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Sandbox Physics

M055 · Rigid body / torque geometry

Gyroscope Precession Bench

Orbit a supported flywheel in three dimensions. Release it freely, prepare either steady-precession branch, reverse the spin or remove gravity. Follow the angular-momentum and torque vectors, measure nutation, and compare the full dynamics with the slow-precession estimate.

Interactive modelGyroscope Precession Bench
Reviewed timePending\text{Pending}
Axis tiltPending\text{Pending}
Instantaneous precessionPending\text{Pending}
Nutation ratePending\text{Pending}
Physical axial spinPending\text{Pending}
Slow-precession estimatePending\text{Pending}
Exact slow branchPending\text{Pending}
Prepared initial statePending\text{Pending}
Rotational kinetic energyPending\text{Pending}
Gravitational potentialPending\text{Pending}
Total energyPending\text{Pending}
Vertical angular momentumPending\text{Pending}
Axial angular momentumPending\text{Pending}
Gravitational torquePending\text{Pending}
Pivot reaction magnitudePending\text{Pending}
Vertical pivot reactionPending\text{Pending}
Max scaled energy defectPending\text{Pending}
Max scaled momentum defectPending\text{Pending}
Axis difference: two stepsPending\text{Pending}
Upright spin thresholdPending\text{Pending}
Spin / upright thresholdPending\text{Pending}

Physics tutorial

Gyroscope Precession Bench

BackgroundOrbit a supported flywheel in three dimensions. Release it freely, prepare either steady-precession branch, reverse the spin or remove gravity. Follow the angular-momentum and torque vectors, measure nutation, and compare the full dynamics with the slow-precession estimate.

Why it mattersA shared rigid-body law explains both an instrument flywheel and a nutating top. Looking only at the spinning texture hides the important motion of its axis.

Start with the essentials

Focus question
Why does a spinning wheel precess instead of simply falling?
One-sentence intuition
Gravity changes world angular momentum through a sideways torque; energy, its vertical component and its body-axis projection remain conserved.

Core mathematical model

Geometry and pivot inertia

I3=12mR2,I⊥=14mR2+mℓ2I_3=\frac12mR^2,\qquad I_\perp=\frac14mR^2+m\ell^2

Homogeneous thin disk on a massless shaft, with the distance measured from the captured pivot to the disk centre. Displayed disk thickness, support and hub are schematic and massless.

Full vector dynamics

L˙=mgℓ ez×n,n˙=L×nI⊥\dot{\mathbf L}=mg\ell\,\mathbf e_z\times\mathbf n,\qquad\dot{\mathbf n}=\frac{\mathbf L\times\mathbf n}{I_\perp}

All quantities refer to the fixed pivot. The symmetry axis and world angular momentum evolve together; the spin angle is reconstructed with a quaternion.

Angular velocity and axial spin

ω=LI⊥+(1I3−1I⊥)(L⋅n)n,ω3=ψ˙+ϕ˙cos⁡θ\boldsymbol\omega=\frac{\mathbf L}{I_\perp}+\left(\frac1{I_3}-\frac1{I_\perp}\right)(\mathbf L\cdot\mathbf n)\mathbf n,\qquad\omega_3=\dot\psi+\dot\phi\cos\theta

Physical axial spin is not the Euler spin-angle rate. Confusing them changes the exact steady-precession condition.

Two steady-precession branches

I⊥cos⁡θ Ω2−I3ω3Ω+mgℓ=0I_\perp\cos\theta\,\Omega^2-I_3\omega_3\Omega+mg\ell=0

A real root prepares constant tilt only with zero initial nutation. A kick perturbs that solution. At a horizontal axis the equation becomes linear.

Slow-precession approximation

Ωslow≃mgℓI3ω3\Omega_{\mathrm{slow}}\simeq\frac{mg\ell}{I_3\omega_3}

This estimate drops the transverse-inertia term. It becomes accurate when precession is small relative to axial spin, and is undefined at zero spin.

Three conserved quantities

E=I⊥2(θ˙2+ϕ˙2sin⁡2θ)+I3ω322+mgℓcos⁡θ,Lz=const,L⋅n=constE=\frac{I_\perp}{2}(\dot\theta^2+\dot\phi^2\sin^2\theta)+\frac{I_3\omega_3^2}{2}+mg\ell\cos\theta,\qquad L_z=\mathrm{const},\quad\mathbf L\cdot\mathbf n=\mathrm{const}

Potential energy is zero when the centre is at pivot height. Maximum errors are scaled by nonzero physical energy and momentum scales; they remain meaningful when an invariant itself vanishes.

Upright stability threshold

∣I3ω3∣>2I⊥mgℓ|I_3\omega_3|>2\sqrt{I_\perp mg\ell}

This is a local linear stability condition. Exactly vertical rotation remains an exact solution even below threshold; use a small tilt to reveal its instability.

Captured-pivot reaction

R=mℓn¨+mgez\mathbf R=m\ell\ddot{\mathbf n}+mg\mathbf e_z

The ideal bearing can supply a negative vertical reaction. A freely resting tip would need a separate unilateral contact and collision model.

Common difficulties

The momentum always follows the shaft

Typical misconceptionThe shaft direction and angular momentum are identical at every spin.

Better mental modelTransverse inertia contributes whenever the axis moves. Compare the blue arrow with the shaft in the low-spin or fast-branch preset.

Exactly sleeping proves stability

Typical misconceptionA vertical top that remains vertical must be stable.

Better mental modelThe exact unperturbed solution survives below threshold. Compare small-tilt presets to test the response to a perturbation.

A pivot is a free tabletop tip

Typical misconceptionA negative vertical reaction can be supplied by an ordinary surface.

Better mental modelThis bearing captures the point. A tabletop cannot pull the tip; sliding, detachment and collisions are outside the model.

Run the experiment

  1. 01

    Prepare and orbit

    Select free release, inspect overview, top and side, then drag the orange tilt handle.

    What to observe: The new initial axis drives both the physical scene and its quantitative record.
  2. 02

    Compare steady branches

    Select slow and fast branches; leave the nutation kick at zero. Reverse the spin.

    What to observe: Constant tilt is a prepared solution; a release without the required precession generally nutates.
  3. 03

    Test the approximation or sleeping state

    Reduce spin and inspect the approximation; for the conical top compare stable and unstable small-tilt presets.

    What to observe: A slow-precession approximation can fail, and a local stability threshold needs a perturbation to become visible.
  4. 04

    Check the numerical record

    Review the full record and inspect conservation defects and the difference between two step sizes. Use fine phase steps for rotor markings.

    What to observe: A small numerical discrepancy supports the retained model; it does not certify omitted friction or tabletop contact physics.