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

1. Setup: why we go to the body frame

WHY a body frame? In the lab (space) frame, $\mathbf I$ changes every instant because the body re-orients. That's a nightmare. If we pick axes fixed in the body along its principal axes, then $\mathbf I = \mathrm{diag}(I_1,I_2,I_3)$ is constant. The price: the body frame rotates, so time derivatives get an extra term.

2. Deriving Euler's equations from scratch

Start from Newton/Euler for rotation in the inertial frame: $\left(\frac{d\vec L}{dt}\right)_{\text{space}} = \vec N \quad(\text{external torque})$

Apply the transport theorem to $\vec L$ : $\left(\frac{d\vec L}{dt}\right)_{\text{body}} + \vec\omega\times\vec L = \vec N$

Why this step? It converts a hard space-frame derivative into a body-frame derivative (where $\mathbf I$ is constant) plus a correction.

In the body principal frame $L_i = I_i\,\omega_i$ , and since $I_i$ are constants: $\left(\frac{d\vec L}{dt}\right)_{\text{body}} = (I_1\dot\omega_1,\, I_2\dot\omega_2,\, I_3\dot\omega_3)$

Now compute the cross product component-wise. With $\vec\omega=(\omega_1,\omega_2,\omega_3)$ , $\vec L=(I_1\omega_1,I_2\omega_2,I_3\omega_3)$ : $(\vec\omega\times\vec L)_1 = \omega_2 L_3 - \omega_3 L_2 = (I_3-I_2)\,\omega_2\omega_3$

Putting it together and setting $\vec N=0$ (torque-free):

3. Two conserved quantities (the constants of motion)

Even with no external torque, in the body frame $\vec\omega$ moves. But two scalars are conserved:

Derivation of energy conservation: $\dfrac{dT}{dt} = I_1\omega_1\dot\omega_1 + I_2\omega_2\dot\omega_2 + I_3\omega_3\dot\omega_3$ . Substitute Euler's equations: $= \omega_1\omega_2\omega_3\big[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)\big] = 0.$ Why this step? The bracket telescopes to zero — energy is conserved without any external work, exactly as expected.

WHY two surfaces? In $\omega$ -space, $2T=$ const is an ellipsoid (energy ellipsoid) and $L^2=$ const is another ellipsoid. The actual motion of $\vec\omega$ lives on their intersection curves — these are the polhodes.

Figure — Torque-free rotation — Euler's equations, asymmetric top

4. Stability of rotation — the tennis-racket theorem

Take the asymmetric top: $I_1 < I_2 < I_3$ (all different). Spin nearly about one principal axis and ask: does a tiny disturbance grow or stay small?

Spin about axis 3 (largest $I$ ): Let $\omega_3\approx\Omega$ (large), $\omega_1,\omega_2$ tiny. From Euler 1 and 2, differentiate and substitute: $\dot\omega_1 = \frac{I_2-I_3}{I_1}\Omega\,\omega_2,\qquad \dot\omega_2 = \frac{I_3-I_1}{I_2}\Omega\,\omega_1$ $\Rightarrow \ddot\omega_1 = \frac{(I_2-I_3)(I_3-I_1)}{I_1 I_2}\Omega^2\,\omega_1$ Why this step? Differentiating the first and plugging in the second gives a single 2nd-order ODE $\ddot\omega_1 = k\,\omega_1$ .

For axis 3 (largest): $(I_2-I_3)<0$ , $(I_3-I_1)>0$ → $k<0$ → oscillation (stable). ✅
For axis 1 (smallest): both factors give $k<0$ → stable. ✅
For axis 2 (intermediate): $(I_3-I_2)>0$ and $(I_2-I_1)>0$ arrange so $k>0$ → exponential growth → UNSTABLE. ❌

5. Worked examples

Worked example Example 1 — Symmetric top (free precession)

A symmetric top has $I_1=I_2\equiv I_\perp \ne I_3$ . Find the motion. Step: Euler 3 gives $I_3\dot\omega_3 = (I_1-I_2)\omega_1\omega_2 = 0$ → $\omega_3=$ const. Why this step? Equal transverse moments kill the RHS, so the spin component along the symmetry axis is frozen. Step: $\dot\omega_1 = \dfrac{I_\perp - I_3}{I_\perp}\omega_3\,\omega_2 \equiv -\Omega_b\,\omega_2$ , $\dot\omega_2 = +\Omega_b\,\omega_1$ , with body-precession rate $\Omega_b = \dfrac{I_3-I_\perp}{I_\perp}\omega_3$ . Why this step? This is circular motion: $\omega_1,\omega_2$ rotate at constant rate $\Omega_b$ → $\vec\omega$ sweeps a cone about axis 3 in the body frame. Earth does this (Chandler wobble, period ~ a year ideal, observed ~433 days).

Common mistake Steel-manning the classic errors

Mistake A: "No torque ⇒ $\vec\omega$ is constant." Why it feels right: For linear motion, no force ⇒ constant velocity. By analogy you expect no torque ⇒ constant $\vec\omega$ . The fix: It's $\vec L$ (not $\vec\omega$ ) that's conserved in space. Since $\vec L=\mathbf I\vec\omega$ and $\mathbf I$ depends on orientation, a constant $\vec L$ generally means a changing $\vec\omega$ . Only for a sphere are they always parallel.

Mistake B: "Euler's equations have external torque, so they don't conserve energy." Why it feels right: The RHS looks like a "driving" term. The fix: The RHS terms are internal (the $\vec\omega\times\vec L$ coupling), not external work. We proved $dT/dt=0$ above; they telescope away.

Mistake C: "All three principal axes are equally stable." Why it feels right: By symmetry of the equations they look interchangeable. The fix: The sign of $(I_a-I_b)(I_b-I_c)$ differs for the intermediate axis. Sign, not symmetry, decides stability.

Recall Feynman: explain to a 12-year-old

Imagine spinning a book that's taped shut. Spin it flat (like a frisbee) — easy and steady. Spin it about its long thin axis — also steady. But try to flip it end-over-end about the middle way — it refuses to stay; it flips itself over and over! Nothing pushes it. The reason: a spinning thing wants to keep its "twirl arrow" pointing the same way in space, but the book is shaped unevenly, so to keep that arrow steady the book has to keep re-tilting itself — and for the middle axis those re-tilts pile up instead of canceling.

Flashcards

Why use the body frame instead of the space frame for rotation?

In the body principal frame the inertia tensor

\mathbf I=\mathrm{diag}(I_1,I_2,I_3)

is constant; in the space frame it changes as the body reorients.

State the transport theorem.

(d\vec A/dt)_{\text{space}} = (d\vec A/dt)_{\text{body}} + \vec\omega\times\vec A

Write Euler's equation for component 1 (torque-free).

I_1\dot\omega_1 = (I_2-I_3)\,\omega_2\omega_3

What conserved quantity is

\vec L

in space vs

\vec\omega

\vec L

is conserved (no torque);

\vec\omega

is generally NOT constant because

\vec L=\mathbf I\vec\omega

and

\mathbf I

depends on orientation.

Name the two conserved scalars in torque-free motion.

Rotational energy

2T=\sum I_i\omega_i^2

and

L^2=\sum I_i^2\omega_i^2

What geometric objects represent these conserved scalars in

\omega

-space?

Two ellipsoids (energy ellipsoid and momentum ellipsoid); motion = their intersection curves (polhodes).

Which principal axes give stable rotation for an asymmetric top?

Axes of largest and smallest moment of inertia (intermediate is unstable).

What is the name of the intermediate-axis instability?

Tennis-racket / Dzhanibekov effect.

For a symmetric top (

I_1=I_2

), what is the body precession rate?

\Omega_b=\dfrac{I_3-I_\perp}{I_\perp}\,\omega_3

, and

\omega_3

is constant.

Why is energy conserved despite the nonzero RHS of Euler's equations?

dT/dt=\omega_1\omega_2\omega_3[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)]=0

— the bracket telescopes; RHS terms are internal, not external work.

What growth/oscillation criterion governs small disturbances near an axis?

\ddot\omega_1=k\omega_1

with

k=\frac{(I_b-I_c)(I_c-I_a)}{I_aI_b}\Omega^2

;

k<0

stable (oscillates),

k>0

unstable (grows).

Connections

Inertia tensor and principal axes — where $\mathbf I=\mathrm{diag}(I_1,I_2,I_3)$ comes from.
Angular momentum in rotating frames — source of the transport theorem.
Rigid body dynamics — Euler angles — converting body $\vec\omega$ to lab orientation.
Symmetric top and gyroscopic precession — the $I_1=I_2$ special case (with torque).
Stability analysis and linearization — the $\ddot x=kx$ method reused here.
Noether's theorem — why $\vec L$ conservation follows from rotational symmetry of free space.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, torque-free rotation ka matlab hai koi bahar se torque nahi lag raha — jaise space mein ghoomta hua satellite ya hawa mein uchhala phone. Aap soch sakte ho ki agar koi torque nahi to spin axis fix rehna chahiye, par twist yeh hai: conserve hota hai angular momentum $\vec L$ , na ki $\vec\omega$ . Kyunki $\vec L=\mathbf I\vec\omega$ , aur jaise body ghoomti hai uska inertia "feel" badalta rehta hai, isliye $\vec\omega$ body ke andar wobble kar sakta hai. Isiliye hum body frame mein baithte hain jahan $I_1,I_2,I_3$ constant ho jaate hain — bas iski keemat yeh hai ki frame khud ghoom raha hai, to derivative mein ek extra $\vec\omega\times\vec L$ term aata hai. Yahi se nikalti hain Euler's equations.

Euler ki teen equations ka pattern simple hai: $I_1\dot\omega_1=(I_2-I_3)\omega_2\omega_3$ , aur cyclically baaki do. Matlab ek axis ka spin badalta hai baaki do omegas ke product se, aur woh bhi inertia ke difference se weighted. Agar saare $I$ equal (sphere) to RHS zero — koi wobble nahi. Do cheezein hamesha conserve hoti hain: energy $2T=\sum I_i\omega_i^2$ aur $L^2=\sum I_i^2\omega_i^2$ . Inko $\omega$ -space mein do ellipsoid samjho; actual motion dono ke intersection (polhode) pe chalti hai.

Sabse mazedaar baat asymmetric top ( $I_1<I_2<I_3$ ) ki stability hai. Largest aur smallest axis ke around spin karo to motion stable rehti hai (chhoti si disturbance bas oscillate karti hai). Lekin intermediate axis ke around spin karo to disturbance exponentially badhti hai — body palat jaati hai! Isi ko tennis-racket ya Dzhanibekov effect kehte hain. Apna phone middle axis pe flip karke dekho, woh hamesha ek extra half-turn le leta hai. Yaad rakhne ka mantra: "BIG and SMALL are solid, MIDDLE is a muddle."

Yeh sab important kyun hai? Satellites ki attitude control, Earth ki Chandler wobble, gymnasts aur divers ka mid-air twist — sab isi physics se chalte hain. Ek baar Eu

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

1. Setup: why we go to the body frame

2. Deriving Euler's equations from scratch

3. Two conserved quantities (the constants of motion)

4. Stability of rotation — the tennis-racket theorem

5. Worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Analytical Mechanics

Connections