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

2.1.23 · Physics › Analytical Mechanics

1. Setup: hum body frame mein kyun jaate hain

Body frame kyun? Lab (space) frame mein, $\mathbf{I}$ har pal badalta rehta hai kyunki body re-orient hoti rehti hai. Yeh ek nightmare hai. Agar hum axes body mein fixed rakhen uske principal axes ke saath, toh $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$ constant ho jaata hai. Iska price: body frame rotate karta hai, isliye time derivatives mein ek extra term aa jaata hai.

2. Euler's equations scratch se derive karna

Inertial frame mein rotation ke liye Newton/Euler se shuru karo: ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{space}}=\vec{N}(\text{external torque})$

$\vec{L}$ par transport theorem apply karo: ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{body}}+\vec{\omega }\times \vec{L}=\vec{N}$

Yeh step kyun? Yeh ek mushkil space-frame derivative ko body-frame derivative (jahan $\mathbf{I}$ constant hai) mein convert karta hai, plus ek correction.

Body principal frame mein $L_i=I_i{\omega }_i$, aur kyunki $I_i$ constants hain: ${\left(\frac{d\vec{L}}{dt}\right)}_{\text{body}}=(I_1{\dot{\omega }}_1,I_2{\dot{\omega }}_2,I_3{\dot{\omega }}_3)$

Ab cross product component-wise compute karo. $\vec{\omega }=({\omega }_1,{\omega }_2,{\omega }_3)$, $\vec{L}=(I_1{\omega }_1,I_2{\omega }_2,I_3{\omega }_3)$ ke saath: $(\vec{\omega }\times \vec{L}{)}_1={\omega }_2{L}_3-{\omega }_3{L}_2=(I_3-I_2){\omega }_2{\omega }_3$

Sab jod ke aur $\vec{N}=0$ set karke (torque-free):

3. Do conserved quantities (constants of motion)

Bina external torque ke bhi, body frame mein $\vec{\omega }$ move karta hai. Lekin do scalars conserved rehte hain:

Energy conservation ka derivation: $\frac{dT}{dt}=I_1{\omega }_1{\dot{\omega }}_1+I_2{\omega }_2{\dot{\omega }}_2+I_3{\omega }_3{\dot{\omega }}_3$. Euler's equations substitute karo: $={\omega }_1{\omega }_2{\omega }_3[(I_2-I_3)+(I_3-I_1)+(I_1-I_2)]=0.$ Yeh step kyun? Bracket telescope karke zero ho jaata hai — energy conserved hai bina kisi external work ke, bilkul expect ke mutabik.

Do surfaces kyun? $\omega$-space mein, $2T=$const ek ellipsoid hai (energy ellipsoid) aur $L^2=$const ek aur ellipsoid hai. $\vec{\omega }$ ki actual motion unke intersection curves par hoti hai — yahi polhodes hain.

4. Rotation ki stability — tennis-racket theorem

Asymmetric top lo: $I_1<I_2<I_3$ (sab alag). Ek principal axis ke aas-paas spin karo aur pucho: kya ek tiny disturbance badhti hai ya choti rehti hai?

Axis 3 ke baare mein spin (sabse bada $I$): Maano ${\omega }_3\approx \Omega$ (bada), ${\omega }_1,{\omega }_2$ bahut chote hain. Euler 1 aur 2 se, differentiate karo aur substitute karo: ${\dot{\omega }}_1=\frac{I_2-I_3}{I_1}\Omega {\omega }_2,{\dot{\omega }}_2=\frac{I_3-I_1}{I_2}\Omega {\omega }_1$ $\Rightarrow {\ddot{\omega }}_1=\frac{(I_2-I_3)(I_3-I_1)}{I_1{I}_2}{\Omega }^2{\omega }_1$ Yeh step kyun? Pehle ko differentiate karke doosre ko plug in karne se ek single 2nd-order ODE milti hai ${\ddot{\omega }}_1=k{\omega }_1$.

• Axis 3 ke liye (sabse bada): $(I_2-I_3)<0$, $(I_3-I_1)>0$ → $k<0$ → oscillation (stable). ✅
• Axis 1 ke liye (sabse chota): dono factors $k<0$ dete hain → stable. ✅
• Axis 2 ke liye (intermediate): $(I_3-I_2)>0$ aur $(I_2-I_1)>0$ is tarah arrange hote hain ki $k>0$ → exponential growth → UNSTABLE. ❌

5. Worked examples

Flashcards

Connections

• Inertia tensor and principal axes — jahan se $\mathbf{I}=\mathrm{diag}(I_1,I_2,I_3)$ aata hai.
• Angular momentum in rotating frames — transport theorem ka source.
• Rigid body dynamics — Euler angles — body $\vec{\omega }$ ko lab orientation mein convert karna.
• Symmetric top and gyroscopic precession — $I_1=I_2$ special case (torque ke saath).
• Stability analysis and linearization — $\ddot{x}=kx$ method yahan reuse hota hai.
• Noether's theorem — kyun $\vec{L}$ conservation free space ki rotational symmetry se follow karta hai.

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