Worked examples — Rigid body dynamics — Euler angles, Euler's equations of motion

Before anything, a one-line reminder of the tool we lean on the whole way down.

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The scenario matrix

Every problem this topic throws is one of these cells. The last column says which worked example nails it.

#	Case class	What changes	Example
A	Symmetric top, torque-free ($I_1=I_2\ne I_3$, $\mathbf{\tau }=0$)	body-frame precession, sign of $I_3-I_1$	Ex 1, Ex 2
B	Fully asymmetric, torque-free ($I_1<I_2<I_3$)	stability of each axis, sign flips	Ex 3
C	Degenerate: sphere ($I_1=I_2=I_3$)	all coupling vanishes	Ex 4
D	Zero input: spin exactly on a principal axis	$\dot{\mathbf{\omega }}=0$, no wander	Ex 4
E	Torque driven, single axis (${\tau }_3\ne 0$)	pure angular acceleration	Ex 5
F	Gyroscopic / steady precession (real-world word problem)	balance torque vs $\mathbf{\omega }\times \mathbf{L}$	Ex 6
G	Euler-angle rates → $\mathbf{\omega }$ (projection trap)	non-orthogonal axes, $\sin \theta ,\cos \theta$	Ex 7
H	Exam twist: prolate vs oblate sign ($I_3>I_1$ vs $I_3<I_1$)	direction of precession reverses	Ex 8
I	Limiting value (fast-spin limit)	which term dominates	Ex 2 (part b)

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Ex 1 — Cell A: torque-free symmetric top, basic precession rate

Step 1. Write equation 3 with ${\tau }_3=0$ and $I_1=I_2$: $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0\Rightarrow {\omega }_3=\text{const}.$ Why this step? If the spin-axis rate weren't constant, "precession about it" wouldn't be a fixed-rate motion — we must confirm ${\omega }_3$ is frozen first.

Step 2. Substitute the constant ${\omega }_3$ into equations 1 and 2. With $I_1=I_2$ they become ${\dot{\omega }}_1=-\Omega {\omega }_2,{\dot{\omega }}_2=+\Omega {\omega }_1,\Omega \equiv \frac{I_3-I_1}{I_1}{\omega }_3.$ Why this step? This pair is the equation of a point going round a circle at angular rate $\Omega$ — that IS the wobble.

Step 3. Plug numbers: $\Omega =\frac{4\times 10^{-3}-2\times 10^{-3}}{2\times 10^{-3}}\times 50=\frac{2\times 10^{-3}}{2\times 10^{-3}}\times 50=50\text{rad/s}.$

Look at the figure: the tip of the transverse angular-velocity vector (pale-yellow arrow) traces the blue circle about the body-3 axis at rate $\Omega$, while the constant ${\omega }_3$ (pink) points straight up.

Recall Forecast answer

Same sign as ${\omega }_3$ (because $I_3>I_1$, so $\Omega >0$), and here it happens to equal ${\omega }_3$ because $I_3=2I_1$ exactly. For a thin disk this is the "$2\times$" relation in disguise.

Verify: Units: $\frac{{\text{kg⋅m}}^2}{{\text{kg⋅m}}^2}\times \text{rad/s}=\text{rad/s}$. ✓ Sign: $I_3>I_1\Rightarrow \Omega >0$, an oblate body, precession same sense as spin. ✓

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Ex 2 — Cell A + Cell I: Earth's Chandler-type wobble and the fast-spin limit

Step 1 (a). From Ex 1, $\Omega =\frac{I_3-I_1}{I_1}{\omega }_3=\frac{1}{305}\times \frac{2\pi }{1\text{day}}$. Why this step? Same symmetric-top formula; only the number $\frac{I_3-I_1}{I_1}$ changed.

Step 2 (a). Period $T=\frac{2\pi }{\Omega }=305\text{days}$. Why this step? Period is $2\pi$ divided by rate; the $2\pi$'s cancel, leaving $305$ spin-periods = 305 days.

Step 3 (b). Equation 1 with $\tau =0$: $I_1{\dot{\omega }}_1=(I_2-I_3){\omega }_2{\omega }_3$. As ${\omega }_3\to \infty$, the right side (linear in ${\omega }_3$) dwarfs any term not multiplied by ${\omega }_3$. Why this step? Identifying the dominant term tells us the wobble frequency scales linearly with spin — spin faster, wobble faster.

Recall Reality check on the real Chandler wobble

The observed wobble is ~433 days, not 305 — because Earth is not perfectly rigid; it deforms. See Chandler wobble. The rigid-body prediction is exactly our 305-day answer, and the discrepancy is itself informative.

Verify: $T=305$ days from $2\pi /\Omega$. Units: $\Omega$ in rad/day $\Rightarrow T$ in days. ✓ Limit: right-hand side $\propto {\omega }_3$, all else independent of ${\omega }_3$. ✓

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Ex 3 — Cell B: fully asymmetric body, the intermediate-axis (tennis-racket) test

Step 1. Spin nearly about axis 1: ${\omega }_1\approx$ const large, ${\omega }_2,{\omega }_3$ tiny. Linearize equations 2 and 3 (drop products of two small quantities): $I_2{\dot{\omega }}_2=(I_3-I_1){\omega }_3{\omega }_1,I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2.$ Why this step? Small-perturbation analysis: keep only terms linear in the small components to see if they grow or oscillate.

Step 2. Differentiate the first, substitute the second: ${\ddot{\omega }}_2=\frac{(I_3-I_1)(I_1-I_2)}{I_2{I}_3}{\omega }_1^2{\omega }_2\equiv -k{\omega }_2.$ Why this step? Reducing two first-order equations to one second-order equation reveals the sign of $k$: $k>0\Rightarrow$ oscillation (stable), $k<0\Rightarrow$ exponential growth (unstable).

Step 3. Evaluate the coefficient $C_i\equiv (I_j-I_i)(I_i-I_k)$ for each axis (same ${\omega }^2/(I_j{I}_k)>0$ factor sits outside, so only this product's sign matters). Let $k=-C_i\times (\text{positive})$, so stable $\iff C_i<0$:

• Axis 1 (smallest): $(I_2-I_1)(I_1-I_3)=(1)(-2)=-2<0$ → stable.
• Axis 2 (middle): $(I_3-I_2)(I_2-I_1)=(1)(1)=+1>0$ → unstable.
• Axis 3 (largest): $(I_1-I_3)(I_3-I_2)=(-2)(1)=-2<0$ → stable.

Recall Forecast answer

Axis 2, the intermediate axis. This is the [tennis-racket / Dzhanibekov effect] — flip your phone spinning about its middle axis and it flips over. See also Angular momentum in rotating frames.

Verify: Products: axis1 $=-2$, axis2 $=+1$, axis3 $=-2$. Exactly one positive → exactly one unstable axis. ✓

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Ex 4 — Cells C & D: sphere, and spin exactly on a principal axis

Step 1 (C). Every coupling factor $(I_i-I_j)=0$, so all three equations read $I{\dot{\omega }}_i=0$. Why this step? The sphere is the ultimate degenerate case — no principal axis is special, so nothing can select a wobble direction.

Step 2 (C). ${\dot{\omega }}_1={\dot{\omega }}_2={\dot{\omega }}_3=0\Rightarrow \mathbf{\omega }=$ const forever. Any axis is a stable spin axis.

Step 3 (D). With ${\omega }_1={\omega }_2=0$: equation 1 has ${\omega }_2{\omega }_3=0$, equation 2 has ${\omega }_3{\omega }_1=0$, equation 3 has ${\omega }_1{\omega }_2=0$. So ${\dot{\omega }}_1={\dot{\omega }}_2={\dot{\omega }}_3=0$. Why this step? This is the zero-input cell: pure principal-axis spin has no gyroscopic coupling because the two other components that would multiply are zero. The motion is a fixed point of the equations.

Verify: (C) all $(I_i-I_j)=0$ so RHS $=0$. ✓ (D) each RHS product contains at least one of ${\omega }_1,{\omega }_2$, both zero, so RHS $=0$. ✓

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Ex 5 — Cell E: single-axis torque, pure angular acceleration

Step 1. Since ${\omega }_1={\omega }_2=0$, equation 3 loses its coupling term: $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2+{\tau }_3={\tau }_3.$ Why this step? We must check the coupling vanishes rather than assume it — here ${\omega }_1{\omega }_2=0$ kills it, so we're left with a clean $F=ma$-style law.

Step 2. Integrate: ${\dot{\omega }}_3={\tau }_3/I_3=2/0.5=4{\text{rad/s}}^2$; with zero start, ${\omega }_3=4t$. Why this step? Constant angular acceleration integrates to a straight-line spin-up.

Step 3. At $t=3$: ${\omega }_3=4\times 3=12\text{rad/s}$.

Recall Forecast answer

No coupling — this is the one case where the naive $\tau =I\dot{\omega }$ is correct, precisely because we spin on a principal axis with no transverse rate.

Verify: ${\omega }_3=12\text{rad/s}$. Units: $\frac{\text{N⋅m}}{{\text{kg⋅m}}^2}\cdot \text{s}=\frac{{\text{kg⋅m}}^2/{\text{s}}^2}{{\text{kg⋅m}}^2}\cdot \text{s}=\text{1/s}=\text{rad/s}$. ✓

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Ex 6 — Cell F: real-world gyroscope, steady precession

Step 1. Gravity torque about the pivot: $\tau =mgd$, horizontal, perpendicular to the spin axis. Why this step? The torque is what the $\mathbf{\omega }\times \mathbf{L}$ machinery must balance in steady precession.

Step 2. For steady precession the spin angular momentum $L=I_3{\omega }_3$ turns at rate ${\Omega }_p$, so $|\dot{\mathbf{L}}|={\Omega }_{p}L$. Setting $\tau ={\Omega }_{p}L$: ${\Omega }_p=\frac{mgd}{I_3{\omega }_3}.$ Why this step? This is Euler's $\mathbf{\tau }=\mathbf{\omega }\times \mathbf{L}$ balance in disguise — the torque doesn't speed the wheel up, it swings its axis. See Gyroscope precession.

Step 3. Numbers: ${\Omega }_p=\frac{0.4\times 9.8\times 0.05}{1.0\times 10^{-2}\times 200}=\frac{0.196}{2.0}=0.098\text{rad/s}.$

Recall Forecast answer

Slower. ${\Omega }_p\propto 1/{\omega }_3$: a fast top precesses lazily, a dying top precesses wildly — exactly what you see as a real top slows down.

Verify: ${\Omega }_p=0.098\text{rad/s}$. Units: $\frac{\text{kg}\cdot {\text{m/s}}^2\cdot \text{m}}{{\text{kg⋅m}}^2\cdot \text{1/s}}=\frac{{\text{kg⋅m}}^2/{\text{s}}^2}{{\text{kg⋅m}}^2/\text{s}}=\text{1/s}$. ✓

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Ex 7 — Cell G: Euler-angle rates → $\mathbf{\omega }$ (the projection trap)

Step 1. Use the 3-1-3 projection formulas from the parent note: ${\omega }_1=\dot{\varphi }\sin \theta \sin \psi +\dot{\theta }\cos \psi ,{\omega }_2=\dot{\varphi }\sin \theta \cos \psi -\dot{\theta }\sin \psi ,{\omega }_3=\dot{\varphi }\cos \theta +\dot{\psi }.$ Why this step? The three Euler axes ($Z$, line of nodes, body-$z$) are not perpendicular, so we cannot just read off components — we must project, which is what these $\sin \theta ,\cos \theta$ factors do.

Step 2. Put $\theta =60^{\circ }$ ($\sin =\frac{\sqrt{3}}{2}\approx 0.8660$, $\cos =\frac{1}{2}$), $\psi =0$ ($\sin \psi =0,\cos \psi =1$), $\dot{\theta }=0$: ${\omega }_1=3(0.8660)(0)+0=0,{\omega }_2=3(0.8660)(1)-0=2.598,{\omega }_3=3(0.5)+5=\mathrm{6.5.}$

Step 3. So $\mathbf{\omega }=(0,2.598,6.5)\text{rad/s}$.

Recall Forecast answer

No — ${\omega }_3=6.5$, not $5$. The precession $\dot{\varphi }$ leaks into the body-3 axis through the $\dot{\varphi }\cos \theta =3\times 0.5=1.5$ term. Treating rates as components is [!mistake]-2 from the parent note.

Verify: $({\omega }_1,{\omega }_2,{\omega }_3)=(0,3\sqrt{3}/2,6.5)$; $3\sqrt{3}/2\approx 2.598$. ✓ See Lagrangian of the symmetric top for where these feed the kinetic energy.

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Ex 8 — Cell H: exam twist, prolate vs oblate flips the precession sense

Step 1 (oblate). $\Omega =\frac{1.5-1}{1}\times 10=+5\text{rad/s}$. Why this step? $I_3>I_1$ makes the factor positive — precession same sense as spin (like a frisbee).

Step 2 (prolate). $\Omega =\frac{0.6-1}{1}\times 10=-4\text{rad/s}$. Why this step? $I_3<I_1$ makes the factor negative — precession opposite sense to spin (like a spinning pencil / cigar). The sign of $I_3-I_1$ is the whole story.

Step 3. Summary: oblate $\Omega =+5$, prolate $\Omega =-4$.

Recall Forecast answer

The prolate (cigar-shaped) top: its transverse $\mathbf{\omega }$ circles the body-3 axis backwards relative to its own spin.

Verify: oblate $+5$, prolate $-4$; signs $\mathrm{sign}(I_3-I_1)$ are $+$ and $-$. ✓

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Full-throttle question to leave with: