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

3.4.6 · D4 · Physics › Rocket Flight Mechanics › Mass properties — CG location, inertia tensor changing with

Recurring reference rocket:

Dry mass $m_{d} = 1000 kg$ fixed station $x_{d} = 6 m$ par.
Full propellant $m_{p 0} = 4000 kg$ , centroid $x_{p} = 2 m$ (fixed treat karo jab tak koi problem aur na kahe).

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

Figure s01 ko kaise padhein: horizontal white arrow $x$ -axis hai (nose $0$ par, aft positive, metres mein label kiya). Cyan lens fuselage hai. White dot $x = 6$ par dry mass hai; cyan dot $x = 2$ par propellant centroid hai; amber diamond $x = 2.8$ par full-load CG hai. Figure ka ek hi message hai: CG dono bodies ke beech baitha hai, bhaari wale ki taraf khicha hua (4000 kg fuel), isliye woh tank ke paas se shuru hota hai aur baad mein dry mass ki taraf aage kheenchta jayega.

Level 1 — Recognition

Goal: kya aap sahi formula pick kar sakte ho aur usse sahi se padh sakte ho?

Exercise 1.1

Do point-like sub-bodies ke liye center of gravity ki definition words aur symbols mein batao, aur identify karo ki isme konsi quantity time-dependent hai ek burning rocket ke liye.

Recall Solution

CG ek mass-weighted average position hai: $x_{cg} = \frac{m _{d} x _{d} + m _{p} x _{p}}{m _{d} + m _{p}} .$ Har symbol ka matlab: $m_{d}, x_{d}$ = dry mass aur uska station (dono fixed); $m_{p}, x_{p}$ = propellant mass aur uska centroid. Time-dependent quantity hai == $m_{p} (t)$ ==, yaani bacha hua propellant mass (aur, ek refined model mein, uska centroid $x_{p} (t)$ bhi). Jab $m_{p}$ girta hai, $x_{p}$ waale term ko diya gaya weight shrink hota hai, toh average $x_{d}$ ki taraf slide karta hai.

Exercise 1.2

Scalar parallel-axis theorem likhiye aur har symbol ko label karo. Konsa term zero hota hai jab aap inertia CG ke baare mein hi compute karte ho?

Recall Solution

$I_{P} = I_{cg} + M d^{2} .$

$I_{cg}$ = CG se guzarne wali axis ke baare mein moment of inertia.
$M$ = body ki total mass.
$d$ = CG axis se naye parallel axis tak ki perpendicular distance.
$I_{P}$ = us shifted axis ke baare mein moment of inertia.

Jab nayi axis CG axis hi ho, $d = 0$ , toh added term $M d^{2} = 0$ aur $I_{P} = I_{cg}$ . Distance hi woh cheez hai jo aapko inertia ka "cost" deti hai; CG par khadhe rehne ka koi cost nahi.

Level 2 — Application

Goal: real numbers ko ek tool mein plug karo aur clean answer lo.

Exercise 2.1

$x_{cg}$ nikalo ignition par ( $m_{p} = 4000$ ) aur 25% burnt point par ( $m_{p} = 3000$ ).

Recall Solution

Ignition par: $x_{cg} = \frac{1000 ( 6 ) + 4000 ( 2 )}{1000 + 4000} = \frac{6000 + 8000}{5000} = \frac{14000}{5000} = 2.8 m .$ 25% burnt par ( $m_{p} = 3000$ ): $x_{cg} = \frac{1000 ( 6 ) + 3000 ( 2 )}{1000 + 3000} = \frac{6000 + 6000}{4000} = \frac{12000}{4000} = 3.0 m .$ CG sirf ek-chauthai fuel jalane ke baad pehle hi $0.2 m$ aage khisak gaya hai ( $6 m$ par dry mass ki taraf).

Exercise 2.2

Point-mass model use karte hue (har body ki apni $I^{c g}$ ignore karo), 25% burnt point par ( $m_{p} = 3000$ , $x_{cg} = 3.0$ ) pitch inertia $I_{y y}$ nikalo.

Recall Solution

CG se offsets $x_{cg} = 3.0$ par: $D_{d} = ∣6 - 3.0∣ = 3.0 m, D_{p} = ∣2 - 3.0∣ = 1.0 m .$ $I_{y y} = m_{d} D_{d}^{2} + m_{p} D_{p}^{2} = 1000 (3.0)^{2} + 3000 (1.0)^{2} = 9000 + 3000 = 12000 kg\cdotm^{2} .$ Dhyan do ki dono bodies CG ke opposite sides par baithe hain, toh dono positively contribute karte hain — inertia distances squared add karta hai, sign-blind.

Level 3 — Analysis

Goal: sirf kisi quantity ki value nahi, ye reason karo ki woh kaise change hoti hai.

Exercise 3.1

CG $2.8 m$ (full) se $6.0 m$ (burnout) tak gaya. Dikhao ki ye migration zyada late burn mein hoti hai, early mein nahi. Pehli aadhi propellant mass ke doran aur doosri aadhi ke doran CG shift compare karo.

Recall Solution

Teen points par CG compute karo ( $m_{p} = 4000, 2000, 0$ ):

$m_{p} = 4000$ : $x_{cg} = 2.8 m$ .
$m_{p} = 2000$ : $x_{cg} = \frac{6000 + 4000}{3000} = 3.333 m$ .
$m_{p} = 0$ : $x_{cg} = 6.0 m$ .

Fuel ki pehli aadhi ( $4000 \to 2000$ ): shift $= 3.333 - 2.8 = 0.533 m$ . Doosri aadhi ( $2000 \to 0$ ): shift $= 6.0 - 3.333 = 2.667 m$ .

Doosri aadhi mein CG 5× zyada doooor jaata hai. Kyun: denominator $m_{d} + m_{p}$ burnout ke paas chhota ho jaata hai, toh bacha hua har kilogram fuel average par bahut zyada leverage rakhta hai. CG bilkul end mein dry mass ki taraf "bhaag" jaata hai — theek tab jab autopilot sabse zyada sensitive hota hai. Figure s02 mein amber curve dekho.

Figure s02 ko kaise padhein: horizontal axis propellant remaining $m_{p}$ kg mein hai — aur yeh reversed hai, toh time left jaata hai jaise fuel drains hota hai (right par full, left par empty). Vertical axis CG position $x_{cg}$ metres mein hai. Amber curve $x_{cg} (m_{p})$ hai; $m_{p} = 4000, 2000, 0$ par marked dots $2.8, 3.33, 6.0 m$ read karte hain. Dhyan do ki curve right par almost flat hai aur left ki taraf tezi se steep hoti hai — woh steepening hi Exercise 3.1 ka "CG late burn mein bhaagta hai" result hai: fuel ke equal chunks burnout ke paas CG ko bahut zyada move karte hain.

Exercise 3.2

Burnout par point-mass model $I_{y y} = 0$ deta hai (fuel khatam, dry mass CG par baitha hai). Explain karo kyun real rocket ka burnout par nonzero pitch inertia hota hai, aur parent note ki woh single quantity identify karo jo usse provide karti hai.

Recall Solution

Point model mein har body ek dimensionless dot hai. Burnout par ek hi body bacha hai dry structure, aur CG dry structure ki hi location hai, toh $D_{d} = 0$ aur $m_{d} D_{d}^{2} = 0$ . Lekin ek real dry structure ek extended body hai — ek lamba tube jisme mass spread out hai — toh uski apni intrinsic == $I_{d}^{c g}$ == hoti hai (uski apne centroid ke baare mein inertia). Parent se composite formula us term ko rakhta hai: $I_{y y}^{burnout} = I_{d}^{c g} + m_{d} = 0 D_{d}^{2} = I_{d}^{c g} .$ Toh rocket $I_{d}^{c g}$ par bottom out karta hai, kabhi truly zero nahi. Point model sirf woh term drop karta hai jo ab dominate kar raha hai.

Level 4 — Synthesis

Goal: ek problem mein CG, Steiner, aur extended-body inertia combine karo.

Exercise 4.1

Ab dry body ko real size do: ise uniform rod of length $L = 2 m$ model karo jo $x_{d} = 6$ par centered ho, toh uski apni pitch inertia $I_{d}^{c g} = \frac{1}{12} m_{d} L^{2}$ hogi. Propellant ko $x_{p} = 2$ par point-like rakhna. Full-load ( $m_{p} = 4000$ ) pitch inertia $I_{y y}$ vehicle CG ke baare mein nikalo.

Recall Solution

Step 1 — CG (point model se unchanged, kyunki humne sirf dry body ki shape badli, uska centroid nahi): $x_{cg} = 2.8 m$ . Step 2 — dry rod ki apni inertia: $I_{d}^{c g} = \frac{1}{12} (1000) (2)^{2} = \frac{1}{12} (4000) = 333.33 kg\cdotm^{2} .$ Step 3 — har body ko CG tak Steiner karo: $D_{d} = ∣6 - 2.8∣ = 3.2, D_{p} = ∣2 - 2.8∣ = 0.8.$ $I_{y y} = [I_{d}^{c g} + m_{d} D_{d}^{2}] + [0 + m_{p} D_{p}^{2}]$ $= [333.33 + 1000 (10.24)] + [4000 (0.64)] = (333.33 + 10240) + 2560 = 13133.33 kg\cdotm^{2} .$ Pure point model ke $12800$ se compare karein, extended rod ne exactly apna $I_{d}^{c g} = 333.33$ add kiya. Ye extra step kyun matter karta hai: shape term full load par chhota hai (kyunki $m_{d} D_{d}^{2}$ Steiner term usse dwarf kar deta hai), lekin burnout par woh akela survivor ban jaata hai — toh aap ise general model mein drop nahi kar sakte.

Exercise 4.2

Maano propellant tank ke bottom se upar ki taraf jalta hai, toh uska centroid drift karta hai. Ise model karo: half-burn par ( $m_{p} = 2000$ ) bacha hua fuel centroid $x_{p} = 1.5 m$ par aa gaya hai (na ki $2$ ). $x_{cg}$ aur point-mass $I_{y y}$ recompute karo, aur fixed-centroid answer se compare karo.

Recall Solution

Moving-centroid CG: $x_{cg} = \frac{1000 ( 6 ) + 2000 ( 1.5 )}{3000} = \frac{6000 + 3000}{3000} = \frac{9000}{3000} = 3.0 m .$ (Fixed-centroid model ne $3.333 m$ diya tha; fuel aage drift karne se CG bhi aage khicha.) Offsets: $D_{d} = ∣6 - 3.0∣ = 3.0$ , $D_{p} = ∣1.5 - 3.0∣ = 1.5$ . $I_{y y} = 1000 (3.0)^{2} + 2000 (1.5)^{2} = 9000 + 4500 = 13500 kg\cdotm^{2} .$ Ye alag kyun hai: centroid move karne dene se CG location aur moment arms dono change ho gaye. Fixed-centroid assumption CG ko $0.333 m$ galat jagah rakhti — itna kaafi hai ki TVC gains ki scheduling galat ho jaye. Real vehicles $x_{p} (t)$ ko tank geometry aur fill level se track karte hain.

Level 5 — Mastery

Goal: time-varying, rate, aur off-diagonal reasoning — real flight problem.

Exercise 5.1

Fuel constant rate $\overset{m}{˙} = - 50 kg/s$ par $m_{p 0} = 4000$ se jalta hai. Fixed-centroid point model use karte hue, CG velocity $\overset{x}{˙}_{cg}$ nikalo ignition par ( $m_{p} = 4000$ ) aur burnout ke paas ( $m_{p} = 200$ ). Ratio interpret karo.

Recall Solution

$x_{d} = 6$ , $x_{p} = 2$ , $m_{d} = 1000$ ke saath likhte hain $x_{cg} (m_{p}) = \frac{6000 + 2 m _{p}}{1000 + m _{p}}$ . $m_{p}$ ke respect mein differentiate karo (quotient rule): $\frac{d x _{cg}}{d m _{p}} = \frac{2 ( 1000 + m _{p} ) - ( 6000 + 2 m _{p} ) ( 1 )}{( 1000 + m _{p} ) ^{2}} .$ Numerator dhyan se dekho: top ko expand karo $2000 + 2 m_{p} - 6000 - 2 m_{p}$ . Do $\pm 2 m_{p}$ terms equal aur opposite hain, toh exactly cancel ho jaate hain, sirf constants $2000 - 6000 = - 4000$ bachte hain: $\frac{d x _{cg}}{d m _{p}} = \frac{- 4000}{( 1000 + m _{p} ) ^{2}} .$ Yahan calculus kyun? Hume rate chahiye jis par CG move karta hai, aur rate exactly wahi hai jo ek derivative measure karta hai — $x_{cg}$ ka slope versus fuel remaining. Phir time par chain-rule: $\overset{x}{˙}_{cg} = \frac{d x _{cg}}{d m _{p}} \overset{m}{˙} = \frac{- 4000}{( 1000 + m _{p} ) ^{2}} \times (- 50) = \frac{200000}{( 1000 + m _{p} ) ^{2}} .$ $m_{p} = 4000$ par: $\overset{x}{˙}_{cg} = \frac{200000}{( 5000 ) ^{2}} = \frac{200000}{25 000 000} = 0.008 m/s .$ $m_{p} = 200$ par: $\overset{x}{˙}_{cg} = \frac{200000}{( 1200 ) ^{2}} = \frac{200000}{1 440 000} = 0.1389 m/s .$ Ratio $\approx 17.4$ : CG burnout ke paas ignition se nearly 17× tezi se bhaagta hai, same fuel burn rate par. Yahi reason hai ki gain-scheduled autopilots burn ke end mein sabse tezi se update karte hain.

Exercise 5.2

Ek single canted vernier engine page ke coordinate system mein $(x, y) = (5, 0.4) m$ par baitha hai ( $y = + 0.4$ matlab $0.4 m$ centreline se positive- $y$ side ki taraf), mass $m_{e} = 20 kg$ (point mass). $x_{cg} = 2.8$ par vehicle CG ke baare mein product of inertia $I_{x y} = - \int x y d m$ mein uska contribution compute karo ( $y_{cg} = 0$ lo). Nonzero $I_{x y}$ kyun matter karta hai?

Recall Solution

CG ke relative position: $x^{'} = 5 - 2.8 = + 2.2 m$ , $y^{'} = 0.4 - 0 = + 0.4 m$ . Dono coordinates positive hain, toh ye mass $(x^{'}, y^{'})$ plane ke quadrant I mein baitha hai. Ek point mass ke liye product of inertia hai $I_{x y} = - m_{e} x^{'} y^{'} = - (20) (2.2) (0.4) = - 17.6 kg\cdotm^{2} .$ Ye kyun matter karta hai: off-diagonal $I_{x y} \neq = 0$ matlab inertia tensor diagonal nahi hai, toh ek pure pitch torque thodi si roll response produce karta hai aur vice versa — axes cross-coupled hain. Ek controller jo decoupled axes assume karta hai woh galat corrections command karega. Real vehicles ya toh layout balance karte hain $I_{x y}$ ko khatam karne ke liye ya coupling ko explicitly model karte hain.

Common mistake L5 trap — product of inertia ko hamesha negative maanna (quadrant ignore karna)

Ye sahi kyun lagta hai: aapne abhi $I_{x y} = - 17.6$ compute kiya aur formula mein leading minus hai, toh answer "obviously" negative hai. Fix: $I_{x y} = - m x^{'} y^{'}$ ka sign $x^{'}$ aur $y^{'}$ ke signs saath milke set hota hai, yaani CG ke relative mass kis quadrant mein hai:

Quadrant I ( $x^{'} > 0, y^{'} > 0$ ) ya III ( $x^{'} < 0, y^{'} < 0$ ): product $x^{'} y^{'} > 0$ , toh $I_{x y} < 0$ .
Quadrant II ( $x^{'} < 0, y^{'} > 0$ ) ya IV ( $x^{'} > 0, y^{'} < 0$ ): product $x^{'} y^{'} < 0$ , toh $I_{x y} > 0$ — woh sign flip kar leta hai.

Concretely, agar wahi engine doosri side par mounted ho $y = - 0.4$ par (quadrant IV), toh $I_{x y} = - (20) (2.2) (- 0.4) = + 17.6 kg\cdotm^{2}$ milega: same magnitude, opposite cross-coupling direction. Aur kisi bhi axis par exactly baitha mass ( $x^{'} = 0$ ya $y^{'} = 0$ ) $I_{x y} = 0$ contribute karta hai. Sign kabhi assume mat karo — quadrant padho.

Related vault topics: Parallel-Axis Theorem · Rigid Body Rotational Dynamics · Thrust Vector Control · Gain Scheduling in Autopilots · Propellant Slosh Dynamics · Tsiolkovsky Rocket Equation.