Rigid body dynamics — Euler angles, Euler's equations of motion: Visual walkthrough

2.1.21 · D2 · Physics › Analytical Mechanics › Rigid body dynamics — Euler angles, Euler's equations of mot

Intuition Yahan se kya milega

Is page ke end tak aap ne Euler's equations of motion apne haathon se build kar li hongi, ek picture ek time, sirf ek sentence se shuru karke — "ek push jo twist kare woh spinning ko badal deta hai." Koi bhi symbol draw karne se pehle nahi aata. Agar aapne kabhi cross product ya rotating frame nahi dekha, Step 1 se shuru karo aur pictures par trust karo.

Hamara target — jis cheez ko hum dhundh rahe hain — yeh box hai:

I_1\dot\omega_1 - (I_2-I_3)\,\omega_2\omega_3 &= \tau_1\\ I_2\dot\omega_2 - (I_3-I_1)\,\omega_3\omega_1 &= \tau_2\\ I_3\dot\omega_3 - (I_1-I_2)\,\omega_1\omega_2 &= \tau_3 \end{aligned}$$ Abhi yeh gibberish lag raha hai. Hum har ek letter ko earn karenge. Poori story [[2.1.21 Rigid body dynamics — Euler angles, Euler's equations of motion (Hinglish)|parent topic]] mein dekho.

Step 1 — Spinning kya hai, aur hum ise measure kaise karte hain?

KYA. Socho ek solid brick empty space mein tumble kar rahi hai. Kisi bhi instant mein woh apne center se guzarne wali kisi line ke baare mein — ek axis — kisi speed se ghoom rahi hai. "Kaun si line" aur "kitni fast" ko hum ek single arrow mein bundle karte hain jise angular velocity kehte hain, likha jaata hai $ω$ (Greek "omega").

ARROW KYUN? Kyunki ek spin ko do pieces of information chahiye: ek direction (woh axis jiske baare mein woh ghoomti hai) aur ek size (kitni fast). Ek arrow exactly yahi do cheezein carry karta hai — uski direction hi axis hai, uski length hi radians per second mein turning rate hai. Hum arrow ko axis ke saath right-hand rule use karke point karte hain: apni right fingers ko us taraf curl karo jis tarah body ghoomti hai, aapka thumb $ω$ ke along point karega.

PICTURE. Amber arrow spin axis hai; cyan circle body hai jo uske around sweep kar rahi hai.

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

Step 2 — Dekhne ke do tarike: lab aur bug

KYA. Tumbling brick ko dekhne ke do viewpoints hain:

Space frame $(X, Y, Z)$ — tum, lab mein still khade ho, ise tumble karte dekh rahe ho.
Body frame $(x, y, z)$ — ek bug brick par chipka hua, uske saath ride kar raha hai, uske saath ghoom raha hai.

DO FRAMES KYUN? Kyunki brick ka mass distribution bug ko constant lagta hai aur tumhe constantly changing. Ek moment jo hum abhi milenge — inertia tensor — bug ke liye frozen hai aur tumhare liye ek shifting nightmare. Toh bug ka frame algebra ko clean karta hai. Catch yeh hai: bug ka frame rotating hai, aur hum Step 5 mein us cheez ki price chakanay wale hain.

PICTURE. Same brick, do coordinate crosses: white $(X, Y, Z)$ fixed, cyan $(x, y, z)$ body ke saath tilted.

Step 3 — Inertia tensor, aur hum principal axes par align kyun karte hain

KYA. Spinning mein ek "reluctance" hoti hai: kuch bodies ek axis ke baare mein ghoomne ko doosre se zyada resist karti hain. Woh reluctance inertia tensor $I$ hai. Yeh woh machine hai jo spin arrow $ω$ ko angular momentum arrow $L$ mein turn karti hai ("stored spinning-motion ki matra"): $L = I ω .$

$L$ $ω$ se alag direction mein kyun point karta hai? Ek lopsided body ke liye, $I$ output ko tilt kar deta hai: tum ek line ke baare mein spin karte ho lekin stored angular momentum ek heavier direction ki taraf lean karta hai. Generally $I$ ek full 3×3 table of nine numbers hoti hai (messy). Lekin har rigid body ke teen special perpendicular axes hote hain — principal axes — jinke along spinning $L$ ko $ω$ ke parallel produce karta hai. Body frame $(x, y, z)$ ko unke along choose karo aur table sirf teen numbers pe collapse ho jaati hai apni diagonal par: $I = I_{1} 00 0 I_{2} 0 00 I_{3}, L_{i} = I_{i} ω_{i} .$

KYUN BOTHER KARO? Kyunki $L_{i} = I_{i} ω_{i}$ (teen clean scalar products) har baar full matrix multiply se better hai. Yeh ek choice hi Euler's equations ke short hone ki wajah hai. Zyada detail Inertia tensor and principal axes mein.

PICTURE. Left: ek general body — $ω$ (amber) aur $L$ (cyan) alag-alag directions mein point karte hain. Right: ek principal axis ke baare mein spin — dono arrows line up ho jaate hain.

Step 4 — Rotation ka Newton's law

KYA. Ek twisting push torque hota hai, likha jaata hai $τ$ (Greek "tau"). Rotation ka fundamental law — $F = m a$ ka rotational cousin — kehta hai: torque angular momentum ke rate of change ke equal hota hai, fixed lab frame mein measure kiya gaya. $τ = (\frac{d L}{d t})_{space} .$

"SPACE" SUBSCRIPT KYUN? Kyunki yeh law sirf ek inertial (non-accelerating) frame mein — lab mein — true hai. Chhota symbol $(\frac{d}{d t})_{space}$ padhta hai "rate of change jaisa lab observer dekhta hai." Yeh distinction abhi enormously matter karne wali hai.

PICTURE. Ek wrench torque $τ$ apply karta hai; angular-momentum arrow $L$ thodi si time mein $τ$ ki direction mein grow karta hai.

Step 5 — Body ride karne ki price: transport theorem

KYA. Hum chahte hain bug ke frame mein kaam karna (jahan $I_{i}$ constant hain), lekin law lab frame mein rehta hai. Toh hume "lab vector change ko kaise dekhta hai" aur "bug vector change ko kaise dekhta hai" ke beech ka bridge chahiye. Woh bridge transport theorem hai: $(\frac{d A}{d t})_{space} = (\frac{d A}{d t})_{body} + ω \times A .$

EXTRA $ω \times A$ TERM KYUN? Socho ek arrow $A$ brick par paint kiya gaya hai aur bug ke liye bilkul nahi hilta. Lab observer phir bhi use swing karte hue dekhta hai, kyunki poori brick ghoom rahi hai. Woh swinging ek chhota sideways shift hai, spin axis $ω$ aur arrow $A$ dono ke perpendicular — exactly jo cross product $ω \times A$ produce karta hai.

CROSS PRODUCT SPECIFICALLY KYUN? Hume ek operation chahiye jo (a) zero de jab $A$ spin axis ke along ho (axis par koi point nahi hilda) aur (b) sabse bada ho jab $A$ seedha bahar stick kare (woh point sabse fast swing karta hai). Cross product exactly yahi karta hai: $∣ ω \times A ∣ = ∣ ω ∣∣ A ∣ sin α$ , zero jab angle $α = 0$ , maximal $α = 9 0^{\circ}$ par, aur sideways pointing (dono ke perpendicular).

PICTURE. Ek arrow spinning body par fixed; bug use still dekhta hai, lekin lab uski tip ko ek circle trace karte hue dekhta hai — tip ki velocity $ω \times A$ hai (amber), grey radius ke perpendicular.

Step 6 — $L$ ko bridge se guzaro

KYA. Transport theorem ko $A = L$ ke saath apply karo, phir Step 4 ke Newton's rotation law mein substitute karo: $τ = (\frac{d L}{d t})_{body} + ω \times L .$

KYUN. Yeh ek line mein poora trick hai. Left side lab law hai. Right par, pehla term ab bug ke frame mein measure kiya gaya hai, jahan $L = I ω$ constant $I_{i}$ ke saath — toh yeh cleanly differentiate hoga. Doosra term unavoidable rotating-frame surcharge hai.

PICTURE. Ek flow diagram: lab rate "bug rate" + "gyroscopic swing" mein split hota hai; neeche, $L$ (cyan) aur uski swing $ω \times L$ (amber) arrows ke roop mein draw kiye gaye hain.

Step 7 — Teen components read off karo

KYA. Step 6 ko teen principal axes ke along likhte hain. Do ingredients:

Body frame mein, $L_{i} = I_{i} ω_{i}$ har $I_{i}$ constant ke saath, toh $(\frac{d L _{i}}{d t})_{body} = I_{i} \overset{ω}{˙}_{i},$ jahan $\overset{ω}{˙}_{i}$ (omega with a dot) matlab hai "number $ω_{i}$ kitni fast change ho raha hai."
Cross product, pehla component: $(ω \times L)_{1} = ω_{2} L_{3} - ω_{3} L_{2} = ω_{2} (I_{3} ω_{3}) - ω_{3} (I_{2} ω_{2}) = (I_{3} - I_{2}) ω_{2} ω_{3} .$

$ω_{2} L_{3} - ω_{3} L_{2}$ KYUN? Yeh ek cross product ke pehle slot ki definition hai: yeh doosre do axes (2 aur 3) ko mix karta hai, kabhi apna nahi. $1 \to 2 \to 3 \to 1$ cycle karne se baaki do slots milte hain.

PICTURE. Cyclic wheel $1 \to 2 \to 3 \to 1$ jo dikhata hai kaunse indices har component ko feed karte hain, sign pattern ke saath.

Assemble karte hue — $τ = (d L / d t)_{body} + ω \times L$ ka har component saath mein rakhte hain:

I_1\dot\omega_1 - (I_2-I_3)\,\omega_2\omega_3 &= \tau_1\\ I_2\dot\omega_2 - (I_3-I_1)\,\omega_3\omega_1 &= \tau_2\\ I_3\dot\omega_3 - (I_1-I_2)\,\omega_1\omega_2 &= \tau_3 \end{aligned}}$$ > [!formula] Euler's equations, term by term (equation 1 lo) > $$\underbrace{I_1}_{\text{axis 1 ke baare mein reluctance}}\ \underbrace{\dot\omega_1}_{\text{axis-1 spin kitni fast change hoti hai}}\ -\ \underbrace{(I_2-I_3)}_{\text{shape asymmetry}}\ \underbrace{\omega_2\omega_3}_{\text{doosre do spins, multiplied}}\ =\ \underbrace{\tau_1}_{\text{axis 1 ke baare mein applied twist}}$$ > Move kiya gaya cross-product piece $-(I_2-I_3)\omega_2\omega_3$ ban gaya (equals sign cross karte waqt sign flip hota hai). Hum is box ko page ke top se dhundh rahe the — yeh ab fully earned hai. --- ## Step 8 — Degenerate & limiting cases (koi gap mat chhodho) **KYA & KYUN.** Ek derivation jis par tum trust kar sako usse apne edge cases survive karne chahiye. Yahan sab kuch hai. **Case A — sphere, sab equal ($I_1=I_2=I_3$).** Har difference $I_i-I_j=0$, toh sab coupling terms mar jaate hain: $$I\dot\omega_i = \tau_i.$$ Koi torque nahi, $\boldsymbol\omega$ frozen hai — **koi bhi** axis ek stable spin axis hai. Yeh ek case hai jahan naive "$\boldsymbol\tau=\mathbf I\dot{\boldsymbol\omega}$" actually sahi hai. **Case B — exactly ek principal axis par spin, koi torque nahi.** Maano $\omega_2=\omega_3=0$. Tab equation 1 deta hai $I_1\dot\omega_1 = 0$: spin hamesha ke liye hold karta hai. Ek body jo perfectly ek principal axis ke baare mein spin ki gayi hai woh kabhi nahi bhatak ti. **Case C — symmetric top, torque-free ($I_1=I_2\ne I_3$, $\boldsymbol\tau=0$).** Equation 3 apna coupling kho deta hai ($I_1-I_2=0$), toh $\omega_3=$ const. Equations 1,2 ban jaate hain, constant $\Omega \equiv \dfrac{I_3-I_1}{I_1}\omega_3$ ke saath, $$\dot\omega_1 = -\Omega\,\omega_2,\qquad \dot\omega_2 = +\Omega\,\omega_1,$$ jinka solution $\omega_1=A\cos\Omega t,\ \omega_2=A\sin\Omega t$ transverse spin ko symmetry axis ke around rate $\Omega$ par **circle** karta hai — free precession, Earth ka [[Chandler wobble]]. **Case D — intermediate-axis (tennis-racket) instability.** $I_1<I_2<I_3$ ke liye, middle axis 2 ke baare mein nearly spin karo aur linearize karo: chhote deviations $\ddot q = +k^2 q$ follow karte hain jahan $k^2>0$, toh woh **exponentially grow** karte hain — flipping-phone effect. Largest ya smallest axis ke baare mein sign flip ho jaata hai $-k^2$ → stable oscillation. **PICTURE.** Char mini-panels, ek har case ke liye: sphere (arrows kahin bhi), locked principal spin, circling transverse spin, aur tumbling middle-axis flip. ![[deepdives/dd-physics-2.1.21-d2-s08.png]] > [!mistake] Equals sign cross karte waqt sign bhoolna mat > $\boldsymbol\omega\times\mathbf L$ *right* par $+(I_3-I_2)\omega_2\omega_3$ ke saath baitha hai. Ise left par $I_1\dot\omega_1$ ke saath move karne par woh $-(I_2-I_3)\omega_2\omega_3$ ban jaata hai. Dono boxed forms equal hain — bas dekho term kaun si side par hai. --- ## Ek-picture summary ![[deepdives/dd-physics-2.1.21-d2-s09.png]] Sab kuch ek blueprint par: spin arrow $\boldsymbol\omega$ (Step 1) principal axes par inertia machine ko feed karta hai $\mathbf L$ banane ke liye (Step 3); lab law $\boldsymbol\tau=(d\mathbf L/dt)_{\text{space}}$ (Step 4) transport bridge se guzarta hai (Step 5–6) aur bug ki clean rate plus gyroscopic swing mein split ho jaata hai (Step 7), Euler's boxed equations par land karta hai. > [!recall]- Feynman retelling — poora walk plain words mein > Socho ek brick space mein tumble kar rahi hai. Uski spin describe karne ke liye hum ek arrow, $\boldsymbol\omega$, use karte hain: yeh axis ke along point karta hai aur uski length batati hai kitni fast (Step 1). Ab, do log dekhte hain: tum lab mein, aur ek bug brick par chipka hua (Step 2). Bug ki life easier hai, kyunki bug ke liye brick ki "har direction mein heaviness" — inertia — kabhi nahi badalti, aur teen special axes ke along yeh sirf teen plain numbers $I_1,I_2,I_3$ ban jaati hai jo spin ko stored spinning-motion $\mathbf L$ mein turn karti hain (Step 3). Rotation ki rulebook kehti hai ek twist $\boldsymbol\tau$ $\mathbf L$ ko badalta hai — lekin woh rule sirf lab observer ke liye kaam karta hai (Step 4). Rule ko bug ki desk par laane ke liye hume ek bridge chahiye: jab bhi tum spinning viewpoint mein switch karte ho, kisi bhi arrow ko extra swing $\boldsymbol\omega\times\mathbf L$ milti hai, kyunki poora frame uske neeche ghoom raha hai (Step 5). $\mathbf L$ ko us bridge ke across push karo (Step 6), teen axes read off karo is fact use karke ki cross product hamesha *doosre do* directions ko mix karta hai (Step 7), aur Euler's teen equations nikal aate hain. Unke andar surprise: chahe koi push na kare, multiplied spins $\omega_j\omega_k$ times shape-difference $I_j-I_k$ spin ko badalta rehta hai — yahi reason hai ki ek tumbling phone apne middle axis ke baare mein flip karta hai lekin ends ke baare mein nahi (Step 8). > [!recall]- Quick self-check > Body frame algebra ko clean kyun karta hai? ::: Kyunki wahan inertia tensor constant hai (diagonal, $I_i$ fixed), lab ke unlike jahan yeh har instant badalti hai. > $\boldsymbol\omega\times\mathbf L$ term kahan se aata hai? ::: Transport theorem — rotating frame mein fixed ek vector phir bhi lab observer ke liye swing karta hai, aur woh swing $\boldsymbol\omega\times$ vector hai. > Sab coupling terms ko kya khatam karta hai? ::: Equal moments of inertia ($I_1=I_2=I_3$, ek sphere) — har $I_i-I_j=0$. > Equation 1 mein, $(I_2-I_3)\omega_2\omega_3$ physically kya represent karta hai? ::: Gyroscopic coupling — axes 2 aur 3 ke baare mein spins axis 1 ke baare mein change drive kar rahe hain, zero torque ke saath bhi present hai.