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

2.1.23 · D2 · Physics › Analytical Mechanics › Torque-free rotation — Euler's equations, asymmetric top

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Step 1 — "Spin arrow" $\vec{\omega }$ kya hota hai?

KYA HAI. Jab koi cheez rotate karti hai, toh har instant mein woh space mein ek line ke baare mein turn karti hai — ek axis. Hum us axis ke saath ek arrow draw karte hain. Uski length yeh batati hai ki hum kitni tezi se spin kar rahe hain (radians per second), uski direction woh axis hai, right-hand rule se pointed (ungli us direction mein curl karo jis direction mein yeh turn karta hai, thumb arrow deta hai). Hum is arrow ko $\vec{\omega }$ ("omega") kehte hain. Upar wala chota arrow sirf yeh batata hai ki "yeh ek direction waali cheez hai," na ki sirf ek number.

YEH OBJECT KYO. Ek rotation ko teen facts chahiye: kaun sa axis, kitni tezi se, kaun sa sense. Ek arrow in teeno ko pack karta hai. Isliye hum ek angle ki jagah ek vector use karte hain.

PICTURE. Neeche, ek flat spinning disc. Purple arrow $\vec{\omega }$ seedha disc ke upar point kar raha hai. Dono baar zyada tezi se spin karo → arrow dono baar itna lamba.

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Step 2 — Body ke paas teen stiffness numbers hote hain: principal moments

KYA HAI. Ek rigid body ko spin karne mein resistance hoti hai. Kitni resistance hai yeh depend karta hai ki tum kis axis ke baare mein spin kar rahe ho. Ek achhe shape wale body ke liye teen khaas perpendicular axes hote hain — principal axes — teen "resistance-to-spin" numbers $I_1,I_2,I_3$ ke saath, jo principal moments of inertia hain. Bada $I$ = mass us axis se door spread hua hai = spin up karna mushkil hai.

YEH AXES KYO. Kisi bhi doosre axis ke saath resistance teeno ko mix kar deti hai (ek ugly $3\times 3$ table). Principal axes ke saath yeh cleanly alag ho jaata hai, ek axis mein ek number. Yahi wajah hai ki hum apna coordinate frame body se chipka lete hain — dekho Inertia tensor and principal axes.

PICTURE. Ek brick (matchbox) jisme teeno axes drawn hain: long-thin axis (chhota $I$), flat axis (bada $I$), aur beech wala axis. Arrows ko is tarah size diya gaya hai ki har $I$ ka hint mile.

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Step 3 — $\vec{L}$ aur $\vec{\omega }$ alag directions mein point karte hain

KYA HAI. $\vec{\omega }$ ke har component ko ek alag number $I_i$ se multiply karo aur result ek naya arrow $\vec{L}$ hai jo kahin aur point karta hai. Sirf tab jab saare $I_i$ equal hon (ek sphere) dono arrows ek line mein aate hain.

YEH KYO MATTER KARTA HAI. Empty space mein bina kisi torque ke, yeh $\vec{L}$ hai jo direction mein frozen rehta hai — $\vec{\omega }$ nahi. Toh agar $\vec{\omega }\ne \vec{L}$, toh $\vec{L}$ ko fixed rakhna $\vec{\omega }$ ko move karne par majboor karta hai. Yahi moving-$\vec{\omega }$ wobble hai. Yahi parent note mein Mistake A hai.

PICTURE. Wahi brick, ek aisi axis ke baare mein spinning jo teen special axon mein se ek nahi hai. Purple $\vec{\omega }$ aur magenta $\vec{L}$ alag ho jaate hain kyunki stretch factors $I_1,I_2,I_3$ alag hain.

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Step 4 — Ek arrow seem to change karne ke do reasons: transport theorem

KYA HAI. Hum apne axes tumbling body se chipkaye rakhte hain. Us spinning frame ke andar se koi bhi arrow $\vec{A}$ dekhne par, woh do alag reasons se change hota dikh sakta hai.

KYO. Reason ek: arrow genuinely badal raha hai (bada ho raha hai, tilt ho raha hai). Reason do: chahe arrow real space mein frozen ho, hamare apne axes uske paas se guzar jaate hain, toh lagta hai ki woh move hua. Dono ko add karna padega.

PICTURE. Ek frozen arrow (grey) room mein still baitha hai. Hamare rotating axes (violet) guzar jaate hain. Hamaari jagah se arrow peeche swing karta dikh raha hai — yahi apparent swing $\vec{\omega }\times \vec{A}$ term hai.

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Step 5 — $\vec{L}$ ko theorem mein feed karo

KYA HAI. Rotation ke liye space-frame Newton kehta hai ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{space}}=\vec{N}$, external torque. Torque-free ka matlab hai $\vec{N}=\vec{0}$. Ab Step 4 use karke awkward space-derivative ko body-derivative se swap karo.

KYO. Body frame mein stiffness numbers $I_i$ constant hain, isliye ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{body}}=(I_1{\dot{\omega }}_1,I_2{\dot{\omega }}_2,I_3{\dot{\omega }}_3)$ — clean. Saari mess tidy cross-product term mein chali jaati hai.

PICTURE. Ek flow box: real-space law → transport apply karo → $\vec{N}=0$ set karo → body-frame law jo components ko couple karta hai.

$\underset{\text{body-frame change of }\vec{L}}{\underbrace{(I_1{\dot{\omega }}_1,I_2{\dot{\omega }}_2,I_3{\dot{\omega }}_3)}}+\underset{\text{frame-swing term}}{\underbrace{\vec{\omega }\times \vec{L}}}=\underset{\text{no torque}}{\underbrace{\vec{0}}}$

Har dotted symbol ${\dot{\omega }}_i$ ka matlab hai "us spin component ke change ki rate." Ab hume sirf cross product chahiye.

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Step 6 — Cross product, term by term work out karo

KYA HAI. Cross product $\vec{\omega }\times \vec{L}$ ke teen components hain. $\vec{\omega }=({\omega }_1,{\omega }_2,{\omega }_3)$ aur $\vec{L}=(I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3)$ use karte hue, pehla component hai $(\vec{\omega }\times \vec{L}{)}_1=\underset{{\omega }_2{I}_3{\omega }_3}{\underbrace{{\omega }_2{L}_3}}-\underset{{\omega }_3{I}_2{\omega }_2}{\underbrace{{\omega }_3{L}_2}}=(I_3-I_2){\omega }_2{\omega }_3.$

KYO. Cross product ke first slot mein hamesha doosre do slots (2 aur 3) use hote hain, khud nahi. Isliye axis 1 ka fate ${\omega }_2$ aur ${\omega }_3$ se decide hota hai — yahi coupling hai. Difference $I_3-I_2$ isliye appear karta hai kyunki dono terms mein ${\omega }_2{\omega }_3$ hai; subtract karne par sirf stiffness ka gap bachta hai.

PICTURE. $\vec{\omega }\times \vec{L}$ ke liye determinant grid, jisme row-1 entries highlight hain aur common ${\omega }_2{\omega }_3$ ka cancellation dikhaya gaya hai.

Us term ko right side par move karne par ($I_1{\dot{\omega }}_1=-(\vec{\omega }\times \vec{L}{)}_1=(I_2-I_3){\omega }_2{\omega }_3$) aur $1\to 2\to 3\to 1$ cycle karne par:

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Step 7 — Degenerate cases: sphere aur symmetric top

KYA HAI. Simple bodies par machine test karo.

• Sphere $I_1=I_2=I_3$: har right side mein $(I_a-I_b)=0$ factor hai, isliye ${\dot{\omega }}_i=0$. Spin arrow body mein frozen hai — kabhi wobble nahi.
• Symmetric top $I_1=I_2=I_{\perp }\ne I_3$: Euler-3 deta hai $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0$, isliye ${\omega }_3$ constant hai. Doosre do obey karte hain ${\dot{\omega }}_1=-{\Omega }_b{\omega }_2,{\dot{\omega }}_2=+{\Omega }_b{\omega }_1$ with ${\Omega }_b=\frac{I_3-I_{\perp }}{I_{\perp }}{\omega }_3$ — pure circular motion, isliye $\vec{\omega }$ axis 3 ke baare mein ek cone sweep karta hai.

YEH KYO DIKHATE HAIN. Ek formula jise tum easy cases par tod nahi sakte, woh untrustworthy hai. Yeh Symmetric top and gyroscopic precession se bhi connect hote hain (free precession, Earth's Chandler wobble).

PICTURE. Left: sphere, arrows sab RHS par zero, $\vec{\omega }$ still. Right: symmetric top, $({\omega }_1,{\omega }_2)$ ek circle trace kar raha hai isliye $\vec{\omega }$ ek cone trace karta hai.

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Step 8 — Intermediate-axis instability, ek sign mein

KYA HAI. Asymmetric top $I_1<I_2<I_3$. Kareeb-kareeb ek axis ke baare mein spin karo, uska bada spin $\Omega$ raho, aur doosre do ko tiny rehne do. Do Euler equations ko combine karne par ek single equation milti hai ${\ddot{\omega }}_1=k{\omega }_1$ with $k=\frac{(I_2-I_3)(I_3-I_1)}{I_1{I}_2}{\Omega }^2(\text{for spin about axis 3}).$

KYO. $k$ ka sign destiny hai. Agar $k<0$ toh solution $\cos ,\sin$ hai — ek bounded wobble (stable). Agar $k>0$ toh solution $e^{\pm \sqrt{k}t}$ hai — yeh grow karta hai (unstable).

• Axis 3 (sabse bada): $(I_2-I_3)<0$, $(I_3-I_1)>0$ → $k<0$ → stable.
• Axis 1 (sabse chhota): dono gaps negative → product positive → $k<0$ → stable.
• Axis 2 (middle): $(I_3-I_2)>0$ aur $(I_2-I_1)>0$ → $k>0$ → unstable — yeh tumble karta hai.

PICTURE. Teen phase pictures: closed loops (stable) axes 1 aur 3 ke liye, aur ek saddle (runaway) axis 2 ke liye — Dzhanibekov / tennis-racket effect. Dekho Stability analysis and linearization.

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Ek-picture summary

Do frozen scalars — energy $2T=I_1{\omega }_1^2+I_2{\omega }_2^2+I_3{\omega }_3^2$ aur $L^2=I_1^2{\omega }_1^2+I_2^2{\omega }_2^2+I_3^2{\omega }_3^2$ — har ek $\omega$-space mein ek surface kaatta hai (do nested ellipsoids). Spin arrow ka tip unke intersection curve par trapped hai, jo polhode hai. Long/short axes ke paas intersections chote closed loops hain (stable wobble); middle axis ke paas woh do great saddle-curves ban jaate hain jo $\vec{\omega }$ ko poora carry kar lete hain (the tumble). Ek figure poori story rakhta hai.

Recall Poore walkthrough ki Feynman retelling

Ek spinning cheez ek "spin arrow" carry karti hai jo apne axis ke saath point karta hai (Step 1). Kyunki object ko ek taraf spin karna doosri taraf se zyada heavy hai, teen khaas stiffness numbers hote hain (Step 2), aur spin arrow ko in unequal numbers se multiply karne par ek doosra arrow milta hai — momentum arrow — jo alag direction mein point karta hai (Step 3). Space mein, bina kuch push kiye, yeh momentum arrow hi frozen rehna chahiye. Lekin kyunki woh spin arrow ke saath line up nahi karta, momentum ko still rakhna spin arrow ko swing karte rehne par majboor karta hai. Bookkeeping karne ke liye hum tumbling body par baith jaate hain; us seat se ek arrow move karta dikh sakta hai dono isliye ki woh genuinely move karta hai aur isliye ki hamaari seat turn kar rahi hai (Step 4). Us language mein "no push" likhna (Step 5) aur cross product grind karna (Step 6) Euler ke teen equations deta hai: har axis ka spin-up = doosre do stiffnesses ka gap $\times$ doosre do spins. Ball par koi gap nahi, isliye woh kabhi wobble nahi karta; symmetric top par spin arrow sirf ek cone mein circle karta hai (Step 7). Aur ek lopsided body ke liye, ek chota sign sab decide karta hai: sabse bade ya sabse chhote axis ke baare mein spin karo aur disturbances harmlessly circle karti hain, lekin middle axis ke baare mein spin karo aur woh blow up karti hain — phone khud-ba-khud end over end flip ho jaata hai (Step 8). Yeh sab do energy-and-momentum ellipsoids ke crossing par live karta hai (summary).