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

2.1.21 · Physics › Analytical Mechanics

Intuition Bada picture

Ek rigid body ke 6 degrees of freedom hote hain: 3 uske center of mass ki position ke liye, aur 3 uske orientation ke liye. Translation easy hai (CM ke liye Newton's law). Mushkil part hai rotation. Do sawal:

Hum orientation describe kaise karein? → Euler angles $(ϕ, θ, ψ)$ batate hain ki fixed lab frame se body frame mein 3 successive rotations se kaise jaayein.
Orientation time ke saath kaise change hoti hai? → Euler's equations torque ko angular velocity se relate karte hain, body frame mein likhe jaate hain jahaan inertia tensor constant (diagonal) hota hai.

WHY is tarah split karte hain? Lab frame mein inertia tensor har pal change hota hai (body tumble karti hai), toh $L = I ω$ ek nightmare hai. Body frame mein $I$ frozen aur principal axes ke along diagonal hota hai — clean. Price yeh hai: body frame rotate karta hai, toh $d / d t$ mein ek extra $ω \times$ term aa jaata hai.

1. Euler Angles — orientation describe karna

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

Euler angles se angular velocity

Derivation

$ω$ ko $\dot{ϕ}, \dot{θ}, \dot{ψ}$ se banana Har Euler rate ek specific axis ke baare mein ek rotation hai. Total angular velocity vector sum hai: $ω = \dot{ϕ} \hat{Z} + \dot{θ} \overset{n}{^} + \dot{ψ} \overset{z}{^}$ jahaan $\hat{Z}$ space- $z$ hai, $\overset{n}{^}$ line of nodes hai, $\overset{z}{^}$ body- $z$ hai.

WHY sirf add karte hain? Angular velocities vectors ki tarah add hoti hain (infinitesimal rotations commute karte hain). HOW body frame mein components paana: rotations use karke har axis ko body coordinates mein express karo. Standard result (3-1-3 convention):

\omega_1 &= \dot\phi\sin\theta\sin\psi + \dot\theta\cos\psi\\ \omega_2 &= \dot\phi\sin\theta\cos\psi - \dot\theta\sin\psi\\ \omega_3 &= \dot\phi\cos\theta + \dot\psi \end{aligned}}$$ $\omega_3$ notice karo: spin $\dot\psi$ plus precession ka body axis pe projection $\dot\phi\cos\theta$. Yahi term fast top ke liye bhi survive karti hai.

2. Rotating-frame derivative (key lemma)

3. Euler's Equations of Motion

Derivation

$τ = d L / d t$ se Euler's equations tak Start (first principles): CM ke baare mein (ya ek fixed point ke), torque rate of change of angular momentum ke barabar hai inertial (space) frame mein: $τ = (\frac{d L}{d t})_{space}$

Step 1 — body frame mein switch karo transport theorem use karke: $τ = (\frac{d L}{d t})_{body} + ω \times L$ Yeh step kyun? Body frame mein inertia tensor constant hai, toh hum cleanly differentiate kar sakte hain.

Step 2 — principal axes use karo, jahaan $I = diag (I_{1}, I_{2}, I_{3})$ hai toh $L_{i} = I_{i} ω_{i}$ hai aur $I_{i}$ constant hain: $(\frac{d L _{i}}{d t})_{body} = I_{i} \overset{ω}{˙}_{i}$ Yeh step kyun? Principal axes choose karne se $I$ diagonalize ho jaata hai — saara off-diagonal mess khatam.

Step 3 — cross product expand karo $(ω \times L)_{1} = ω_{2} L_{3} - ω_{3} L_{2} = (I_{3} - I_{2}) ω_{2} ω_{3}$ , etc.

Result — Euler's equations:

I_1\dot\omega_1 - (I_2-I_3)\,\omega_2\omega_3 &= \tau_1\\ I_2\dot\omega_2 - (I_3-I_1)\,\omega_3\omega_1 &= \tau_2\\ I_3\dot\omega_3 - (I_1-I_2)\,\omega_1\omega_2 &= \tau_3 \end{aligned}}$$

Worked example Torque-free symmetric top (

$I_{1} = I_{2} \neq = I_{3}$ , $τ = 0$ ) Yeh case kyun? Yeh sabse clean non-trivial solution hai aur Earth ki free precession, ek frisbee, ek tumbling satellite ko model karta hai.

Equation 3: $I_{3} \overset{ω}{˙}_{3} = (I_{1} - I_{2}) ω_{1} ω_{2} = 0 \Rightarrow ω_{3} = const$ . Kyun? $I_{1} = I_{2}$ teesra coupling term kill kar deta hai — symmetry axis ke baare mein spin constant hai.

Equations 1,2 ban jaati hain, $Ω \equiv \frac{I _{3} - I _{1}}{I _{1}} ω_{3}$ ke saath (ek constant): $\overset{ω}{˙}_{1} = - Ω ω_{2}, \overset{ω}{˙}_{2} = + Ω ω_{1}$ Kyun? $ω_{3}$ ko constant substitute karne se coupled pair simple harmonic rotation mein convert ho jaati hai.

Solution: $ω_{1} = A cos Ω t, ω_{2} = A sin Ω t$ . $ω$ ka transverse part body-3 axis ke baare mein rate $Ω$ pe circles karta hai — yeh hai body-frame free precession. Earth ke liye: $Ω \approx \frac{C - A}{A} ω_{3} \Rightarrow$ Chandler-wobble-type period.

4. Common mistakes

Recall Feynman: ek 12-year-old ko explain karo

Socho ek spinning top. Kis taraf point kar raha hai yeh batane ke liye tumhe teen twists chahiye: ise clock hand ki tarah ghuma do ( $ϕ$ ), ise tipa do ( $θ$ ), aur ise apne point pe spin karo ( $ψ$ ). Woh teen "twist ki matra" hain Euler angles. Ab, yeh move kaise karta hai? Agar tum ek chota sa bug ho jo TOP PE sawaar ho, toh top tumhare liye hamesha same dikhta hai (uska weight hamesha same tarah balanced hai), toh math bug ke liye ground pe khade aadmi se simpler hai. Euler's equations bug ki rulebook hai. Funny part: jab bhi koi top ko push nahi karta, bug phir bhi apna spin axis swing hote feel karta hai — yeh isliye hai kyunki bug ki poori duniya rotate kar rahi hai. Woh swinging hi reason hai ki phone "wrong way" spin hokar toss karne pe khud ko flip kar leta hai.

Connections

Inertia tensor and principal axes — kyun hum $I$ diagonalize kar sakte hain.
Angular momentum in rotating frames — transport theorem.
Lagrangian of the symmetric top — same physics $L (ϕ, θ, ψ)$ se.
Gyroscope precession — $ω \times L$ term ka direct application.
Coriolis and centrifugal forces — same rotating-frame derivative idea.
Chandler wobble — Earth ki torque-free precession.

Flashcards

Ek free rigid body ke kitne degrees of freedom hote hain, aur kaise split hote hain?

6 total: 3 translational (CM position) + 3 rotational (orientation, e.g. Euler angles).

Teen Euler angles ke naam aur unka physical meaning batao (3-1-3).

ϕ

precession (space-Z ke baare mein),

θ

nutation (line of nodes ke baare mein),

ψ

spin (body-z ke baare mein).

Euler's equations body frame mein kyun likhi jaati hain, space frame mein kyun nahi?

Kyunki body frame mein inertia tensor constant (aur principal axes pe diagonal) hota hai; space mein body ke tumble karne se woh change hota rehta hai.

Transport theorem state karo.

(d A / d t)_{s p a ce} = (d A / d t)_{b o d y} + ω \times A

Axis 1 ke liye Euler's equation likho.

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

Gyroscopic coupling terms kya hain aur kab vanish hote hain?

(I_{i} - I_{j}) ω_{j} ω_{k}

terms; yeh spherical top (

I_{1} = I_{2} = I_{3}

) ke liye ya ek single principal axis ke baare mein spin karne pe vanish hote hain.

Torque-free symmetric top ke liye

ω_{3}

kya hai?

Constant, kyunki teesri Euler equation

I_{3} \overset{ω}{˙}_{3} = 0

deti hai.

Body frame mein symmetric top ki free precession rate?

Ω = \frac{I _{3} - I _{1}}{I _{1}} ω_{3}

Tennis-racket theorem mein kaun sa principal axis unstable rotation deta hai?

Intermediate-moment axis (

I_{2}

jab

I_{1} < I_{2} < I_{3}

(\dot{ϕ}, \dot{θ}, \dot{ψ}) = (ω_{1}, ω_{2}, ω_{3})

kyun nahi hota?

Euler rotation axes mutually orthogonal nahi hain; tumhe unhe body axes pe project karna padta hai (isliye

sin θ, cos θ

factors aate hain).

Euler-angle rates ke terms mein

ω_{3}

express karo (3-1-3).

ω_{3} = \dot{ϕ} cos θ + \dot{ψ}