Torque-free rotation — Euler's equations, asymmetric top: Worked examples

This page is the drill ground for the parent topic. We do not re-derive Euler's equations here — we exercise them on every kind of input they can receive. If a symbol looks unfamiliar, it was built on the parent note; the quick reminders below re-anchor them.

Recall Quick symbol re-anchor (read if any notation feels new)

$I_{1}, I_{2}, I_{3}$ — the three principal moments of inertia: how hard the body resists spinning about each of its three special body-glued axes. Built from the Inertia tensor and principal axes.
$ω_{1}, ω_{2}, ω_{3}$ — the three components of the spin arrow $ω$ measured along those body axes.
$Ω$ (capital omega) — the large, near-constant spin rate about the axis we are perturbing. It is a specific value of one $ω_{i}$ , not a new object.
$ω_{⊥}$ — the transverse spin vector, shorthand for the two-component pair $(ω_{1}, ω_{2})$ living in the plane perpendicular to axis 3. Its length is $∣ ω_{⊥} ∣ = ω_{1}^{2} + ω_{2}^{2}$ .
A dot means "rate of change per second": $\overset{ω}{˙}_{1} = d ω_{1} / d t$ .
$T$ — the rotational kinetic energy; we usually track $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2}$ . It is conserved with no external work.
$L = I ω$ — the angular momentum. Its length-squared $L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2}$ is conserved in torque-free motion. See Angular momentum in rotating frames.
$k$ — the dimensionless stability constant, defined in Ex 3 by writing the linearized wobble equation as $\overset{ω}{¨}_{1} = k Ω^{2} ω_{1}$ . So $k$ is pure number (it does NOT contain $Ω^{2}$ ); the $Ω^{2}$ is written separately every time. Its sign decides stability: $k < 0$ ⇒ safe oscillation, $k > 0$ ⇒ runaway tumble.

The scenario matrix

Every problem this topic can hand you falls into one of these cells. The examples below hit each one at least once. (The stability constant $k$ referenced in cells C–E is defined in Example 3 — treat it here as a placeholder for "the pure-number sign that decides stability.")

Cell	What makes it that case	Why it needs its own treatment	Example
A. All $I$ equal (sphere)	$I_{1} = I_{2} = I_{3}$	RHS $\to 0$ : every derivative vanishes, degenerate limit	Ex 1
B. Two $I$ equal (symmetric top)	$I_{1} = I_{2} \neq = I_{3}$	One equation dies, other two give steady precession	Ex 2
C. All $I$ distinct, spin near LARGEST axis	$I_{1} < I_{2} < I_{3}$ , spin $\approx$ axis 3	Sign of $k < 0$ ⇒ stable oscillation ( $k$ defined in Ex 3)	Ex 3
D. All $I$ distinct, spin near SMALLEST axis	spin $\approx$ axis 1	Sign of $k < 0$ ⇒ stable (the other stable one)	Ex 4
E. All $I$ distinct, spin near INTERMEDIATE axis	spin $\approx$ axis 2	Sign of $k > 0$ ⇒ exponential tumble	Ex 5
F. Conservation cross-check	any motion	Verify $2 T$ and $L^{2}$ are actually constant on the trajectory	Ex 6
G. Zero / one-component spin (degenerate)	$ω$ along a single axis exactly	Pure principal-axis spin: RHS all zero, no wobble ever	Ex 7
H. Real-world word problem	Earth's Chandler wobble	Turning a physics number into a real observable period	Ex 8
I. Exam twist (sign trap)	axis ordering scrambled	Tests whether you track the sign, not the axis label	Ex 9

Example 1 — Cell A: the sphere (fully degenerate)

Worked example Statement

A uniform solid sphere has $I_{1} = I_{2} = I_{3} = I$ . It is set spinning with $ω = (2, - 3, 1)$ rad/s in its body frame. Describe $ω$ one hour later.

Forecast: Guess before reading — does the spin arrow drift, wobble, or stay frozen?

Plug into all three Euler equations. $I \overset{ω}{˙}_{1} = (I - I) ω_{2} ω_{3} = 0, I \overset{ω}{˙}_{2} = 0, I \overset{ω}{˙}_{3} = 0.$ Why this step? Every coefficient is a difference of two equal moments, so it is exactly $0$ . The engine has no coupling to drive anything.
Integrate. $\overset{ω}{˙}_{1} = \overset{ω}{˙}_{2} = \overset{ω}{˙}_{3} = 0 \Rightarrow ω = (2, - 3, 1)$ forever. Why this step? Zero derivative means the value never changes — in the body frame the spin is frozen.

Verify: For a sphere $I = I 1$ , so $L = I ω$ is parallel to $ω$ . In space $L$ is conserved (no torque), hence $ω$ is too. Consistent. One hour later: still $(2, - 3, 1)$ rad/s.

Example 2 — Cell B: symmetric top (steady precession)

Worked example Statement

A discus-like symmetric top has $I_{1} = I_{2} = I_{⊥} = 2$ , $I_{3} = 5$ (kg·m²). Its spin along the symmetry axis is $ω_{3} = 4$ rad/s, with small transverse start $ω_{1} (0) = 0.1$ , $ω_{2} (0) = 0$ . Find the body precession rate $Ω_{b}$ and the period.

Forecast: Will the transverse components decay, blow up, or circle at constant rate?

Euler 3 first. $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = 0$ , so $ω_{3} = 4$ stays constant. Why this step? Equal transverse moments make that coefficient zero, freezing the axial spin — the cleanest quantity to lock down first.
Reduce Euler 1 & 2. With $I_{1} = I_{2} = I_{⊥}$ : $\overset{ω}{˙}_{1} = \frac{I _{⊥} - I _{3}}{I _{⊥}} ω_{3} ω_{2}, \overset{ω}{˙}_{2} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} ω_{1} .$ Define $Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} = \frac{5 - 2}{2} \cdot 4 = 6$ rad/s. Why this step? Both right sides share the same magnitude $Ω_{b}$ with opposite signs — the signature of pure rotation of the pair $(ω_{1}, ω_{2})$ .
Solve the circle. Recall from the re-anchor box that $ω_{⊥} = (ω_{1}, ω_{2})$ is the transverse spin vector. With $\overset{ω}{˙}_{1} = - Ω_{b} ω_{2}, \overset{ω}{˙}_{2} = Ω_{b} ω_{1}$ , the tip of $ω_{⊥}$ rotates at rate $Ω_{b}$ . Period $T_{prec} = 2 π / Ω_{b} = 2 π /6 \approx 1.047$ s. Why this step? These are the equations of a point going round a circle at angular speed $Ω_{b}$ — that's exactly the free precession cone.

The figure below shows this. The blue closed loop is the path traced by the transverse spin $ω_{⊥} = (ω_{1}, ω_{2})$ ; the orange arrows are the spin vector at eight instants (all the same length — that is why the loop is a perfect circle), and the red label marks the constant turning rate $Ω_{b} = 6$ rad/s. Read it as: the tip of the transverse spin marches steadily around a circle and never spirals in or out.

Verify: $∣ ω_{⊥} ∣^{2} = ω_{1}^{2} + ω_{2}^{2}$ should stay $= 0. 1^{2} = 0.01$ . Since $\frac{d}{d t} (ω_{1}^{2} + ω_{2}^{2}) = 2 ω_{1} \overset{ω}{˙}_{1} + 2 ω_{2} \overset{ω}{˙}_{2} = 2 ω_{1} (- Ω_{b} ω_{2}) + 2 ω_{2} (Ω_{b} ω_{1}) = 0$ . The transverse magnitude is constant — a circle, exactly as the figure shows. $Ω_{b} = 6$ rad/s, $T_{prec} \approx 1.047$ s. ✅

Example 3 — Cell C: spin near the largest axis (stable)

Worked example Statement

Asymmetric top $I_{1} = 1, I_{2} = 2, I_{3} = 3$ . Spin mainly about axis 3 with $Ω = 10$ rad/s and tiny $ω_{1}, ω_{2}$ . Find the stability constant $k$ and the wobble frequency.

Forecast: Sign of $k$ : positive (tumble) or negative (gentle wobble)?

Linearize. Treat $ω_{3} \approx Ω$ constant, keep $ω_{1}, ω_{2}$ first-order: $\overset{ω}{˙}_{1} = \frac{I _{2} - I _{3}}{I _{1}} Ω ω_{2}, \overset{ω}{˙}_{2} = \frac{I _{3} - I _{1}}{I _{2}} Ω ω_{1} .$ Why this step? Products of two small quantities are negligible; this is linearization about the pure-spin state.
Combine into one ODE and DEFINE $k$ . Differentiate the first, insert the second: $\overset{ω}{¨}_{1} = k (pure number) \frac{( I _{2} - I _{3} ) ( I _{3} - I _{1} )}{I _{1} I _{2}} Ω^{2} ω_{1} .$ The stability constant $k$ is exactly the bracketed dimensionless coefficient — it does not include $Ω^{2}$ , which we always write separately: $k = \frac{( I _{2} - I _{3} ) ( I _{3} - I _{1} )}{I _{1} I _{2}} = \frac{( 2 - 3 ) ( 3 - 1 )}{1 \cdot 2} = \frac{( - 1 ) ( 2 )}{2} = - 1.$ Why this step? Keeping $k$ a pure number ( $k = - 1$ here) lets us read stability from its sign alone; the physical growth/wobble rate then comes from $k Ω^{2}$ .
Read off the wobble. With $k = - 1$ : $\overset{ω}{¨}_{1} = k Ω^{2} ω_{1} = - Ω^{2} ω_{1}$ , simple harmonic. Frequency $ω_{wob} = ∣ k ∣ Ω = 1 \cdot 10 = 10$ rad/s. Why this step? A negative $k$ is the equation of a spring — bounded, safe oscillation. Axis 3 (largest $I$ ) is stable. ✅

Verify: Both factors: $(I_{2} - I_{3}) = - 1 < 0$ and $(I_{3} - I_{1}) = + 2 > 0$ , product negative ⇒ $k = - 1 < 0$ . Wobble frequency $∣ k ∣ Ω = 10$ rad/s, period $\approx 0.628$ s. Stable, as the tennis-racket rule predicts.

Example 4 — Cell D: spin near the smallest axis (also stable)

Worked example Statement

Same body $I_{1} = 1, I_{2} = 2, I_{3} = 3$ , but now spin mainly about axis 1 with $Ω = 10$ rad/s. Show it is stable and give the frequency.

Forecast: Two stable axes exist. Is the smallest one really stable, or is that only the largest?

Cyclic-shift the stability constant. For a big spin about axis 1, the linearized ODE for $ω_{2}$ is $\overset{ω}{¨}_{2} = k Ω^{2} ω_{2}, k = \frac{( I _{3} - I _{1} ) ( I _{1} - I _{2} )}{I _{2} I _{3}} (pure number) .$ Why this step? The role of "the axis you spin about" cycles through the equations; we just relabel $(1, 2, 3)$ correctly and keep $k$ dimensionless as in Ex 3.
Compute $k$ . $k = \frac{( 3 - 1 ) ( 1 - 2 )}{2 \cdot 3} = \frac{( 2 ) ( - 1 )}{6} = - \frac{1}{3} \approx - 0.333.$ Why this step? Plugging the numbers evaluates the sign directly — no $Ω^{2}$ inside $k$ .
Read off the wobble. $k < 0$ ⇒ oscillation, frequency $∣ k ∣ Ω = 1/3 \cdot 10 \approx 5.77$ rad/s. Stable. ✅

Verify: $(I_{3} - I_{1}) = + 2 > 0$ , $(I_{1} - I_{2}) = - 1 < 0$ ⇒ product negative ⇒ $k = - 1/3 < 0$ . Both the largest and smallest axes give $k < 0$ : exactly the "BIG and SMALL are SOLID" rule. Frequency $\approx 5.77$ rad/s.

Example 5 — Cell E: spin near the intermediate axis (tumble)

Worked example Statement

Same body $I_{1} = 1, I_{2} = 2, I_{3} = 3$ , spin mainly about axis 2 (the intermediate) with $Ω = 10$ rad/s. Find $k$ and the e-folding time of a disturbance.

Forecast: This is the Dzhanibekov / tennis-racket axis. Guess the sign of $k$ before computing.

Stability constant for axis 2. $\overset{ω}{¨}_{1} = k Ω^{2} ω_{1}, k = \frac{( I _{3} - I _{2} ) ( I _{2} - I _{1} )}{I _{1} I _{3}} (pure number) .$ Why this step? Same linearization machine, now with axis 2 as the "big spin," and $k$ still a dimensionless number.
Compute $k$ . $k = \frac{( 3 - 2 ) ( 2 - 1 )}{1 \cdot 3} = \frac{( 1 ) ( 1 )}{3} = \frac{1}{3} \approx 0.333.$ Why this step? Both factors are positive here, so their product is positive — the crucial sign flip.
Read off the growth. $k > 0$ ⇒ $\overset{ω}{¨}_{1} = + k Ω^{2} ω_{1}$ , solution $\propto e^{k Ω t}$ . Growth rate $k Ω = 1/3 \cdot 10 \approx 5.77 s^{- 1}$ ; e-folding time $τ = 1/ (k Ω) \approx 0.173$ s. Why this step? A positive $k$ turns the spring into a runaway — the disturbance multiplies by $e$ every $0.173$ s. Tumble confirmed. ❌

The figure below contrasts this against Example 3. The red curve is this case (axis 2): a disturbance that grows without bound like $e^{k Ω t}$ ; the vertical dashed line marks the e-folding time $τ \approx 0.173$ s where it has multiplied by $e$ . The blue curve is Example 3's axis-3 case: a bounded, gently oscillating wobble that never runs away. One picture, two fates decided purely by the sign of $k$ .

Verify: $(I_{3} - I_{2}) = + 1 > 0$ and $(I_{2} - I_{1}) = + 1 > 0$ ⇒ product positive ⇒ $k = 1/3 > 0$ . Contrast Ex 3–4 where one factor was negative. Growth rate $k Ω \approx 5.77 s^{- 1}$ , $τ \approx 0.173$ s. This is the "MIDDLE is a MUDDLE" cell — matching the red curve above.

Example 6 — Cell F: the two conservation laws hold on the motion

Worked example Statement

Take $I_{1} = 1, I_{2} = 2, I_{3} = 3$ and the instantaneous spin $ω = (3, 1, 2)$ rad/s. Compute $2 T$ and $L^{2}$ , then show their time derivatives are zero using Euler's equations. (Recall from the re-anchor box: $T$ is the rotational kinetic energy, $L = I ω$ is the angular momentum.)

Forecast: Even though each $ω_{i}$ changes in time, do the two scalars really stay pinned?

Evaluate the constants. $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2} = 1 (9) + 2 (1) + 3 (4) = 9 + 2 + 12 = 23.$ $L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} = 1 (9) + 4 (1) + 9 (4) = 9 + 4 + 36 = 49.$ Why this step? These are the two ellipsoids ( $2 T =$ const and $L^{2} =$ const in $ω$ -space) whose intersection $ω$ must ride.
Differentiate energy and substitute Euler. $\frac{d}{d t} (2 T) = 2 (I_{1} ω_{1} \overset{ω}{˙}_{1} + I_{2} ω_{2} \overset{ω}{˙}_{2} + I_{3} ω_{3} \overset{ω}{˙}_{3}) .$ Now $I_{i} \overset{ω}{˙}_{i}$ are the Euler right-hand sides, so $\frac{d}{d t} (2 T) = 2 [ω_{1} (I_{2} - I_{3}) ω_{2} ω_{3} + ω_{2} (I_{3} - I_{1}) ω_{3} ω_{1} + ω_{3} (I_{1} - I_{2}) ω_{1} ω_{2}]$ $= 2 ω_{1} ω_{2} ω_{3} [(I_{2} - I_{3}) + (I_{3} - I_{1}) + (I_{1} - I_{2})] = 0.$ Why this step? Substituting $I_{i} \overset{ω}{˙}_{i}$ from Euler's equations pulls out the common factor $ω_{1} ω_{2} ω_{3}$ , and the three moment-differences telescope to $0$ — energy needs no external work.
Differentiate $L^{2}$ and substitute Euler — shown in full. $\frac{d}{d t} L^{2} = 2 (I_{1}^{2} ω_{1} \overset{ω}{˙}_{1} + I_{2}^{2} ω_{2} \overset{ω}{˙}_{2} + I_{3}^{2} ω_{3} \overset{ω}{˙}_{3}) = 2 (I_{1} ω_{1} \cdot I_{1} \overset{ω}{˙}_{1} + \dots) .$ Insert each $I_{i} \overset{ω}{˙}_{i}$ (the Euler RHS): $= 2 [I_{1} ω_{1} (I_{2} - I_{3}) ω_{2} ω_{3} + I_{2} ω_{2} (I_{3} - I_{1}) ω_{3} ω_{1} + I_{3} ω_{3} (I_{1} - I_{2}) ω_{1} ω_{2}] .$ Factor out $ω_{1} ω_{2} ω_{3}$ : $= 2 ω_{1} ω_{2} ω_{3} [I_{1} (I_{2} - I_{3}) + I_{2} (I_{3} - I_{1}) + I_{3} (I_{1} - I_{2})] .$ Expand the bracket: $I_{1} I_{2} - I_{1} I_{3} + I_{2} I_{3} - I_{2} I_{1} + I_{3} I_{1} - I_{3} I_{2} = 0$ (each product appears once with a $+$ and once with a $-$ ). Hence $\frac{d}{d t} L^{2} = 0$ . Why this step? $L$ has fixed length in space, so $L^{2}$ must be constant — and here we see why algebraically: every $I_{a} I_{b}$ pair cancels its twin.

Verify: $2 T = 23$ (joules) and $L^{2} = 49$ (kg²·m⁴·s⁻²). Both $\frac{d}{d t} (2 T) = 0$ and $\frac{d}{d t} L^{2} = 0$ are confirmed symbolically in the VERIFY block.

Example 7 — Cell G: exact single-axis spin (degenerate input)

Worked example Statement

Asymmetric top $I_{1} = 1, I_{2} = 2, I_{3} = 3$ . Spin exactly about axis 2: $ω = (0, Ω, 0)$ with $Ω = 7$ rad/s and no perturbation. What does it do?

Forecast: Axis 2 is unstable — but with a perfectly clean start, does it still tumble?

Plug the exact state in. $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3} = (2 - 3) (Ω) (0) = 0,$ $I_{2} \overset{ω}{˙}_{2} = (I_{3} - I_{1}) ω_{3} ω_{1} = (3 - 1) (0) (0) = 0, I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = (1 - 2) (0) (Ω) = 0.$ Why this step? Every right side contains a factor of a zero component, so all derivatives vanish.
Conclude. $ω = (0, 7, 0)$ is a fixed point — mathematically it spins forever about axis 2. Why this step? A pure principal-axis spin is always an equilibrium of Euler's equations, regardless of stability.
Reconcile with instability. The equilibrium is unstable: any real $ω_{1}$ or $ω_{3}$ , however tiny, grows as $e^{k Ω t}$ (Ex 5). Physics always has that seed, so real bodies tumble; the idealized point does not. Why this step? Fixed point $\neq =$ stable — this is the distinction Ex 5 quantified.

Verify: All three $\overset{ω}{˙}_{i} = 0$ at $(0, 7, 0)$ . Also $2 T = I_{2} Ω^{2} = 2 (49) = 98$ and $L^{2} = I_{2}^{2} Ω^{2} = 4 (49) = 196$ ; both trivially constant since $ω$ doesn't move. Degenerate case handled.

Example 8 — Cell H: real-world word problem (Chandler wobble)

Worked example Statement

Earth is nearly a symmetric top with $\frac{I _{3} - I _{⊥}}{I _{⊥}} \approx \frac{1}{305}$ (the "dynamical ellipticity"). It spins once per sidereal day, $ω_{3} = 2 π / (86164 s)$ . Predict the free-precession (Euler/Chandler) period in the body frame.

Forecast: Days? A month? About a year? Guess the order of magnitude.

Use the symmetric-top precession rate from Ex 2: $Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} = \frac{ω _{3}}{305} .$ Why this step? Earth's tiny equatorial bulge makes $I_{3} \neq = I_{⊥}$ ; that difference sets the wobble rate.
Convert to a period. The wobble period is $T_{wob} = \frac{2 π}{Ω _{b}} = 305 \cdot \frac{2 π}{ω _{3}} = 305 \times (1 sidereal day) \approx 305 days .$ Why this step? $2 π / ω_{3}$ is one day, so the wobble period is just $305$ days — a direct, memorable result.
Compare to reality. The rigid prediction is ~305 days; the observed Chandler wobble is ~433 days. The gap is because Earth is not perfectly rigid (elastic crust + fluid core lengthen the period). Why this step? Because our derivation assumed a perfectly rigid body; comparing model to data isolates exactly the assumption that fails and shows what the idealized model omits — the essential habit of checking a prediction against measurement.

Verify: $T_{wob} = 305$ sidereal days $= 305 \times 86164 s \approx 2.628 \times 1 0^{7} s \approx 304.2$ solar days. Order-of-magnitude "about a year" forecast is right; ~305 d rigid vs ~433 d observed.

Example 9 — Cell I: exam sign-trap (scrambled axis labels)

Worked example Statement

A tricky exam gives $I_{1} = 5, I_{2} = 1, I_{3} = 3$ (note: not sorted!). The body spins mainly about axis 3. Is axis 3 stable? Don't trust the label — check the sign.

Forecast: "Axis 3 is largest ⇒ stable" — is that reasoning safe here?

Rank the moments. Sorted: $I_{2} = 1 < I_{3} = 3 < I_{1} = 5$ . So axis 3 is the intermediate moment, despite its label "3". Why this step? Stability depends on whether the moment is smallest/largest/middle — the value, not the index.
Compute $k$ for spin about axis 3. For a big spin about axis 3, the pure-number stability constant (from Ex 3's pattern) is $k = \frac{( I _{2} - I _{3} ) ( I _{3} - I _{1} )}{I _{1} I _{2}} = \frac{( 1 - 3 ) ( 3 - 5 )}{5 \cdot 1} = \frac{( - 2 ) ( - 2 )}{5} = \frac{4}{5} = 0.8.$ Why this step? Both factors are negative, so their product is positive ⇒ $k > 0$ ⇒ instability, exactly like the intermediate case in Ex 5.
Conclude. $k = 0.8 > 0$ : axis 3 is UNSTABLE here. The label "largest index" fooled you; the intermediate value wins. Growth rate $= k Ω = 0.8 Ω \approx 0.894 Ω$ . Why this step? This is the trap in Mistake C from the parent note — the sign of the product $(I_{a} - I_{b}) (I_{b} - I_{c})$ decides, never the axis name.

Verify: With $Ω$ generic, growth rate $= Ω 0.8 \approx 0.894 Ω s^{- 1}$ . Positive $k = 0.8$ confirms tumble. The only stable axes here are the ones carrying $I = 1$ (axis 2) and $I = 5$ (axis 1).

Recall Self-test — cover the answers

Which axis of $I = (1, 2, 3)$ is unstable, and by what sign rule? ::: The intermediate ( $I_{2} = 2$ ); unstable because $(I_{3} - I_{2}) (I_{2} - I_{1}) > 0$ gives $k > 0$ . For a symmetric top $I_{1} = I_{2} = 2, I_{3} = 5, ω_{3} = 4$ , what is the body precession rate? ::: $Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} = \frac{3}{2} \cdot 4 = 6$ rad/s. Does the stability constant $k$ contain $Ω^{2}$ ? ::: No — $k$ is a pure number; the physical rate is $k Ω^{2}$ , with $Ω^{2}$ written separately. Why does an exact single-axis spin never tumble mathematically? ::: Every Euler RHS has a factor of a zero component, so all $\overset{ω}{˙}_{i} = 0$ — it's a fixed point (though unstable).