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

2.1.21 · D5 · Physics › Analytical Mechanics › Rigid body dynamics — Euler angles, Euler's equations of mot

Symbols ka ek reminder pehle, taaki koi line surprise na kare:

$(ϕ, θ, ψ)$ — teen Euler angles: space- $Z$ ke baare mein precession, line of nodes ke baare mein nutation (tilt), body- $z$ ke baare mein spin.
$ω$ — angular velocity vector; iske body-frame components hain $(ω_{1}, ω_{2}, ω_{3})$ .
$L = I ω$ — angular momentum; $I$ hai inertia tensor (dekho Inertia tensor and principal axes).
$I_{1}, I_{2}, I_{3}$ — principal moments of inertia ( $I$ ki diagonal jab hum principal axes par baith jaate hain).
$τ$ — applied torque.

Woh geometry jise ye traps refer karte hain

Traps se pehle, woh picture dekho jise ye pura page baar baar point karta hai. Do frames ek origin share karte hain: navy mein draw kiya hua fixed space frame $(X, Y, Z)$ , aur magenta mein draw kiya hua tumbling object se glued body frame $(x, y, z)$ . Teen Euler angles woh teen twists hain jo ek ko doosre mein le jaate hain.

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

Kuch traps jo projection factors invoke karte hain woh folklore nahi hain — yeh raha pura kinematic map, jo tum upar ke figure se read off kar sakte ho (har Euler-rate arrow body axes par project hua). Ek piece jo picture sirf implicitly dikhati hai woh hai $\overset{n}{^}$ body coordinates mein likha hua, isliye hum use explicitly state karte hain:

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

Pehle, teen axis unit vectors body frame $(1, 2, 3)$ mein express kiye gaye: $\overset{z}{^} = (0, 0, 1), \overset{n}{^} = (cos ψ, - sin ψ, 0), \hat{Z} = (sin θ sin ψ, sin θ cos ψ, cos θ) .$ Line of nodes $\overset{n}{^}$ body- $(1, 2)$ plane mein hai lekin spin $ψ$ se rotated hai, isliye uske $cos ψ, - sin ψ$ components hain; $\hat{Z}$ tilt $θ$ se body- $z$ se lean karta hai, jisse uska $cos θ$ vertical part aur $sin θ$ horizontal part milta hai.

Ab vector sum $ω = \dot{ϕ} \hat{Z} + \dot{θ} \overset{n}{^} + \dot{ψ} \overset{z}{^}$ banao aur components read off karo:

\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}$$ $\sin\theta$ isliye aata hai kyunki $\hat Z$ body-$z$ se angle $\theta$ par lean karta hai, isliye sirf uska perpendicular part $\dot\phi\sin\theta$ $(1,2)$-plane tak pahunchta hai; $\cos\theta$ hai $\hat Z$ ka projection *onto* body-$z$. Extra $\sin\psi,\cos\psi$ isliye aate hain kyunki spin ne $(1,2)$ axes ko $\psi$ se rotate kar diya hai.

True or false — justify karo

TF1. "Euler angles ko exactly teen numbers chahiye kyunki 3D orientation ek 3-dimensional space mein rehti hai."

True — rotation group 3-dimensional hai, isliye koi bhi orientation teen independent parameters se fix hoti hai; Euler ka choice bas un teeno ko geometrically readable banata hai (sweep, tilt, spin).

TF2. "

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

mutually perpendicular axes ke along teen vectors ka sum hai."

False — space-

Z

, line of nodes

\overset{n}{^}

, aur body-

z

generally ek dusre ke relative tilted hain (sirf

\overset{n}{^} ⊥ \hat{Z}

aur

\overset{n}{^} ⊥ \overset{z}{^}

), aur yahi exact reason hai ki unhe project karne se boxed

ω_{1}, ω_{2}, ω_{3}

mein

sin θ, cos θ

factors aate hain.

TF3. "Torque-free body ke liye,

ω

hamesha constant hota hai."

False — sirf

L

space frame mein conserved hai;

ω

wander kar sakta hai kyunki gyroscopic terms

(I_{i} - I_{j}) ω_{j} ω_{k}

\overset{ω}{˙}_{i} \neq = 0

banate hain jab tak spin principal axis ke baare mein na ho.

TF4. "Body frame mein inertia tensor constant hai."

True — body ki mass distribution body se glued axes ke relative frozen hai, isliye

I

wahan change nahi hota; yahi constancy woh poori reason hai jis wajah se hum body frame mein dynamics karte hain.

TF5. "Perfectly spherical rigid body (

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

) mein koi gyroscopic coupling nahi hoti."

True — har coupling term mein ek factor

(I_{i} - I_{j})

hota hai, jo tab zero ho jaata hai jab saare moments equal hote hain, isliye koi bhi axis equilibrium hai aur free rotation steady hai.

TF6. "Euler's equations

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

kisi bhi body-fixed axes mein hold karte hain jो tum chaaho."

False — ye assume karte hain ki axes principal hain, taaki

I

diagonal ho; non-principal body axes mein off-diagonal inertia terms wapas aa jaate hain aur clean boxed form toot jaati hai.

TF7. "Largest moment of inertia ki axis ke baare mein rotation stable hai."

True — torque-free equations ko linearize karne se largest aur smallest

I

dono ke liye restoring (oscillatory) coefficient milta hai; sirf intermediate axis exponential growth deta hai.

TF8. "Transport theorem ka

ω \times A

term ek real physical force hai."

False — yeh ek kinematic correction hai jo do observers ke time-derivatives ko relate karta hai, force nahi; forces/torques tab hi enter karte hain jab hum ise

L

par apply karte hain aur

τ = (d L / d t)_{space}

set karte hain.

TF9. "Symmetric top ki torque-free motion mein symmetry axis ke baare mein spin component

ω_{3}

constant hota hai."

True —

I_{1} = I_{2}

ke saath teesra coupling term

(I_{1} - I_{2}) ω_{1} ω_{2}

zero ho jaata hai, isliye

I_{3} \overset{ω}{˙}_{3} = 0

TF10. "Body-frame free precession aur lab observer jo wobble dekhta hai unki rate same hoti hai."

False — body frame mein transverse

ω

body-

z

ke baare mein

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

rate se circle karta hai; lekin space observer

L

fixed dekhta hai, aur geometry force karti hai ki symmetry axis

L

ke baare mein

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

rate se cone kare, jo generally

Ω

se different hai.

Error pakdo

SE1. "Kyunki torque angular momentum ka rate of change hai, isliye

τ = I \dot{ω}

directly."

Error yeh hai:

τ = (d L / d t)_{space}

, aur body frame mein transport karne se

(d L / d t)_{body} + ω \times L

milta hai;

ω \times L

isliye aata hai kyunki body axes khud rotate karte hain, isliye ek body-fixed

L

bhi space mein ghoomta hai. Use drop karna galti hai.

SE2. "Euler rates angular-velocity components hain:

ω_{1} = \dot{ϕ}

ω_{2} = \dot{θ}

ω_{3} = \dot{ψ}

Error yeh hai: teen Euler axes non-orthogonal hain, isliye har

ω_{i}

teeno rates ko upar boxed formula mein listed

sin / cos

projection factors ke saath mix karta hai — e.g.

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

, sirf

\dot{ψ}

nahi, kyunki

\dot{ϕ} \hat{Z}

ka body-

z

par

cos θ

shadow hota hai.

SE3. "Torque-free asymmetric body ke liye hum

τ_{i} = 0

set karte hain isliye saare

\overset{ω}{˙}_{i} = 0

aur motion steady spin hai."

Error yeh hai: torques ko zero set karna right-hand couplings ko zero nahi karta;

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

abhi bhi

ω_{1}

drive karta hai jab tak do components already zero na hon.

SE4. "Principal axes choose karna sirf neatness ke liye hai; physics kisi bhi axes use karne se identical hai."

Error yeh hai: principal axes

I

ko diagonalize karte hain, jo allow karta hai

L_{i} = I_{i} ω_{i}

constant

I_{i}

ke saath; dusre axes mein

L

component-per-component

ω

ke parallel nahi hota aur equations genuinely form mein differ karti hain.

SE5. "

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

mein hum fast spinning top ke liye

\hat{Z}

term drop kar sakte hain kyunki

\dot{ψ}

dominate karta hai."

Error yeh hai: fast top bhi precess karta hai, aur

\dot{ϕ} cos θ

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

ke andar baitha hai; precession drop karne se exactly woh coupling kho jaati hai jo top ko gravity ke under precess karti hai.

SE6. "Euler's equations sirf orientation angles describe karte hain, spin rates nahi."

Error yeh hai: Euler's equations angular-velocity components

ω_{i} (t)

govern karte hain; orientation angles paane ke liye tumhe additionally boxed formula se kinematic relations

ω_{i} (\dot{ϕ}, \dot{θ}, \dot{ψ})

integrate karni hoti hain.

SE7. "Kyunki

L

conserved hai jab

τ = 0

, body ki spin axis space mein fixed rehti hai."

Error yeh hai:

L

fixed hona matlab

ω

fixed nahi hota — woh ek hi line par point karte hain sirf tab jab principal axis ke baare mein spin ho; warna spin axis fixed

L

ke baare mein cone karta hai.

Why questions

W1. "Hum rotational dynamics rotating body frame mein kyun karte hain, fixed lab frame mein kyun nahi?"

Kyunki lab frame mein

I

har instant change karta rehta hai jab body tumble karti hai,

L = I ω

ko intractable bana deta hai; body frame mein

I

frozen aur diagonal hai, ek extra

ω \times

term ki price par.

W2. "Transport theorem ek cross product specifically kyun introduce karta hai?"

Ek infinitesimal rotation har point ko

ω \times r

move karta hai; rotating body mein fixed ek vector isliye space observer ko

ω \times A

rate se change hota dikhta hai, aur cross product woh operation hai jo "axis ke perpendicular rotate karo" encode karta hai.

W3. "Intermediate-axis spin unstable kyun ho jaata hai jabki baaki do stable hain (tennis-racket theorem)?"

Growth coefficient mein

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

-type products hote hain; intermediate axis ke liye ek factor sign flip karta hai, oscillation ko exponential growth mein badal deta hai, isliye koi bhi tiny perturbation blow up ho jaata hai.

W4. "Symmetric-top solution mein

ω_{3}

special kyun hai?"

I_{1} = I_{2}

ke saath

ω_{3}

ki equation apna coupling term kho deti hai, isliye

ω_{3}

ek conserved constant hai jo phir fixed frequency

Ω

ki tarah kaam karta hai jo

(ω_{1}, ω_{2})

ki circular motion drive karta hai.

W5. "Angular velocities different rotations se simply vectors ki tarah add kyun ki ja sakti hain?"

Kyunki infinitesimal limit mein har rotation

R \approx 1 + ϵ Ω

hai, aur non-commuting piece second order hai (

Ω_{a} Ω_{b} - Ω_{b} Ω_{a}

, size

ϵ^{2}

ka) — isliye first order mein generators add karte hain;

d t

se divide karne par

ω

ek genuine vector sum milta hai. Finite rotations commute nahi karte, lekin unke per-unit-time rates karte hain.

W6. "Gyroscope precess kyun karta hai girne ki bajaye?"

Gravity torque

L

L

ke perpendicular change karta hai (

d L / d t = τ

ke zariye), isliye spin axis neeche tip hone ki bajaye sideways ghoomti hai — gyroscopic coupling Euler's equations mein explicit hai, Gyroscope precession mein expand kiya gaya.

W7. "Chandler wobble in same equations se kyun arise hoti hai?"

Earth almost ek symmetric top hai jisme

I_{3} > I_{1}

; torque-free Euler's equations

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

par body-frame free precession predict karte hain, jo observed Chandler wobble ka theoretical seed hai.

W8. "Hum centrifugal/Coriolis forces ki jagah rotating-frame derivative ke saath body-frame angular momentum kyun use karte hain?"

Woh pseudo-forces (dekho Coriolis and centrifugal forces) same rotating-frame kinematics ka point-particle face hain; rigid-body rotation ke liye compact bookkeeping hai ek akela

ω \times L

term.

Edge cases

E1. "Spherical top,

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

, ke liye Euler's equations mein kya hota hai?"

Saare coupling terms zero ho jaate hain,

I \overset{ω}{˙}_{i} = τ_{i}

rehta hai; free rotation kisi bhi axis ke baare mein steady hai, aur

L

har time exactly

ω

ke parallel hota hai.

E2. "Agar body exactly ek principal axis ke baare mein spin kare bina kisi perturbation ke?"

Tab teen mein se do

ω_{i}

zero hain, saare products

ω_{j} ω_{k}

vanish ho jaate hain, isliye torque-free motion perfectly steady hai — lekin yeh knife-edge hai; stability depend karti hai kis axis par (E4).

E3. "Symmetric-top formula kya deta hai jab

I_{3} = I_{1}

(sphere approach karte hue)?"

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

, isliye free-precession period infinity jaati hai — practically koi wobble nahi, jo spherical case ke no coupling ke saath consistent hai.

E4. "

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

ke liye, kaun se axes stable steady spin dete hain aur kaunsa nahi?"

Axes 1 (min) aur 3 (max) stable hain — perturbations oscillate karti hain; axis 2 (intermediate) unstable hai — perturbations exponentially grow karti hain aur body tumble karne lagti hai.

E5. "

θ = 0

par (top bilkul seedha khada) Euler-angle-to-

ω

formulas ka kya hota hai?"

Line of nodes undefined ho jaati hai aur

ϕ, ψ

aligned

Z

aur

z

axes ke baare mein same rotation describe karte hain — ek coordinate singularity (gimbal lock); physics fine hai lekin

(ϕ, θ, ψ)

chart degenerate ho jaata hai.

E6. "

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}

mein

θ = π /2

par kya hota hai?"

Precession ka body-axis spin mein contribution drop out ho jaata hai (

cos \frac{π}{2} = 0

), isliye

ω_{3} = \dot{ψ}

akela — ek clean limit, singularity nahi, kyunki line of nodes yahan abhi bhi well defined hai.

E7. "

θ = π

par kya hota hai (body-

z

space-

Z

ke opposite point kare)?"

Same gimbal-lock trouble jaise

θ = 0

par:

\overset{z}{^}

\hat{Z}

ke anti-parallel hone se,

ϕ

aur

ψ

dono same line ke baare mein rotate karte hain (ab inverted), isliye sirf combination

ϕ - ψ

determine hota hai — chart phir se degenerate ho jaata hai, aur

sin θ = 0

transverse projection ko exactly

θ = 0

ki tarah collapse karta hai.

E8. "Agar applied torque hamesha

ω

ke parallel ho, toh kya body bas ek fixed axis ke along spin up karti hai?"

Generally nahi — coupling terms abhi bhi angular velocity axes ke beech redistribute karte hain jab tak

ω

principal axis ke along na ho, isliye

ω

ke along torque necessarily axis fixed nahi rakhta.

Recall Traps ka one-line summary

Symbols geometry se earn hote hain (non-orthogonal Euler axes → $sin θ, cos θ$ projection factors), $τ = I \dot{ω}$ ek jhooth hai kyunki $I$ lab frame mein tumble karta hai (tum $ω \times L$ term miss karte ho), coupling terms $(I_{i} - I_{j}) ω_{j} ω_{k}$ zero torque par bhi survive karte hain isliye free body ka spin axis wander kar sakta hai, aur teen principal axes mein se intermediate wala hai jiska steady spin unstable hai.