Question bank — Torque-free rotation — Euler's equations, asymmetric top

False. Space mein jo conserve hota hai woh $\vec{L}$ hai, $\vec{\omega }$ nahi; kyunki $\vec{L}=\mathbf{I}\vec{\omega }$ hai aur body ki orientation lab mein $\mathbf{I}$ ko change karti hai, isliye $\vec{\omega }$ generally move karta hai jabki $\vec{L}$ frozen rehta hai.

False. $\vec{L}$ space frame mein constant hota hai; body frame mein woh move karta hai kyunki body ke axes uske neeche rotate karte hain — yahi cheez Euler's equations mein $\vec{\omega }\times \vec{L}$ track karta hai.

False. Yeh $\omega$-space mein do alag ellipsoids hain (semi-axes $1/\sqrt{I_i}$ versus $1/I_i$ ki tarah scale hote hain); real motion unke intersection, yani polhode, par rehti hai.

False. Har Euler right-hand side mein $(I_2-I_3)$ jaisa ek factor hota hai; jab sab moments equal hoon toh yeh vanish ho jaate hain, isliye $\dot{\vec{\omega }}=0$ aur $\vec{\omega }$ body mein frozen hai — koi wobble nahi.

True. $\Omega$ unperturbed spin rate ke saath, linearized growth constant $k=\frac{(I_3-I_2)(I_2-I_1)}{I_1{I}_3}{\Omega }^2$ do positive factors ka product hai, isliye $k>0$ exponential growth deta hai (Dzhanibekov / tennis-racket effect).

True (diagonal form mein jo di gayi hai). Principal axes ke bahar $\mathbf{I}$ diagonal nahi hai, $L_i\ne I_i{\omega }_i$, aur messy off-diagonal terms aate hain; clean cyclic form ke liye principal axes zaroori hain.

Causal claim ke roop mein False. $T$ aur $\vec{L}$ dono independently $\vec{N}=0$ se follow karte hain; energy conservation directly Euler's equations ko $\dot{T}$ mein substitute karke prove hoti hai, bina kabhi $\vec{L}$ conservation invoke kiye.

True. Euler-3 padhta hai $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0$ jab $I_1=I_2$ ho, isliye ${\omega }_3$ frozen hai aur transverse part uske around precess karta hai.

False. Yeh ek linear instability hai: koi bhi nonzero perturbation, chahe kitni bhi choti ho, exponentially $e^{\sqrt{k}t}$ ki tarah grow karti hai jab tak dominate na kar le.

Galti $\dot{\vec{L}}=0$ se $\dot{\vec{\omega }}=0$ ki taraf jump karne mein hai. Yeh tabhi equal hote hain jab $\mathbf{I}$ constant ho aur $\vec{L}\parallel \vec{\omega }$ ho; non-spherical body ke liye dono nahi hold hote, isliye $\vec{\omega }$ move karta hai.

Right-hand side ke terms $(I_i-I_j){\omega }_j{\omega }_k$ $\vec{\omega }\times \vec{L}$ transport term se internal geometric coupling hain, external work nahi. $\vec{N}=0$ ke saath system phir bhi torque-free hai.

Cyclic symmetry sirf labels move karti hai, lekin stability $(I_a-I_b)(I_b-I_c)$ ke sign par depend karti hai, jo sirf intermediate axis ke liye flip hoti hai. Form ki symmetry ≠ behaviour ki symmetry.

Fixed magnitudes par direction badalna koi energy nahi leta. Concretely $\dot{T}=\sum I_i{\omega }_i{\dot{\omega }}_i$; Euler's equations substitute karne par milta hai ${\omega }_1{\omega }_2{\omega }_3[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)]$, aur bracket term-by-term cancel ho jaata hai — $+I_2-I_3+I_3-I_1+I_1-I_2=0$ — isliye $\dot{T}=0$ exactly.

Nahi — ek symmetric top ki free precession torque-free hoti hai; transverse moment $I_{\perp }\equiv I_1=I_2$ ke saath, spin ${\omega }_3$ constant rehta hai aur transverse spin body rate ${\Omega }_b=\frac{I_3-I_{\perp }}{I_{\perp }}{\omega }_3$ par circle karta hai, poori tarah Earth ki apni asymmetry se driven.

Ulta hai — axis-3 spin ka matlab hai ${\omega }_3\approx \Omega$ (unperturbed spin rate) woh bada term hai jise hum constant rakhte hain, jabki ${\omega }_1,{\omega }_2$ woh small perturbations hain jinhe hum track karte hain.

Kyunki body frame mein inertia tensor ek constant diagonal $\mathrm{diag}(I_1,I_2,I_3)$ hota hai; lab mein yeh har instant badalta hai jaise body reorient karti hai, jo $\dot{\vec{L}}=\mathbf{I}\dot{\vec{\omega }}+\dot{\mathbf{I}}\vec{\omega }$ ko intractable bana deta.

Ek vector do reasons se change ho sakta hai: woh sach mein change hota hai (body-frame derivative) ya humare rotating axes ek fixed vector ke past sweep karte hain, use change hota dikhaate hain; $\vec{\omega }\times \vec{A}$ woh apparent change hai. Dekho Angular momentum in rotating frames.

Coupling $\vec{\omega }\times \vec{L}$ se aati hai, jo do vectors ka cross product hai jinmein se har ek $\omega$ mein linear hai; result quadratic hai, isliye har equation baaki do components ke ek product se fed hoti hai.

Do ellipsoids ke intersection par, max aur min inertia axes ke paas polhodes chote closed loops hain, isliye $\vec{\omega }$ nearby orbit karta hai — ek bounded, oscillatory (stable) motion, runaway growth nahi. Dekho Stability analysis and linearization.

Intermediate-inertia axis ke around spinning $k>0$ deta hai, isliye infinitesimal wobble exponentially grow karta hai; phone rapidly orientation swap karta hai — Dzhanibekov effect tumhare haath mein.

Kyunki axis-3 spin ke paas do small equations ${\omega }_1,{\omega }_2$ mein linear hain: ${\dot{\omega }}_1=a{\omega }_2$ aur ${\dot{\omega }}_2=b{\omega }_1$ constant coefficients $a,b$ ke saath (constant kyunki ${\omega }_3\approx \Omega$ fixed rakha jaata hai aur ${\omega }_1{\omega }_2$, second-order-small hone ki wajah se, drop ho jaata hai). ${\dot{\omega }}_1=a{\omega }_2$ differentiate karne par milta hai ${\ddot{\omega }}_1=a{\dot{\omega }}_2=ab{\omega }_1$, isliye ${\omega }_2$ eliminate ho jaata hai aur $k=ab$. Elimination tabhi kaam karti hai jab coefficients constants hon — yeh $\Omega$ ke (first order tak) fixed hone par rely karta hai.

$\vec{L}=\mathbf{I}\vec{\omega }$, $\vec{\omega }$ ke parallel tabhi hota hai jab $\vec{\omega }$ ek principal axis ke along ho ($\mathbf{I}$ ka eigenvector); ek generic $\vec{\omega }$ alag $I_i$ wale axes mix karta hai, $\vec{L}$ ko away tilt kar deta hai.

Har right-hand side vanish ho jaata hai, isliye $\dot{\vec{\omega }}=0$: spin arrow body mein frozen hai aur (kyunki $\vec{L}\parallel \vec{\omega }$ hamesha) space mein bhi frozen hai.

Woh wahan hamesha ke liye ruk jaati hai — ek principal axis Euler's equations ka fixed point hai (saare products ${\omega }_j{\omega }_k$ jo ise kick karte zero hain), stable ya unstable axis ke hisaab se.

Yeh do curves mein degenerate ho jaata hai jo unstable axis par cross karte hain; ek body jo exactly us par launch ho woh asymptotically middle axis ke paas pahunch jaayegi, lekin koi bhi deviation use around bhej deta hai, jo do stable families ke beech boundary mark karta hai.

Growth constant $k\propto (I_2-I_1)\to 0$, isliye instability kamzor hoti jaati hai aur vanish ho jaati hai; exact symmetry par koi unique "intermediate" axis nahi hoti aur motion steady free precession ban jaati hai.

Tab $k\propto {\Omega }^2=0$: koi coupling nahi, koi wobble nahi, koi instability nahi — rest par ek body rest par hi rehti hai, $\dot{\vec{\omega }}=0$ ke consistent.

Body precession rate ${\Omega }_b\propto {\omega }_3=0$, isliye ${\omega }_1,{\omega }_2$ frozen hain: body simply ek fixed transverse axis ke baare mein bina kisi precession ke steadily spin karti hai. Dekho Symmetric top and gyroscopic precession.

Space ki rotational invariance — Noether's theorem se, space ke rotations ke under symmetry angular momentum $\vec{L}$ ka conservation guarantee karti hai.