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

2.1.21 · D3 · Physics › Analytical Mechanics › Rigid body dynamics — Euler angles, Euler's equations of mot

Shuru karne se pehle, ek line ka reminder us tool ka jo hum poore time use karenge.

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The scenario matrix

Is topic ki har problem inhi cells mein se ek hoti hai. Last column batata hai ki kaun sa worked example usse solve karta hai.

#	Case class	Kya badalta hai	Example
A	Symmetric top, torque-free ($I_1=I_2\ne I_3$, $\mathbf{\tau }=0$)	body-frame precession, $I_3-I_1$ ka sign	Ex 1, Ex 2
B	Fully asymmetric, torque-free ($I_1<I_2<I_3$)	har axis ki stability, sign flips	Ex 3
C	Degenerate: sphere ($I_1=I_2=I_3$)	saari coupling khatam	Ex 4
D	Zero input: exactly principal axis par spin	$\dot{\mathbf{\omega }}=0$, koi wander nahi	Ex 4
E	Torque driven, single axis (${\tau }_3\ne 0$)	pure angular acceleration	Ex 5
F	Gyroscopic / steady precession (real-world word problem)	torque vs $\mathbf{\omega }\times \mathbf{L}$ ka balance	Ex 6
G	Euler-angle rates → $\mathbf{\omega }$ (projection trap)	non-orthogonal axes, $\sin \theta ,\cos \theta$	Ex 7
H	Exam twist: prolate vs oblate sign ($I_3>I_1$ vs $I_3<I_1$)	precession ki direction ulat jaati hai	Ex 8
I	Limiting value (fast-spin limit)	kaun sa term dominate karta hai	Ex 2 (part b)

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Ex 1 — Cell A: torque-free symmetric top, basic precession rate

Step 1. Equation 3 ko ${\tau }_3=0$ aur $I_1=I_2$ ke saath likhte hain: $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2=0\Rightarrow {\omega }_3=\text{const}.$ Yeh step kyun? Agar spin-axis rate constant nahi hoti, toh "uske baare mein precession" fixed-rate motion nahi hoti — pehle confirm karna zaroori hai ki ${\omega }_3$ frozen hai.

Step 2. Constant ${\omega }_3$ ko equations 1 aur 2 mein substitute karo. $I_1=I_2$ ke saath yeh ban jaate hain: ${\dot{\omega }}_1=-\Omega {\omega }_2,{\dot{\omega }}_2=+\Omega {\omega }_1,\Omega \equiv \frac{I_3-I_1}{I_1}{\omega }_3.$ Yeh step kyun? Yeh pair ek point ke equation hai jo angular rate $\Omega$ par circle mein ghoom raha hai — yahi wobble hai.

Step 3. Numbers daalo: $\Omega =\frac{4\times 10^{-3}-2\times 10^{-3}}{2\times 10^{-3}}\times 50=\frac{2\times 10^{-3}}{2\times 10^{-3}}\times 50=50\text{rad/s}.$

Figure dekho: transverse angular-velocity vector (pale-yellow arrow) ki tip body-3 axis ke baare mein blue circle trace karti hai rate $\Omega$ par, jabki constant ${\omega }_3$ (pink) seedha upar point karta hai.

Recall Forecast answer

${\omega }_3$ ke same sign mein (kyunki $I_3>I_1$, isliye $\Omega >0$), aur yahan yeh ${\omega }_3$ ke barabar hota hai kyunki exactly $I_3=2I_1$ hai. Thin disk ke liye yeh "$2\times$" relation disguise mein hai.

Verify: Units: $\frac{{\text{kg⋅m}}^2}{{\text{kg⋅m}}^2}\times \text{rad/s}=\text{rad/s}$. ✓ Sign: $I_3>I_1\Rightarrow \Omega >0$, ek oblate body, precession spin ke same sense mein. ✓

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Ex 2 — Cell A + Cell I: Earth ki Chandler-type wobble aur fast-spin limit

Step 1 (a). Ex 1 se, $\Omega =\frac{I_3-I_1}{I_1}{\omega }_3=\frac{1}{305}\times \frac{2\pi }{1\text{day}}$. Yeh step kyun? Same symmetric-top formula; sirf number $\frac{I_3-I_1}{I_1}$ badla.

Step 2 (a). Period $T=\frac{2\pi }{\Omega }=305\text{days}$. Yeh step kyun? Period hota hai $2\pi$ divided by rate; $2\pi$ cancel ho jaate hain, $305$ spin-periods = 305 days bachta hai.

Step 3 (b). Equation 1 with $\tau =0$: $I_1{\dot{\omega }}_1=(I_2-I_3){\omega }_2{\omega }_3$. Jab ${\omega }_3\to \infty$, right side (${\omega }_3$ mein linear) kisi bhi aisi term ko dwarf karta hai jo ${\omega }_3$ se multiply nahi hui. Yeh step kyun? Dominant term identify karne se pata chalta hai ki wobble frequency linearly spin ke saath scale karti hai — faster spin karo, faster wobble hoga.

Recall Real Chandler wobble par reality check

Observed wobble ~433 days hai, 305 nahi — kyunki Earth perfectly rigid nahi hai; yeh deform hoti hai. Dekho Chandler wobble. Rigid-body prediction bilkul humara 305-day answer hai, aur yeh discrepancy apne aap mein informative hai.

Verify: $T=305$ days from $2\pi /\Omega$. Units: $\Omega$ in rad/day $\Rightarrow T$ in days. ✓ Limit: right-hand side $\propto {\omega }_3$, baaki sab ${\omega }_3$ se independent. ✓

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Ex 3 — Cell B: fully asymmetric body, intermediate-axis (tennis-racket) test

Step 1. Axis 1 ke baare mein nearly spin: ${\omega }_1\approx$ const large, ${\omega }_2,{\omega }_3$ tiny. Equations 2 aur 3 ko linearize karo (do chhoti quantities ka product drop karo): $I_2{\dot{\omega }}_2=(I_3-I_1){\omega }_3{\omega }_1,I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2.$ Yeh step kyun? Small-perturbation analysis: sirf chhote components mein linear terms rakho yeh dekhne ke liye ki woh badhte hain ya oscillate karte hain.

Step 2. Pehle wale ko differentiate karo, doosre ko substitute karo: ${\ddot{\omega }}_2=\frac{(I_3-I_1)(I_1-I_2)}{I_2{I}_3}{\omega }_1^2{\omega }_2\equiv -k{\omega }_2.$ Yeh step kyun? Do first-order equations ko ek second-order equation mein reduce karne se $k$ ka sign milta hai: $k>0\Rightarrow$ oscillation (stable), $k<0\Rightarrow$ exponential growth (unstable).

Step 3. Har axis ke liye coefficient $C_i\equiv (I_j-I_i)(I_i-I_k)$ evaluate karo (bahar ${\omega }^2/(I_j{I}_k)>0$ factor hota hai, isliye sirf is product ka sign matter karta hai). Maano $k=-C_i\times (\text{positive})$, toh stable $\iff C_i<0$:

• Axis 1 (smallest): $(I_2-I_1)(I_1-I_3)=(1)(-2)=-2<0$ → stable.
• Axis 2 (middle): $(I_3-I_2)(I_2-I_1)=(1)(1)=+1>0$ → unstable.
• Axis 3 (largest): $(I_1-I_3)(I_3-I_2)=(-2)(1)=-2<0$ → stable.

Recall Forecast answer

Axis 2, intermediate axis. Yahi [tennis-racket / Dzhanibekov effect] hai — apna phone middle axis ke baare mein spin karake flip karo aur yeh palta jaata hai. Dekho bhi Angular momentum in rotating frames.

Verify: Products: axis1 $=-2$, axis2 $=+1$, axis3 $=-2$. Exactly ek positive → exactly ek unstable axis. ✓

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Ex 4 — Cells C & D: sphere, aur exactly principal axis par spin

Step 1 (C). Har coupling factor $(I_i-I_j)=0$ hai, isliye teeno equations padhti hain $I{\dot{\omega }}_i=0$. Yeh step kyun? Sphere ultimate degenerate case hai — koi bhi principal axis special nahi, isliye koi wobble direction select nahi ho sakti.

Step 2 (C). ${\dot{\omega }}_1={\dot{\omega }}_2={\dot{\omega }}_3=0\Rightarrow \mathbf{\omega }=$ const forever. Koi bhi axis stable spin axis hai.

Step 3 (D). ${\omega }_1={\omega }_2=0$ ke saath: equation 1 mein ${\omega }_2{\omega }_3=0$ hai, equation 2 mein ${\omega }_3{\omega }_1=0$ hai, equation 3 mein ${\omega }_1{\omega }_2=0$ hai. Isliye ${\dot{\omega }}_1={\dot{\omega }}_2={\dot{\omega }}_3=0$. Yeh step kyun? Yeh zero-input cell hai: pure principal-axis spin mein koi gyroscopic coupling nahi hoti kyunki baaki do components jo multiply hote woh zero hain. Yeh motion equations ka ek fixed point hai.

Verify: (C) saare $(I_i-I_j)=0$ isliye RHS $=0$. ✓ (D) har RHS product mein ${\omega }_1,{\omega }_2$ mein se kam se kam ek hai, dono zero hain, isliye RHS $=0$. ✓

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Ex 5 — Cell E: single-axis torque, pure angular acceleration

Step 1. Kyunki ${\omega }_1={\omega }_2=0$ hai, equation 3 apna coupling term kho deti hai: $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2+{\tau }_3={\tau }_3.$ Yeh step kyun? Hume coupling vanish hona check karna chahiye assume karne ki jagah — yahan ${\omega }_1{\omega }_2=0$ use khata hai, isliye hum ek clean $F=ma$-style law se bache hain.

Step 2. Integrate karo: ${\dot{\omega }}_3={\tau }_3/I_3=2/0.5=4{\text{rad/s}}^2$; zero start ke saath, ${\omega }_3=4t$. Yeh step kyun? Constant angular acceleration ek straight-line spin-up mein integrate hoti hai.

Step 3. $t=3$ par: ${\omega }_3=4\times 3=12\text{rad/s}$.

Recall Forecast answer

Koi coupling nahi — yeh woh ek case hai jahan naive $\tau =I\dot{\omega }$ sahi hai, precisely isliye kyunki hum principal axis par koi transverse rate ke bina spin kar rahe hain.

Verify: ${\omega }_3=12\text{rad/s}$. Units: $\frac{\text{N⋅m}}{{\text{kg⋅m}}^2}\cdot \text{s}=\frac{{\text{kg⋅m}}^2/{\text{s}}^2}{{\text{kg⋅m}}^2}\cdot \text{s}=\text{1/s}=\text{rad/s}$. ✓

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Ex 6 — Cell F: real-world gyroscope, steady precession

Step 1. Pivot ke baare mein gravity torque: $\tau =mgd$, horizontal, spin axis ke perpendicular. Yeh step kyun? Yahi torque hai jise steady precession mein $\mathbf{\omega }\times \mathbf{L}$ machinery ko balance karna hota hai.

Step 2. Steady precession ke liye spin angular momentum $L=I_3{\omega }_3$ rate ${\Omega }_p$ par turn karta hai, isliye $|\dot{\mathbf{L}}|={\Omega }_{p}L$. $\tau ={\Omega }_{p}L$ set karo: ${\Omega }_p=\frac{mgd}{I_3{\omega }_3}.$ Yeh step kyun? Yeh Euler ka $\mathbf{\tau }=\mathbf{\omega }\times \mathbf{L}$ balance disguise mein hai — torque wheel ko speed up nahi karta, yeh uski axis ko swing karta hai. Dekho Gyroscope precession.

Step 3. Numbers: ${\Omega }_p=\frac{0.4\times 9.8\times 0.05}{1.0\times 10^{-2}\times 200}=\frac{0.196}{2.0}=0.098\text{rad/s}.$

Recall Forecast answer

Slower. ${\Omega }_p\propto 1/{\omega }_3$: ek fast top lazily precess karta hai, ek dying top wildly — bilkul wahi jo aap real top slowdown hote waqt dekhte ho.

Verify: ${\Omega }_p=0.098\text{rad/s}$. Units: $\frac{\text{kg}\cdot {\text{m/s}}^2\cdot \text{m}}{{\text{kg⋅m}}^2\cdot \text{1/s}}=\frac{{\text{kg⋅m}}^2/{\text{s}}^2}{{\text{kg⋅m}}^2/\text{s}}=\text{1/s}$. ✓

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Ex 7 — Cell G: Euler-angle rates → $\mathbf{\omega }$ (projection trap)

Step 1. Parent note ke 3-1-3 projection formulas use karo: ${\omega }_1=\dot{\varphi }\sin \theta \sin \psi +\dot{\theta }\cos \psi ,{\omega }_2=\dot{\varphi }\sin \theta \cos \psi -\dot{\theta }\sin \psi ,{\omega }_3=\dot{\varphi }\cos \theta +\dot{\psi }.$ Yeh step kyun? Teen Euler axes ($Z$, line of nodes, body-$z$) perpendicular nahi hain, isliye hum sirf components read off nahi kar sakte — hume project karna padega, yahi kaam yeh $\sin \theta ,\cos \theta$ factors karte hain.

Step 2. $\theta =60^{\circ }$ daalo ($\sin =\frac{\sqrt{3}}{2}\approx 0.8660$, $\cos =\frac{1}{2}$), $\psi =0$ ($\sin \psi =0,\cos \psi =1$), $\dot{\theta }=0$: ${\omega }_1=3(0.8660)(0)+0=0,{\omega }_2=3(0.8660)(1)-0=2.598,{\omega }_3=3(0.5)+5=\mathrm{6.5.}$

Step 3. Toh $\mathbf{\omega }=(0,2.598,6.5)\text{rad/s}$.

Recall Forecast answer

Nahi — ${\omega }_3=6.5$ hai, 5 nahi. Precession $\dot{\varphi }$ body-3 axis mein $\dot{\varphi }\cos \theta =3\times 0.5=1.5$ term ke through leak karta hai. Rates ko components ki tarah treat karna parent note ka [!mistake]-2 hai.

Verify: $({\omega }_1,{\omega }_2,{\omega }_3)=(0,3\sqrt{3}/2,6.5)$; $3\sqrt{3}/2\approx 2.598$. ✓ Kinetic energy mein yeh kahan feed hote hain, uske liye dekho Lagrangian of the symmetric top.

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Ex 8 — Cell H: exam twist, prolate vs oblate precession ka sense flip karta hai

Step 1 (oblate). $\Omega =\frac{1.5-1}{1}\times 10=+5\text{rad/s}$. Yeh step kyun? $I_3>I_1$ factor ko positive banata hai — precession spin ke same sense mein (frisbee ki tarah).

Step 2 (prolate). $\Omega =\frac{0.6-1}{1}\times 10=-4\text{rad/s}$. Yeh step kyun? $I_3<I_1$ factor ko negative banata hai — precession spin ke opposite sense mein (spinning pencil / cigar ki tarah). $I_3-I_1$ ka sign hi poori kahani hai.

Step 3. Summary: oblate $\Omega =+5$, prolate $\Omega =-4$.

Recall Forecast answer

Prolate (cigar-shaped) top: uska transverse $\mathbf{\omega }$ body-3 axis ke around apne khud ke spin ke ulte direction mein circle karta hai.

Verify: oblate $+5$, prolate $-4$; signs $\mathrm{sign}(I_3-I_1)$ hain $+$ aur $-$. ✓

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