Euler's Equations of Motion

In 1985, cosmonaut Vladimir Dzhanibekov unscrewed a wing-nut aboard Salyut 7 and watched it spin off the bolt, flip 180° in mid-air, spin steadily for a few seconds, then flip again — metronomically, as if haunted. You can stage the same séance tonight: toss a phone spinning about its long axis and it comes back clean; about the axis through the screen, clean again; end-over-end about the short axis and it always corkscrews. Nothing is touching these objects. The flips are written into three differential equations that Leonhard Euler derived in the 1750s. Euler's equations are what Newton's $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ becomes when you insist on describing rotation from the body's own point of view — the one frame where the mass distribution, and hence the inertia tensor, stays put. The price of that convenience is a fictitious term $\boldsymbol{\omega}\times\mathbf{L}$ , and that term makes the equations nonlinear: the inertia differences $I_i-I_j$ let each pair of spin components torque the third, even with no external torque at all. Out of that nonlinearity falls the entire zoo of rigid-body behavior: the polhode curves that angular velocity traces on a conserved-energy ellipsoid, the serene precession of a thrown frisbee, the tennis-racket instability behind Dzhanibekov's flipping nut, and the reason the first American satellite ended up tumbling end over end. These three lines of mathematics are the dynamical heart of everything a spinning body can do — and this article derives them, integrates them, and breaks a satellite with them.

Key takeaways

Euler's equations are the rotational analogue of Newton's $\mathbf{F}=m\mathbf{a}$ , written in the rotating body frame aligned with the principal axes of the inertia tensor.
They read $I_1\dot{\omega}_1-(I_2-I_3)\,\omega_2\omega_3=\tau_1$ and cyclic permutations: the inertia differences $I_i-I_j$ couple the angular-velocity components nonlinearly.
The nonlinearity is pure kinematics — the cost of taking the time derivative of $\mathbf{L}$ in a frame that is itself rotating, via the transport-theorem term $\boldsymbol{\omega}\times\mathbf{L}$ .
For a free body ( $\boldsymbol{\tau}=0$ ) they conserve both $L^2$ and the kinetic energy $T$ ; the angular velocity is confined to their intersection curve, the polhode.
They predict the intermediate-axis (tennis-racket / Dzhanibekov) instability: spin about the middle moment of inertia grows exponentially at rate $\sigma=\omega_0\sqrt{(I_2-I_1)(I_3-I_2)/(I_1I_3)}$ , while the extreme axes are stable.
Combined with the Euler-angle kinematics they give the full attitude evolution of any rigid body — the working equations of spacecraft attitude control.
Key idea: Write $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ in the body frame: $\left(\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}\right)_{\text{body}}+\boldsymbol{\omega}\times\mathbf{L}=\boldsymbol{\tau}$ . Aligning with principal axes diagonalises $\mathbf{L}=(I_1\omega_1,I_2\omega_2,I_3\omega_3)$ , and the cross product injects the inertia differences that couple the components.

Where this lives today

Every spacecraft attitude-control system integrates equation (4) in real time. Reaction wheels and magnetorquers supply $\boldsymbol{\tau}$ ; a quaternion carries the orientation alongside the Euler dynamics; and the stability theorem is baked into design rules — spin-stabilised craft spin about the maximum-inertia axis, or accept active nutation damping if mission geometry forces a minor-axis spin. Detumbling a freshly deployed CubeSat (the "B-dot" controller that bleeds $\boldsymbol{\omega}$ using Earth's magnetic field) is Euler's equations plus a dissipative torque, and the major-axis rule of Worked Example 2 governs everything from Explorer 1's 1958 surprise to the slow stabilisation of tumbling space debris, whose light-curves observatories fit with torque-free Euler solutions to infer their spin states. The same equations, quantised, run molecular physics: the rotational Hamiltonian of an asymmetric-top molecule (water, for one) is $\hat H=\sum_i \hat L_i^2/2I_i$ , and microwave spectroscopists live on its eigenvalue ladder. And the Dzhanibekov effect has become the internet's favourite physics party trick — T-handles flipping in ISS footage — a 260-year-old linearisation going viral.

The physical picture

Stand on the spinning body. From your vantage the mass distribution is frozen around you — the inertia tensor never changes, which is precisely why this frame is worth the trouble. But Newton's law lives in the inertial lab, so when you import $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ into your rotating perch you pick up a bookkeeping term, exactly the $\boldsymbol{\omega}\times\mathbf{L}$ that makes the equations nonlinear. The inertia differences $I_i-I_j$ are the levers through which one spin component torques another: energy and angular momentum slosh between the body axes with no outside help. If the body were a perfect sphere ( $I_1=I_2=I_3$ ), every coupling term would vanish and $I\dot{\boldsymbol{\omega}}=\boldsymbol{\tau}$ — a sphere tumbles as blandly as a point mass obeys $\mathbf{F}=m\mathbf{a}$ . Every interesting rigid-body effect, from a tossed phone flipping to a gyroscope holding its axis, is a consequence of the inertia tensor being anisotropic. Your phone has three unequal moments, and the middle one is a trap:

Figure

Three axes of a phone — spin about the extreme moments is stable, the intermediate axis flips

There is a beautiful global picture of why. With no torque, two quantities are locked: the kinetic energy $2T=\sum I_i\omega_i^2$ (an ellipsoid in $\boldsymbol{\omega}$ -space) and the squared angular momentum $L^2=\sum I_i^2\omega_i^2$ (another ellipsoid — or, in $\mathbf{L}$ -space, a sphere). The angular velocity must live on the intersection of the two surfaces, a closed curve called the polhode. Near the axes of largest and smallest inertia those intersections are tiny closed loops — the spin circulates harmlessly. Near the intermediate axis the curves are the arms of a saddle: a trajectory starting arbitrarily close to the axis rides the separatrix to the far side of the body and back. That single geometric fact is the tennis-racket theorem:

Figure

Polhode curves on the angular-momentum sphere — closed loops around the extreme axes, a saddle at the intermediate axis

Hold that picture — sphere, loops, saddle — and everything that follows is just calculus catching up with geometry.

From Newton to Euler

Everything starts from one operator identity, the same transport theorem that generates the Coriolis force. For any vector $\mathbf{A}$ , an inertial observer and an observer riding the body disagree about its rate of change by exactly the sweep of the rotating basis:

\left(\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\right)_{\text{inertial}}=\left(\frac{\mathrm{d}\mathbf{A}}{\mathrm{d}t}\right)_{\text{body}}+\boldsymbol{\omega}\times\mathbf{A} \qquad (1)

where: $\boldsymbol{\omega}$ — the angular velocity of the body (and of the body-fixed frame); the subscripts mark the frame in which components are differentiated. Applied to $\mathbf{A}=\boldsymbol{\omega}$ itself the correction $\boldsymbol{\omega}\times\boldsymbol{\omega}$ vanishes: the angular acceleration $\dot{\boldsymbol{\omega}}$ is the one vector both observers agree on — a small miracle we will lean on repeatedly. The object we want to differentiate is the angular momentum. In a general frame $\mathbf{L}=\mathsf{I}\boldsymbol{\omega}$ involves a time-dependent inertia tensor; in the body frame aligned with the principal axes (see Principal Axes and Principal Moments of Inertia) the tensor is constant and diagonal, and the angular momentum has the transparent component form

\mathbf{L}=I_1\omega_1\,\hat{\mathbf{e}}_1+I_2\omega_2\,\hat{\mathbf{e}}_2+I_3\omega_3\,\hat{\mathbf{e}}_3 \qquad (2)

where: $\hat{\mathbf{e}}_i$ — the body-fixed principal axes; $I_1,I_2,I_3$ — the principal moments, taken about the centre of mass or about a fixed point of the body. This is the entire motivation for the body frame: three constants $I_i$ instead of six functions $I_{ab}(t)$ .

Derivation AFrom the transport theorem to Euler's equations

We are deriving the equations of motion for the body-frame components $\omega_i(t)$ , starting from the inertial-frame law $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ and nothing else. The strategy: transport the derivative into the principal-axis frame with identity (1), where the inertia tensor freezes and the derivative term becomes trivial, then expand the leftover cross product component by component. The result is exact — no small-angle or slow-spin approximation enters. The point to watch is the reference point: torque and inertia moments must be taken about the same point, either the centre of mass or a point of the body fixed in space; mixing them is the classic setup error.

Step 1. Apply (1) to $\mathbf{L}$ and impose Newton's law $\left(\mathrm{d}\mathbf{L}/\mathrm{d}t\right)_{\text{inertial}}=\boldsymbol{\tau}$ :

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

Why this is legitimate: the law $\dot{\mathbf{L}}=\boldsymbol{\tau}$ is a statement about the inertial derivative; equation (1) merely re-expresses that same derivative in body components. No physics has changed — only bookkeeping.

Step 2. Evaluate the body-frame derivative using (2). In the body frame the $\hat{\mathbf{e}}_i$ are constant by construction and the $I_i$ are constant because the mass distribution is rigid, so only the components move:

\left(\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}\right)_{\text{body}}=\big(I_1\dot\omega_1,\;I_2\dot\omega_2,\;I_3\dot\omega_3\big)

Why the frame was chosen: this single line is the payoff. In the lab frame this term would carry $\dot{\mathsf{I}}\boldsymbol{\omega}$ with a full time-dependent tensor.

Step 3. Expand the cross product in the same components, using $L_i=I_i\omega_i$ :

(\boldsymbol{\omega}\times\mathbf{L})_1=\omega_2L_3-\omega_3L_2=(I_3-I_2)\,\omega_2\omega_3

Why an inertia difference appears: if all moments were equal, $\mathbf{L}$ would be parallel to $\boldsymbol{\omega}$ and the cross product would die. The coupling exists only because the body responds unequally about its different axes.

Step 4. Assemble component 1 and generate the rest by cyclic permutation $1\to2\to3\to1$ :

\begin{aligned}I_1\dot{\omega}_1+(I_3-I_2)\,\omega_2\omega_3&=\tau_1\\ I_2\dot{\omega}_2+(I_1-I_3)\,\omega_3\omega_1&=\tau_2\\ I_3\dot{\omega}_3+(I_2-I_1)\,\omega_1\omega_2&=\tau_3\end{aligned} \qquad (3)

where: $\tau_i$ — principal-axis components of the external torque. These are Euler's equations: three coupled, first-order, quadratically nonlinear ODEs for the body-frame spin components. Written with the sign convention $I_1\dot\omega_1-(I_2-I_3)\omega_2\omega_3=\tau_1$ they are the form you will meet in Goldstein. Two quick sanity checks: setting $I_1=I_2=I_3$ kills every coupling term and returns $I\dot{\boldsymbol{\omega}}=\boldsymbol{\tau}$ , as expected for a sphere; and each term in (3) carries units of moment of inertia times angular acceleration, i.e. torque.

For numerical work the three equations compress into a single vector line, solved for the angular acceleration:

\dot{\boldsymbol{\omega}}=\mathsf{I}^{-1}\Big(\boldsymbol{\tau}-\boldsymbol{\omega}\times(\mathsf{I}\boldsymbol{\omega})\Big) \qquad (4)

where: $\mathsf{I}=\mathrm{diag}(I_1,I_2,I_3)$ in the principal frame. This is the form every spacecraft simulator integrates, timestep after timestep. Note what it does not give you: the orientation of the body in space. Euler's equations evolve $\boldsymbol{\omega}$ in body components only; to know where the body actually points you must simultaneously integrate the kinematic relations between $\boldsymbol{\omega}$ and the Euler angles (see Angular Velocity in Terms of Euler Angles) or a quaternion.

Which equation answers which question

you want	use	notes
evolve the body-frame spin under torque	Euler's equations (3)	valid only in the principal body frame
one-line form for a numerical integrator	eq. (4)	pair with quaternion kinematics for attitude
conserved quantities when $\boldsymbol{\tau}=0$	eqs. (5), (6)	the polhode construction
is a given steady spin stable?	eq. (7)	sign of $(I_1-I_2)(I_1-I_3)$ decides
how fast does an unstable spin blow up?	eq. (8)	e-folding rate $\sigma\propto\omega_0$
symmetric body $I_1=I_2$ shortcut	$\Omega=\frac{I_3-I_1}{I_1}\omega_3$	full story in the free-symmetric-top article

Torque-free motion: two integrals and the polhode

Switch off the torque. The equations remain nonlinear, yet the motion is completely integrable, because two functions of $\boldsymbol{\omega}$ refuse to change. The first is the rotational kinetic energy (see Rotational Kinetic Energy),

2T=I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2 \qquad (5)

where: $T$ — kinetic energy of rotation about the reference point; in $\boldsymbol{\omega}$ -space this is an ellipsoid with semi-axes $\propto1/\sqrt{I_i}$ . The second is the squared magnitude of the angular momentum,

L^2=I_1^2\omega_1^2+I_2^2\omega_2^2+I_3^2\omega_3^2 \qquad (6)

where: $L=|\mathbf{L}|$ — constant because with $\boldsymbol{\tau}=0$ the vector $\mathbf{L}$ is fixed in space; its body-frame components move, but its length cannot.

Go deeper · Derivation B · 1 minBoth integrals follow from the equations themselves

Derivation BBoth integrals follow from the equations themselves

The goal is to prove $\dot T=0$ and $\mathrm{d}(L^2)/\mathrm{d}t=0$ directly from the torque-free Euler equations, with no appeal to the lab frame. The strategy is mechanical: differentiate each candidate integral, substitute $I_i\dot\omega_i$ from (3) with $\tau_i=0$ , and watch the sum telescope. The manipulation is worth doing once by hand because the same telescoping structure — cyclic differences summing to zero — is what makes the rigid body one of the rare integrable nonlinear systems.

Step 1. Differentiate the energy: $\dot T=\sum_i I_i\omega_i\dot\omega_i$ . Substituting $I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3$ and its cyclic partners,

\dot T=\omega_1\omega_2\omega_3\Big[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)\Big]=0

Why it telescopes: each moment appears once with a plus and once with a minus around the cycle — the bracket is identically zero for any $I_i$ whatsoever.

Step 2. Differentiate the momentum: $\tfrac12\,\mathrm{d}(L^2)/\mathrm{d}t=\sum_i I_i^2\omega_i\dot\omega_i$ . The same substitution gives

\tfrac12\,\frac{\mathrm{d}(L^2)}{\mathrm{d}t}=\omega_1\omega_2\omega_3\Big[I_1(I_2-I_3)+I_2(I_3-I_1)+I_3(I_1-I_2)\Big]=0

Why this bracket also dies: expand it — every product $I_iI_j$ appears exactly twice with opposite signs. Again true for arbitrary moments.

Step 3. Read the geometry. Fix the constants $T$ and $L$ : equation (5) confines $\boldsymbol{\omega}$ to an ellipsoid, equation (6) to a second ellipsoid; the trajectory is their intersection curve, the polhode. Equivalently, plot the body-frame components of $\mathbf{L}$ : they live on the sphere $L^2=\text{const}$ cut by the energy surface $\sum L_i^2/2I_i=T$ — the sphere-and-loops picture in the figure above. Both brackets are identically zero for arbitrary $(I_1,I_2,I_3)$ , as can be confirmed by expanding them symbolically — every moment appears with equal and opposite coefficients.

Two consistency checks the polhode hands you for free. Spin exactly about a principal axis and the "curve" degenerates to a point — steady rotation, exactly as (3) says. And for the symmetric top $I_1=I_2$ the polhode becomes a circle around the symmetry axis: $\omega_3$ stays constant while the transverse components turn at the fixed rate

\Omega=\frac{I_3-I_1}{I_1}\,\omega_3 \qquad (7)

where: $\Omega$ — the body-frame precession rate of the transverse spin. That one formula unfolds into the rolling body-cone/space-cone geometry, the wobble of a frisbee, and the Chandler wobble of the Earth — the complete story, including where the cones actually roll, is the next article, Free Symmetric Top and Body-Cone/Space-Cone Precession. Here we only note that it drops out of (3) in two lines.

The tennis-racket theorem

The polhode picture shows the intermediate axis is special; now we prove it, quantitatively, with the growth rate attached.

Go deeper · Derivation C · 2 minStability of spin about each principal axis

Derivation CStability of spin about each principal axis

We are testing whether steady rotation about a principal axis survives a small knock — the linear-stability analysis whose answer is the tennis-racket theorem. Order the moments $I_1<I_2<I_3$ and consider spin about a general axis $k$ at rate $\omega_0$ , perturbed by small transverse components. The strategy: keep the torque-free equations, discard everything second order in the perturbations, and see whether the survivors oscillate or explode. The step to watch is the very first: you must check that the unperturbed motion is an exact solution before linearising around it — for a principal axis it is, for any other axis it is not, which is why "spin about an arbitrary axis" is not even a candidate steady state.

Step 1. Take spin about $\hat{\mathbf{e}}_1$ : $\boldsymbol{\omega}=(\omega_0,0,0)$ . All three coupling products $\omega_i\omega_j$ vanish, so $\dot{\boldsymbol{\omega}}=0$ : an exact steady solution of (3) with $\boldsymbol{\tau}=0$ . Why only principal axes work: for $\mathbf{L}$ to stay fixed in the body while also fixed in space, $\boldsymbol{\omega}$ must be parallel to $\mathbf{L}$ — which happens precisely on principal axes.

Step 2. Perturb: $\boldsymbol{\omega}=(\omega_0,\eta_2,\eta_3)$ with $|\eta_i|\ll\omega_0$ . The first equation gives $I_1\dot\omega_1=(I_2-I_3)\eta_2\eta_3$ — quadratic in small quantities — so $\omega_1=\omega_0$ stays constant to first order. The other two equations, linearised:

I_2\dot\eta_2=(I_3-I_1)\,\omega_0\,\eta_3,\qquad I_3\dot\eta_3=(I_1-I_2)\,\omega_0\,\eta_2

Why we may treat $\omega_0$ as frozen: its drift is second order, so over the first-order dynamics it acts as a constant coefficient.

Step 3. Differentiate the first, substitute the second, and the pair decouples into a single second-order equation:

\ddot\eta_2=\frac{(I_3-I_1)(I_1-I_2)}{I_2I_3}\,\omega_0^2\,\eta_2

Why the numerator is a product of two comparisons: each factor asks how $I_1$ ranks against one of the other moments — the dynamics cares only about where the spin axis sits in the ordering, not about absolute sizes.

Step 4. Read off the three verdicts from the sign of the coefficient. If $I_1$ is the smallest moment: $(I_3-I_1)>0$ and $(I_1-I_2)<0$ — coefficient negative, $\ddot\eta_2=-|c|\,\eta_2$ , bounded oscillation: stable. If $I_1$ is the largest: $(I_3-I_1)<0$ , $(I_1-I_2)>0$ — negative again: stable. If $I_1$ is the intermediate moment: both factors positive — coefficient positive, solutions $e^{\pm\sigma t}$ : exponential growth. Equivalently, for spin about axis $s$ the linearised eigenvalues obey $\lambda^2=\frac{(I_s-I_a)(I_b-I_s)}{I_aI_b}\,\omega_0^2$ with $a,b$ the other two axes — imaginary for the extreme axes, real for the middle one. Why the perturbations circulate rather than damp in the stable case: there is no dissipation anywhere in the equations; stability here means the knocked spin axis traces a small closed loop (a tiny polhode) around the principal axis forever.

Step 5. For spin about the intermediate axis $\hat{\mathbf{e}}_2$ at rate $\omega_0$ , the e-folding rate is

\sigma=\omega_0\,\sqrt{\frac{(I_2-I_1)(I_3-I_2)}{I_1I_3}} \qquad (8)

where: $\sigma$ — exponential growth rate of the transverse perturbations; the perturbation doubles every $\ln 2/\sigma$ . Faster spin flips sooner in angle-count terms: $\sigma$ scales linearly with $\omega_0$ . The sign analysis is consistent with the polhode geometry — the intermediate axis is a saddle point — and a direct numerical integration of (3) for $I=(20,30,40)\,\mathrm{kg\,m^2}$ with $\omega=(0.01,1.5,0.01)\,\mathrm{rad/s}$ shows e-folding at $\sigma=0.53\,\mathrm{s^{-1}}$ , in agreement with equation (8).

The instability does not tear the body apart — it flips it. The trajectory leaves the saddle along the separatrix, swings around the polhode sphere, and returns to the neighbourhood of the reversed axis: the wing-nut turns 180°, spins calmly, and flips back. In the ideal rigid body this repeats forever with a fixed period; the period depends logarithmically on how precisely you hit the axis, which is why every real toss flips at its own tempo but always flips.

Limiting cases

Case	Result	Meaning
spherical top, $I_1=I_2=I_3$	couplings vanish; $I\dot{\boldsymbol{\omega}}=\boldsymbol{\tau}$	free motion is constant $\boldsymbol{\omega}$ ; the body is dynamically a point mass
symmetric top, $I_1=I_2\neq I_3$ , $\boldsymbol{\tau}=0$	$\omega_3$ const; transverse spin turns at $\Omega=\frac{I_3-I_1}{I_1}\omega_3$	body-cone precession; polhode is a circle
asymmetric top, $I_1<I_2<I_3$ , $\boldsymbol{\tau}=0$	Jacobi elliptic functions $\mathrm{cn},\mathrm{sn},\mathrm{dn}$	fully nonlinear yet integrable; polhode loops and separatrix
spin about extreme axis	$\lambda^2<0$ : bounded oscillation	stable — the classic rifle-bullet and frisbee spins
spin about intermediate axis	$\lambda=\pm\sigma$ , eq. (8)	tennis-racket / Dzhanibekov flips
$\omega_2\omega_3$ products small (near-axis spin)	equations linearise	gyroscopic small-oscillation theory

Common trapIn the body frame L is constant, so dL/dt = tau looks wrong — resolving this is the whole equation

The seductive blunder: "the inertia tensor is constant in the body frame, $\mathbf{L}=\mathsf{I}\boldsymbol{\omega}$ , so surely $\mathrm{d}\mathbf{L}/\mathrm{d}t=\mathsf{I}\dot{\boldsymbol{\omega}}=\boldsymbol{\tau}$ ." That misses the transport term: Newton's law equates the torque to the inertial derivative, and the body-frame derivative differs from it by $\boldsymbol{\omega}\times\mathbf{L}$ — which is generally not zero, because $\mathbf{L}$ and $\boldsymbol{\omega}$ are not parallel unless the spin is along a principal axis. Three further traps. The $\dot\omega_i$ in (3) are derivatives of body-frame components (though, uniquely for $\boldsymbol{\omega}$ , the assembled vector $\dot{\boldsymbol{\omega}}$ is frame-agnostic since $\boldsymbol{\omega}\times\boldsymbol{\omega}=0$ ). The torque and the moments must refer to the same point — centre of mass or a fixed point, never one of each. And Euler's equations alone do not know where the body points in space: they evolve $\boldsymbol{\omega}$ , not the orientation, which requires the separate Euler-angle or quaternion kinematics.

Worked examples

1 · The phone flip, with numbers (confidence builder)

A phone of mass $m=0.20\ \mathrm{kg}$ measures $16\ \mathrm{cm}\times7.5\ \mathrm{cm}\times0.8\ \mathrm{cm}$ . It is tossed spinning end-over-end about its width axis at $2$ revolutions per second. Find the principal moments, identify the unstable axis, and estimate how long a wobble takes to double.

Step 1. Model the phone as a uniform rectangular slab with $a=0.160\ \mathrm{m}$ , $b=0.075\ \mathrm{m}$ , $c=0.008\ \mathrm{m}$ . The slab moments about the three symmetry axes are $I=\tfrac{m}{12}(\text{sum of the two transverse dimensions squared})$ : about the long axis $I_1=\tfrac{0.20}{12}(0.075^2+0.008^2)=9.5\times10^{-5}\ \mathrm{kg\,m^2}$ ; about the width axis $I_2=\tfrac{0.20}{12}(0.160^2+0.008^2)=4.28\times10^{-4}\ \mathrm{kg\,m^2}$ ; about the face normal $I_3=\tfrac{0.20}{12}(0.160^2+0.075^2)=5.20\times10^{-4}\ \mathrm{kg\,m^2}$ .

Step 2. Order them: $I_1<I_2<I_3$ , so the width axis — the end-over-end flip axis — carries the intermediate moment. That is the unstable one, exactly the axis your thumbs naturally launch.

Step 3. Spin rate $\omega_0=2\ \mathrm{rev/s}=4\pi=12.6\ \mathrm{rad/s}$ . In equation (8) the common factor $10^{-4}\ \mathrm{kg\,m^2}$ cancels between numerator and denominator:

\sigma=12.6\,\sqrt{\frac{(4.28-0.95)\times(5.20-4.28)}{0.95\times5.20}}=12.6\times0.79=9.9\ \mathrm{s^{-1}}

Step 4. Doubling time $t_2=\ln2/\sigma=\boxed{0.070\ \mathrm{s}}$ . A half-second toss spans about $5$ e-foldings — a $1°$ imperfection at launch grows by a factor $e^{9.9\times0.5}\approx140$ , far past the linear regime: the phone will corkscrew. To toss it clean you would need launch alignment of a few hundredths of a degree, which human thumbs do not deliver.

2 · A satellite spun about its middle axis (exam level)

A satellite has principal moments $I_1=20$ , $I_2=30$ , $I_3=40\ \mathrm{kg\,m^2}$ and is accidentally deployed spinning at $\omega=1.5\ \mathrm{rad/s}$ about $\hat{\mathbf{e}}_2$ . (a) Find the e-folding time of the tumble. (b) The satellite carries flexible antennas that slowly dissipate kinetic energy while $\mathbf{L}$ is conserved. Find the spin state it finally settles into and the energy shed on the way.

Step 1. Growth rate from (8): $\sigma=1.5\sqrt{\dfrac{(30-20)(40-30)}{20\times40}}=1.5\sqrt{0.125}=0.53\ \mathrm{s^{-1}}$ , so the e-folding time is $\boxed{1.9\ \mathrm{s}}$ — the tumble develops within seconds.

Step 2. The conserved magnitude is $L=I_2\omega=30\times1.5=45\ \mathrm{kg\,m^2/s}$ . Kinetic energy of the initial state: $T_2=L^2/2I_2=45^2/60=33.8\ \mathrm{J}$ .

Step 3. At fixed $L$ , spin about a principal axis $I$ costs $T=L^2/2I$ — energy decreases as $I$ increases. The minimum-energy state compatible with the conserved $L$ is pure spin about the largest moment: $T_3=L^2/2I_3=2025/80=25.3\ \mathrm{J}$ , at rate $\omega_3=L/I_3=45/40=1.125\ \mathrm{rad/s}$ .

Step 4. Any internal dissipation therefore drags the polhode toward $\hat{\mathbf{e}}_3$ : the satellite ends in a flat spin about its maximum-inertia axis at $1.125\ \mathrm{rad/s}$ , having dissipated $T_2-T_3=\boxed{8.4\ \mathrm{J}}$ into its structure. Note the ideal Euler equations conserve $T$ — the migration is driven entirely by the non-rigidity of the real satellite, with the rigid-body theory supplying the start and end points. This "major-axis rule" is why Explorer 1, a pencil-shaped craft spun about its long (minimum-inertia) axis, was tumbling end-over-end within hours of its 1958 launch: its whip antennas flexed, $T$ fell at fixed $L$ , and the only place to go was the flat spin.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · St. Petersburg & Berlin academies

Euler wrote the mechanics of the rigid body almost single-handedly. In 1750 he recognised that $\mathbf{F}=m\mathbf{a}$ , applied to every particle of a continuum, splits into two independent laws — linear momentum and angular momentum — and by 1758 he had found the trick that tamed rotation: refer everything to axes carried by the body, where the inertia integrals freeze. The memoir that announced it introduced, in one sweep, the principal axes, the moments of inertia, and the three equations that bear his name; the 1765 treatise Theoria motus corporum solidorum seu rigidorum built the whole edifice, including the torque-free solution for the symmetric top. He did much of this while going blind — by 1771 he was dictating calculations to assistants, and his productivity famously increased. The angles, the equations, the buckling load, the fluid equations, $e$ , $\gamma$ , and the bridges of Königsberg: mechanics is written in his handwriting.

History

1750

— Euler splits Newton's law into the linear and angular momentum principles for extended bodies

1758

— Euler derives the body-frame rotation equations in his memoir on the motion of solid bodies about a variable axis

1765

— Theoria motus corporum solidorum publishes the complete rigid-body theory, principal axes and all

1834

— Poinsot gives the geometric solution: the inertia ellipsoid rolling on the invariable plane, tracing polhode and herpolhode

1888

— Kovalevskaya discovers the third and last integrable heavy top, winning the Prix Bordin

1985

— Dzhanibekov watches a wing-nut flip periodically aboard Salyut 7, the intermediate-axis instability made vivid in microgravity

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 9 (Angular Momentum, Part II) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §5.5-5.6 · Landau & Lifshitz, Mechanics, 3rd ed., §36-37 · Taylor, Classical Mechanics, Ch. 10 · Arnold, Mathematical Methods of Classical Mechanics, Ch. 6 The Chandler-wobble flattening $\tfrac{I_3-I_1}{I_1}\approx 1/305$ is a standard geophysical value. Worked-example numbers on this page (phone moments $0.95/4.28/5.20\times10^{-4}\,\mathrm{kg\,m^2}$ , $\sigma=9.9\,\mathrm{s^{-1}}$ , doubling $0.070$ s; satellite $\sigma=0.53\,\mathrm{s^{-1}}$ , flat spin $1.125$ rad/s, $8.4$ J shed) follow directly from the boxed formulas.

Builds on: Angular Momentum of a Rigid Body · Principal Axes and Principal Moments of Inertia · Moment of Inertia Tensor · Rotational Kinetic Energy Leads to: Free Symmetric Top and Body-Cone/Space-Cone Precession · Classification of Rigid Bodies: Spherical, Symmetric, and Asymmetric Tops · Euler Angles · Angular Velocity in Terms of Euler Angles

Why can't I just apply dL/dt = τ directly in the lab frame and skip Euler's equations?

You can — it is the same physics — but the price is the inertia tensor. In the lab frame $\mathbf{L}=\mathsf{I}(t)\,\boldsymbol{\omega}$ with all six independent components of $\mathsf{I}$ oscillating as the body turns, so $\dot{\mathbf{L}}=\dot{\mathsf{I}}\boldsymbol{\omega}+\mathsf{I}\dot{\boldsymbol{\omega}}$ buries the dynamics inside a time-dependent matrix you would have to track anyway by integrating the orientation. The body frame trades that mess for three constant numbers $I_1,I_2,I_3$ plus one algebraic term $\boldsymbol{\omega}\times\mathbf{L}$ . For numerical work the modern compromise is exactly Euler's: integrate (4) in the body frame and carry a quaternion to map back to the lab.

Is the angular acceleration in Euler's equations the body-frame one or the lab-frame one?

Both — and this is special to $\boldsymbol{\omega}$ . The transport theorem applied to $\boldsymbol{\omega}$ itself reads $(\mathrm{d}\boldsymbol{\omega}/\mathrm{d}t)_{\text{inertial}}=(\mathrm{d}\boldsymbol{\omega}/\mathrm{d}t)_{\text{body}}+\boldsymbol{\omega}\times\boldsymbol{\omega}$ , and the cross product of a vector with itself vanishes. So $\dot{\boldsymbol{\omega}}$ is the one derivative on which the two frames agree, and the $\dot\omega_i$ in (3) can be read either way. No other dynamical vector in the problem has this property — $\dot{\mathbf{L}}$ certainly does not, which is the whole content of the equations.

Do Euler's equations require principal axes, or is that just a convenience?

Any body-fixed axes work, but then $\mathbf{L}=\mathsf{I}\boldsymbol{\omega}$ with a full symmetric matrix (constant, at least), and the component equations carry products of inertia in every term. Choosing the principal frame diagonalises $\mathsf{I}$ (see Principal Axes and Principal Moments of Inertia) and collapses the algebra to the clean cyclic form. There is one more subtlety worth knowing: the reference point must be either the centre of mass or a point of the body fixed in space — those are the two cases in which $\dot{\mathbf{L}}=\boldsymbol{\tau}$ holds without correction terms.

Why is it exactly the middle axis that is unstable and not, say, the smallest?

Because the stability coefficient is a product of two comparisons. Linearising about spin axis $s$ gives eigenvalues $\lambda^2\propto(I_s-I_a)(I_b-I_s)$ , where $a,b$ are the other two axes. If $I_s$ is the largest or the smallest, the two factors have opposite signs, $\lambda^2<0$ , and the perturbation circulates in a bounded ellipse. Only when $I_s$ sits between the other two do both factors carry the same sign, making $\lambda$ real — exponential growth. Geometrically it is the polhode: the constant- $T$ and constant- $L$ surfaces intersect in closed loops around the extreme axes but cross in a saddle at the intermediate one.

If the ideal equations conserve T, why does every real spinning body end up rotating about its largest moment of inertia?

Because no real body is perfectly rigid. Flexing antennas, sloshing propellant, even elastic hysteresis dissipate kinetic energy internally, while the total angular momentum — which no internal force can touch — stays fixed. At fixed $L$ , spin about axis $I$ costs $T=L^2/2I$ , so the accessible minimum is spin about $I_{\max}$ . Euler's rigid-body theory supplies the endpoints and the timescale hierarchy; the migration itself lives outside the rigid idealisation. Explorer 1 (1958) is the canonical cautionary tale, and modern spacecraft either spin about the major axis or pay for active damping.

What actually happens on the separatrix — does the Dzhanibekov flip take infinite time?

On the exact separatrix, yes: the trajectory approaches the reversed intermediate axis asymptotically and never arrives — the mathematical flip time is infinite. Any real toss starts a small distance $\epsilon$ off the axis, and the period of the resulting near-separatrix polhode loop scales as $\ln(1/\epsilon)$ : finite, but exquisitely sensitive to the launch. That logarithm is why Dzhanibekov's wing-nut flipped with an almost fixed rhythm (its $\epsilon$ was set once, by the thread of the bolt) while your phone flips differently on every toss. Between flips the body lingers near the axis, spinning deceptively calmly — the slow drift near a saddle.

Where do the Jacobi elliptic functions everyone mentions actually enter?

In the general asymmetric free top, $I_1<I_2<I_3$ with no symmetry to exploit. Using the two integrals (5)–(6) to eliminate $\omega_1$ and $\omega_3$ turns the middle Euler equation into $\dot\omega_2^2=(\text{quartic in }\omega_2)$ , whose solution is the elliptic function $\mathrm{sn}$ ; the other components ride along as $\mathrm{cn}$ and $\mathrm{dn}$ . Landau–Lifshitz §37 carries this out in two pages. The symmetric top is the degenerate case where the elliptic modulus goes to zero and $\mathrm{sn},\mathrm{cn}$ collapse to sine and cosine — which is why that case reduces to the circular precession of equation (7).

Euler's equations give me ω(t) in the body frame — how do I recover what the body actually looks like from the lab?

You integrate a second, kinematic layer. The orientation is a rotation (three Euler angles, or better a unit quaternion), and its rate is slaved to the body-frame $\boldsymbol{\omega}$ through the kinematic relations — for Euler angles, the formulas in Angular Velocity in Terms of Euler Angles; for quaternions, $\dot q=\tfrac12\,q\otimes(0,\boldsymbol{\omega})$ . A rigid-body simulator therefore integrates six-plus-one numbers: three dynamical ( $\boldsymbol{\omega}$ from Euler) and four kinematic (the quaternion, kept unit-normalised). Quaternions are preferred in software because the Euler-angle map degenerates at $\theta=0$ — the gimbal-lock singularity.