horizontal component of $\mathbf{L}$, whose length is $L\sin\theta$, turns through an angle $\mathrm{d}\varphi$, so the change in $\mathbf{L}$ has magnitude $L\sin\theta\,\mathrm{d}\varphi$. Setting that equal to $\tau\,\mathrm{d}t = Mgl\sin\theta\,\mathrm{d}t$ gives

$$\dot\varphi = \frac{Mgl}{L} = \frac{Mgl}{I_3\omega_3},$$

with the two sines cancelling, which is why the precession rate does not depend on how far the top is tilted. It is inversely proportional to the spin: a faster top precesses more slowly, which is precisely backwards from what anybody guesses and is the sharpest test of whether the argument above has been understood.

The rate that is one root of two #

That formula is right and it is not the whole answer, because the derivation quietly assumed the axis moves only in azimuth — that the tilt does not change. Allowing the tilt to move and asking for a solution in which it does not gives a quadratic rather than a linear equation:

$$I_1\cos\theta\,\dot\varphi^2 - I_3\omega_3\dot\varphi + Mgl = 0.$$

The extra term is the centrifugal contribution from the precession itself, which was dropped by assuming the precession is slow. Solve it properly and there are two rates at which a top can precess steadily at a given tilt.

The slow root’s leading term is $Mgl/I_3\omega_3$, as it must be, and the agreement is not exact: at 3000 rpm the true root is 3.269 radians a second where the simple formula gives 3.122, a discrepancy of four and a half per cent that grows rapidly as the spin comes down. The fast root at the same spin is 69 radians a second — twenty-one times faster — and it is a perfectly good solution of the equations that nobody ever sets up.

Where does the fast root come from physically? At a high precession rate the axis is being swung round so quickly that the centrifugal effect of that swinging is itself comparable with gravity, and the top is held up by its own precession rather than by the coupling to its spin. It is the same solution branch as a conical pendulum: a body on a string, swung fast enough, holds a large angle with no spin involved at all. The two roots are therefore two different mechanisms wearing the same name, and the quadratic that produces them is the statement that at a given tilt they can both balance the same weight.

The discriminant carries a statement neither root does. Real roots require

$$I_3^2\omega_3^2 \ge 4I_1 Mgl\cos\theta,$$

so below a certain spin there is no steady precession at any rate whatever. For the small top in these figures at a tilt of 30°, that threshold is 1245 revolutions a minute. Below it the axis has no choice but to fall, and the familiar demonstration simply stops working — which is the everyday observation that a top has to be spun fast enough, given a number.

The threshold moves with the tilt, and the direction it moves in is the useful part.

What a released top actually does #

Here is the part the textbook picture leaves out. Steady precession requires the axis to be moving sideways at exactly one of those two rates at the moment it is let go. Nobody arranges that. A top released from rest has $\dot\varphi = 0$, which is not a root of the quadratic, so it is not in steady precession — and what it does instead is fall.

The fall is real and it is visible. Releasing a top at 30° from rest, the axis drops to 33.3° before coming back up, and it repeats that dip at the nutation frequency — much faster than the precession — while the whole pattern drifts round. Averaged over a nutation cycle the drift is the slow root, which is why the simple formula describes what a casual observer sees: the cusps are small and quick, and the eye integrates them away.

Two things about the cusps are worth stating carefully. The first is why the top falls at all if angular momentum is conserved — it is not conserved here, because gravity exerts a torque, and the vertical component of $\mathbf{L}$ is conserved while its magnitude is not. The second is that the fall is what supplies the precession: as the axis drops, the torque does work, and the resulting sideways motion is the axis converting that into azimuthal rotation. Nothing pushes the top sideways. It is dropped, and dropping is how a spinning body turns.

There is an energy statement underneath all this that is worth having. The quantity conserved during the motion is not the kinetic energy alone but the sum of the kinetic energy and the height of the centre of mass, and the tilt oscillates between two values in exactly the way a bead oscillates between two turning points of a potential well. The upper turning point is the release angle, which is why the path always comes back to the same tilt and why the cusps lie on a circle rather than drifting; the lower turning point is set by how much the top can fall before the azimuthal motion it has acquired is fast enough to carry it back up. Everything the drawn paths do is that one-dimensional oscillation, seen from above and combined with a drift.

A real top’s nutation dies away, and the reason is friction at the pivot, which the equations here do not contain. A top that has been running for a few seconds is genuinely in something close to steady precession, not because it was launched into it but because the dissipation has damped the nutation out — which is a different and much better argument than “it precesses steadily”, and it explains why the cusps are easy to see in the first second and hard to see afterwards.

The same equation somewhere else #

The structure $\mathrm{d}\mathbf{L}/\mathrm{d}t = \boldsymbol{\tau}$ with $\boldsymbol{\tau}$ perpendicular to $\mathbf{L}$ is not about tops.

A current loop is a magnetic moment, and a magnetic moment in an external field feels a torque $\boldsymbol{\mu} \times \mathbf{B}$ — perpendicular to the moment, exactly as the weight’s torque is perpendicular to a top’s angular momentum. If the loop carries angular momentum as well as magnetic moment, which it does whenever the current is made of moving mass or of spin, the response is the same one: precession, at a rate that does not depend on the tilt.

The correspondence is exact enough to be used in both directions. A classical top’s precession rate depends on its spin, and a nucleus’s does not — the magnetic moment and the angular momentum are proportional for a given species, so the ratio that sets the rate is a property of the particle rather than of how fast it happens to be going. That is what makes magnetic resonance a spectroscopy: every proton in a sample precesses at the same rate in the same field, and a shift of a few parts per million in that rate reports on the chemical environment the proton sits in.

That precession is called Larmor precession when the moment belongs to an orbiting electron and Larmor frequency when it belongs to a proton in a magnetic-resonance scanner, and the rate is $\gamma B$ with $\gamma$ the ratio of magnetic moment to angular momentum. The whole of magnetic resonance imaging is a measurement of that rate, one nucleus at a time, in a field that has been made to vary across a body so that the rate encodes position. The equation being solved is the one on this page.

A related cancellation is worth noticing beside it. The rate at which a charged particle circles in a magnetic field does not depend on its speed at all, non-relativistically, because a faster particle travels a proportionally larger circle and the two factors cancel. That is structurally the same surprise as a top’s precession rate not depending on its tilt, and it has the same cause: one factor of a quantity coming out of the force and another out of the geometry, in opposite senses.

One thing that is not the same deserves separating out, because it wears the same word. An orbit under a force law departing slightly from the inverse square precesses — the ellipse’s long axis creeps round — and there is no spin in it, no torque about the centre, and no angular momentum being turned. What precesses is a direction that returns to itself slightly late. Mercury’s perihelion advance is that; the axis of a top is not, and keeping the two apart is worth the effort, because the word is doing different work in each.

The couple that comes back #

The reaction the figures cannot show is worth putting a number on, because it is the reason gyroscopic effects appear in the design of machines that have nothing to do with tops.

Turn a spinning rotor’s axis and the rotor resists, with a couple of magnitude $I_3\omega_3\Omega$ — its angular momentum times the rate at which the axis is being turned, at right angles to both. For a substantial marine or aircraft rotor of a thousand kilogram square metres at three thousand revolutions a minute, being turned at a tenth of a radian a second, that is over thirty kilonewton-metres, applied to whatever the bearings are bolted to.

Which is a real load in three familiar places. A ship’s turbine rotor pitches the hull when the ship turns; a single-engined aircraft’s propeller produces a yaw when the aircraft pitches, which is why such aircraft are trimmed asymmetrically and why the effect is on the pre-flight briefing; and a large fan or flywheel in a vehicle has to have its bearings sized for a load that appears only when the vehicle manoeuvres.

Run deliberately, the same couple is a thruster. A control moment gyroscope is a spinning rotor mounted on a gimbal, and torquing the gimbal precesses the rotor, which delivers a large couple to the spacecraft it is bolted to — far more torque per unit of mass and power than accelerating a reaction wheel, because the torque comes from the spin the rotor already has rather than from changing it. Large spacecraft are attitude-controlled that way for exactly that reason.

The limitation is the one the geometry imposes. A gimballed rotor can be precessed only so far before its axis reaches an orientation where further torque in the wanted direction is unavailable — a singularity in the arrangement — and steering a cluster of them around those configurations is a control problem in its own right. The device delivers its torque from an existing angular momentum rather than creating one, and that is a constraint as well as an advantage.

The bicycle, which is not this #

The most widely repeated application of everything above is that a bicycle stays up because its wheels are gyroscopes. It is worth going through, because it is wrong and the experiment that shows it is wrong is a good one.

A bicycle’s front wheel does carry angular momentum, and leaning the bicycle does precess it in a direction that turns the wheel into the lean — which is a real effect and does contribute to the steering geometry that keeps the machine upright. The question is whether it is necessary, and the way to find out is to cancel it.

In 1970 that was done directly: a second wheel was mounted alongside the front one, clear of the ground, spun the opposite way at the same rate, so that the two angular momenta cancelled exactly and the machine had no net gyroscopic effect at the front at all. It remained rideable. The rider had to work slightly harder, and the bicycle would not stay up on its own when pushed away riderless — but the everyday claim, that a bicycle is held up by its wheels’ gyroscopic action, does not survive the experiment.

Nor does the usual second candidate, the trail — the distance by which the steering axis meets the ground ahead of the contact patch, which makes the front wheel castor. A bicycle was later built with neither gyroscopic effect nor trail, using a distribution of masses instead, and it was self-stable over a range of speeds.

What that leaves is the honest answer: self-stability comes from a combination of effects, no one of which is necessary, and which one dominates depends on the machine. That is unsatisfying next to a clean single-cause explanation, and it is the case. The gyroscopic story survives because the demonstration with a spinning wheel on a handle is so striking that it feels like it must be the explanation for the nearest familiar thing.

The Earth as a top #

The largest instance of this essay’s equation is worth naming with its number, because it was measured before anybody knew what it was.

The Earth is not a sphere: it bulges at the equator by about one part in three hundred. The Sun and Moon pull on the near side of that bulge harder than on the far side, and because the equatorial plane is tilted to the orbital planes, the resulting couple is at right angles to the spin axis. That is a top’s situation exactly — a spinning body with a torque perpendicular to its angular momentum — and the axis precesses.

The rate follows from the same triangle. Angular momentum divided into torque, with the tilt’s sines cancelling as before, and the answer is one turn in about 25,772 years. The Moon supplies roughly twice the torque the Sun does, despite being vastly less massive, because the torque falls as the cube of the distance.

Hipparchus noticed it around 130 BC by comparing his own star positions with records a century and a half older, and finding that the longitudes had all shifted by about two degrees. He had no mechanism and could not have had one; what he had was two catalogues and the sense to compare them. The explanation is Newton’s, in the Principia, and it is one of the first things the theory of gravitation was used for after the orbits themselves.

Where the model stops #

The top is assumed symmetric. Everything here takes $I_1 = I_2$ — two equal transverse moments — which is what makes the spin about the symmetry axis a constant of the motion. An asymmetric body under a torque is a much harder problem and has no steady solution of this kind at all.

An asymmetric body does something else even with no torque at all. Its angular velocity moves anyway, because the moment of inertia is a tensor and the angular velocity and the angular momentum need not be parallel — the free motion traces closed paths on the sphere of angular-velocity directions, stable near two of the principal axes and unstable near the third. A gyroscope avoids all of it by being built symmetric, which is a design choice rather than a law, and the essay below this one is about what happens when it is not made.

The pivot is assumed to be a point that does not move. A real pivot has friction, which damps the nutation and also slowly drags the top; and it has a finite size, so the contact point moves as the top leans, which is what makes a spinning top rise — the phenomenon of a sleeping top, where a tilted top gradually stands upright, is entirely a friction effect and is invisible to the equations here.

Gravity is taken as uniform and the top as rigid. Both are excellent approximations for a laboratory top and neither is exact. The first fails for the largest gyroscope anybody uses, which is the Earth: the Sun and Moon exert a torque on its equatorial bulge, and the resulting precession takes 25,772 years to go round once — the same equation, the same triangle, and a rate slow enough that it was noticed as a discrepancy in star positions before anyone knew what was doing it.

What the pictures cannot show #

The figures draw the axis and nothing else. The disc’s own rotation — the thing that makes all of this happen — is invisible in every one of them, because a symmetric disc looks the same at every stage of its own spin. What is drawn is a slow motion that would be a fall in the absence of a fast one that cannot be seen.

Nor do they show the reaction at the pivot. A precessing top exerts a torque back on whatever holds it, and it is not small: turning a gyroscope by hand is the standard demonstration that the force needed is at right angles to the direction it is being pushed. That reaction is the whole of what a gyrocompass, a ship’s stabiliser and a rate gyroscope in an aircraft are built to feel, and it is missing from these pictures entirely.

Where the ladder goes next #

The rotation ladder has gone from torque and the lever arm, through the moment of inertia and the axis that will not hold, to angular momentum as the thing conserved, and now to what a torque does to it. The rungs after it are the full Euler-angle treatment with the effective potential drawn in the tilt variable, where the nutation appears as oscillation between two turning points; the sleeping top and its stability condition, which is the same discriminant evaluated at zero tilt; the gyrocompass, where a constrained gyroscope on a rotating Earth finds north on its own; and the rigid body in three dimensions with no symmetry at all.

The habit worth carrying away is about what an equation of motion actually promises. $\mathrm{d}\mathbf{L}/\mathrm{d}t = \boldsymbol{\tau}$ does not say what a body does; it says how a vector changes, and what the body does depends on what that vector was. The same torque applied to the same object produces a fall or a horizontal sweep depending on nothing but whether the object was already spinning — and a picture of the forces, which is identical in the two cases, cannot tell them apart.