Rigid body dynamics — Euler angles, Euler's equations of motion: Question bank

Reminders of the cast of symbols before we start, so no line surprises you:

$(ϕ, θ, ψ)$ — the three Euler angles: precession about space- $Z$ , nutation (tilt) about the line of nodes, spin about body- $z$ .
$ω$ — the angular velocity vector; its body-frame components are $(ω_{1}, ω_{2}, ω_{3})$ .
$L = I ω$ — angular momentum; $I$ is the inertia tensor (see Inertia tensor and principal axes).
$I_{1}, I_{2}, I_{3}$ — the principal moments of inertia (the diagonal of $I$ once we sit on principal axes).
$τ$ — the applied torque.

The geometry these traps refer to

Before the traps, look at the picture the whole page keeps pointing back to. Two frames share an origin: the fixed space frame $(X, Y, Z)$ drawn in navy, and the body frame $(x, y, z)$ glued to the tumbling object drawn in magenta. The three Euler angles are the three twists that carry one into the other.

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

The projection factors invoked by several traps below are not folklore — here is the full kinematic map, which you should read off the figure above (each Euler-rate arrow projected onto the body axes). The one piece the picture only shows implicitly is $\overset{n}{^}$ written in body coordinates, so we state it explicitly:

Formula Euler rates → body-frame angular velocity (3-1-3)

First, the three axis unit vectors expressed in the body frame $(1, 2, 3)$ : $\overset{z}{^} = (0, 0, 1), \overset{n}{^} = (cos ψ, - sin ψ, 0), \hat{Z} = (sin θ sin ψ, sin θ cos ψ, cos θ) .$ The line of nodes $\overset{n}{^}$ lies in the body- $(1, 2)$ plane but rotated by the spin $ψ$ , hence its $cos ψ, - sin ψ$ components; $\hat{Z}$ leans off body- $z$ by the tilt $θ$ , giving its $cos θ$ vertical part and $sin θ$ horizontal part.

Now form the vector sum $ω = \dot{ϕ} \hat{Z} + \dot{θ} \overset{n}{^} + \dot{ψ} \overset{z}{^}$ and read off components:

\omega_1 &= \dot\phi\sin\theta\sin\psi + \dot\theta\cos\psi\\ \omega_2 &= \dot\phi\sin\theta\cos\psi - \dot\theta\sin\psi\\ \omega_3 &= \dot\phi\cos\theta + \dot\psi \end{aligned}$$ The $\sin\theta$ appears because $\hat Z$ leans at angle $\theta$ off body-$z$, so only its perpendicular part $\dot\phi\sin\theta$ reaches the $(1,2)$-plane; the $\cos\theta$ is $\hat Z$'s projection *onto* body-$z$. The extra $\sin\psi,\cos\psi$ come from the spin having rotated the $(1,2)$ axes by $\psi$.

True or false — justify

TF1. "Euler angles need exactly three numbers because 3D orientation lives in a 3-dimensional space."

True — the rotation group is 3-dimensional, so any orientation is fixed by three independent parameters; Euler's choice just makes those three geometrically readable (sweep, tilt, spin).

TF2. "

ω = \dot{ϕ} \hat{Z} + \dot{θ} \overset{n}{^} + \dot{ψ} \overset{z}{^}

is a sum of three vectors along mutually perpendicular axes."

False — space-

Z

, the line of nodes

\overset{n}{^}

, and body-

z

are generally tilted relative to each other (only

\overset{n}{^} ⊥ \hat{Z}

and

\overset{n}{^} ⊥ \overset{z}{^}

), which is exactly why projecting them produces the

sin θ, cos θ

factors in the boxed

ω_{1}, ω_{2}, ω_{3}

above.

TF3. "For a torque-free body,

ω

is always constant."

False — only

L

is conserved in the space frame;

ω

can wander because the gyroscopic terms

(I_{i} - I_{j}) ω_{j} ω_{k}

make

\overset{ω}{˙}_{i} \neq = 0

unless the spin is about a principal axis.

TF4. "In the body frame the inertia tensor is constant."

True — the body's mass distribution is frozen relative to axes glued to the body, so

I

doesn't change there; that constancy is the whole reason we do dynamics in the body frame.

TF5. "A perfectly spherical rigid body (

I_{1} = I_{2} = I_{3}

) has no gyroscopic coupling."

True — every coupling term carries a factor

(I_{i} - I_{j})

, which vanishes when all moments are equal, so any axis is an equilibrium and free rotation is steady.

TF6. "Euler's equations

I_{i} \overset{ω}{˙}_{i} - (I_{j} - I_{k}) ω_{j} ω_{k} = τ_{i}

hold in any body-fixed axes you like."

False — they assume the axes are principal, so

I

is diagonal; in non-principal body axes off-diagonal inertia terms reappear and the clean boxed form breaks.

TF7. "Rotation about the axis of largest moment of inertia is stable."

True — linearizing the torque-free equations gives a restoring (oscillatory) coefficient for both the largest and smallest

I

; only the intermediate axis gives exponential growth.

TF8. "The transport theorem's

ω \times A

term is a real physical force."

False — it is a kinematic correction relating two observers' time-derivatives, not a force; forces/torques only enter when we later apply it to

L

and set

τ = (d L / d t)_{space}

TF9. "In torque-free motion of a symmetric top the spin component

ω_{3}

about the symmetry axis is constant."

True — with

I_{1} = I_{2}

the third coupling term

(I_{1} - I_{2}) ω_{1} ω_{2}

vanishes, so

I_{3} \overset{ω}{˙}_{3} = 0

TF10. "Body-frame free precession and the wobble a lab observer sees have the same rate."

False — in the body frame the transverse

ω

circles body-

z

Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3}

; but a space observer sees

L

fixed, and geometry forces the symmetry axis to cone about

L

at rate

\dot{ϕ} = \frac{I _{3} ω _{3}}{I _{1} c o s θ}

, which differs from

Ω

in general.

Spot the error

SE1. "Since torque is rate of change of angular momentum,

τ = I \dot{ω}

directly."

The error:

τ = (d L / d t)_{space}

, and transporting to the body frame gives

(d L / d t)_{body} + ω \times L

; the

ω \times L

arises because the body axes themselves rotate, so a body-fixed

L

still turns in space. Dropping it is the mistake.

SE2. "The Euler rates are the angular-velocity components:

ω_{1} = \dot{ϕ}

ω_{2} = \dot{θ}

ω_{3} = \dot{ψ}

The error: the three Euler axes are non-orthogonal, so each

ω_{i}

mixes all three rates with the

sin / cos

projection factors listed in the boxed formula above — e.g.

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

, not just

\dot{ψ}

, because

\dot{ϕ} \hat{Z}

has a

cos θ

shadow on body-

z

SE3. "For a torque-free asymmetric body we set

τ_{i} = 0

so all

\overset{ω}{˙}_{i} = 0

and motion is steady spin."

The error: setting the torques to zero does not zero the right-hand couplings;

I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3}

still drives

ω_{1}

unless two components already vanish.

SE4. "Choosing principal axes is just for neatness; the physics is identical to using any axes."

The error: principal axes diagonalize

I

, which is what lets

L_{i} = I_{i} ω_{i}

with constant

I_{i}

; in other axes

L

is not parallel-per-component to

ω

and the equations genuinely differ in form.

SE5. "In

ω = \dot{ϕ} \hat{Z} + \dot{θ} \overset{n}{^} + \dot{ψ} \overset{z}{^}

we can drop the

\hat{Z}

term for a fast spinning top since

\dot{ψ}

dominates."

The error: even a fast top precesses, and

\dot{ϕ} cos θ

sits inside

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

; dropping precession loses the very coupling that makes a top precess under gravity.

SE6. "Euler's equations describe only the orientation angles, not the spin rates."

The error: Euler's equations govern the angular-velocity components

ω_{i} (t)

; to get the orientation angles you must additionally integrate the kinematic relations

ω_{i} (\dot{ϕ}, \dot{θ}, \dot{ψ})

from the boxed formula.

SE7. "Because

L

is conserved when

τ = 0

, the body's spin axis stays fixed in space."

The error:

L

fixed does not mean

ω

fixed — they point along the same line only when spinning about a principal axis; otherwise the spin axis cones around the fixed

L

Why questions

W1. "Why do we do rotational dynamics in the rotating body frame rather than the fixed lab frame?"

Because in the lab frame

I

changes every instant as the body tumbles, making

L = I ω

intractable; in the body frame

I

is frozen and diagonal, at the price of one extra

ω \times

term.

W2. "Why does the transport theorem introduce a cross product specifically?"

An infinitesimal rotation moves each point by

ω \times r

; a vector fixed in the rotating body therefore appears to a space observer to change at rate

ω \times A

, and cross product is the operation encoding "rotate perpendicular to the axis."

W3. "Why does the intermediate-axis spin go unstable while the other two are stable (tennis-racket theorem)?"

The growth coefficient contains

(I_{2} - I_{1}) (I_{1} - I_{3})

-type products; for the intermediate axis one factor flips sign, turning oscillation into exponential growth, so any tiny perturbation blows up.

W4. "Why is

ω_{3}

special in the symmetric-top solution?"

With

I_{1} = I_{2}

the equation for

ω_{3}

loses its coupling term, so

ω_{3}

is a conserved constant that then acts as the fixed frequency

Ω

driving the circular motion of

(ω_{1}, ω_{2})

W5. "Why can angular velocities from different rotations simply be added as vectors?"

Because in the infinitesimal limit each rotation is

R \approx 1 + ϵ Ω

, and the non-commuting piece is second order (

Ω_{a} Ω_{b} - Ω_{b} Ω_{a}

, of size

ϵ^{2}

) — so to first order the generators add; dividing by

d t

gives

ω

as a genuine vector sum. Finite rotations do not commute, but their per-unit-time rates do.

W6. "Why does a gyroscope precess instead of just falling over?"

The gravity torque changes

L

perpendicular to

L

(via

d L / d t = τ

), so the spin axis turns sideways rather than tipping down — the gyroscopic coupling made explicit in Euler's equations, expanded in Gyroscope precession.

W7. "Why does the Chandler wobble arise from these same equations?"

Earth is a nearly-symmetric top with

I_{3} > I_{1}

; torque-free Euler's equations predict body-frame free precession at

Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3}

, the theoretical seed of the observed Chandler wobble.

W8. "Why must we use body-frame angular momentum with the rotating-frame derivative rather than centrifugal/Coriolis forces?"

Those pseudo-forces (see Coriolis and centrifugal forces) are the point-particle face of the same rotating-frame kinematics; for rigid-body rotation the compact bookkeeping is the single

ω \times L

term.

Edge cases

E1. "What happens to Euler's equations for a spherical top,

I_{1} = I_{2} = I_{3} = I

All coupling terms vanish, leaving

I \overset{ω}{˙}_{i} = τ_{i}

; free rotation is steady about any axis, and

L

is exactly parallel to

ω

at all times.

E2. "What if the body spins exactly about a principal axis with no perturbation?"

Then two of the three

ω_{i}

are zero, all products

ω_{j} ω_{k}

vanish, so torque-free motion is perfectly steady — but this is a knife-edge; stability depends on which axis (E4).

E3. "What does the symmetric-top formula give when

I_{3} = I_{1}

(approaching a sphere)?"

Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3} \to 0

, so the free-precession period goes to infinity — there is effectively no wobble, consistent with the spherical case having no coupling.

E4. "For

I_{1} < I_{2} < I_{3}

, which axes give stable steady spin and which does not?"

Axes 1 (min) and 3 (max) are stable — perturbations oscillate; axis 2 (intermediate) is unstable — perturbations grow exponentially and the body tumbles.

E5. "What happens to the Euler-angle-to-

ω

formulas at

θ = 0

(top standing perfectly upright)?"

The line of nodes becomes undefined and

ϕ, ψ

describe the same rotation about the aligned

Z

and

z

axes — a coordinate singularity (gimbal lock); the physics is fine but the

(ϕ, θ, ψ)

chart degenerates.

E6. "What happens at

θ = π /2

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

The precession's contribution to the body-axis spin drops out (

cos \frac{π}{2} = 0

), so

ω_{3} = \dot{ψ}

alone — a clean limit, not a singularity, since the line of nodes is still well defined here.

E7. "What happens at

θ = π

(body-

z

flipped to point opposite space-

Z

)?"

Same gimbal-lock trouble as

θ = 0

: with

\overset{z}{^}

anti-parallel to

\hat{Z}

, both

ϕ

and

ψ

rotate about the same line (now inverted), so only the combination

ϕ - ψ

is determined — the chart degenerates again, and

sin θ = 0

collapses the transverse projection just as at

θ = 0

E8. "If the applied torque is always parallel to

ω

, does the body just spin up along a fixed axis?"

Not generally — the coupling terms still redistribute angular velocity among the axes unless

ω

lies along a principal axis, so a torque along

ω

need not keep the axis fixed.

Recall One-line summary of the traps

Symbols are earned by geometry (non-orthogonal Euler axes → the $sin θ, cos θ$ projection factors), $τ = I \dot{ω}$ is a lie because $I$ tumbles in the lab frame (you miss the $ω \times L$ term), the coupling terms $(I_{i} - I_{j}) ω_{j} ω_{k}$ survive even at zero torque so a free body's spin axis can wander, and among the three principal axes it is the intermediate one whose steady spin is unstable.