Reaction wheel on a free body

The motor torque acts only in the wheel’s shape equation. It pushes μ around in the body frame, but the momentum map J = gμ stays fixed unless an external torque acts.

Drag to orbit

J (spatial) ω wheel

External torque

Motor torque u0.00

‖J‖: –
Energy: –
Wheel rate φ̇: –

Body momentum μ, body axes 123

Momentum map J = gμ, spatial axes 123

Same vertical scale as μ. With no external torque, these lines stay flat whatever the motor does.

Wheel momentum p p

What is being integrated

The configuration is g ∈ SO(3) for the body and the wheel angle φ relative to the body. The group is SO(3) and the shape space is the circle. Here I is the locked inertia (body plus wheel, wheel not spinning), I_w is the wheel’s axial inertia, and â is the wheel axis in the body frame.

Reduced Lagrangian: l(ω, φ̇) = ½ ω·Iω + I_w φ̇ (â·ω) + ½ I_w φ̇²
Body momentum: μ = ∂l/∂ω = Iω + I_w φ̇ â
Wheel momentum: p = ∂l/∂φ̇ = I_w (â·ω + φ̇)
Euler–Poincaré: μ̇ = μ × ω + τ
Shape equation: ṗ = u
Recover ω: ω = (I − I_w ââ^T)⁻¹ (μ − p â)
Reconstruction: ġ = g ω̂
Momentum map: J = gμ, with J̇ = gτ

Parameters: I = diag(3, 2, 1.2), I_w = 0.08, â along body axis 3. The external torque is τ = (0.06, 0, 0) in the body frame. Integrated with RK4 at a fixed step of 0.004.