M055 · Torque on a spinning rigid body

Gyroscope Precession Bench

Why does a spinning wheel precess instead of simply falling? Because gravity's torque on the wheel points sideways, not down. A torque changes angular momentum in its own direction, so the wheel's large spin angular momentum swings around the vertical instead of tipping over. Energy, the angular momentum about the vertical and the spin along the axle all stay constant during the motion.

Release a spinning flywheel held at one pivot, start it on either steady-precession branch, reverse its spin or switch gravity off. Follow the angular-momentum and torque arrows in three dimensions, measure the nutation and compare the full motion with the slow-precession estimate.

Reviewed time $\text{Pending}$

Axis tilt $\text{Pending}$

Instantaneous precession $\text{Pending}$

Nutation rate $\text{Pending}$

Physical axial spin $\text{Pending}$

Slow-precession estimate $\text{Pending}$

Exact slow branch $\text{Pending}$

Prepared initial state $\text{Pending}$

Rotational kinetic energy $\text{Pending}$

Gravitational potential $\text{Pending}$

Total energy $\text{Pending}$

Vertical angular momentum $\text{Pending}$

Axial angular momentum $\text{Pending}$

Gravitational torque $\text{Pending}$

Pivot reaction magnitude $\text{Pending}$

Vertical pivot reaction $\text{Pending}$

Max scaled energy defect $\text{Pending}$

Max scaled momentum defect $\text{Pending}$

Axis difference: two steps $\text{Pending}$

Upright spin threshold $\text{Pending}$

Spin / upright threshold $\text{Pending}$

Set up, measure, then verify

Set the initial angular momentum

Show physical vectorsEach kind of arrow is normalized on its own; the readouts give the actual magnitudes.Body mass

m

1.00\,\mathrm{kg}

Body radius

R

0.20\,\mathrm{m}

Pivot to centre distance

\ell

0.22\,\mathrm{m}

Physical axial spin

\omega_3

100.00\,\mathrm{rad/s}

Initial axis tilt

\theta_0

75.00\,\mathrm{{}^\circ}

Free-release precession

\dot\phi_0

0.00\,\mathrm{rad/s}

Initial nutation rate

\dot\theta_0

0.00\,\mathrm{rad/s}

Initial state preparation

s

0.00\,

Record duration

T

10.00\,\mathrm{s}

Uniform gravity

g

9.81\,\mathrm{m/s^2}

Reviewed fraction

t/T

0.00\,\%

Drag the orange tilt handle below. The initialization setting chooses the start: 0 is a free release, 1 the slow steady-precession branch and 2 the fast branch. A nutation kick disturbs the slow branch; the fast branch always has zero kick and follows its exact solution. If no real branch exists, the free-release settings apply and the status line says so.

The wheel is a uniform thin disk on a massless shaft, with the distance measured from the fixed pivot to the disk's centre. The drawn disk thickness, support and hub are only illustrations and carry no mass. The model is an ideal symmetric rigid body with one point fixed, in uniform gravity, and the full equations are integrated without assuming a small tilt or fast spin. Spin here means the component of angular velocity along the wheel's axis, which differs from the rate of the Euler spin angle. The bearing can pull as well as push, so this is not a tip that slides or bounces freely on a table. Friction, air drag, deformation, collisions with the floor or support, and spin decay are left out. The full-record plots include samples calculated ahead of the current time. At high spin the screen frame rate can make the rotation look wrong (aliasing); pause and use the fine-step button to follow the phase. The upright stability threshold applies only to infinitesimal tilts away from vertical spinning.

Physics tutorial

Gyroscope Precession Bench

BackgroundA spinning flywheel is held at one fixed pivot, and you can turn the view around it in three dimensions. Release it freely, start it on either steady-precession branch, reverse the spin or switch gravity off. Follow the angular-momentum and torque arrows, measure the nutation, and compare the full motion with the slow-precession estimate.

Why it mattersThe same rigid-body law explains the flywheel in an instrument and a top that wobbles as it spins. If you watch only the spinning surface, you miss what matters: the axis itself is moving.

Start with the essentials

Focus question: Why does a spinning wheel precess instead of simply falling?
One-sentence intuition: Gravity's sideways torque changes the angular momentum in the fixed frame, yet the energy, the vertical part of the angular momentum and its part along the axis all stay constant.

Core mathematical model

Moments of inertia about the pivot

I_3=\frac12mR^2,\qquad I_\perp=\frac14mR^2+m\ell^2

The wheel is a uniform thin disk on a massless shaft, and the distance is measured from the fixed pivot to the disk's centre. The drawn disk thickness, support and hub are only illustrations and carry no mass.

Equations of motion in vector form

\dot{\mathbf L}=mg\ell\,\mathbf e_z\times\mathbf n,\qquad\dot{\mathbf n}=\frac{\mathbf L\times\mathbf n}{I_\perp}

Every quantity is taken about the fixed pivot. The symmetry axis and the angular momentum in the fixed frame change together, and the spin angle is tracked with a quaternion.

Angular velocity and spin about the axis

\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

The physical spin about the axis is not the rate of the Euler spin angle. Mixing the two up gives the wrong condition for exact steady precession.

Two steady-precession branches

I_\perp\cos\theta\,\Omega^2-I_3\omega_3\Omega+mg\ell=0

A real root gives a constant tilt only if the wheel starts with no nutation. A kick disturbs that solution. When the axis is horizontal, the equation becomes linear.

Slow-precession approximation

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

This estimate drops the term with the transverse moment of inertia. It is accurate when the precession is slow compared with the spin about the axis, and it is undefined at zero spin.

Three conserved quantities

E=\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 disk's centre is level with the pivot. The largest errors are divided by nonzero energy and angular-momentum scales of the system, so they still mean something when a conserved quantity is itself zero.

Upright stability threshold

|I_3\omega_3|>2\sqrt{I_\perp mg\ell}

This is a linear stability condition for small tilts near vertical. Below the threshold, perfectly vertical spinning is still an exact solution; add a small tilt to see that it is unstable.

Force from the pivot

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

The ideal bearing can pull down as well as push up. A tip resting freely on a surface would need a separate model for one-sided contact and collisions.

Common difficulties

Angular momentum always points along the shaft

Typical misconceptionThe shaft and the angular momentum point the same way at every spin rate.

Better mental modelWhenever the axis moves, the transverse moment of inertia adds to the angular momentum. Compare the blue arrow with the shaft at low spin or on the fast branch.

A top that stays upright must be stable

Typical misconceptionIf a vertical top stays vertical, it must be stable.

Better mental modelBelow the threshold, the undisturbed vertical solution still exists. Compare runs that start from a small tilt to see how the top responds to a disturbance.

The pivot works like a tip on a table

Typical misconceptionAn ordinary table could supply the downward pull that the pivot provides.

Better mental modelHere a bearing holds the tip in place. A table cannot pull the tip down, and sliding, lifting off and collisions are outside the model.

Run the experiment

01
Set up and look around
Select free release, look at the overview, top and side views, then drag the orange tilt handle.
What to observe: The new starting axis changes both the scene and the recorded numbers.
02
Compare the steady branches
Select the slow and fast branches with the nutation kick left at zero. Then reverse the spin.
What to observe: The tilt stays constant only when the wheel starts exactly on that solution. Released without the right precession, it usually nutates.
03
Test the approximation and the upright top
Reduce the spin and check the approximation. For the conical top, compare the stable and unstable small-tilt presets.
What to observe: The slow-precession approximation can fail, and the upright stability threshold shows up only when you disturb the top.
04
Check the numerical accuracy
Review the full record and look at the conservation errors and the difference between two step sizes. Use fine phase steps to follow the rotor markings.
What to observe: Small numerical errors show that the model is solved accurately. They say nothing about friction or table contact, which the model leaves out.