Myla's notebook

The Gyroscopic Motion

21 May, 2025

Quick Guide to Gyroscope Dynamics

A symmetric top ( $I_{1} = I_{2} \neq I_{3}$ ) under gravity is described by

$ϕ (t)$ : tilt from horizontal
$θ (t)$ : precession around vertical
constant spin rate $ω$ about its symmetry axis

1. Full equations

\ddot{ϕ} = \dot{θ} \cos ϕ (- \dot{θ} \sin ϕ + \frac{I_{3}}{I_{1}} ω) - \frac{m g l}{I_{1}} \cos ϕ

\ddot{θ} = \frac{\dot{ϕ}}{\cos ϕ} (2 \dot{θ} \sin ϕ - \frac{I_{3}}{I_{1}} ω)

2. Small‐tilt approximation ( $ϕ ≪ 1$ )

Set $\sin ϕ \approx ϕ$ , $\cos ϕ \approx 1$ . Then

\dot{θ} = - \frac{I_{3}}{I_{1}} ω ϕ + Ω_{0}

\ddot{ϕ} + κ^{2} ϕ = C with κ^{2} = Ω_{0}^{2} + (\frac{I_{3}}{I_{1}} ω)^{2}, C = \frac{I_{3}}{I_{1}} ω Ω_{0} - \frac{m g l}{I_{1}}

Mean nutation angle:

ϕ_{m e a n} = \frac{C}{κ^{2}}

3. Azimuthal drift per nutation

One period $T = 2 π / κ$ gives

Δ θ = \int_{0}^{T} \dot{θ} d t = 2 π \frac{m g l I_{1}}{I_{3}^{2} ω^{2}} = 2 π ϕ_{m e a n}

4. Steady precession

Constant tilt ( $\dot{ϕ} = 0$ , $\ddot{θ} = 0$ ) requires

I_{1} \sin ϕ Ω^{2} - I_{3} ω Ω + m g l = 0

Solutions (“slow” and “fast” modes):

Ω = \frac{I_{3} ω \pm \sqrt{(I_{3} ω)^{2} - 4 I_{1} m g l \sin ϕ}}{2 I_{1} \sin ϕ}

Exists if $(I_{3} ω)^{2} \geq 4 I_{1} m g l \sin ϕ$ .
Slow root is the usual top precession; fast root is high‐energy.