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

2.1.21 · D4 · Physics › Analytical Mechanics › Rigid body dynamics — Euler angles, Euler's equations of mot

Poore note mein, teen principal moments of inertia $I_{1}, I_{2}, I_{3}$ hain (har principal axis ke baare mein "spinning ki resistance"), body-frame angular-velocity components $ω_{1}, ω_{2}, ω_{3}$ hain (body kitni tez har principal axis ke baare mein ghoomti hai, body par sawar ek kide ke nazariye se), aur torque components $τ_{1}, τ_{2}, τ_{3}$ hain. Jo torque-free Euler equations hum use karte hain woh hain

I_{1} \overset{ω}{˙}_{1} - (I_{2} - I_{3}) ω_{2} ω_{3} = 0, I_{2} \overset{ω}{˙}_{2} - (I_{3} - I_{1}) ω_{3} ω_{1} = 0, I_{3} \overset{ω}{˙}_{3} - (I_{1} - I_{2}) ω_{1} ω_{2} = 0.

Yahan $\overset{ω}{˙}_{i}$ ka matlab hai " $ω_{i}$ ki time ke saath rate of change," dot Newton ka shorthand hai $d / d t$ ke liye.

Level 1 — Recognition

L1.1 Teen Euler angles ko naam do

Standard 3-1-3 rotation order likhо aur plain words mein batao ki $ϕ, θ, ψ$ mein se har ek ek spinning top ke saath physically kya karta hai.

Recall Solution

Order: space $Z$ -axis ke baare mein $ϕ$ se rotate karo, phir nayi $x$ -axis (line of nodes) ke baare mein $θ$ se, phir final body $z$ -axis ke baare mein $ψ$ se. Toh $R = R_{z} (ψ) R_{x} (θ) R_{z} (ϕ)$ .

$ϕ$ = precession: kaise tilted top vertical ke aas-paas sweep karta hai, jaise ghadi ki suyi.
$θ$ = nutation: top ki axis ka vertical se jhukav.
$ψ$ = spin: top apne point par kitni tez ghoomti hai.

L1.2 Moment condition padhna

Ek body hai jiske liye $I_{1} = I_{2} = I_{3}$ hai (ek uniform sphere). Koi torque nahi ke saath, har Euler equation $\overset{ω}{˙}_{i}$ ke baare mein kya kehti hai?

Recall Solution

Har coupling coefficient $(I_{i} - I_{j})$ zero hai, isliye teeno equations $I_{i} \overset{ω}{˙}_{i} = 0$ mein collapse ho jaati hain, yaani $\overset{ω}{˙}_{1} = \overset{ω}{˙}_{2} = \overset{ω}{˙}_{3} = 0$ . Angular velocity kisi bhi axis ke baare mein constant hai — ek free sphere jis bhi axis par tumne set kiya tha usi par hamesha ke liye ghoomti rahegi. Koi wobble possible nahi hai.

L1.3 Surviving spin ko identify karo

Ek symmetric body $I_{1} = I_{2} \neq = I_{3}$ ke liye, zero torque ke saath, $ω$ ka kaun sa single component guaranteed constant hai, aur kis equation se?

Recall Solution

Equation 3: $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = 0$ kyunki $I_{1} = I_{2}$ . Isliye $ω_{3} = const$ — symmetry axis ke baare mein spin frozen hai.

Level 2 — Application

L2.1 Ek leaning, spinning top ki angular velocity

Ek symmetric top ke instantaneous Euler angles $θ = 6 0^{\circ}$ , $ψ = 0^{\circ}$ hain aur rates $\dot{ϕ} = 4 rad/s$ , $\dot{θ} = 0$ , $\dot{ψ} = 10 rad/s$ hain. Body-frame components $ω_{1}, ω_{2}, ω_{3}$ compute karo.

Recall Solution

3-1-3 formulas use karo. $ψ = 0$ ke saath: $sin ψ = 0, cos ψ = 1$ . $θ = 6 0^{\circ}$ ke saath: $sin θ = \frac{3}{2} \approx 0.8660$ , $cos θ = \frac{1}{2}$ . $ω_{1} = \dot{ϕ} sin θ sin ψ + \dot{θ} cos ψ = 4 (0.8660) (0) + 0 = 0.$ $ω_{2} = \dot{ϕ} sin θ cos ψ - \dot{θ} sin ψ = 4 (0.8660) (1) - 0 = 3.464 rad/s .$ $ω_{3} = \dot{ϕ} cos θ + \dot{ψ} = 4 (0.5) + 10 = 12 rad/s .$ Toh $ω = (0, 3.464, 12) rad/s$ .

L2.2 Ek symmetric body ki free-precession rate

Ek torque-free symmetric body ke liye $I_{1} = I_{2} = 2.0 kg\cdotm^{2}$ , $I_{3} = 3.0 kg\cdotm^{2}$ , aur constant spin $ω_{3} = 6.0 rad/s$ hai. Body-frame precession rate $Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3}$ aur period $T = 2 π /∣Ω∣$ nikalo.

Recall Solution

$Ω = \frac{3.0 - 2.0}{2.0} (6.0) = \frac{1.0}{2.0} (6.0) = 3.0 rad/s .$ $T = \frac{2 π}{3.0} = 2.094 s .$ Transverse part $(ω_{1}, ω_{2})$ body-3 axis ke baare mein ek circle trace karta hai har $2.09$ s mein ek baar.

L2.3 Transverse circle ki radius

L2.2 wali body ke liye transverse solution hai $ω_{1} = A cos Ω t, ω_{2} = A sin Ω t$ . $t = 0$ par body mein $ω_{1} (0) = 0.4 rad/s$ , $ω_{2} (0) = 0$ hai. $A$ aur $t = T /4$ par $ω_{2}$ nikalo.

Recall Solution

$t = 0$ par: $ω_{1} (0) = A cos 0 = A = 0.4 rad/s$ , aur $ω_{2} (0) = A sin 0 = 0$ ✓. $t = T /4$ par, $Ω t = Ω \cdot \frac{2 π}{4Ω} = \frac{π}{2}$ , toh $ω_{1} = A cos \frac{π}{2} = 0$ aur $ω_{2} = A sin \frac{π}{2} = A = 0.4 rad/s$ . Transverse vector ki tip ne quarter turn rotate ki hai: axis 1 ki direction se axis 2 ki direction mein, magnitude preserved. Neeche circle dekho.

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

Level 3 — Analysis

L3.1 Symmetric-top harmonic pair derive karo

$ω_{1}, ω_{2}$ ke liye $I_{1} = I_{2}$ ke saath do torque-free equations se shuru karke, dikhao ki woh $\overset{ω}{˙}_{1} = - Ω ω_{2}, \overset{ω}{˙}_{2} = + Ω ω_{1}$ mein reduce ho jaati hain, aur isliye $ω_{1}^{2} + ω_{2}^{2}$ constant hai.

Recall Solution

$I_{1} = I_{2}$ aur $τ = 0$ ke saath:

I_1\dot\omega_2=(I_3-I_1)\omega_3\omega_1 .$$ $I_1$ se divide karo aur $\Omega=\dfrac{I_3-I_1}{I_1}\omega_3$ define karo (constant, kyunki $\omega_3$ constant hai): $$\dot\omega_1=-\Omega\omega_2,\qquad \dot\omega_2=+\Omega\omega_1 .$$ **Conserved magnitude:** $\frac{d}{dt}(\omega_1^2+\omega_2^2)=2\omega_1\dot\omega_1+2\omega_2\dot\omega_2 =2\omega_1(-\Omega\omega_2)+2\omega_2(\Omega\omega_1)=0.$ Toh transverse speed $A=\sqrt{\omega_1^2+\omega_2^2}$ fixed hai — tip ek **circle** par move karta hai, jo L2.3 confirm karta hai.

L3.2 Tennis-racket coefficient

$I_{1} < I_{2} < I_{3}$ wali body ke liye, axis 2 ke baare mein almost spin karte hue small $ω_{1}, ω_{3}$ ke saath, torque-free equations ko linearize karo taaki dikhao ki $\overset{ω}{¨}_{1} \propto k ω_{1}$ hai aur $k$ ka sign determine karo.

Recall Solution

Torque-free: $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3}$ aur $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2}$ . $ω_{2} \approx const$ (bada) treat karo, $ω_{1}, ω_{3}$ small hain. Pehli ko differentiate karo aur teesri substitute karo: $I_{1} \overset{ω}{¨}_{1} = (I_{2} - I_{3}) ω_{2} \overset{ω}{˙}_{3} = (I_{2} - I_{3}) ω_{2} \cdot \frac{( I _{1} - I _{2} ) ω _{1} ω _{2}}{I _{3}} .$ $\overset{ω}{¨}_{1} = \frac{( I _{2} - I _{3} ) ( I _{1} - I _{2} )}{I _{1} I _{3}} ω_{2}^{2} ω_{1} \equiv k ω_{1} .$ Signs: $I_{2} - I_{3} < 0$ aur $I_{1} - I_{2} < 0$ , product $> 0$ , aur $\frac{ω _{2}^{2}}{I _{1} I _{3}} > 0$ , toh $k > 0$ . $\overset{ω}{¨}_{1} = + k ω_{1}$ with $k > 0$ ke exponentially growing solutions $e^{k t}$ hain → axis 2 unstable hai. Yahi tumbling middle-axis (Dzhanibekov) effect hai.

L3.3 Stable axes confirm karo

L3.2 ka linearization axis 1 (smallest $I$ ) ke baare mein spin ke liye repeat karo aur dikhao ki coefficient negative hai (oscillatory, stable).

Recall Solution

Axis 1 ke baare mein spin: $ω_{1} \approx const$ bada, $ω_{2}, ω_{3}$ small. $I_{2} \overset{ω}{˙}_{2} = (I_{3} - I_{1}) ω_{3} ω_{1}, I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} .$ Pehli differentiate karo, doosri sub karo: $I_{2} \overset{ω}{¨}_{2} = (I_{3} - I_{1}) ω_{1} \overset{ω}{˙}_{3} = (I_{3} - I_{1}) ω_{1} \cdot \frac{( I _{1} - I _{2} ) ω _{1} ω _{2}}{I _{3}},$ $\overset{ω}{¨}_{2} = \frac{( I _{3} - I _{1} ) ( I _{1} - I _{2} )}{I _{2} I _{3}} ω_{1}^{2} ω_{2} .$ Signs: $I_{3} - I_{1} > 0$ , $I_{1} - I_{2} < 0$ , product $< 0$ . Toh $\overset{ω}{¨}_{2} = - c ω_{2}$ with $c > 0$ → $c$ par oscillation → stable. Axis 3 (largest $I$ ) bhi isi tarah kaam karta hai (dono factors phir se negative coefficient dete hain). Sirf intermediate axis unstable hai. Diagram dekho.

Level 4 — Synthesis

L4.1 Frisbee versus Earth — precession compare karo

(a) Ek frisbee (oblate) ka $I_{3} = 2 I_{1}$ hai. (b) Earth almost spherical hai jisme $\frac{I _{3} - I _{1}}{I _{1}} = \frac{C - A}{A} \approx \frac{1}{305}$ hai aur yeh din mein ek baar spin karta hai. Har ek ke liye body-frame precession period spin period $T_{spin} = 2 π / ω_{3}$ ki units mein compute karo.

Recall Solution

$Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3}$ , toh free-precession period hai $T_{Ω} = \frac{2 π}{∣Ω∣} = \frac{2 π}{ω _{3}} \cdot \frac{I _{1}}{∣ I _{3} - I _{1} ∣} = T_{spin} \cdot \frac{I _{1}}{∣ I _{3} - I _{1} ∣} .$ (a) Frisbee: $I_{3} - I_{1} = 2 I_{1} - I_{1} = I_{1}$ , toh $\frac{I _{1}}{∣ I _{3} - I _{1} ∣} = 1$ . Isliye $T_{Ω} = T_{spin}$ — yeh har spin mein ek baar wobble karta hai (ek fast, obvious tumble). (b) Earth: $\frac{I _{1}}{I _{3} - I _{1}} = 305$ , toh $T_{Ω} = 305 days \approx 0.84 yr$ . Yeh rigid-body Chandler prediction (~305 d) hai; real Chandler wobble ~433 d hai kyunki Earth elastic/fluid hai, perfectly rigid nahi.

L4.2 Ek free body ki energy aur angular-momentum constraints

Ek torque-free rigid body dono rotational kinetic energy $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2}$ aur $L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2}$ conserve karta hai. $I_{1} = 1, I_{2} = 2, I_{3} = 3$ (SI) aur initial $ω = (2, 1, 0)$ ke liye, dono invariants compute karo aur batao ki motion ek energy ellipsoid aur ek momentum sphere ke intersection par hai ya nahi.

Recall Solution

$2 T = 1 (2^{2}) + 2 (1^{2}) + 3 (0^{2}) = 4 + 2 + 0 = 6 \Rightarrow T = 3 J (per unit... using SI numbers) .$ $L^{2} = 1^{2} (2^{2}) + 2^{2} (1^{2}) + 3^{2} (0^{2}) = 4 + 4 + 0 = 8 \Rightarrow L = 8 = 2.828.$ $ω$ -space mein, constant $2 T$ ek ellipsoid hai aur constant $L^{2}$ ek aur quadric hai; actual motion unke intersection curve (the "polhode") tak confined hai. Kyunki dono zero torque ke saath conserved hain, $ω (t)$ kabhi us curve ko nahi chhodega — yeh tennis-racket theorem ka geometric backbone hai.

L4.3 Body type se free precession ki direction

$Ω$ ke sign ka use karke dikhao ki oblate body ka transverse $ω$ spin ke same rotational sense mein circles karta hai, jabki prolate body ka opposite mein circles karta hai.

Recall Solution

$\overset{ω}{˙}_{1} = - Ω ω_{2}, \overset{ω}{˙}_{2} = + Ω ω_{1}$ integrate hoke $ω_{1} = A cos Ω t, ω_{2} = A sin Ω t$ dete hain. Tip $+ 1$ -axis se $+ 2$ -axis ki taraf advance karta hai jab $Ω > 0$ (counterclockwise, wahi sense jo spin $ω_{3} > 0$ right-hand rule se define karta hai), aur reverse karta hai jab $Ω < 0$ .

Oblate ( $I_{3} > I_{1}$ ): $Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3} > 0$ → spin ke same sense mein.
Prolate ( $I_{3} < I_{1}$ ): $Ω < 0$ → opposite sense. Yeh sign observable hai: yeh ek tumbling frisbee (co-rotating wobble) ko ek tumbling pencil (counter-rotating wobble) se alag karta hai.

Level 5 — Mastery

L5.1 Scratch se fast-top steady-precession condition banao

Ek symmetric top ( $I_{1} = I_{2} \equiv I_{t}$ , spin moment $I_{3}$ ) large $ω_{3}$ ke saath gravity ke under spin karta hai, weight $m g$ , center of mass fixed pivot se $ℓ$ doori par, tilt $θ$ constant rakha gaya hai. $τ = ∣ torque ∣ = m g ℓ sin θ$ aur steady slow precession ke liye gyroscopic relation $τ = \dot{ϕ} (I_{3} ω_{3}) sin θ$ use karke, precession rate $\dot{ϕ}$ derive karo aur $m g ℓ = 0.20 N\cdotm$ , $I_{3} = 4.0 \times 1 0^{- 3} kg\cdotm^{2}$ , $ω_{3} = 150 rad/s$ ke liye evaluate karo.

Recall Solution

Physics: pivot ke baare mein gravity torque ka magnitude $m g ℓ sin θ$ hai, spin axis ke perpendicular. Fast top ke liye angular momentum spin se dominate hota hai, $L \approx I_{3} ω_{3} \overset{z}{^}$ . $L$ ke perpendicular torque sirf $L$ ko sideways rotate kar sakta hai (length nahi badal sakta) → steady precession. Sideways rate: $L$ ka horizontal projection $I_{3} ω_{3} sin θ$ length ka hai aur rate $\dot{ϕ}$ se sweep karta hai, toh $∣ \dot{L} ∣ = \dot{ϕ} I_{3} ω_{3} sin θ$ . Torque ke equal set karo: $m g ℓ sin θ = \dot{ϕ} I_{3} ω_{3} sin θ \Rightarrow \dot{ϕ} = \frac{m g ℓ}{I _{3} ω _{3}} .$ $sin θ$ cancel ho jaata hai — steady precession rate tilt se independent hai (fast-top approximation ke liye). Number: $\dot{ϕ} = \frac{0.20}{( 4.0 \times 1 0 ^{- 3} ) ( 150 )} = \frac{0.20}{0.60} = 0.333 rad/s .$ Full slow/fast branch analysis ke liye Gyroscope precession dekho.

L5.2 Elliptic reasoning se general torque-free period (bounding argument)

L4.2 ki free body ke liye ( $I = (1, 2, 3)$ , $ω_{0} = (2, 1, 0)$ ), motion polhode par oscillate karta hai. Dikhao ki $ω_{3}$ energy aur momentum se bounded maximum tak pahunch sakta hai, aur woh maximum $ω_{3}^{m a x}$ nikalo.

Recall Solution

Motion ke along, $2 T = 6$ aur $L^{2} = 8$ (L4.2 se), aur $ω_{1}, ω_{2}$ real hain toh $ω_{1}^{2} \geq 0, ω_{2}^{2} \geq 0$ . $ω_{3}$ ke terms mein $ω_{1}^{2}, ω_{2}^{2}$ ke liye do invariants solve karo: $I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} = 2 T - I_{3} ω_{3}^{2} = 6 - 3 ω_{3}^{2},$ $I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} = L^{2} - I_{3}^{2} ω_{3}^{2} = 8 - 9 ω_{3}^{2} .$ $I_{1} = 1, I_{2} = 2$ ke saath: $ω_{1}^{2} + 2 ω_{2}^{2} = 6 - 3 ω_{3}^{2}$ aur $ω_{1}^{2} + 4 ω_{2}^{2} = 8 - 9 ω_{3}^{2}$ . Subtract karo: $2 ω_{2}^{2} = (8 - 9 ω_{3}^{2}) - (6 - 3 ω_{3}^{2}) = 2 - 6 ω_{3}^{2} \Rightarrow ω_{2}^{2} = 1 - 3 ω_{3}^{2} .$ Phir $ω_{1}^{2} = (6 - 3 ω_{3}^{2}) - 2 ω_{2}^{2} = 6 - 3 ω_{3}^{2} - 2 (1 - 3 ω_{3}^{2}) = 4 + 3 ω_{3}^{2}$ . $ω_{2}^{2} \geq 0$ require karo: $1 - 3 ω_{3}^{2} \geq 0 \Rightarrow ω_{3}^{2} \leq \frac{1}{3}$ , toh $ω_{3}^{m a x} = \frac{1}{3} = 0.5774 rad/s .$ ( $ω_{1}^{2} = 4 + 3 ω_{3}^{2} > 0$ hamesha, toh yeh kabhi limit nahi karta.) Body ka axis 3 ke baare mein spin $- 0.577$ aur $+ 0.577$ ke beech oscillate karta hai — polhode ek closed loop hai, L4.2 ki picture se match karta hai.

L5.3 Consistency check: kya $ω_{0}$ curve par hai?

Verify karo ki initial condition $ω_{0} = (2, 1, 0)$ , L5.2 mein derive kiye gaye invariants ke saath consistent hai.

Recall Solution

L5.2 expressions mein $ω_{3} = 0$ daalo: $ω_{2}^{2} = 1 - 0 = 1 \Rightarrow ω_{2} = 1$ ✓ aur $ω_{1}^{2} = 4 + 0 = 4 \Rightarrow ω_{1} = 2$ ✓. Exactly $ω_{0}$ . Initial point polhode ke $ω_{3} = 0$ crossing par baitha hai — maximum transverse spin, minimum axial spin ka moment. Self-consistent.

Recall One-line self-test

Spin period ke terms mein free precession period ::: $T_{Ω} = T_{spin} \frac{I _{1}}{∣ I _{3} - I _{1} ∣}$ Fast-top precession rate ::: $\dot{ϕ} = \frac{m g ℓ}{I _{3} ω _{3}}$ (note $sin θ$ cancels) $I_{1} < I_{2} < I_{3}$ ke liye kaun sa axis unstable hai? ::: intermediate axis, $I_{2}$