Gyroscopes and Precession

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.

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 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.

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.

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.

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).