2.1.23 · D3Torque-free rotation — Euler's equations, asymmetric top
Yeh page parent topic ka drill ground hai. Yahan hum Euler's equations dobara derive nahi karte — balki unhe har tarah ke input par exercise karte hain. Agar koi symbol unfamiliar lage, toh woh parent note mein build hua tha; neeche ke quick reminders unhe dobara anchor karte hain.
Recall Quick symbol re-anchor (padho agar koi notation naya lage)
I 1 , I 2 , I 3 principal moments of inertia : body kitni mushkil se apne teen special body-glued axes ke baare mein spin resist karti hai. Inertia tensor and principal axes se build hua.ω 1 , ω 2 , ω 3 ω un body axes ke saath measure kiye jaate hain.Ω ω i specific value hai, koi naya object nahi.ω ⊥ transverse spin vector , do-component pair ( ω 1 , ω 2 ) ∣ ω ⊥ ∣ = ω 1 2 + ω 2 2 Ek dot matlab "rate of change per second": ω ˙ 1 = d ω 1 / d t
T rotational kinetic energy ; hum usually 2 T = I 1 ω 1 2 + I 2 ω 2 2 + I 3 ω 3 2 L = I ω angular momentum . Iska length-squared L 2 = I 1 2 ω 1 2 + I 2 2 ω 2 2 + I 3 2 ω 3 2 Angular momentum in rotating frames .k dimensionless stability constant , Ex 3 mein defined — linearized wobble equation ko ω ¨ 1 = k Ω 2 ω 1 k pure number hai (isme Ω 2 Ω 2 sign stability decide karta hai: k < 0 k > 0
Is topic ke har problem ko in cells mein se kisi ek mein daala ja sakta hai. Neeche ke examples har ek ko kam se kam ek baar hit karte hain. (Cells C–E mein jo stability constant k reference hai woh Example 3 mein defined hai — yahan use "woh pure-number sign jo stability decide karta hai" ka placeholder maano.)
Cell
Kya cheez ise woh case banati hai
Kyun iska alag treatment chahiye
Example
A. Saare I I 1 = I 2 = I 3 RHS → 0
Ex 1
B. Do I I 1 = I 2 = I 3 Ek equation khatam, doosre do steady precession dete hain
Ex 2
C. Saare I I 1 < I 2 < I 3 ≈ k < 0 k Ex 3
D. Saare I spin ≈
k < 0 doosra stable wala)Ex 4
E. Saare I spin ≈
k > 0 Ex 5
F. Conservation cross-check koi bhi motion
Verify karo ki 2 T L 2
Ex 6
G. Zero / one-component spin (degenerate) ω Pure principal-axis spin: RHS sab zero, wobble kabhi nahi
Ex 7
H. Real-world word problem Earth ka Chandler wobble
Ek physics number ko ek real observable period mein convert karna
Ex 8
I. Exam twist (sign trap) axis ordering scrambled
Check karta hai ki tum sign track karte ho, axis label nahi
Ex 9
Ek uniform solid sphere mein I 1 = I 2 = I 3 = I ω = ( 2 , − 3 , 1 ) ω
Forecast: Padhne se pehle andaza lagao — kya spin arrow drift karega, wobble karega, ya frozen rahega?
Teeno Euler equations mein plug karo.
I ω ˙ 1 = ( I − I ) ω 2 ω 3 = 0 , I ω ˙ 2 = 0 , I ω ˙ 3 = 0. Yeh step kyun? Har coefficient do equal moments ka difference hai, isliye exactly 0 Integrate karo. ω ˙ 1 = ω ˙ 2 = ω ˙ 3 = 0 ⇒ ω = ( 2 , − 3 , 1 ) Yeh step kyun? Zero derivative matlab value kabhi nahi badlegi — body frame mein spin frozen hai.
Verify: Sphere ke liye I = I 1 L = I ω ω parallel hai. Space mein L ω ( 2 , − 3 , 1 )
Ek discus-jaisa symmetric top hai jisme I 1 = I 2 = I ⊥ = 2 I 3 = 5 ω 3 = 4 ω 1 ( 0 ) = 0.1 ω 2 ( 0 ) = 0 Body precession rate Ω b
Forecast: Kya transverse components decay karenge, blow up karenge, ya constant rate par circle karenge?
Pehle Euler 3. I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 = 0 ω 3 = 4 Yeh step kyun? Equal transverse moments woh coefficient zero kar dete hain, axial spin ko freeze kar dete hain — pehle lock down karne ki sabse clean quantity.Euler 1 & 2 reduce karo. I 1 = I 2 = I ⊥ ω ˙ 1 = I ⊥ I ⊥ − I 3 ω 3 ω 2 , ω ˙ 2 = I ⊥ I 3 − I ⊥ ω 3 ω 1 . Ω b = I ⊥ I 3 − I ⊥ ω 3 = 2 5 − 2 ⋅ 4 = 6 Yeh step kyun? Dono right sides opposite signs ke saath same magnitude Ω b ( ω 1 , ω 2 ) Circle solve karo. Re-anchor box se yaad karo ki ω ⊥ = ( ω 1 , ω 2 ) transverse spin vector hai. ω ˙ 1 = − Ω b ω 2 , ω ˙ 2 = Ω b ω 1 ω ⊥ Ω b T prec = 2 π / Ω b = 2 π /6 ≈ 1.047 Yeh step kyun? Yeh equations ek point ke angular speed Ω b free precession cone hai.
Neeche ki figure yeh dikhati hai. Blue closed loop woh path hai jo transverse spin ω ⊥ = ( ω 1 , ω 2 ) Ω b = 6
Verify: ∣ ω ⊥ ∣ 2 = ω 1 2 + ω 2 2 = 0. 1 2 = 0.01 d t d ( ω 1 2 + ω 2 2 ) = 2 ω 1 ω ˙ 1 + 2 ω 2 ω ˙ 2 = 2 ω 1 ( − Ω b ω 2 ) + 2 ω 2 ( Ω b ω 1 ) = 0 Ω b = 6 T prec ≈ 1.047
Asymmetric top I 1 = 1 , I 2 = 2 , I 3 = 3 Ω = 10 ω 1 , ω 2 Stability constant k
Forecast: k
Linearize karo. ω 3 ≈ Ω ω 1 , ω 2 ω ˙ 1 = I 1 I 2 − I 3 Ω ω 2 , ω ˙ 2 = I 2 I 3 − I 1 Ω ω 1 . Yeh step kyun? Do small quantities ka product negligible hai; yeh pure-spin state ke baare mein linearization hai.Ek ODE mein combine karo aur k Pehle differentiate karo, doosra insert karo:
ω ¨ 1 = k ( pure number ) I 1 I 2 ( I 2 − I 3 ) ( I 3 − I 1 ) Ω 2 ω 1 . Stability constant k exactly woh bracketed dimensionless coefficient hai — isme Ω 2 nahi hai, jo hum hamesha alag likhte hain:
k = I 1 I 2 ( I 2 − I 3 ) ( I 3 − I 1 ) = 1 ⋅ 2 ( 2 − 3 ) ( 3 − 1 ) = 2 ( − 1 ) ( 2 ) = − 1. Yeh step kyun? k k = − 1 k Ω 2 Wobble read off karo. k = − 1 ω ¨ 1 = k Ω 2 ω 1 = − Ω 2 ω 1 ω wob = ∣ k ∣ Ω = 1 ⋅ 10 = 10 Yeh step kyun? Negative k I stable hai. ✅
Verify: Dono factors: ( I 2 − I 3 ) = − 1 < 0 ( I 3 − I 1 ) = + 2 > 0 k = − 1 < 0 ∣ k ∣ Ω = 10 ≈ 0.628
Same body I 1 = 1 , I 2 = 2 , I 3 = 3 axis 1 ke baare mein spin Ω = 10
Forecast: Do stable axes hote hain. Kya sabse chhhota wala sach mein stable hai, ya sirf sabse bada?
Stability constant ko cyclic-shift karo. Axis 1 ke baare mein bade spin ke liye, ω 2 ω ¨ 2 = k Ω 2 ω 2 , k = I 2 I 3 ( I 3 − I 1 ) ( I 1 − I 2 ) ( pure number ) . Yeh step kyun? "Jis axis ke baare mein spin kar rahe ho" ka role equations mein cycle karta hai; hum sirf ( 1 , 2 , 3 ) k k k = 2 ⋅ 3 ( 3 − 1 ) ( 1 − 2 ) = 6 ( 2 ) ( − 1 ) = − 3 1 ≈ − 0.333. Yeh step kyun? Numbers plug karne se directly sign evaluate hota hai — k Ω 2 Wobble read off karo. k < 0 ∣ k ∣ Ω = 1/3 ⋅ 10 ≈ 5.77 Stable. ✅
Verify: ( I 3 − I 1 ) = + 2 > 0 ( I 1 − I 2 ) = − 1 < 0 k = − 1/3 < 0 k < 0 ≈ 5.77
Same body I 1 = 1 , I 2 = 2 , I 3 = 3 axis 2 (intermediate) ke baare mein spin Ω = 10 k
Forecast: Yeh Dzhanibekov / tennis-racket axis hai. Compute karne se pehle k
Axis 2 ke liye stability constant.
ω ¨ 1 = k Ω 2 ω 1 , k = I 1 I 3 ( I 3 − I 2 ) ( I 2 − I 1 ) ( pure number ) . Yeh step kyun? Same linearization machine, ab axis 2 "big spin" ke roop mein, aur k k k = 1 ⋅ 3 ( 3 − 2 ) ( 2 − 1 ) = 3 ( 1 ) ( 1 ) = 3 1 ≈ 0.333. Yeh step kyun? Dono factors yahan positive hain, toh unka product positive hai — yahi crucial sign flip hai.Growth read off karo. k > 0 ω ¨ 1 = + k Ω 2 ω 1 ∝ e k Ω t k Ω = 1/3 ⋅ 10 ≈ 5.77 s − 1 τ = 1/ ( k Ω ) ≈ 0.173 Yeh step kyun? Positive k 0.173 e
Neeche ki figure ise Example 3 se contrast karti hai. Red curve yeh case hai (axis 2): ek disturbance jo e k Ω t τ ≈ 0.173 e k
Verify: ( I 3 − I 2 ) = + 1 > 0 ( I 2 − I 1 ) = + 1 > 0 k = 1/3 > 0 k Ω ≈ 5.77 s − 1 τ ≈ 0.173
Lo I 1 = 1 , I 2 = 2 , I 3 = 3 ω = ( 3 , 1 , 2 ) 2 T L 2 T L = I ω
Forecast: Bhaale har ω i
Constants evaluate karo.
2 T = I 1 ω 1 2 + I 2 ω 2 2 + I 3 ω 3 2 = 1 ( 9 ) + 2 ( 1 ) + 3 ( 4 ) = 9 + 2 + 12 = 23. L 2 = I 1 2 ω 1 2 + I 2 2 ω 2 2 + I 3 2 ω 3 2 = 1 ( 9 ) + 4 ( 1 ) + 9 ( 4 ) = 9 + 4 + 36 = 49. Yeh step kyun? Yeh do ellipsoids hain (2 T = L 2 = ω ω Energy differentiate karo aur Euler substitute karo.
d t d ( 2 T ) = 2 ( I 1 ω 1 ω ˙ 1 + I 2 ω 2 ω ˙ 2 + I 3 ω 3 ω ˙ 3 ) . I i ω ˙ i d t d ( 2 T ) = 2 [ ω 1 ( I 2 − I 3 ) ω 2 ω 3 + ω 2 ( I 3 − I 1 ) ω 3 ω 1 + ω 3 ( I 1 − I 2 ) ω 1 ω 2 ] = 2 ω 1 ω 2 ω 3 [ ( I 2 − I 3 ) + ( I 3 − I 1 ) + ( I 1 − I 2 ) ] = 0. Yeh step kyun? Euler's equations se I i ω ˙ i ω 1 ω 2 ω 3 0 L 2 d t d L 2 = 2 ( I 1 2 ω 1 ω ˙ 1 + I 2 2 ω 2 ω ˙ 2 + I 3 2 ω 3 ω ˙ 3 ) = 2 ( I 1 ω 1 ⋅ I 1 ω ˙ 1 + … ) . I i ω ˙ i = 2 [ I 1 ω 1 ( I 2 − I 3 ) ω 2 ω 3 + I 2 ω 2 ( I 3 − I 1 ) ω 3 ω 1 + I 3 ω 3 ( I 1 − I 2 ) ω 1 ω 2 ] . ω 1 ω 2 ω 3 = 2 ω 1 ω 2 ω 3 [ I 1 ( I 2 − I 3 ) + I 2 ( I 3 − I 1 ) + I 3 ( I 1 − I 2 ) ] . I 1 I 2 − I 1 I 3 + I 2 I 3 − I 2 I 1 + I 3 I 1 − I 3 I 2 = 0 + − d t d L 2 = 0 Yeh step kyun? L L 2 kyun algebraically dekhte hain: har I a I b
Verify: 2 T = 23 L 2 = 49 d t d ( 2 T ) = 0 d t d L 2 = 0
Asymmetric top I 1 = 1 , I 2 = 2 , I 3 = 3 ω = ( 0 , Ω , 0 ) Ω = 7 koi perturbation nahi. Yeh kya karta hai?
Forecast: Axis 2 unstable hai — lekin perfectly clean start ke saath, kya yeh phir bhi tumble karega?
Exact state plug in karo.
I 1 ω ˙ 1 = ( I 2 − I 3 ) ω 2 ω 3 = ( 2 − 3 ) ( Ω ) ( 0 ) = 0 , I 2 ω ˙ 2 = ( I 3 − I 1 ) ω 3 ω 1 = ( 3 − 1 ) ( 0 ) ( 0 ) = 0 , I 3 ω ˙ 3 = ( I 1 − I 2 ) ω 1 ω 2 = ( 1 − 2 ) ( 0 ) ( Ω ) = 0. Yeh step kyun? Har right side mein ek zero component ka factor hai, toh saare derivatives vanish karte hain.Conclude karo. ω = ( 0 , 7 , 0 ) Yeh step kyun? Pure principal-axis spin hamesha Euler's equations ka equilibrium hota hai, stability ke bawajood.Instability se reconcile karo. Equilibrium unstable hai: koi bhi real ω 1 ω 3 e k Ω t Yeh step kyun? Fixed point =
Verify: Teeno ω ˙ i = 0 ( 0 , 7 , 0 ) 2 T = I 2 Ω 2 = 2 ( 49 ) = 98 L 2 = I 2 2 Ω 2 = 4 ( 49 ) = 196 ω
Earth almost ek symmetric top hai jisme I ⊥ I 3 − I ⊥ ≈ 305 1 ω 3 = 2 π / ( 86164 s )
Forecast: Days? Ek mahina? Lagbhag ek saal? Order of magnitude guess karo.
Ex 2 se symmetric-top precession rate use karo:
Ω b = I ⊥ I 3 − I ⊥ ω 3 = 305 ω 3 . Yeh step kyun? Earth ka tiny equatorial bulge I 3 = I ⊥ Period mein convert karo. Wobble period hai
T wob = Ω b 2 π = 305 ⋅ ω 3 2 π = 305 × ( 1 sidereal day ) ≈ 305 days . Yeh step kyun? 2 π / ω 3 305 Reality se compare karo. Rigid prediction ~305 days hai; observed Chandler wobble ~433 days hai. Gap isliye hai kyunki Earth perfectly rigid nahi hai (elastic crust + fluid core period lengthen karte hain).
Yeh step kyun? Kyunki hamari derivation perfectly rigid body assume karti hai; model ko data se compare karna exactly woh assumption isolate karta hai jo fail hoti hai aur dikhata hai ki idealized model kya chhod deta hai — prediction ko measurement ke against check karne ki essential habit.
Verify: T wob = 305 = 305 × 86164 s ≈ 2.628 × 1 0 7 s ≈ 304.2
Ek tricky exam deta hai I 1 = 5 , I 2 = 1 , I 3 = 3 sorted nahi! ). Body mainly axis 3 ke baare mein spin kar rahi hai. Kya axis 3 stable hai? Label par trust mat karo — sign check karo.
Forecast: "Axis 3 sabse bada hai ⇒ stable" — kya yeh reasoning yahan safe hai?
Moments rank karo. Sorted: I 2 = 1 < I 3 = 3 < I 1 = 5 intermediate moment hai, bhaale uska label "3" ho.
Yeh step kyun? Stability depend karta hai ki moment smallest/largest/middle hai ya nahi — value par, index par nahi.Axis 3 ke baare mein spin ke liye k Axis 3 ke baare mein bade spin ke liye, pure-number stability constant (Ex 3 ke pattern se) hai
k = I 1 I 2 ( I 2 − I 3 ) ( I 3 − I 1 ) = 5 ⋅ 1 ( 1 − 3 ) ( 3 − 5 ) = 5 ( − 2 ) ( − 2 ) = 5 4 = 0.8. Yeh step kyun? Dono factors negative hain, toh unka product positive hai ⇒ k > 0 Conclude karo. k = 0.8 > 0 UNSTABLE hai. "Largest index" label ne tumhe fool kiya; intermediate value jeetti hai. Growth rate = k Ω = 0.8 Ω ≈ 0.894 Ω Yeh step kyun? Yahi trap hai parent note ke Mistake C mein — product ( I a − I b ) ( I b − I c )
Verify: Generic Ω = Ω 0.8 ≈ 0.894 Ω s − 1 k = 0.8 I = 1 I = 5
Recall Self-test — answers cover karoI = ( 1 , 2 , 3 ) I 2 = 2 ( I 3 − I 2 ) ( I 2 − I 1 ) > 0 k > 0 I 1 = I 2 = 2 , I 3 = 5 , ω 3 = 4 Ω b = I ⊥ I 3 − I ⊥ ω 3 = 2 3 ⋅ 4 = 6 k Ω 2 k k Ω 2 Ω 2 ω ˙ i = 0
Mnemonic One-line takeawayI ( I a − I b ) ( I b − I c ) label ek decoy hai.