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

2.1.23 · D4 · Physics › Analytical Mechanics › Torque-free rotation — Euler's equations, asymmetric top

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Euler's equations (torque-free) — equations of motion, ek har body axis ke liye: $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3}, I_{2} \overset{ω}{˙}_{2} = (I_{3} - I_{1}) ω_{3} ω_{1}, I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2}$ Yahan $I_{1}, I_{2}, I_{3}$ principal moments of inertia hain (teen fixed numbers jo describe karte hain ki mass har body axis ke around kaise spread hai), aur $ω_{1}, ω_{2}, ω_{3}$ spin vector $ω$ ke components hain un axes ke along. Dot ka matlab hai "time mein rate of change". Do conserved scalars: $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2} (energy), L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} (angular momentum squared)$ Stability sign axis $b$ ke around nearly spin ke liye (jahan $a, c$ baaki do hain): $k = \frac{( I _{a} - I _{b} ) ( I _{b} - I _{c} )}{I _{a} I _{c}} Ω^{2} < 0 \Rightarrow stable (oscillate), > 0 \Rightarrow unstable (grow) .$

Related depth yahan milegi: Inertia tensor and principal axes, Stability analysis and linearization, aur Symmetric top and gyroscopic precession.

Level 1 — Recognition

Problem 1.1

Ek rigid body ka $I_{1} = I_{2} = I_{3} = 4 kg\cdotm^{2}$ hai (ek uniform ball). Ise $ω = (3, - 1, 2) rad/s$ se spin diya gaya aur koi external torque nahi hai. Us instant par $\dot{ω}$ kya hai?

Recall Solution 1.1

Kya notice karna hai: teeno moments equal hain. Har Euler equation dekho — har right-hand side par moments ka difference hota hai jaise $(I_{2} - I_{3})$ . Jab saare $I_{i}$ equal hों, to har aisa difference zero hoga. $I_{1} \overset{ω}{˙}_{1} = (I_{2} - I_{3}) ω_{2} ω_{3} = (4 - 4) ω_{2} ω_{3} = 0 \Rightarrow \overset{ω}{˙}_{1} = 0,$ aur isi tarah $\overset{ω}{˙}_{2} = \overset{ω}{˙}_{3} = 0$ . Answer: $\dot{ω} = (0, 0, 0)$ . Spin kabhi nahi badalta — ek sphere mein koi wobble nahi hoti kyunki axes ko couple karne ke liye koi moment-difference nahi hota.

Problem 1.2

$I_{1} = 1, I_{2} = 2, I_{3} = 3$ wale body ke liye, kaun sa ek principal axis hai jiske around spin karna unstable hai? Bas naam batao.

Recall Solution 1.2

Moments ko order karo: sabse chhota $I_{1}$ hai, sabse bada $I_{3}$ hai, intermediate $I_{2}$ hai. Tennis-racket theorem kehta hai ki intermediate axis ke around rotation unstable hoti hai. Answer: axis 2 (wala $I_{2} = 2$ axis).

Level 2 — Application

Problem 2.1

Ek symmetric top ka $I_{1} = I_{2} = I_{⊥} = 2 kg\cdotm^{2}$ aur $I_{3} = 5 kg\cdotm^{2}$ hai. Yeh $ω_{3} = 10 rad/s$ se spin karta hai. Body precession rate $Ω_{b}$ nikalo (woh rate jis par $ω_{1}, ω_{2}$ axis 3 ke around rotate karte hain).

Recall Solution 2.1

Kyun kaam karta hai: $I_{1} = I_{2}$ ke saath, Euler ki teesri equation deti hai $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = 0$ , to $ω_{3}$ frozen hai. Pehli do equations phir yeh ban jaati hain: $\overset{ω}{˙}_{1} = - Ω_{b} ω_{2}, \overset{ω}{˙}_{2} = + Ω_{b} ω_{1}, Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3} .$ Yeh pair $(ω_{1}, ω_{2})$ ki uniform circular motion hai. $Ω_{b} = \frac{5 - 2}{2} \times 10 = \frac{3}{2} \times 10 = 15 rad/s .$ Answer: $Ω_{b} = 15 rad/s$ — $ω$ axis 3 ke around ek cone sweep karta hai fifteen radians per second.

Problem 2.2

$I_{1} = 1, I_{2} = 2, I_{3} = 3$ wali body ka ek instant par $ω = (2, 0, 3) rad/s$ hai. $2 T$ aur $L^{2}$ compute karo aur batao ki yeh baad ke saare time mein kya rahenge.

Recall Solution 2.2

Conserved scalars mein seedha plug karo. $2 T = I_{1} ω_{1}^{2} + I_{2} ω_{2}^{2} + I_{3} ω_{3}^{2} = 1 (2^{2}) + 2 (0^{2}) + 3 (3^{2}) = 4 + 0 + 27 = 31 J (i.e. 2 T) .$ $L^{2} = I_{1}^{2} ω_{1}^{2} + I_{2}^{2} ω_{2}^{2} + I_{3}^{2} ω_{3}^{2} = 1 (4) + 4 (0) + 9 (9) = 4 + 0 + 81 = 85.$ Kyun constant hai: humne prove kiya tha ki $d T / d t = 0$ (moment-differences telescope karte hain) aur $L$ space mein fixed hai to $L^{2}$ change nahi ho sakta. Dono har samay hold karte hain. Answer: $2 T = 31 J$ , $L^{2} = 85 kg^{2} m^{4} / s^{2}$ , hamesha ke liye.

Level 3 — Analysis

Problem 3.1

$I_{1} = 1, I_{2} = 2, I_{3} = 3$ ke liye, ek body axis 1 ke around nearly $Ω = 8 rad/s$ se spin karti hai. Stability constant $k$ aur small disturbances ki oscillation frequency compute karo.

Recall Solution 3.1

Axis 1 ke around spin karte hue, "baaki do" hain $2, 3$ . Linearised equation hai $\overset{ω}{¨}_{2} = k ω_{2}$ jahan $k = \frac{( I _{2} - I _{1} ) ( I _{1} - I _{3} )}{I _{2} I _{3}} Ω^{2} = \frac{( 2 - 1 ) ( 1 - 3 )}{2 \cdot 3} (8)^{2} = \frac{( 1 ) ( - 2 )}{6} \times 64 = - \frac{128}{6} = - 21.33.$ Kyunki $k < 0$ hai, $k = - ν^{2}$ likho to solution $cos (ν t)$ hai: yeh oscillate karta hai, isliye stable. $ν = ∣ k ∣ = 21.33 = 4.62 rad/s .$ Answer: $k = - 21.33$ , small wobbles $ν \approx 4.62 rad/s$ par oscillate karti hain — axis 1 (smallest $I$ ) stable hai. ✅

Problem 3.2

Same body, lekin axis 2 ke around nearly $Ω = 8 rad/s$ se spin. $k$ aur exponential growth rate compute karo; ek disturbance $e$ factor tak grow hone mein kitna time lagega?

Recall Solution 3.2

Ab "baaki do" $1, 3$ hain: $k = \frac{( I _{1} - I _{2} ) ( I _{2} - I _{3} )}{I _{1} I _{3}} Ω^{2} = \frac{( 1 - 2 ) ( 2 - 3 )}{1 \cdot 3} (64) = \frac{( - 1 ) ( - 1 )}{3} \times 64 = \frac{64}{3} = 21.33 > 0.$ Positive $k$ ka matlab hai solution $e^{k t}$ hai — exponential growth, unstable. Growth rate hai: $k = 21.33 = 4.62 s^{- 1}, t_{e} = \frac{1}{k} = 0.217 s .$ Answer: $k = + 21.33$ , e-folding time $\approx 0.22 s$ — axis 2 (intermediate $I$ ) unstable hai. ❌ Yahi Dzhanibekov flip hai.

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

Level 4 — Synthesis

Problem 4.1

$I_{1} = 1, I_{2} = 2, I_{3} = 4$ wali body $ω = (0, 0, 5) rad/s$ (axis 3 ke around pure spin) se start karti hai. Dikhao ki agar yeh $ω = (ϵ, 0, 5)$ tiny $ϵ$ ke saath start kare, to $ω$ ki trajectory ek thin closed loop par rehti hai, aur axis 3 ke around wobble frequency nikalo.

Recall Solution 4.1

Strategy: axis 3 yahan sabse bada $I$ hai, isliye hum stability expect karte hain (closed loop). Confirm karo linearise karke: $ω_{3} \approx Ω = 5$ constant rakho, $ω_{1}, ω_{2}$ small. $\overset{ω}{˙}_{1} = \frac{I _{2} - I _{3}}{I _{1}} Ω ω_{2}, \overset{ω}{˙}_{2} = \frac{I _{3} - I _{1}}{I _{2}} Ω ω_{1} .$ Pehle ko differentiate karo, doosra substitute karo: $\overset{ω}{¨}_{1} = \frac{( I _{2} - I _{3} ) ( I _{3} - I _{1} )}{I _{1} I _{2}} Ω^{2} ω_{1} = \frac{( 2 - 4 ) ( 4 - 1 )}{1 \cdot 2} (25) ω_{1} = \frac{( - 2 ) ( 3 )}{2} (25) ω_{1} = - 75 ω_{1} .$ $k = - 75 < 0$ ⇒ oscillation, isliye $ω$ axis 3 ke around ek small ellipse trace karta hai (ek polhode loop) — yeh kabhi nahi bhaagta, jo thin closed loop confirm karta hai. Wobble frequency: $ω_{wobble} = ∣ k ∣ = 75 = 8.66 rad/s .$ Answer: stable closed loop, wobble frequency $\approx 8.66 rad/s$ . Do conserved surfaces $2 T$ aur $L^{2}$ axis 3 ke paas ek tight closed polhode mein intersect karte hain.

Problem 4.2

Problem 2.1 wale symmetric top ke liye ( $I_{⊥} = 2, I_{3} = 5, ω_{3} = 10$ ), transverse spin ka amplitude $ω_{⊥} = ω_{1}^{2} + ω_{2}^{2} = 1 rad/s$ hai. Nikalo (a) body frame mein $ω$ jo cone axis 3 ke around sweep karta hai uska half-angle, aur (b) conserved $L^{2}$ .

Recall Solution 4.2

(a) Spin vector axis 3 ke saath angle $α$ banata hai jahan $tan α = \frac{ω _{⊥}}{ω _{3}} = \frac{1}{10} = 0.1 \Rightarrow α = arctan (0.1) = 5.7 1^{\circ} .$ $tan$ kyun? Us right triangle par jiske legs $ω_{⊥}$ (tilt ke opposite side) aur $ω_{3}$ (axis ke along adjacent side) hain, tangent = opposite/adjacent exactly woh ratio hai jo encode karta hai ki $ω$ axis se kitna tip karta hai. (b) $ω_{1}^{2} + ω_{2}^{2} = 1$ aur $I_{1} = I_{2} = 2$ ke saath: $L^{2} = I_{⊥}^{2} (ω_{1}^{2} + ω_{2}^{2}) + I_{3}^{2} ω_{3}^{2} = 2^{2} (1) + 5^{2} (100) = 4 + 2500 = 2504.$ Answer: (a) $α \approx 5.7 1^{\circ}$ ; (b) $L^{2} = 2504 kg^{2} m^{4} / s^{2}$ .

Level 5 — Mastery

Problem 5.1

Ek body ka $I_{1} = 1, I_{2} = 2, I_{3} = 3$ hai. Ise $2 T = 31$ aur $L^{2} = 85$ diya gaya hai (Problem 2.2 wali state). Separatrix — woh borderline polhode jo unstable axis 2 se guzarti hai — tab hoti hai jab $L^{2} = 2 T I_{2}$ . Check karo ki yeh exact state separatrix par hai, uske upar hai (energy ellipsoid axis 3 ko enclose karta hai), ya neeche hai (axis 1 ko enclose karta hai), aur physically interpret karo.

Recall Solution 5.1

Strategy — ratio $L^{2} / (2 T)$ ek "effective $I$ " ki tarah kaam karta hai. Ise compute karo aur moments se compare karo. $\frac{L ^{2}}{2 T} = \frac{85}{31} = 2.742.$ Yeh $I_{2} = 2$ aur $I_{3} = 3$ ke beech pada hai. Interpretation: kyunki $L^{2} / (2 T) > I_{2}$ hai, do ellipsoids ka intersection ek polhode hai jo axis 3 (sabse bade $I$ wale axis) ke around loop karti hai, axis 2 ke around nahi. To yeh state stable side par hai, large axis ko wrap karti hui. Separatrix ke liye $L^{2} / (2 T) = I_{2} = 2$ exactly chahiye hoga, yani $L^{2} = 2 T \cdot I_{2} = 31 \times 2 = 62$ . Hamare paas $85 \neq = 62$ hai, isliye hum separatrix par nahi hain; hum safely uske upar hain (axis-3 loop). Answer: $L^{2} / (2 T) = 2.742$ jo $I_{2}$ aur $I_{3}$ ke beech hai ⇒ polhode axis 3 ke around circle karti hai, stable side; separatrix value hoti $L^{2} = 62$ .

Problem 5.2

Design check (Dzhanibekov timer). Space mein ek wingnut ko $I_{1} = 1, I_{2} = 2, I_{3} = 3 kg\cdotm^{2}$ model kiya gaya hai, unstable axis 2 ke around $Ω = 12 rad/s$ se tossa gaya hai aur ek unavoidable seed disturbance $ω_{1} (0) = 0.01 rad/s$ hai. Estimate karo ki disturbance $ω_{1} = 1 rad/s$ tak pahunchne mein kitna time lagega (roughly "flip-scale"). Linear growth $ω_{1} (t) = ω_{1} (0) e^{k t}$ use karo.

Recall Solution 5.2

Step 1 — growth rate. Axis 2 ke liye: $k = \frac{( I _{1} - I _{2} ) ( I _{2} - I _{3} )}{I _{1} I _{3}} Ω^{2} = \frac{( - 1 ) ( - 1 )}{1 \cdot 3} (144) = \frac{144}{3} = 48, k = 48 = 6.928 s^{- 1} .$ Step 2 — exponential invert karo. Humein $ω_{1} (t) / ω_{1} (0) = 1/0.01 = 100$ chahiye: $e^{k t} = 100 \Rightarrow t = \frac{l n 100}{k} = \frac{4.6052}{6.928} = 0.6647 s .$ Log kyun? Exponential growth $e^{k t}$ jawab deta hai "time $t$ ke baad kitna bada"; natural log lena ise invert karta hai — yeh reverse sawaal ka jawab deta hai "ek given size tak pahunchne mein kitna time lagega". Yahi $ln$ undo karta hai. Answer: roughly $0.66 s$ toss se visible flip tak — wingnut ek second ke andar tumble kar jaata hai, jo famous ISS footage se match karta hai.

Recall Self-test index (page band karo, memory se jawab do)

Kaun sa axis unstable hai aur kyun ::: Intermediate- $I$ axis, kyunki product $(I_{a} - I_{b}) (I_{b} - I_{c})$ wahan positive hota hai, $k > 0$ aur exponential growth deta hai. Torque-free motion mein kaun se do scalars conserved hain ::: $2 T = \sum I_{i} ω_{i}^{2}$ aur $L^{2} = \sum I_{i}^{2} ω_{i}^{2}$ . Symmetric-top body precession rate ka formula ::: $Ω_{b} = \frac{I _{3} - I _{⊥}}{I _{⊥}} ω_{3}$ . Zero torque ke baad bhi $ω$ constant kyun nahi hota ::: Kyunki $L = I ω$ conserved hai, $ω$ nahi; ek lopsided (non-sphere) $I$ constant $L$ ko ek moving $ω$ force karta hai.