Mass properties — CG location, inertia tensor changing with propellant depletion: Visual walkthrough

3.4.6 · D2 · Physics › Rocket Flight Mechanics › Mass properties — CG location, inertia tensor changing with

Shuru karne se pehle, ek promise: hum koi bhi symbol use nahi karenge jab tak woh pehle ek picture na ho. Isliye hum sabse primitive cheez se shuru karte hain — ek heavy dot jo ek stick par kahin baitha hai.

Step 1 — Stick par do heavy dots

KYA. Rocket ko flat lita lo aur edge-on dekho. Curves aur tanks bhool jao. Yeh actually sirf do lumps of mass hain jo ek line par do jagah baithe hain: dry structure (engine + payload + skin) aur propellant (fuel). Rocket ki length ke saath ek ruler draw karo — us ruler ki reading ko station kaho, $x$ likho, nose se metres mein measure karo.

$m_{d}$ = dry part ki mass (kilograms mein ek number).
$x_{d}$ = woh station (ruler mark) jahan dry part akele balance karta hai.
$m_{p}$ = propellant ki mass.
$x_{p}$ = woh station jahan propellant akele balance karta hai.

Bas itna hi cast of characters hai.

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

YEH YAHAAN SE KYU shuru karein. Parent note ka daraauna integral $\int r d m$ asliyat mein kuch nahi hai sirf "bahut saare tiny dots ko add karo" ke alawa. Agar hum do dots master kar lein, toh integral wahi kaam ek million dots ke saath karta hai. Pehle atom master karo.

Step 2 — "Balance point" ka matlab kya hai?

KYA. Stick ke neeche ek knife-edge (pivot) rakho aur tab tak slide karo jab tak stick kisi bhi taraf tip na kare. Woh pivot station centre of gravity hai, $x_{cg}$ likha jaata hai. "Tip nahi karta" ka matlab hai ki left lump ki turning effort right lump ki turning effort ko exactly cancel kar deti hai.

Ek lump ki turning effort = uski mass $\times$ uska lever arm (woh pivot se kitni door baitha hai). Pivot ke left wala lump ek taraf tip karta hai; right wala doosri taraf. Balance = dono cancel ho jaate hain:

$dry mass m_{d} lever arm, signed (x_{d} - x_{cg}) + fuel mass m_{p} lever arm, signed (x_{p} - x_{cg}) = 0$

Har term hai $mass \times pivot se signed distance$ — ek turning effort.
Distance signed hai: negative agar lump pivot ke left mein hai, positive agar right mein. Wahi sign dono efforts ko cancel karne deta hai.
Puri sum $0$ hona mathematically "yeh balance karta hai" likhne ka tarika hai.

YEH equation kyun, koi aur kyun nahi. Humne CG ko ek formula se define nahi kiya aur umeed ki ki balance karega — humne balance demand kiya (turning efforts ka sum $= 0$ ) aur ab algebra ko bataane denge ki pivot kahan hona chahiye. Yahi honest order hai.

Step 3 — Pivot ke liye solve karo: CG formula saamne aata hai

KYA. Balance equation lo aur sirf $x_{cg}$ isolate karo. Multiply out karo:

$m_{d} x_{d} - m_{d} x_{cg} + m_{p} x_{p} - m_{p} x_{cg} = 0.$

Do $x_{cg}$ pieces ko ek side gather karo:

$m_{d} x_{d} + m_{p} x_{p} = total mass M (m_{d} + m_{p}) x_{cg} .$

Total mass $M = m_{d} + m_{p}$ se divide karo:

$x_{cg} = \frac{m _{d} x _{d} + m _{p} x _{p}}{m _{d} + m _{p}}$

Top line $m_{d} x_{d} + m_{p} x_{p}$ : har station apne lump ki heaviness ke hisaab se weighted hai. Ek heavy lump answer ko strongly apni station ki taraf kheenchta hai.
Bottom line $m_{d} + m_{p}$ : total mass, toh result ek proper average hai (weights ka sum 1 hai).
Result $x_{cg}$ : ek single station — mass-weighted average position.

YEH aisa kyun dikhta hai. Yeh ek weighted average hai. Agar fuel dry part se bahut zyada heavy hai, toh average fuel ke upar almost baithta hai. Yeh wahi formula hai jo parent note ka $r_{cg} = \frac{1}{M} \sum m_{i} r_{i}$ hai — humne ise sirf do dots ke liye haath se derive kiya.

Step 4 — CG ko march karte dekho jab fuel jalta hai

KYA. Ab $m_{p}$ ko zero ki taraf drop karne do aur formula dobara run karo. Parent note ke numbers use karo: $m_{d} = 1000 kg$ at $x_{d} = 6 m$ , full fuel $m_{p 0} = 4000 kg$ at $x_{p} = 2 m$ .

moment	fuel $m_{p}$	$x_{cg}$
liftoff	$4000$	$\frac{1000 ( 6 ) + 4000 ( 2 )}{5000} = 2.8 m$
half-burnt	$2000$	$\frac{1000 ( 6 ) + 2000 ( 2 )}{3000} = 3.33 m$
burnout	$0$	$\frac{1000 ( 6 )}{1000} = 6 m$

PICTURE. Pivot $2.8 m$ (fuel tank ke andar) se slide karke poori tarah $6 m$ tak (dry structure par) pahunch jaata hai. Yeh us mass ko chase karta hai jo peechhe rah jaata hai.

YEH nose ki taraf kyun move karta hai, tail ki taraf kyun nahi. Log expect karte hain ki CG exhaust ke saath peechhe jaayega. Lekin formula sirf woh mass jaanta hai jo abhi bhi board par hai. Jab $m_{p} \to 0$ , top line apna $m_{p} x_{p}$ term kho deta hai aur pura average $x_{d}$ par collapse ho jaata hai. Jo bacha hua mass hai woh jeetta hai. Yeh parent note ki classic mistake kill karta hai.

Step 5 — Balancing se twisting ki taraf: doosra number kyun chahiye

KYA. Balancing batata hai kahan push karna hai. Yeh nahi batata ki rocket spin karne par kitna resist karta hai. Do rockets ek hi CG share kar sakte hain lekin ek whip around karta hai aur doosra barely budge karta hai. Woh number jo "resistance to spin" measure karta hai woh moment of inertia hai, $I$ likha jaata hai (pitch axis ke liye, $I_{y y}$ ).

Spin axis se $d$ distance par baithay ek single dot of mass $m$ ke liye:

$I = m d^{2} .$

$m$ : dot ki mass — zyada mass, spin karna mushkil.
$d$ : axis se distance jiske baare mein tum spin kar rahe ho. Yahi pivot hai jo humne abhi find kiya, $x_{cg}$ .
$d^{2}$ : squared! Distance mass se zyada matter karta hai. Arm double karo aur inertia chaar guna ho jaati hai.

Square kyun, aur yeh tool kyun. Jab rocket ek tiny angle rotate karta hai, axis se door ka dot ek lamba arc sweep karta hai aur bahut speed pick karta hai; axis ke paas ka dot barely move karta hai. Spin ki kinetic energy $\frac{1}{2} m v^{2}$ ke saath add hoti hai jahan $v = d ω$ , isliye $d$ do baar aata hai — ek baar speed ke liye, ek baar arc ke liye. Wahi squared distance exactly isliye hai ki axis se door fuel rakho toh rocket sluggish hota hai, aur drain karo toh rocket suddenly nimble ho jaata hai. (Full tensor form aur $L = I ω$ ki kahani Rigid Body Rotational Dynamics mein hai.)

Step 6 — Dots add karo, lekin sirf ek common pivot ke baare mein

KYA. Rocket ke paas do lumps hain, toh do $m d^{2}$ terms add karo — har distance ek hi pivot $x_{cg}$ se measure ki gayi ho:

$I_{y y} = dry lump about CG m_{d} (x_{d} - x_{cg})^{2} + fuel lump about CG m_{p} (x_{p} - x_{cg})^{2}$

Har bracket $(x - x_{cg})$ us lump ka arm hai balance point se.
Squared, isliye dono terms positive hain — spin resistance kabhi subtract nahi hoti.
Woh add hote hain kyunki ek shared axis ke baare mein inertia simply additive hai.

"Same pivot" kyun non-negotiable hai. Tum ek lump ki nose ke baare mein measured resistance ko doosre ki tail ke baare mein measured se add nahi kar sakte — yeh apples aur oranges add karna hai. Woh rule jo inertia ko lump ke apne centre se shared pivot par move karne deta hai woh parallel-axis (Steiner) theorem hai, $I_{about pivot} = I_{own centre} + m d^{2}$ . Point-like lumps ke liye "own centre" part $0$ hai, sirf $m d^{2}$ bachta hai. (Deeper: Parallel-Axis Theorem.)

Liftoff numbers plug in karo ( $x_{cg} = 2.8$ ):

$I_{y y} = 1000 (6 - 2.8)^{2} + 4000 (2 - 2.8)^{2} = 1000 (10.24) + 4000 (0.64) = 12800 kg\cdotm^{2} .$

Step 7 — Inertia collapse dekho, aur autopilot ko react kyun karna padta hai

KYA. $I_{y y}$ ko fuel drain hone ke saath re-run karo, har baar Step 4 ka naya $x_{cg}$ use karte hue.

moment	$x_{cg}$	$I_{y y}$ (point model)
liftoff	$2.8$	$12800 kg\cdotm^{2}$
half-burnt	$3.33$	$1000 (6 - 3.33)^{2} + 2000 (2 - 3.33)^{2} \approx 10667 kg\cdotm^{2}$
burnout	$6$	$1000 (0)^{2} = 0$ (point model)

PICTURE. Inertia bar dramatically shrink hoti hai. Point model mein burnout par $0$ hit karti hai kyunki humne pretend kiya tha ki dry structure khud axis par ek dot hai — ek real rocket apni dry-body inertia $I_{d}^{c g}$ rakhta hai, isliye true curve ek small positive floor par flatten hoti hai, zero par nahi.

YEH chapter ka poora point kyun hai. Rotational law kehta hai torque $τ = I \overset{ω}{˙}$ . Agar wahi control torque $τ$ ek aise inertia $I$ se milta hai jo collapse ho gayi hai, toh angular acceleration $\overset{ω}{˙} = τ / I$ blow up ho jaata hai — rocket usi nudge ke liye bahut zyada teezi se spin karta hai. Heavy liftoff rocket ke liye tune kiya hua autopilot ab over-steer karega aur oscillate karega. Isliye controller apna gain schedule $I (t)$ ke against karta hai. Yeh exactly Gain Scheduling in Autopilots hai, seedha is figure se driven, aur yeh Thrust Vector Control ke liye authority budget set karta hai.

Step 8 — Do edge cases jinke liye tayaar rehna chahiye

KYA — degenerate case A: bilkul fuel nahi ( $m_{p} = 0$ ). CG formula ban jaata hai $x_{cg} = \frac{m _{d} x _{d}}{m _{d}} = x_{d}$ , aur fuel inertia term vanish ho jaata hai. Sab kuch bare dry body par reduce ho jaata hai. Achha — formula gracefully degrade karta hai.

KYA — degenerate case B: fuel exactly dry station par ( $x_{p} = x_{d}$ ). Toh dono lumps ek hi station par baithte hain, fuel burn hone par CG kabhi move nahi karta, aur $(x_{p} - x_{cg}) = (x_{d} - x_{cg})$ isliye inertia sirf mass ke saath scale karta hai. Is tarah se built rocket ko koi CG scheduling ki zaroorat nahi hogi — real rockets itne lucky kabhi nahi hote kyunki tanks aur engines alag-alag ends par rehte hain.

YEH kyun cover karte hain. Agar tumhara formula kisi limiting case mein divide-by-zero ya nonsense negative inertia de, toh woh galat hoga. Extremes test karna hi woh tarika hai jisse tum general result par trust karte ho. (Off-axis effects jaise neat symmetry todna wala fuel sloshing Propellant Slosh Dynamics mein handle hote hain; woh mass budget jo burn ko velocity gain se jodhta hai woh Tsiolkovsky Rocket Equation hai.)

Ek-picture summary

Is page par sab kuch, compressed: top track CG dot ko nose-ward slide karte dikhata hai jab fuel bar drain hota hai; bottom track inertia bar ko lock-step mein shrink hote dikhata hai. Ise left-to-right padho jaise "burn ke dauran time."

Recall Feynman: puri walkthrough seedhe shabdon mein

Rocket ko flat lao: yeh ek ruler par do heavy dots hain — dry body aur fuel. Uske neeche ek knife slide karo jab tak balance na ho; woh spot CG hai, aur woh whichever dot heavier hai uske paas utarta hai. Kyunki fuel heavy one hai, CG tank ke paas shuru hota hai. Ab fuel jalao: heavy dot fade hota jaata hai, toh balance point jo bacha hai uski taraf slide karta hai — aage dry body ki taraf, exhaust ki taraf nahi. Doosra sawaal: spin karna kitna mushkil hai? Har dot spin resist karta hai mass times pivot se distance squared se — door ke dots bahut zyada fight karte hain. Dono dots ko ek hi balance point ke baare mein add karo. Jab door, heavy fuel disappear hota hai, woh spin-resistance collapse ho jaati hai, isliye wahi steering nudge ab rocket ko bahut teezi se flip karta hai. Flight computer dono numbers har instant watch karta hai aur apna steering gain neeche karta hai jab rocket light aur twitchy ho jaata hai. Balance point leftover ko chase karta hai; inertia shrink hoti hai — lekin tum hamesha dono ko naye pivot ke baare mein re-measure karte ho.

Active recall

Recall Quick self-test

Kon sa lump CG ke zyada paas hota hai, aur kyun? ::: Heavier lump — yeh ek mass-weighted average hai, isliye ek bada mass average ko apni taraf kheenchta hai. Fuel zero tak drain hone par CG kahan pahunchta hai? ::: Dry station $x_{d}$ par — leftover mass jeetta hai; yeh nose-ward move karta hai, nozzle ki taraf nahi. $I = m d^{2}$ mein distance squared kyun hai? ::: Ek door ka dot ek lamba arc sweep karta hai AUR teezi se move karta hai ( $v = d ω$ ), isliye $d$ do baar aata hai. Har lump ki inertia ek hi pivot ke baare mein kyun measure honi chahiye? ::: Inertia sirf ek common axis ke baare mein additive hai; Steiner ( $I = I_{c g} + m d^{2}$ ) har lump ko us shared CG par move karta hai. Autopilot burnout ke paas apna gain kyun lower karta hai? ::: $\overset{ω}{˙} = τ / I$ ; jab $I$ collapse hoti hai toh wahi torque light rocket ko over-rotate karta hai, isliye gain drop karna padta hai. "Kam fuel" $I (t)$ ke liye kaun se do updates maangta hai? ::: Moved CG recompute karo, phir us naye CG ke baare mein har $m d^{2}$ redo karo.