Question bank — Mass properties — CG location, inertia tensor changing with propellant depletion

False — CG us mass ki taraf drift karta hai jo peeche reh jaati hai, aur zyaadatar rockets mein yahi dry structure (engine + payload) hota hai jo tanks ke aage hota hai. Peeche se fuel jaane ka matlab hai ki average ko peeche kheenchne wali mass kam ho gayi, isliye CG aage jaata hai.

False — yahan inertia mass ke proportional nahi hai. Fuel hatane se CG bhi move karta hai, jo har Steiner arm $d$ ko change karta hai, isliye har bacha hue body ka $M{d}^2$ contribution bhi change ho jaata hai. Aapko CG recompute karna hoga aur parallel-axis shift dobara run karni hogi, sirf scale nahi karna.

True — parallel-axis inertia CG par minimize hoti hai, isliye koi bhi parallel offset $d$ sirf positive quantity $M{d}^2$ add kar sakta hai. Doosri taraf jaane par (CG ki taraf) aap isse subtract karte ho, bas dono axes parallel hon aur $d$ unke beech measure ho.

Roughly true jab tak symmetry bani rahe — lekin real rockets mein asymmetric tank drain, sloshing propellant, aur canted engines se ye khatam ho jaati hai. Ye symmetry tod dete hain aur off-diagonals populate ho jaate hain, roll aur pitch ko cross-couple karte hue. Dekho Propellant Slosh Dynamics.

True — saari propellant jaane ke baad, sirf dry body bacha rehta hai, aur vehicle CG dry-body CG ke saath coincide karta hai (average karne ke liye koi doosra mass nahi). Toh $I_{veh}=I_d^{cg}$ with ek zero Steiner arm. Ek point-mass model galat $I=0$ deta hai kyunki ye dry body ki apni spatial extent ko ignore karta hai.

False — yahi equation definition hai. Hum ek aisa point demand karte hain jiske baare mein mass-weighted position offsets ka sum zero ho (balance point); jaani-pehchani formula ${\mathbf{r}}_{cg}=\frac{1}{M}\sum m_i{\mathbf{r}}_i$ toh bas us condition ko solve karne ka algebra hai.

False — ulta risk aata hai. Kam inertia matlab wahi control torque ek bada angular acceleration produce karta hai, isliye liftoff ke liye tune kiya gaya gain burnout par overshoot ya destabilize kar sakta hai. Isliye gains schedule kiye jaate hain; dekho Gain Scheduling in Autopilots.

Ye ek mass-weighted average hai, plain average nahi. $x_{cg}=\frac{m_d{x}_d+m_p{x}_p}{m_d+m_p}$; ek simple midpoint tabhi sahi hoga jab $m_d=m_p$ ho.

Steiner arm do axes ke beech ki distance hai, yani $d=|x_p-x_{cg}|$, na ki propellant ka absolute station $x_p$. Raw coordinate use karne se position aur offset mein confusion hoti hai.

Inertias sirf ek common point ke baare mein additive hoti hain. Aapko sum karne se pehle har body ko uske apne CG se shared vehicle CG tak Steiner-shift karna hoga; own-CG values directly add karne se saare $M{d}^2$ terms ignore ho jaate hain.

Sirf linear cross-terms ($2d_y\int y^{\prime }dm$, etc.) vanish hoti hain kyunki $\int {\mathbf{r}}^{\prime }dm=0$ CG se measured hoti hai. Quadratic term $M{d}^2$ bachti hai — parallel-axis theorem ka poora content yahi hai.

Constant thrust par bhi, CG migrate hone se moment arm change hoti hai, isliye CG ke baare mein control/misalignment torque change hota hai. TVC (dekho Thrust Vector Control) ko moving CG ke against apna deflection schedule karna hoga.

Sirf tab jab $\mathbf{\omega }$ ek principal axis (ek eigenvector of $\mathbf{I}$) ke saath ho. Nonzero products of inertia ke saath, $\mathbf{L}$ aur $\mathbf{\omega }$ alag directions mein point karte hain, gyroscopic cross-coupling produce karte hue — yahi reason hai ki products of inertia matter karte hain. Dekho Rigid Body Rotational Dynamics.

Kyunki fuel drain hone ke saath CG location aur inertia continuously change hoti hai, aur dono control response set karte hain ($\tau =I\dot{\omega }$, CG ke baare mein moment arms). Liftoff values par frozen ek controller burnout tak mistuned — sluggish ya unstable — ho jaata hai.

Kyunki $\int {\mathbf{r}}^{\prime }dm=0$ by definition CG ka — primed coordinates usse measured hain. Kisi non-CG reference se shift karo aur woh integral nonzero hai, toh cross-term nahi marti.

Kyunki har body ka inertia integral ek specific axis ke baare mein liya jaata hai; $r^2$ weighting us axis tak distance par depend karti hai. Har body ko Steiner-shift karke usi axis par laane ke baad hi integrals ek common measuring stick share karte hain aur legitimately summable ban jaate hain.

Hum ek linear map demand karte hain jo $\mathbf{\omega }$ ko $\mathbf{L}=\int \mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})dm$ bheje. Triple product expand karo aur $\mathbf{\omega }$ ko factor out karo toh exactly wahi bracketed operator force hota hai — yahi map hai, koi arbitrary definition nahi.

Cant us mirror symmetry ko todo deta hai jo $\int xydm$-type integrals ko cancel karti thi. Off-axis mass distribution (aur off-axis thrust line) nonzero off-diagonal $I$ terms introduce karta hai, roll aur pitch responses ko couple karte hue.

$\dot{\omega }=\tau /I$ ke saath, chhota $I$ vehicle ko same commanded torque par zyada violently react karta hai. Loop ko overshoot se bachane ke liye, controller ko unit error par kam torque command karni padti hai — yani lower gain.

Formula $\frac{m_d{x}_d+m_p{x}_p}{m_d+m_p}$, $m_d\to 0$ par $x_p$ tend karta hai — CG poori tarah propellant centroid par hoga, kyunki average karne ke liye kuch nahi bacha.

Ye $x_{cg}=x_d$ par collapse ho jaata hai: CG exactly dry structure ke station par land karta hai, kyunki bacha hua single mass dry body hai. Ye poore CG migration ka endpoint hai.

Haan — formula instantaneous hai. Har moment par aap current $m_p(t)$ aur current $x_p(t)$ plug in karte ho. Subtlety ye hai ki $x_p$ ab time-dependent hai (aur slosh isse dynamically move kar sakta hai), isliye isse har step par re-evaluate karna hoga, fixed nahi rakh sakte.

Tab dono Steiner arms $D_d=D_p=0$, isliye vehicle inertia bas har body ki own-CG inertia ka sum hai bina kisi parallel-axis contribution ke — ek degenerate case jahan fuel burn hone par CG move nahi karta.

Haan, agar bacha hua propellant ka centroid aisa drift kare ki uski mass-weighted pull shrinking $m_p$ se aage nikal jaaye — jaise ek tank jo upar se drain ho raha ho aur uska centroid peeche move ho raha ho. "Bache hue mass ki taraf" waala rule phir bhi hold karta hai; bas bacha hua wahan nahi tha jahan aapne assume kiya tha.

Uss particular spin ke liye $\mathbf{L}$ par kuch contribute nahi karte — $\mathbf{L}$, $\mathbf{\omega }$ ke parallel rehta hai. Lekin ye tensor mein abhi bhi exist karte hain aur jaise hi spin axis principal direction se tilt ho (jaise slosh disturbance ke baad), ye kaam aayenge.

Ye char instincts hain jo fail karte hain: "CG exhaust follow karta hai", "$I\propto m$", "own-CG inertias directly add karo", aur "halka matlab fly karna aasaan hai".