University Physics I · Angular Momentum · 11.5

Gyroscopes and Precession

A torque perpendicular to angular momentum changes where the axis points without changing how fast the wheel spins. That is why a pivoted gyroscope sweeps sideways instead of toppling.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Στ⃗ext = dL⃗/dt is a vector statement, and a vector can change in two ways. A torque along L⃗ changes how fast the body spins. A torque across L⃗ changes only where the spin axis points.

A gyroscope pivoted at one end of its axle is the second case in pure form: its weight acts downward at the centre of mass, so the torque about the pivot is horizontal and square to the axle. Every dL⃗ = τ⃗ dt is added sideways, so the tip of L⃗ travels along a circle instead of stretching along L⃗. The axis sweeps around the vertical at Ω = Mgr/(Iω) — quicker when the torque is larger, slower when the spin is faster.

The spin holds nothing up — the pivot carries the weight. What it redirects is the weight's torque, into a turn instead of a topple.

Simple definition: Precession is the slow, steady turning of a spinning object's axis, driven by a torque that pushes the axis sideways rather than changing how fast it spins.
Example: A bicycle wheel spun up and hung by one end of its axle does not fall. The axle stays roughly level and the whole wheel drifts slowly around the support point.

Vector form of the second lawΣτ⃗ext = dL⃗/dt

Torque sets how L⃗ changes — its direction as well as its size.

τ in N m, L in kg m² s⁻¹

Steering versus spinning upτ⃗ ∥ L⃗ → dL/dt = ττ⃗ ⟂ L⃗ → dφ/dt = τ/L

Along L⃗ the torque changes the spin rate; across L⃗ it only turns the axis.

φ is the axis angle, in rad

Steady precession rateΩ = τ/L = Mgr/(Iω)

More torque precesses faster; more spin precesses slower.

rad s⁻¹; r is pivot to centre of mass

Precession as a rotationdL⃗/dt = Ω⃗ × L⃗

The axis turns about the vertical; reverse the spin and the sweep reverses.

Ω⃗ is vertical through the pivot

Exact steady-precession conditionMgr = Ω(Iω − I⟂Ω cos θ)

Two roots, one slow and one fast; the slow one reduces to Mgr/(Iω).

θ from vertical; I⟂ about the pivot

Fast-spin validity test(Iω)² ≫ 4MgrI⟂cos θ

Comfortably true and Ω = Mgr/(Iω) is the rate; only below the equality does steady precession stop existing.

Both sides in kg² m⁴ s⁻²

Two things a torque can do

Στ⃗ext = dL⃗/dt is a vector equation, so split the torque relative to L⃗. The component along L⃗ changes only the magnitude — the body spins up or slows down. The component across L⃗ changes only the direction — the axis is steered. Pivot a spinning wheel at one end of its axle and only the second component survives: r⃗ runs along the axle, so r⃗ × F⃗ is perpendicular to the axle whatever the weight does. |L⃗| therefore holds steady: the wheel neither speeds up nor slows down while it precesses.

Follow one small step

Put the pivot at the origin, the axle along x̂ with L⃗ = Iω x̂, and gravity along −ẑ. Then τ⃗ = (r x̂) × (−Mg ẑ) = +Mgr ŷ. Over a short dt the wheel gains dL⃗ = τ⃗ dt, which points sideways, not downward. Add it: to first order the new L⃗ has the same length as the old one but points a little towards +ŷ. The axle has swung horizontally, not dropped. Repeat, and the axle sweeps a horizontal circle — counterclockwise seen from above, for an L⃗ pointing away from the pivot. Reverse the spin and the sweep reverses with it.

The steady rate

The tip of L⃗ travels on a circle of radius L, and it moves at |dL⃗/dt| = τ, so the axis turns through dφ = τ dt/L, giving Ω = τ/L = Mgr/(Iω). Take a flywheel with I = 0.30 kg m² spun at ω = 60 rad s⁻¹, total mass 3.0 kg, pivot-to-centre distance r = 0.20 m. Then τ = 3.0 × 9.81 × 0.20 = 5.9 N m, L = 18 kg m² s⁻¹, and Ω = 5.9/18 = 0.33 rad s⁻¹ — one circuit in 19 s. Double the spin and the circuit takes 38 s. Slide a small extra mass outward along the axle and the precession quickens, because Mgr rises while Iω does not.

Tilt makes no difference

Let the axle sit at angle θ from the vertical. The torque drops to Mgr sin θ, but the tip of L⃗ now traces a cone of radius L sin θ rather than a full circle. The two factors cancel: Ω = Mgr sin θ/(L sin θ) = Mgr/(Iω), the same rate as for a level axle. In vector form, dL⃗/dt = Ω⃗ × L⃗ with Ω⃗ vertical through the pivot. The Earth is this problem at scale: solar and lunar torques on its equatorial bulge swing the 23.4° axis around at Ω ≈ 7.7 × 10⁻¹² rad s⁻¹, one turn in about 26 000 years.

Gyroscopic stability

Read Ω = τ/L the other way round. A disturbance delivering angular impulse τΔt turns the axis through only Δφ ≈ τΔt/L. Make L large and any given push barely moves the axis, so the direction of the axle becomes a usable reference. That stability is bought with spin, not with stiffness. It also explains why the response feels wrong in the hand: the axis moves at right angles to the push, so pressing down on the free end of a spinning wheel makes it swing sideways. Rifled bullets, spin-stabilised spacecraft and the gimballed gyroscopes of an inertial navigation unit all trade on the same 1/L.

Where the fast-spin picture ends

The elementary result assumes all of L⃗ lies along the spin axis. It does not: the precession itself contributes I⟂Ω about the vertical, where I⟂ is the moment of inertia about a transverse axis through the pivot. The exact steady-precession condition is Mgr = Ω(Iω − I⟂Ω cos θ), a quadratic with real roots only when (Iω)² ≥ 4MgrI⟂cos θ; below that no steady precession exists and the top falls over. Released from rest, a gyroscope also nutates — the axis dips and recovers at about Iω/I⟂, with amplitude 2I⟂Mgr sin θ/(Iω)². For the flywheel above (I⟂ ≈ 0.27 kg m²) that is 67 rad s⁻¹ and 0.6°, quickly damped by pivot friction, and the average precession is still Mgr/(Iω).

Change one variable at a time

Make the relationship visible.

Interactive model

Spin rate ω60 rad s⁻¹

Tilt θ from vertical90 °

Pivot to centre r0.16 m

Drag the tilt and watch the circle and the torque shrink together while the angle swept in one second refuses to move; then raise the spin and watch that angle shrink, because Ω = Mgr/(Iω).

TORQUE Mgr sinθ3.92 N m

SPIN L = Iω15.00 kg m² s⁻¹

PRECESSION Ω0.262 rad s⁻¹

ONE CIRCUIT24.0 s

Live interpretationTORQUE Mgr sinθ: 3.92 N m. SPIN L = Iω: 15.00 kg m² s⁻¹. PRECESSION Ω: 0.262 rad s⁻¹. ONE CIRCUIT: 24.0 s

Catch the common trap

Explain before calculating.

A gyroscope precesses steadily with its axle horizontal. The same wheel, spin rate and pivot-to-centre distance are then used with the axle tilted to 60° from the vertical, still in the fast-spin regime. What happens to the steady precession rate?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA bicycle wheel of moment of inertia I = 0.14 kg m² is spun to ω = 32 rad s⁻¹ and hung by one end of its axle, with the axle level. Wheel and axle together mass 2.4 kg, and the centre of mass sits 0.18 m from the pivot. Find the steady precession rate and the time for one full sweep.

Torque about the pivot: τ = Mgr = 2.4 kg × 9.81 m s⁻² × 0.18 m = 4.24 N m, horizontal and square to the axle.
Spin angular momentum, along the axle: L = Iω = 0.14 kg m² × 32 rad s⁻¹ = 4.48 kg m² s⁻¹.
Every dL⃗ = τ⃗ dt is added sideways, so the tip of L⃗ turns at Ω = τ/L = 4.24 N m ÷ 4.48 kg m² s⁻¹ = 0.946 rad s⁻¹.
One circuit: T = 2π/Ω = 6.283 ÷ 0.946 rad s⁻¹ = 6.6 s.

AnswerΩ ≈ 0.95 rad s⁻¹; the axle sweeps one horizontal circle in about 6.6 s.

MediumA demonstration gyroscope has I = 0.020 kg m² about its spin axis, total mass 1.5 kg, and centre of mass 0.12 m from the pivot. (a) What spin rate makes the level axle take 12 s to precess once round? (b) The axle is then tilted to 30° from the vertical at that same spin — how long does one circuit take now?

Target precession rate: Ω = 2π/T = 6.283 ÷ 12 s = 0.5236 rad s⁻¹.
Weight torque about the pivot, axle level: τ = Mgr = 1.5 kg × 9.81 m s⁻² × 0.12 m = 1.766 N m.
Rearrange Ω = Mgr/(Iω): Iω = Mgr/Ω = 1.766 N m ÷ 0.5236 rad s⁻¹ = 3.372 kg m² s⁻¹.
ω = 3.372 kg m² s⁻¹ ÷ 0.020 kg m² = 169 rad s⁻¹, about 27 revolutions per second.
(b) Tilting cuts the torque to Mgr sin 30° = 0.883 N m, but the tip of L⃗ now traces a cone of radius L sin 30° instead of a full circle; the two sin 30° factors cancel, leaving Ω = Mgr/(Iω) = 0.524 rad s⁻¹.

Answerω ≈ 1.7 × 10² rad s⁻¹ (about 27 rev s⁻¹); at 30° from the vertical the circuit still takes 12 s.

HardA top has I = 4.0 × 10⁻⁴ kg m² about its spin axis and I⟂ = 3.0 × 10⁻⁴ kg m² about a transverse axis through the pivot. Its mass is 0.080 kg, its centre of mass is 0.030 m from the pivot, and the axle sits at 60° from the vertical. Find the slowest spin that permits steady precession, and compare the exact and elementary rates at ω = 40 rad s⁻¹.

Weight term: Mgr = 0.080 kg × 9.81 m s⁻² × 0.030 m = 0.02354 N m — the sin θ has already cancelled out of Mgr = Ω(Iω − I⟂Ω cos θ).
Steady precession needs real roots: (Iω)² ≥ 4MgrI⟂cos θ = 4 × 0.02354 × 3.0 × 10⁻⁴ × 0.500 = 1.413 × 10⁻⁵ kg² m⁴ s⁻², so Iω ≥ 3.759 × 10⁻³ and ω ≥ 3.759 × 10⁻³ ÷ 4.0 × 10⁻⁴ = 9.4 rad s⁻¹.
At ω = 40 rad s⁻¹, Iω = 4.0 × 10⁻⁴ kg m² × 40 rad s⁻¹ = 0.0160 kg m² s⁻¹, so the elementary rate is Ω = Mgr/(Iω) = 0.02354 ÷ 0.0160 = 1.47 rad s⁻¹.
Exact condition as a quadratic in Ω: I⟂cos θ Ω² − IωΩ + Mgr = 0, i.e. 1.50 × 10⁻⁴ Ω² − 0.0160 Ω + 0.02354 = 0.
Slow root: Ω = [0.0160 − √(2.560 × 10⁻⁴ − 1.413 × 10⁻⁵)] ÷ (2 × 1.50 × 10⁻⁴) = (0.0160 − 0.015552) ÷ 3.00 × 10⁻⁴ = 1.49 rad s⁻¹.
Mgr/(Iω) is therefore 1.4 % low here; the other root, 105 rad s⁻¹, is the fast precession that pivot friction damps away.

AnswerMinimum spin ω ≈ 9.4 rad s⁻¹; at ω = 40 rad s⁻¹ the exact slow rate is Ω ≈ 1.49 rad s⁻¹ against 1.47 rad s⁻¹ from Mgr/(Iω).