Gyroscope

Set the wheel spinning, choose the angle its axle is let go at, and decide whether to give it a sideways push as you let go. The page integrates the motion while you watch and prints how far the wheel falls, how fast it turns, and how far the textbook formula is from what it actually does.

Worked example — Gyroscope

Moment of inertia about the axle (hoop)	0.180 kg·m²
Spin angular momentum	11.31 kg·m²/s
Torque of the weight about the pivot	3.923 N·m
Steady turning rate, slow-precession formula	0.3468 rad/s
One turn at that rate	18.12 s
Simulation against the formula	−0.005%
Sideways turning rate, slowest to fastest	0.0000 to 0.6937 rad/s
Drop below the release, simulated	0.597°

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Gyroscope

A gyroscope here is a spinning wheel whose axle rests on a fixed pivot at one end. With the axle level, its weight exerts a torque τ = mgl about the pivot, horizontal and at right angles to the axle, and torque is the rate of change of angular momentum, so the spin's angular momentum L = I₃ω₃ is turned sideways rather than tipped down. In steady precession the axle turns round the vertical at Ω = mgl/L; let go from rest, it first falls a little and then nutates, bobbing between its starting height and a lowest point about 2I₁mgl/L² radians below it, at about L/(2πI₁) times a second.

What each symbol means

L: angular momentum of the spin about the axle, I₃ω₃
Ω: steady precession rate, mgl/L, the rate at which the axle turns round the vertical
I_3: moment of inertia about the axle; mR² for a hoop
I_1: moment of inertia about a line through the pivot at right angles to the axle, ml² + ½mR²
l: distance from the pivot to the wheel's centre
k: sideways turning rate at release, in units of Ω

Assumes: A rigid wheel treated as a hoop with all its mass on the rim, on a light axle, resting on a frictionless point pivot. Standard gravity, 9.80665 m/s². No air and no friction anywhere, so energy and the angular momentum about the vertical and about the axle are all conserved; the results table prints how well the integrator keeps them.
Breaks when: A real pivot has friction, which damps the wobble and leaves the wheel precessing smoothly a little below where it was let go. The bearing's friction also slows the spin, and as L falls the precession speeds up and the drop deepens, until the wheel falls over.

Let go, a spinning wheel falls before it turns 🖖

Press Let go at 10 rev/s and watch the second panel. Turning sideways means the wheel's centre is moving, and that kinetic energy has to come from somewhere, so the wheel cannot start precessing the moment you let go. It falls first. The axle drops 0.597°, its centre 2.09 mm, and at the bottom of the drop the sideways rate has reached 0.6937 rad/s, twice its average. Then it climbs back to the height it was let go at, where the sideways motion stops dead, and the cycle starts again: the end of the axle draws a row of cusps, 10.59 of them a second. Peraire and Widnall's MIT dynamics notes name this cuspidal motion, and David Tong's Cambridge lectures derive it as exactly what a top does when you spin it and let go. A drop this small is hard to see, and friction at a real pivot smooths the wobble away, which is why a demonstration wheel seems to float. The grey ghost is the same wheel with no spin, hanging straight down after 0.39 s.

Spin it twice as fast and it turns half as fast 🖖

The weight's torque about the pivot, 3.923 N·m with the axle level, changes L by 3.923 kg·m²/s every second, always sideways. The bigger L is, the longer that takes to swing it round, so the turning rate is Ω = mgl/L: one turn in 9.06, 18.12 and 36.23 s at 5, 10 and 20 rev/s. The drop goes as 1/L², a quarter for each doubling: the simulation gives 2.386°, 0.597° and 0.149°. Ω = mgl/L treats L as lying along the axle, which is only true while the spin's angular momentum dwarfs the part the sideways turning adds. The table follows it down: the simulated wheel turns 0.087% slower than the formula at 5 rev/s, 3.026% at 2, and 21.481% at 1, where the wheel swings 39.013° down and behaves more like a pendulum with a twist than a gyroscope.

Push the axle sideways and the wheel climbs 🖖

A shove on the end of the axle is a torque, and like the weight's it moves the tip of L in the direction of the torque, not of the shove. A shove in the direction the wheel is already turning makes an upward torque, so the axle rises. Press Pushed at 2Ω: let go already turning sideways at twice the steady rate, the axle climbs 0.597°, the same distance the unpushed wheel fell, and its mean turning rate is still 0.3468 rad/s. Hold it back with a push of −1 and it drops 1.195°, twice as far. Press Let go turning at Ω and there is no drop and no wobble at all, because for a level axle steady precession at exactly mgl/L is a solution of the full equations at any spin. Set the push to 1 and read the table: at 1 rev/s, where letting go from rest was 21.481% out, the wheel now turns at 3.4684 rad/s, the formula's own number. The low-spin error is the price of letting go from rest. Tong's notes put it as needing "exactly the right push" to make a top precess without bobbing.

Problem solved in full

What does the pivot carry, and how fast must the wheel spin? 5 steps
The wheel spins at 10 rev/s with its axle level, precessing steadily. How hard does the pivot push on the axle, upwards and sideways? And what spin would keep the drop after letting go from rest under 0.1°? Press Let go at 10 rev/s to check the first step.

Open this problem in the tool
1. The spin angular momentum and the steady turning rate are on the page:
  
  $L = I_{3} ω_{3} = 0.180 \times 2 π \times 10 = 11.31 kg m^{2} /s Ω = \frac{m g l}{L} = \frac{3 . 923}{11 . 31} = 0.3468 rad/s$
2. Precessing steadily with the axle level, the centre stays at one height, so it has no vertical acceleration and the pivot pushes up with exactly the weight. The spin takes none of it:
  
  $N_{z} = m g = 2 \times 9.80665 = 19.61 N$
3. The centre goes round a circle of radius l at Ω, so the pivot must also push it towards the stand:
  
  $N_{h} = m l Ω^{2} = 2 \times 0.2 \times 0.346 8^{2} = 0.0481 N$
4. For the second question, turn the fast-top estimate of the drop round, with δ = 0.1° = 0.001745 rad:
  
  $δ = \frac{2 I _{1} m g l}{L ^{2}} \Rightarrow L \geq \frac{2 I _{1} m g l}{δ} = \frac{2 \times 0 . 170 \times 3 . 923}{0 . 001745} = 27.64 kg m^{2} /s$
5. and divide by I₃ and by 2π:
  
  $ω_{3} = \frac{27 . 64}{0 . 180} = 153.6 rad/s = \frac{153 . 6}{2 π} rev/s = 24.44 rev/s$
Answer

The pivot carries 19.61 N upwards, the wheel's whole weight, and 0.0481 N sideways towards the stand, under a quarter of a percent of it; the spin makes the wheel no lighter. Keeping the drop under 0.1° takes 24.44 rev/s, and solving the energy equation exactly at that spin gives a drop of 0.0999997°, so the estimate is good to about 2.5 parts in a million there.

$N_{z} = 19.61 N, N_{h} = 0.0481 N, f \geq 24.44 rev/s (exact drop 0.099999 7^{\circ})$

Learning path

Rotation and angular momentum

References (2)

The heavy symmetric top from its Lagrangian: the two conserved momenta and the energy, the three kinds of nutation, why a top that is spun and simply let go falls first and traces cusps, and why precessing without bobbing needs exactly the right push: D. Tong, "Classical Dynamics", lecture notes for a final-year undergraduate course, University of Cambridge, chapter 3 "The Motion of Rigid Bodies", section 3.6 "The Heavy Symmetric Top".
Steady precession from Euler's equations in Euler angles, the gyroscopic limit in which the precession rate is the weight's moment over the spin angular momentum, and the names unidirectional, looping and cuspidal for the three kinds of unsteady motion: J. Peraire and S. Widnall, "Lecture L30 - 3D Rigid Body Dynamics: Tops and Gyroscopes", 16.07 Dynamics, MIT OpenCourseWare, Fall 2009.

By the 42Math team · Reviewed against published sources. Corrections welcome.