JOVANA
Explore Library Glossary Getting Started Three Levels Fields How it works Mission
Join the mission
All guides

Euler's Equations, the Free Top and the Gyroscope

Newton's law for rotation, written in the frame spinning with the body — how Euler's equations explain the tumbling wrench, the wobble of the Earth, and why a spinning gyroscope precesses instead of falling.

The trouble with the lab frame

Rotation obeys Newton's rotational law: the rate of change of angular momentum equals the torque. In an inertial (lab) frame that reads simply as follows.

\left.\frac{d\mathbf{L}}{dt}\right|_{\text{lab}} = \boldsymbol{\tau}

The rotational Newton's law — exact, but awkward, because in the lab frame the inertia tensor keeps changing as the body reorients.

The awkwardness is that \mathbf{L}=\mathbf{I}\,\boldsymbol{\omega}, but as the body turns, the components of \mathbf{I} in the lab frame keep changing. The cure is to sit in the body-fixed frame — the frame spinning with the object. There the inertia tensor is constant (just the three principal moments), at the price of working in a rotating, non-inertial frame.

The transport theorem and Euler's equations

Any vector seen from a rotating frame has an extra piece in its time derivative: the transport theorem says (d/dt)_{\text{lab}} = (d/dt)_{\text{body}} + \boldsymbol{\omega}\times. Apply it to \mathbf{L}:

\left(\frac{d\mathbf{L}}{dt}\right)_{\text{body}} + \boldsymbol{\omega}\times\mathbf{L} = \boldsymbol{\tau}

The same physical law carried into the rotating body frame; the extra \boldsymbol{\omega}\times\mathbf{L} term is the price of leaving an inertial frame.

Now choose the body's principal axes, so \mathbf{L}=(I_1\omega_1,\,I_2\omega_2,\,I_3\omega_3) with the three moments constant. Writing the equation out component by component gives the celebrated set:

\begin{aligned} I_1\dot{\omega}_1 &= (I_2 - I_3)\,\omega_2\omega_3 + \tau_1 \\ I_2\dot{\omega}_2 &= (I_3 - I_1)\,\omega_3\omega_1 + \tau_2 \\ I_3\dot{\omega}_3 &= (I_1 - I_2)\,\omega_1\omega_2 + \tau_3 \end{aligned}

Euler's equations of rigid-body motion — three coupled, nonlinear equations for the body-frame angular velocity.

The torque-free top: wobble without a push

Set \boldsymbol{\tau}=0 — a body spinning freely in space, or a planet with no external couple. Astonishingly, the axis can still move. For a symmetric top (I_1=I_2\neq I_3), Euler's equations show \omega_3 stays constant while \omega_1 and \omega_2 rotate steadily: the spin axis traces a cone in the body frame. This is torque-free precession, and its rate is

\Omega_{\text{body}} = \frac{I_3 - I_1}{I_1}\,\omega_3

The rate at which the spin axis circles in the body frame of a freely spinning symmetric top — zero only if the body is spherically symmetric (I_3=I_1).

The Earth is a real example. It is slightly oblate (I_3 a touch larger than I_1), so this formula predicts a free wobble of its rotation axis with a period of about 305 days — the 'Euler period'. The wobble is genuinely observed, but with a period of roughly 433 days: the Chandler wobble. The discrepancy is honest physics, not error — the Earth is not perfectly rigid, and its elastic yielding lengthens the period.

The heavy gyroscope: precession under gravity

Now hold the spinning top by a point below its centre, so gravity applies a torque. For a fast top the angular momentum is nearly all along the spin axis, \mathbf{L}\approx I_3\omega_3\,\hat{\mathbf{n}}. Gravity's torque \boldsymbol{\tau}=\mathbf{r}\times M\mathbf{g} points horizontally, perpendicular to \mathbf{L}. A perpendicular torque cannot change the length of \mathbf{L}, only swing its direction — so the axis sweeps sideways around a cone instead of falling.

Gravity's torque is perpendicular to the spin angular momentum, so it swings the axis sideways into steady precession rather than toppling it.

\frac{d\mathbf{L}}{dt} = \boldsymbol{\tau} \;\Longrightarrow\; \Omega_{\text{prec}} = \frac{\tau}{L\sin\theta} = \frac{Mgr}{I_3\,\omega_3}

The steady gyroscopic precession rate — remarkably, it does not depend on the tilt angle \theta, and it slows down as the spin \omega_3 increases.

Real tops also nod up and down — a fast, small tremor called nutation — superimposed on the smooth precession. Friction usually damps the nod away within moments, leaving the steady sweep the formula describes.

Worked example: how fast does a bicycle wheel precess?

  1. Set up an illustrative case. Suppose a bicycle wheel of mass M=2\ \text{kg} and radius r=0.33\ \text{m} is spun at $10$ revolutions per second and hung from one end of its axle, a distance \ell=0.10\ \text{m} from the wheel's centre.
  2. Model it as a hoop. Nearly all the mass sits at the rim, so the spin moment is I_3\approx Mr^2 = 2\times0.33^2 \approx 0.22\ \text{kg·m}^2.
  3. Spin angular momentum. \omega_3 = 2\pi\times10 \approx 63\ \text{rad/s}, so L = I_3\omega_3 \approx 0.22\times63 \approx 14\ \text{kg·m}^2/\text{s}.
  4. Gravitational torque about the support. \tau = Mg\ell = 2\times9.8\times0.10 \approx 2.0\ \text{N·m}.
  5. Precession rate. \Omega = \tau/L \approx 2.0/14 \approx 0.14\ \text{rad/s} — about one slow revolution every 45 seconds. Spin the wheel faster and it precesses slower, which is why a rapidly spinning gyroscope seems almost frozen in place.

The specific numbers are round assumptions, but the scaling \Omega\propto 1/\omega_3 is exact and is the whole principle behind gyrocompasses, camera stabilizers and the reaction wheels that point spacecraft.