Euler's Equations and Torque-Free Precession

Derivation

Euler's Equations and Torque-Free Precession

D-036 Home PU-101 Threads force · matter · symmetry Depends on The Moment-of-Inertia Tensor, Angular Momentum, Torque, and Central Forces

Statement

For a rigid body referred to its principal axes fixed in the body, angular-momentum balance \(\vec{\tau}=\left(\dfrac{d\vec{L}}{dt}\right)_{\text{space}}\) becomes the three Euler equations \(I_i\dot{\omega}_i-(I_j-I_k)\omega_j\omega_k=\tau_i\) (cyclic in \(i,j,k\)). In the torque-free case these predict that a symmetric body's transverse angular velocity precesses about its symmetry axis at a fixed rate \(\Omega=\dfrac{I_\parallel-I_\perp}{I_\perp}\omega_3\), and that steady rotation is stable only about the axes of greatest and least moment of inertia, never the intermediate one.

Why it matters

Newton's law \(\vec\tau=d\vec L/dt\) is exact but unusable in the lab frame for a spinning body, because the inertia tensor there tumbles from instant to instant. Passing to the body frame freezes the inertia tensor into three constants; the price is a gyroscopic coupling term \(\vec\omega\times\vec L\), and that term is exactly what makes tops, gyroscopes, tumbling wrenches, wobbling frisbees and precessing planets behave as they do.

The torque-free predictions are the cleanest test of the whole rigid-body formalism. They explain the Earth's Chandler wobble, the design rule that spin-stabilised satellites must spin about their axis of maximum inertia, and the tennis-racket theorem (Dzhanibekov effect) — a striking prediction with no free small parameter, drawn purely from the algebra of three coupled first-order equations.

Assumptions

The body is rigid.If parts move relative to one another the inertia tensor is time-dependent in the body frame and \(L_i=I_i\omega_i\) with constant \(I_i\) fails; internal dissipation also drains rotational energy at fixed \(\vec L\) and drives the spin toward the maximum-inertia axis. Axes are the principal axes of \(\mathbf{I}\), fixed in the body.Off-principal axes leave products of inertia in \(\mathbf I\), so \(\vec L\) is not simply \((I_1\omega_1,I_2\omega_2,I_3\omega_3)\) and the equations acquire extra cross terms. The reference point is the centre of mass (or a fixed pivot).Otherwise \(\vec L\) and \(\vec\tau\) must be taken about the same moving point, and a term in \(m\ddot{\vec R}_{\text{cm}}\) appears, breaking the clean form. For the precession result, two principal moments are equal (\(I_1=I_2\equiv I_\perp\)).For a fully asymmetric body the torque-free motion is not simple precession but a Jacobi-elliptic-function polhode; the constant-rate circle is replaced by a closed loop of varying speed on the inertia ellipsoid.

Derivation

\[ \vec{\tau}=\left(\frac{d\vec{L}}{dt}\right)_{\text{space}} \]

Angular-momentum theorem about the centre of mass, valid in any inertial (space) frame. A

\[ \left(\frac{d\vec{A}}{dt}\right)_{\text{space}}=\left(\frac{d\vec{A}}{dt}\right)_{\text{body}}+\vec{\omega}\times\vec{A} \]

Transport theorem: for any vector \(\vec A\), its rate in an inertial frame equals its rate in a frame rotating at \(\vec\omega\) plus the carry-along term. Pure kinematics of rotating frames; the two frames share an origin. B

\[ \left(\frac{d\vec{L}}{dt}\right)_{\text{body}}+\vec{\omega}\times\vec{L}=\vec{\tau} \]

Apply Step 2 with \(\vec A=\vec L\) and substitute Step 1. No approximation — this is \(\vec\tau=d\vec L/dt\) re-expressed in the body frame. B

\[ \vec{L}=\mathbf{I}\,\vec{\omega}=\left(I_1\omega_1,\;I_2\omega_2,\;I_3\omega_3\right) \]

In principal axes the inertia tensor is diagonal, so each component of \(\vec L\) is a principal moment times the matching component of \(\vec\omega\). The \(I_i\) are constants because the axes are glued to the rigid body. A

\[ \left(\vec{\omega}\times\vec{L}\right)_1=\omega_2 L_3-\omega_3 L_2=(I_3-I_2)\,\omega_2\omega_3 \]

Take the first component of the cross product and insert Step 4. The two cyclic partners follow by the index rotation \(1\to2\to3\to1\). B

\[ \boxed{\,I_1\dot{\omega}_1+(I_3-I_2)\,\omega_2\omega_3=\tau_1\,}\quad(\text{+ cyclic}) \]

Component 1 of Step 3, using \((d\vec L/dt)_{\text{body},1}=I_1\dot\omega_1\) (constant \(I_1\)) and Step 5. Cyclic permutation gives components 2 and 3. These are Euler's equations. B

\[ \tau_i=0:\qquad I_1\dot{\omega}_1=(I_2-I_3)\,\omega_2\omega_3\quad(\text{+ cyclic}) \]

Set \(\vec\tau=0\) (torque-free motion) and move the coupling term across. From here the free motion is read off. A

\[ I_1=I_2\;\Rightarrow\;I_3\dot{\omega}_3=(I_1-I_2)\omega_1\omega_2=0\;\Rightarrow\;\omega_3=\text{const} \]

Impose symmetry about axis 3. The third Euler equation loses its source term, so the spin component along the symmetry axis is conserved. B

\[ \dot{\omega}_1=-\Omega\,\omega_2,\quad \dot{\omega}_2=+\Omega\,\omega_1,\quad \Omega\equiv\frac{I_3-I_1}{I_1}\,\omega_3 \]

The first two torque-free equations with \(I_1=I_2\) and constant \(\omega_3\) become a linear pair. Their solution \(\omega_1=\omega_\perp\cos\Omega t,\ \omega_2=\omega_\perp\sin\Omega t\) shows the transverse part of \(\vec\omega\) sweeps a circle of fixed radius at rate \(\Omega\) — free precession in the body frame. B

\[ \omega_1=\omega_0+\delta\omega_1,\qquad \ddot{\omega}_2=\frac{(I_3-I_1)(I_1-I_2)}{I_2 I_3}\,\omega_0^{2}\,\omega_2 \]

Now for a general (asymmetric) body, spin steadily about axis 1 and linearise the torque-free equations in the small transverse components. To first order \(\omega_1\approx\omega_0\) is constant; differentiating the second equation and substituting the third gives a single second-order equation for \(\omega_2\) (identical form for \(\omega_3\)). C

\[ K\equiv\frac{(I_3-I_1)(I_1-I_2)}{I_2 I_3}\,\omega_0^{2}:\quad K<0\Rightarrow\text{oscillation (stable)},\quad K>0\Rightarrow\text{growth (unstable)} \]

The sign of the coefficient decides everything. If axis 1 is the largest or smallest moment the factors \((I_3-I_1)\) and \((I_1-I_2)\) have opposite signs, \(K<0\), and perturbations oscillate. If axis 1 is intermediate they share a sign, \(K>0\), and perturbations grow like \(e^{\sqrt{K}\,t}\) — the intermediate-axis instability. C

Result

\[ I_i\dot{\omega}_i-(I_j-I_k)\,\omega_j\omega_k=\tau_i \qquad \Omega=\frac{I_3-I_1}{I_1}\,\omega_3 \qquad K=\frac{(I_3-I_1)(I_1-I_2)}{I_2 I_3}\,\omega_0^{2} \]

Reading. The three Euler equations are Newton's second law for rotation written in axes that turn with the body. Each says: torque about a principal axis equals the direct angular acceleration \(I_i\dot\omega_i\) plus a gyroscopic term coupling the other two spin components through the difference of their moments. With no torque, a symmetric body keeps its spin \(\omega_3\) fixed and its transverse angular velocity circles the symmetry axis at rate \(\Omega\), positive for an oblate body (\(I_3>I_1\), flattened, like the Earth or a disc) and negative for a prolate body (\(I_3<I_1\), elongated). Steady spin is stable about the extreme axes (\(K<0\)) and unstable about the intermediate one (\(K>0\)), which flips the body end over end.

Units check. \(I_1\dot\omega_1\) has units \((\text{kg·m}^2)(\text{s}^{-1})(\text{s}^{-1})=\text{kg·m}^2\text{s}^{-2}=\text{N·m}\), matching \(\tau_1\); the coupling term \((I_3-I_2)\omega_2\omega_3\) is likewise \(\text{kg·m}^2\cdot\text{s}^{-2}=\text{N·m}\). \(\Omega\) is (dimensionless ratio)\(\times\text{s}^{-1}=\text{s}^{-1}\); \(K\) is \(\text{s}^{-2}\), so \(\sqrt{K}\) is a rate, as a growth constant must be.

Limiting cases

Spherical top \(I_1=I_2=I_3\): every coupling coefficient vanishes, \(\Omega=0\), and torque-free \(\vec\omega\) is simply constant — any axis is a permanent rotation axis.
Thin disc / flat plate \(I_3=2I_1\): \(\Omega=(2I_1-I_1)\omega_3/I_1=\omega_3\) — the body-frame precession rate equals the spin rate (the origin of the "wobble at twice the flip" look of a tossed coin in the lab).
Nearly spherical \(I_3\approx I_1\): \(\Omega\to0\), precession is very slow — the Earth (\((C-A)/A\approx1/305\)) precesses freely with a period of hundreds of days despite a one-day spin.
Prolate vs oblate: \(I_3<I_1\) gives \(\Omega<0\), reversing the sense of body precession relative to the oblate \(I_3>I_1\) case.
Torque present, steady precession: reinstating \(\vec\tau\neq0\) and seeking constant-\(\vec\omega\) solutions recovers the gyroscope's forced precession \(\vec\tau=\vec\omega_p\times\vec L\) — the ordinary heavy top is a special case of Euler's equations.

Breaks when

The body is not rigid or dissipates energy. With internal friction (flexing panels, sloshing fuel, tidal yielding) rotational kinetic energy decays at fixed \(\vec L\), so the spin migrates to the minimum-energy state — rotation about the maximum-inertia axis. Minor-axis "spinners" like the 1958 Explorer 1 satellite therefore flatten out within hours; the energy-conserving analysis above cannot see this.
Non-principal or time-varying inertia. Deploying booms or shifting fuel reintroduce products of inertia and \(\dot{\mathbf I}\neq0\) in the body frame, so \(L_i=I_i\omega_i\) with constant \(I_i\) collapses and extra terms appear.
Deformation from the spin itself. A fast-spinning fluid or elastic body bulges, so \(I_i\) depend on \(\vec\omega\) and the equations become nonlinear — this shifts the Earth's free-precession period from the rigid "Euler" \(\approx305\) days to the observed Chandler \(\approx433\) days.
Relativistic or non-classical rotation. The derivation uses \(L_i=I_i\omega_i\) with a fixed Newtonian inertia tensor; near \(\omega R\sim c\), or for a quantum rotor where angular momentum is quantised, the constitutive relation and even the meaning of a body frame change.

Failure modes

Sign slips in the coupling term. Writing \((I_2-I_3)\) where the cyclic convention demands \((I_3-I_2)\) flips the direction of precession and can turn a stable prediction into an unstable one. Fix the order \(1\to2\to3\) once and apply it mechanically.
Treating \(\dot\omega_i\) as the space-frame acceleration. Dropping \(\vec\omega\times\vec L\) and writing \(d\vec L/dt=\mathbf I\dot{\vec\omega}\) erases the entire gyroscopic physics — that term is the whole point.
Confusing body-frame precession \(\Omega\) with space-frame precession. The symmetry axis precesses about the fixed \(\vec L\) at rate \(\dot\phi=L/I_\perp\) in the lab; the two rates differ and are routinely mixed up.
Assuming \(\vec L\parallel\vec\omega\). Except along a principal axis \(\vec L\) and \(\vec\omega\) point differently; taking them parallel erases free precession entirely.
Assuming the intermediate axis is "in between" in stability. It is the most unstable case, a full inversion, not a mild wobble.

Discussion

The deep move in this derivation is the change of frame. In the inertial frame the inertia tensor \(\mathbf I(t)\) tumbles with the body and is hopeless to track; in the body frame it is three fixed numbers. The cost is the transport term \(\vec\omega\times\vec L\), and that term is not a nuisance — it is the physical origin of all gyroscopic phenomena. A spinning top does not fall because the coupling redirects the gravitational torque into a sideways precession rather than a downward acceleration; the same algebra that gives free precession gives forced precession once \(\vec\tau\neq0\).

Geometrically, torque-free motion conserves both \(\vec L\) (a fixed vector in space) and the kinetic energy \(T=\tfrac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)\). In \(\vec\omega\)-space the energy condition is an ellipsoid and the \(L^2\) condition a sphere; the tip of \(\vec\omega\) must lie on both, so it traces their intersection — the polhode. Near the maximum- and minimum-inertia axes the intersections are small closed loops (stable); near the intermediate axis they are crossing separatrices, and a trajectory launched nearby swings all the way to the far side — the geometric face of the \(K>0\) instability. This is Poinsot's picture: the inertia ellipsoid rolls without slipping on a plane fixed in space.

The stability result has hard engineering consequences. Spin-stabilised spacecraft must spin about the axis of maximum moment of inertia, because any real body dissipates energy and, at fixed \(\vec L\), drifts to the lowest-energy spin \(T=L^2/2I_{\max}\) — which is rotation about the maximum-inertia axis. The minor-axis spin, mathematically stable in the rigid analysis, is destroyed by the slightest damping; Explorer 1 demonstrated this in orbit in 1958.

Formally, Euler's equations are the reduction of the free rigid body to the dual of the Lie algebra \(\mathfrak{so}(3)\): they form a Lie–Poisson (Euler–Arnold) system whose Casimir \(|\vec L|^2\) and Hamiltonian \(T\) are the two conserved quantities intersected above. The intermediate-axis instability is then the statement that the middle equilibrium of this integrable system is a hyperbolic (saddle) fixed point while the outer two are elliptic centres — the same normal-form classification that governs a pendulum's up/down equilibria, transplanted onto the rotation group.

Common misconceptions. Free precession is not caused by any torque — it is the torque-free motion of a body whose \(\vec L\) and \(\vec\omega\) are misaligned, needing no external agent. And "the intermediate axis is unstable" does not mean the body flies apart; angular momentum and energy are conserved throughout — the body merely reorients, periodically flipping in the idealised rigid case.

Worked examples

Free precession of the Earth (rigid Euler period)

Model the Earth as a rigid oblate symmetric top with dynamical ellipticity \((C-A)/A=0.00327\) and sidereal spin \(\omega_3=2\pi/86164\ \text{s}=7.292\times10^{-5}\ \text{rad·s}^{-1}\). Find the body-frame precession period. A

\[ \Omega=\frac{C-A}{A}\,\omega_3=0.00327\times7.292\times10^{-5}\ \text{s}^{-1} \]

Insert into \(\Omega=(I_3-I_1)\omega_3/I_1\) with \(I_3=C,\ I_1=A\). Symbols first, then numbers. A

\[ \Omega=2.385\times10^{-7}\ \text{s}^{-1}\ \Rightarrow\ T=\frac{2\pi}{\Omega}=2.635\times10^{7}\ \text{s} \]

Divide \(2\pi\) by the rate. A

\[ T=2.635\times10^{7}\ \text{s}\approx305\ \text{days} \]

Reading. A perfectly rigid Earth would let its rotation pole trace a small circle about the figure axis once every \(\sim305\) days. The observed Chandler wobble is \(\sim433\) days — the difference is the elastic and oceanic yielding flagged in "Breaks when".

Units check. The prefactor \((C-A)/A\) is dimensionless, so \([\Omega]=\text{s}^{-1}\); \(T=2\pi/\Omega\) is in \(\text{s}\). Consistent.

Intermediate-axis flip of a book (tennis-racket theorem)

A uniform rectangular slab of mass \(m=0.50\ \text{kg}\), sides \(a=0.24\ \text{m}\) and \(b=0.16\ \text{m}\), is spun at \(\omega_0=2\ \text{rev·s}^{-1}=12.57\ \text{rad·s}^{-1}\) about its intermediate principal axis. Find the e-folding time of the instability. B

\[ I_x=\frac{mb^2}{12},\qquad I_y=\frac{ma^2}{12},\qquad I_z=\frac{m(a^2+b^2)}{12} \]

Standard thin-plate principal moments; the perpendicular-axis theorem gives \(I_z=I_x+I_y\). A

\[ I_x=1.067\times10^{-3},\quad I_y=2.400\times10^{-3},\quad I_z=3.467\times10^{-3}\ \text{kg·m}^2 \]

Evaluate. The ordering \(I_x<I_y<I_z\) identifies \(y\) as the intermediate axis. A

\[ K=\frac{(I_x-I_y)(I_y-I_z)}{I_z I_x}\,\omega_0^{2} \]

Apply the stability coefficient with the spin axis taken as \(y\) and the transverse pair \(x,z\). Both factors are negative, so their product is positive and \(K>0\) (unstable). B

\[ \frac{K}{\omega_0^{2}}=\frac{(-1.333\times10^{-3})(-1.067\times10^{-3})}{3.699\times10^{-6}}=0.385 \]

Substitute the numbers. B

\[ \lambda=\sqrt{K}=0.620\,\omega_0=7.79\ \text{s}^{-1}\ \Rightarrow\ \tau_e=\frac{1}{\lambda}=0.128\ \text{s} \]

The perturbation grows as \(e^{\lambda t}\); the e-folding time is \(1/\lambda\). B

\[ \tau_e\approx0.13\ \text{s} \]

Reading. A tiny wobble multiplies e-fold every eighth of a second, so within a fraction of a rotation the book flips end over end — exactly what you see tossing a phone spun about its middle axis. Spun about the long or short face axis instead (\(K<0\)) it stays put.

Units check. \([K]=(\text{kg·m}^2)^2\,\text{s}^{-2}/(\text{kg·m}^2)^2=\text{s}^{-2}\), so \([\lambda]=\text{s}^{-1}\) and \([\tau_e]=\text{s}\). Consistent.

Problems

Starting from \(\vec\tau=d\vec L/dt\) and the transport theorem, write out the three torque-free Euler equations for a symmetric body (\(I_1=I_2\)) and show that both \(\omega_3\) and \(\omega_1^2+\omega_2^2\) are conserved.
Solution
The equations are \(I_1\dot\omega_1=(I_1-I_3)\omega_2\omega_3\), \(I_1\dot\omega_2=(I_3-I_1)\omega_3\omega_1\), \(I_3\dot\omega_3=(I_1-I_2)\omega_1\omega_2=0\) since \(I_1=I_2\), so \(\omega_3=\text{const}\). Multiply the first by \(\omega_1\) and the second by \(\omega_2\) and add: \(I_1(\omega_1\dot\omega_1+\omega_2\dot\omega_2)=(I_1-I_3)\omega_3\omega_1\omega_2+(I_3-I_1)\omega_3\omega_1\omega_2=0\). Hence \(\tfrac{d}{dt}(\omega_1^2+\omega_2^2)=0\): the transverse spin magnitude is constant, confirming steady circular precession.
A thin uniform disc (\(I_3=\tfrac12 mR^2\), \(I_1=I_2=\tfrac14 mR^2\)) spins torque-free at \(\omega_3=10\ \text{rad·s}^{-1}\) about its symmetry axis with a slight transverse wobble. Find the body-frame precession rate and period.
Solution
\(\Omega=\dfrac{I_3-I_1}{I_1}\omega_3=\dfrac{\tfrac12-\tfrac14}{\tfrac14}\omega_3=1\times10=10\ \text{rad·s}^{-1}\). Period \(T=2\pi/\Omega=0.628\ \text{s}\). For a thin disc the body-frame precession rate equals the spin rate.
A symmetric satellite has \(I_3=1200\ \text{kg·m}^2\) (symmetry axis) and \(I_1=I_2=800\ \text{kg·m}^2\), spinning at \(\omega_3=0.50\ \text{rad·s}^{-1}\). Find the free-precession period, and state whether this spin axis is the stable choice for a real (dissipative) satellite.
Solution
\(\Omega=\dfrac{(1200-800)(0.50)}{800}=0.25\ \text{rad·s}^{-1}\), so \(T=2\pi/0.25=25.1\ \text{s}\). Since \(I_3\) is the maximum moment, this is the maximum-inertia axis — the energetically lowest spin at fixed \(\vec L\) — so with internal dissipation the satellite is stable spinning this way. Spinning about a transverse (minimum-inertia) axis would be destroyed by damping.
A rigid body has principal moments \(I_1=2\), \(I_2=5\), \(I_3=8\ \text{kg·m}^2\). For steady spin about each axis in turn, evaluate the sign of the stability coefficient and classify each axis.
Solution
Spin about axis 1: \((I_3-I_1)(I_1-I_2)=(6)(-3)=-18<0\) → stable (least axis). Spin about axis 3: \((I_2-I_3)(I_3-I_1)=(-3)(6)=-18<0\) → stable (greatest axis). Spin about axis 2: \((I_2-I_3)(I_1-I_2)... \) i.e. the two relevant differences \((I_3-I_2)(I_2-I_1)=(3)(3)=+9>0\) → unstable (intermediate axis). Confirms: greatest and least stable, intermediate unstable.
A smartphone modelled as a thin slab, \(m=0.17\ \text{kg}\), \(a=0.15\ \text{m}\), \(b=0.075\ \text{m}\), is flung spinning at \(3\ \text{rev·s}^{-1}\) about its intermediate axis. Find the e-folding time of the flip, and comment on whether a human toss (airborne \(\sim0.6\ \text{s}\)) will show it.
Solution
\(I_x=mb^2/12=7.97\times10^{-5}\), \(I_y=ma^2/12=3.19\times10^{-4}\), \(I_z=m(a^2+b^2)/12=3.98\times10^{-4}\ \text{kg·m}^2\); intermediate axis is \(y\). \(\omega_0=3\times2\pi=18.85\ \text{rad·s}^{-1}\). \(K/\omega_0^2=\dfrac{(I_x-I_y)(I_y-I_z)}{I_z I_x}=\dfrac{(-2.39\times10^{-4})(-7.97\times10^{-5})}{3.17\times10^{-8}}=0.600\). Then \(\lambda=\sqrt{0.600}\times18.85=0.775\times18.85=14.6\ \text{s}^{-1}\), so \(\tau_e=1/\lambda=0.069\ \text{s}\). In \(0.6\ \text{s}\) of flight the perturbation grows by \(e^{0.6/0.069}\approx e^{8.7}\approx6000\times\), so the flip is fully visible — the phone tumbles end over end mid-air.