Question bank — 6DOF equations — translational (Newton), rotational (Euler's equations)

False. Newton's law sirf inertial frame mein acceleration ke liye valid hai; body-frame rate ko transport theorem se correct karna padta hai, jo $\vec{\omega }\times \vec{v}$ term add karta hai.

False. Constant speed par agar rocket turn kar raha hai toh velocity ki direction rotate ho rahi hai, isliye $\vec{\omega }\times \vec{v}\ne 0$ aur woh inertial frame mein genuinely accelerate karta hai.

Saamaanyatah False. Sirf $\vec{H}$ (inertial frame mein) aur kinetic energy conserved hoti hai; agar $I_x\ne I_y\ne I_z$ toh coupling terms $p,q,r$ ko energy trade kara dete hain aur body tumble karti hai (Dzhanibekov effect).

True. Ek principal axis ke along $\vec{\omega }\times (\mathbf{I}\vec{\omega })=0$ aur $\vec{M}=0$, isliye $\dot{\vec{\omega }}=0$ — yeh spin stabilization ki buniyad hai.

False. Rigid body se chipki honi ke karan, body frame mein $\mathbf{I}$ constant hai; exactly isliye Euler's equations wahan likhe jaate hain. Yeh $\mathbf{I}$ inertial frame mein express karne par rotate hoti hai aur time-varying ho jaati hai.

False. Euler's law $\vec{M}=\mathbf{I}\dot{\vec{\omega }}{|}_B+\vec{\omega }\times (\mathbf{I}\vec{\omega })$ kisi bhi body axes ke liye valid hai; principal axes chunna sirf $\mathbf{I}$ ko diagonalize karta hai aur clean scalar form deta hai.

Aadha sach. Ye kinematically "CoM ki motion" + "CoM ke baare mein rotation" mein decouple hote hain, lekin physics ke through re-couple ho jaate hain: thrust misalignment torque create karta hai, aur — concretely — jab body pitch karti hai, uski nose velocity vector se door point karti hai, jisse angle of attack $\alpha$ banta hai; aerodynamic lift/drag $\alpha$ ke saath scale karte hain, isliye attitude (ek rotational state) Newton ki equation mein force ko directly reset karta hai.

True. Dono ek hi theorem se aate hain — transport theorem — body components mein store kiye gaye momentum vector par apply kiya gaya; sirf stored vector alag hai ($m\vec{v}$ vs $\mathbf{I}\vec{\omega }$).

$\dot{m}\vec{v}$ term thrust nahi hai; ejected gas momentum le jaati hai, isliye sahi variable-mass form yeh hai: ${\vec{F}}_{ext}+\dot{m}{\vec{v}}_{exh,rel}=m\dot{\vec{v}}$. Sabse clean fix: thrust ko ek external force maano aur constant-mass Newton use karo.

Coupling term $(I_z-I_y)qr$ genuinely wahan hai, lekin ek axisymmetric rocket ke liye $I_y=I_z=I_t$ ise exactly zero bana deta hai, isliye $\dot{p}=0$ exact hai — approximation nahi.

Galat — $\dot{\vec{\omega }}=0$ $\mathbf{I}\dot{\vec{\omega }}$ term ko khatam karta hai, lekin $\vec{\omega }\times (\mathbf{I}\vec{\omega })$ wahan rehta hai aur woh torque ke barabar hai jo off-axis constant spin hold karne ke liye chahiye.

Sirf tab jab $\vec{\omega }$ ek principal axis ke along ho. Saamaanyatah $\mathbf{I}$ ek tensor hai, isliye $\vec{H}$ aur $\vec{\omega }$ alag-alag directions mein point karte hain — yahi coning aur precession ka source hai.

Neglect nahi kiya — yeh exactly zero hai kyunki humne $p=r=0$ se start kiya tha. Pitch cleanly decouple hoti hai sirf uss initial condition ki wajah se, approximation se nahi.

Internal forces 3rd law ke anusaar pairs mein cancel hoti hain; unke torques tab cancel hote hain jab har pair central ho (do particles ko milane wali line ke along kaam kare), toh moment arms match karte hain. Woh (usually-satisfy hone wali) condition hi hai jo ${\vec{F}}_{ext}=\dot{\vec{p}}$ aur ${\vec{M}}_{ext}=\dot{\vec{H}}$ ko valid banati hai — yeh justification hai, koi re-drop karne wali cheez nahi.

Kyunki body frame mein inertia tensor constant hai; inertial frame mein $\mathbf{I}$ rotate karta hai aur uska time-derivative single cross-product term se kahin zyada equation ko cluttered kar deta.

Translation mein inertia ek single scalar $m$ hai; rotation mein resistance ek tensor $\mathbf{I}$ hai jo har axis ke baare mein alag hai aur teen rates $p,q,r$ ko couple karta hai, gyroscopic effects produce karta hai jinka translation mein koi analogue nahi.

$I_y=I_z=I_t$ ke saath transverse equations $\dot{q}=\lambda r,\dot{r}=-\lambda q$ ban jaate hain, jo $(q,r)$ ka pure rotation hai — spin axis frequency $|\lambda |$ par ek clean cone trace karta hai, isliye sounding rockets spin-stabilize karte hain.

Clean constant-mass Newton form rakhne ke liye aur Meshchersky bookkeeping trap se bachne ke liye; ejected exhaust ki momentum ko ek single thrust vector $\vec{T}=-\dot{m}{\vec{u}}_e$ mein package kiya jaata hai.

Kyunki Newton's rotational law hai $\vec{M}=d\vec{H}/dt{|}_I$; $\vec{M}=0$ ke saath yeh $\vec{H}$ hai jo inertial frame mein fixed rehta hai. Body-frame $\vec{\omega }$ phir bhi vary kar sakta hai kyunki $\mathbf{I}$ unhe non-trivially link karta hai. Dekho Angular momentum conservation.

Dono ek hi theorem — transport theorem — ko body components mein store kiye gaye momentum vector par apply karne se aate hain; sirf stored vector alag hai ($m\vec{v}$ vs $\mathbf{I}\vec{\omega }$).

Finite rotations vectors ki tarah add nahi hote, isliye $\vec{\omega }$ ko ek proper kinematic equation mein feed karna padta hai — Euler angles ya quaternions — orientation ko bina singularities ke track karne ke liye.

Saare differences $(I_z-I_y)$ etc. vanish ho jaate hain, isliye $\vec{M}=\mathbf{I}\dot{\vec{\omega }}$ — rotation utna hi simply behave karta hai jitna translation, $\vec{M}=0$ hone par $\vec{\omega }$ constant rehta hai.

Cross term bilkul drop ho jaata hai, sirf $\vec{F}=m\dot{\vec{v}}{|}_B$ bachta hai, jo elementary point-mass Newtonian motion mein reduce ho jaata hai — Ex 1 mein sanity-check limit.

Motion unstable hoti hai: choti perturbations exponentially grow karti hain aur body flip ho jaati hai — intermediate axis theorem. Sabse bade ya sabse chote axis ke baare mein spin stable hoti hai.

$p=p_0$ constant rehta hai, isliye coning frequency $\lambda =\frac{I_x-I_t}{I_t}{p}_0$ bhi fixed rehti hai — ek steady, predictable cone, tri-axial tumbling case se alag.

Iska sign $(I_x-I_t)$ se aata hai: prolate (pencil-jaisi) body ke liye $I_x<I_t$ isliye $\lambda <0$ aur transverse rates ek taraf rotate karte hain; oblate (disk-jaisi) body ke liye $I_x>I_t$ isliye $\lambda >0$ aur woh ulti taraf rotate karte hain. $|\lambda |$ cone ki speed set karta hai, sign batata hai ki coning spin ke relative prograde hai ya retrograde.

Zaroor nahi — jaise fuel drain hota hai CoM shift karta hai, moment arms aur inertia tensor ko badalta hai; strictly body-frame $\mathbf{I}$ aur CoM sirf quasi-constant hain aur burn ke dauran update karne padte hain.

Dono cross terms vanish ho jaate hain, sirf $\vec{M}=\mathbf{I}\dot{\vec{\omega }}$ bachta hai — rest se pure angular acceleration, exactly Ex 3 mein start-of-maneuver ka clean case.