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

3.4.6 · D3 · Physics › Rocket Flight Mechanics › Mass properties — CG location, inertia tensor changing with

Yeh page Mass properties — CG location, inertia tensor changing with propellant depletion ka drill floor hai. Yahan hum koi naya theory nahi seekhte — hum bas yeh pakka karte hain ki koi bhi case tumhe surprise na kar sake: distance ka har sign, har degenerate mass, har limit, ek real word problem, aur ek exam-style twist — sab kuch neeche aata hai, har ek zero se solve kiya gaya hai.

Pehli line se pehle, plain words mein reminders taaki koi bhi symbol unearned na lage:

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The scenario matrix

Is topic ka har problem inhi cells mein se ek mein baithta hai. Neeche aane wale examples mein har ek [Cell N] tag carry karta hai.

#	Cell (the case class)	Isme kya special hai	Example
1	Baseline / mid-burn	ordinary two-body CG, dono masses present	Ex 1
2	Degenerate mass → 0 (burnout)	propellant $m_p=0$: CG dry mass par snap karta hai	Ex 2
3	Degenerate mass → 0 (liftoff heavy)	dry mass propellant ke comparison mein tiny: tank ki taraf limit	Ex 3
4	Moment arm ka sign	dry mass forward vs. propellant aft of CG — $x-x_{\text{cg}}$ ke opposite signs, same squared contribution	Ex 4 (figure)
5	Moving propellant centroid	$x_p(t)$ khud drift karta hai jab tank drain hota hai	Ex 5
6	Inertia with body's own $I^{cg}$ kept	point masses nahi — cylinders/rods jo apni own inertia rakhte hain	Ex 6
7	$I(t)$ ka limiting behaviour	inertia kitni tezi se collapse hoti hai; do times par ratio	Ex 7
8	Real-world word problem	TVC moment-arm / gain consequence	Ex 8
9	Exam twist (off-diagonal / product of inertia)	asymmetric drain $I_{xy}\ne 0$ banata hai	Ex 9

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Ex 1 — Baseline mid-burn CG [Cell 1]

Forecast: compute karne se pehle guess karo — balance point $2\text{ m}$ ke kareeb hoga ya $6\text{ m}$ ke? (Fuel utna hi structure se double hai, toh $2\text{ m}$ ki taraf jhukao.)

1. Har mass ko uske station se multiply karo. $1000\times 6=6000$; $2000\times 2=4000$. Yeh step kyun? Formula ek mass-weighted average hai — bhaare pieces average ko apni taraf kheenchte hain, isliye hum pehle positions ko mass se weight karte hain.
2. Weighted positions aur masses ko add karo. Top $=6000+4000=10000$; bottom $=1000+2000=3000$. Yeh step kyun? Numerator total "moment" hai, denominator total mass hai; unka ratio balance point hai.
3. Divide karo. $x_{\text{cg}}=10000/3000=3.333\dots \text{ m}$ (ek repeating decimal $=\frac{10}{3}\text{ m}$; hum ise reporting ke liye $3.33\text{ m}$ round karte hain, lekin exact $\frac{10}{3}$ rakhte hain jab bhi baad mein square karte hain).

Verify karo: $3.33\text{ m}$ $2$ aur $6$ ke beech hai (ek genuine average hona hi chahiye), aur yeh $2$ ke kareeb hai (bhaari side) — hamare forecast se milta hai. Units: $(\text{kg}\cdot \text{m})/\text{kg}=\text{m}$. ✓

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Ex 2 — Degenerate: propellant gone (burnout) [Cell 2]

Forecast: tank mein kuch nahi bacha, toh sirf dry structure ka mass hai — CG bilkul usee par baithna chahiye.

1. $m_p=0$ substitute karo. $x_{\text{cg}}=\frac{1000(6)+0\cdot 2}{1000+0}=\frac{6000}{1000}$. Yeh step kyun? Zero mass zero moment contribute karta hai aur total mein zero — yeh fraction ke dono parts se cleanly disappear ho jaata hai.
2. Simplify karo. $x_{\text{cg}}=6\text{ m}=x_d$. Yeh step kyun? Surviving moment $6000$ ko surviving mass $1000$ se divide karna poora averaging formula hai jisme sirf ek body bachi hai — toh "average position" sirf us ek body ki position ho sakti hai.

Verify karo: answer exactly $x_d$ hai, jaisa hona chahiye jab sirf ek body bache. Division-by-zero nahi hota kyunki $m_d\ne 0$. ✓ (Danger case se contrast karo: agar DONO masses zero hote toh formula $0/0$ hai — undefined, matlab "koi body nahi, koi balance point nahi." Hamesha check karo total mass $>0$.)

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Ex 3 — Degenerate limit: heavy-propellant liftoff [Cell 3]

Forecast: propellant ab structure ko swamp kar raha hai — balance point almost tank ke upar $2\text{ m}$ par hona chahiye.

1. Weight aur sum karo. Top $=1000(6)+99000(2)=6000+198000=204000$; bottom $=1000+99000=100000$. Yeh step kyun? Same mass-weighted average; humne sirf limit probe karne ke liye extreme mass ratio choose ki.
2. Divide karo. $x_{\text{cg}}=204000/100000=2.04\text{ m}$. Yeh step kyun? Total moment ka total mass se ratio hi balance station hai; division carry out karna do accumulated totals ko ek single position mein convert karta hai jo hum axis par padh sakte hain.
3. Limit lo. Jab $m_p\to \infty$, top mein $m_p{x}_p$ term dominate karta hai aur bottom mein $m_p$ dominate karta hai, isliye $x_{\text{cg}}\to x_p=2\text{ m}$. Yeh step kyun? Top aur bottom ko $m_p$ se divide karo: $x_{\text{cg}}=\frac{(m_d/m_p)x_d+x_p}{(m_d/m_p)+1}\to x_p$.

Verify karo: $2.04\text{ m}$ barely $2\text{ m}$ se aage hai — CG tank se chipa hua hai, limit confirm ho rahi hai. Ex 3 ($2.04$) → Ex 1 ($3.33$) → Ex 2 ($6.00$) ke beech tum dekh sakte ho CG dry mass ki taraf march kar raha hai jaise fuel drain hota hai. ✓

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Ex 4 — Moment arm ke signs (the two-sided case) [Cell 4]

Forecast: $6$ par dry mass CG ke peeche hai (positive offset); $2$ par propellant CG ke aage hai (negative offset). Lekin kyunki Steiner distance ko square karta hai, dono inertia mein positively add karte hain.

1. Dry-mass offset. $x_d-x_{\text{cg}}=6-\frac{10}{3}=\frac{8}{3}=+2.667\text{ m}$ (CG ke aft → positive). Yeh step kyun? Sign batata hai kaunsi side — baad mein torque directions ke liye vital — lekin figure dekho: dry body violet dot hai balance line ke right mein.
2. Propellant offset. $x_p-x_{\text{cg}}=2-\frac{10}{3}=-\frac{4}{3}=-1.333\text{ m}$ (CG ke forward → negative). Magenta dot, line ke left mein. Yeh step kyun? Opposite sign confirm karta hai ki dono bodies CG ke aas-paas hain — isliye kyun balance hota hai.
3. Square aur weight karo (Steiner, point model). $I_{yy}=1000{\left(\frac{8}{3}\right)}^2+2000{\left(\frac{4}{3}\right)}^2=1000(7.111)+2000(1.778)$. Yeh step kyun? $d=|x-x_{\text{cg}}|$, aur squaring sign ko erase kar deta hai — ek body ka twist contribution kabhi care nahi karta ki woh kis side par hai, sirf kitni door hai.
4. Add karo. $I_{yy}=7111.1+3555.6=10666.7{\text{ kg⋅m}}^2\approx 10667{\text{ kg⋅m}}^2$. Yeh step kyun? Same point (yahan CG) ke baare mein measure ki gayi inertias simply sum hoti hain — har body ka pitching ke against resistance seedha vehicle ke total mein add hota hai.

Verify karo: opposite signs ke bawajood dono terms positive hain — exactly squaring ki guarantee hai. Sanity check: kareeb, bhaari propellant yahan door, halke dry mass se kam contribute karta hai, kyunki distance square hota hai aur mass factor ko beat karta hai. Units ${\text{kg⋅m}}^2$. ✓

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Ex 5 — Moving propellant centroid [Cell 5]

Forecast: Ex 1 ke comparison mein (jisne $x_p=2$ fix kiya tha), fuel centroid ko $2.5$ tak slide hone dena CG ko $3.33\text{ m}$ se thoda aur aft pull karna chahiye.

1. Current propellant centroid nikalo. $x_p=2+1.0(0.5)=2.5\text{ m}$. Yeh step kyun? Centroid ab ek constant nahi hai; hume average karne se pehle ise is instant par padhna hoga.
2. Current propellant mass nikalo. $m_p=m_{p0}(1-f)=4000(0.5)=2000\text{ kg}$. Yeh step kyun? $f$ mass aur centroid ko ek saath tie karta hai — dono ek variable ke saath move karte hain.
3. Weighted average. $x_{\text{cg}}=\frac{1000(6)+2000(2.5)}{3000}=\frac{6000+5000}{3000}=\frac{11000}{3000}=3.667\text{ m}$. Yeh step kyun? Same balance-station formula jaise hamesha — lekin ab updated $x_p$ aur $m_p$ feed kiye gaye hain, isliye answer full-tank waale ki jagah mid-burn snapshot reflect karta hai.

Verify karo: $3.667\text{ m}>3.33\text{ m}$ Ex 1 se — sliding centroid ne CG ko aft push kiya, jaise forecast tha. $2.5$ aur $6$ ke beech abhi bhi hai. ✓ (Dekho Propellant Slosh Dynamics jaanne ke liye ki fuel surface — aur hence $x_p$ — perfectly steady kyun nahi hoti.)

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Ex 6 — Inertia keeping each body's own $I^{cg}$ [Cell 6]

Forecast: hume Ex 4 ($10666.7$) ka point-mass answer milega plus do extra "own-inertia" bumps — toh total $10667$ se zyada hoga.

1. Dry rod ki own inertia. $I_d^{cg}=\frac{1}{12}(1000)(4^2)=\frac{16000}{12}=1333.3{\text{ kg⋅m}}^2$. Yeh step kyun? Axis ke saath spread out ek body pitch ke against resist karti hai apne centre ke baare mein bhi — hum ise ignore nahi kar sakte jab yeh ek point nahi hai.
2. Propellant cylinder ki own inertia. $3R^2=3(0.64)=1.92$; $L_p^2=9$; sum $=10.92$. $I_p^{cg}=\frac{1}{12}(2000)(10.92)=\frac{21840}{12}=1820{\text{ kg⋅m}}^2$. Yeh step kyun? Same reason — fuel column ki real length hai, isliye uski intrinsic pitch inertia hai.
3. Har ek ko vehicle CG tak Steiner-shift karo aur point terms add karo (Ex 4 se): $1000{\left(\frac{8}{3}\right)}^2=7111.1$, $2000{\left(\frac{4}{3}\right)}^2=3555.6$. Yeh step kyun? Inertias sirf common point ke baare mein add hoti hain, isliye hum har body ki own inertia ko $M{d}^2$ ke zariye $x_{\text{cg}}$ tak drag karte hain. Dekho Parallel-Axis Theorem.
4. Total. $I_{yy}=1333.3+7111.1+1820+3555.6=13820.0{\text{ kg⋅m}}^2$. Yeh step kyun? Jab har piece — dono own-inertias aur dono Steiner terms — ek hi CG ke reference mein ho, parallel-axis theorem inhe seedha ek vehicle inertia mein add karne deta hai.

Verify karo: $13820>10667$ (Ex 4 point model), exactly do own-inertia bumps $1333.3+1820=3153.3$ se; aur sach mein $10666.7+3153.3=13820.0$. ✓ Units ${\text{kg⋅m}}^2$.

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Ex 7 — Limiting behaviour: $I$ kitni tezi se collapse hoti hai? [Cell 7]

Forecast: aadha fuel gaya lekin inertia aadhe se kam girni chahiye, kyunki remaining dry mass CG se door hai (bada $d^2$).

1. Ratio lo. $I_{yy}(\text{half})/I_{yy}(0)=10667/12800=0.833$. Yeh step kyun? Ratio units strip karta hai aur proportional collapse dikhata hai jo autopilot feel karta hai.
2. Interpret karo. Sirf $50\%$ fuel bacha hai phir bhi $83.3\%$ inertia bachi hai — kyunki surviving dry mass apna lever arm badha chuka hai (CG aft move ho gaya uski taraf). Yeh step kyun? $d^2$ dominate karta hai; near-CG fuel khona $I$ ko barely lower karta hai.
3. Burnout limit. Jab $m_p\to 0$, point model deta hai $I_{yy}\to m_d(x_d-x_d{)}^2=0$; real rocket dry body ki apni $I_d^{cg}=1333.3$ rakhta hai (Ex 6). Yeh step kyun? Burnout par $x_{\text{cg}}=x_d$, isliye dry lever arm vanish ho jaata hai aur sirf intrinsic inertia bachti hai — kabhi truly zero nahi.

Verify karo: $0.833\times 12800=10666.7\approx 10667$ ✓. Aur $0.833>0.5$ confirm karta hai inertia fuel mass se slower girti hai. ✓

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Ex 8 — Real-world word problem: TVC moment arm & gain [Cell 8]

Forecast: burnout par moment arm shrink karta hai lekin inertia bahut zyada shrink karti hai, isliye $\dot{\omega }$ bada hona chahiye — rocket twitchy ho jaata hai. Isi karan gains schedule karne chahiye (Gain Scheduling in Autopilots).

1. Side force. $T\sin \delta =60000\times 0.05234=3140.4\text{ N}$. Yeh step kyun? Canted thrust ka sirf sideways component pitch torque banata hai; axial part bas aage push karta hai.
2. Mid-burn arm & torque. arm $=x_e-x_{\text{cg}}=7-\frac{10}{3}=\frac{11}{3}=3.667\text{ m}$; $\tau =3140.4\times 3.667=11515\text{ N⋅m}$. $\dot{\omega }=11515/10667=1.080{\text{ rad/s}}^2$. Yeh step kyun? Torque = force × lever arm current CG ke baare mein; current inertia se divide karo response ke liye.
3. Burnout arm & torque. arm $=x_e-x_{\text{cg}}=7-6=1\text{ m}$; $\tau =3140.4\times 1=3140.4\text{ N⋅m}$. $\dot{\omega }=3140.4/1333.3=2.355{\text{ rad/s}}^2$.

Verify karo: arm $3.667\to 1$ gira (×0.27) lekin $\dot{\omega }$ badha $1.08\to 2.36$ (×2.18), kyunki inertia ×0.125 giri — inertia jeeti. Ek fixed-gain autopilot jo mid-burn ke liye tuned hai woh burnout par over-react karega. ✓ (Links to Thrust Vector Control aur Rigid Body Rotational Dynamics.)

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Ex 9 — Exam twist: asymmetric drain → nonzero product of inertia [Cell 9]

Forecast: agar tanks equal hote, toh $+y$ aur $-y$ contributions cancel ho jaate aur $I_{xy}=0$ hota. Unequal masses cancellation tod dete hain → nonzero cross-coupling jo pitch commands ko roll mein leak karta hai.

1. Left tank contribution. $m_{L}xy=600(-\frac{4}{3})(+0.9)=-720$. Yeh step kyun? Product of inertia mass ko measure karta hai jo dono symmetry planes se distribute hua hai; har term mass × do offsets hai. Hum exact $-\frac{4}{3}$ rakhte hain taaki numbers cleanly close hों.
2. Right tank contribution. $m_{R}xy=400(-\frac{4}{3})(-0.9)=+480$. Yeh step kyun? Opposite $y$ sign is term ko positive flip karta hai — would-be cancellation ka source.
3. Andar sum karo, phir leading minus sign apply karo. $\sum mxy=-720+480=-240$, isliye $I_{xy}=-(-240)=+240{\text{ kg⋅m}}^2$. Yeh step kyun? Definition explicit minus carry karti hai, isliye negative bracket positive product of inertia ban jaata hai.

Verify karo: $m_R=m_L=600$ set karo: tab $\sum mxy=600(-\frac{4}{3})(0.9)+600(-\frac{4}{3})(-0.9)=-720+720=0$, isliye $I_{xy}=0$ — symmetric case mein koi coupling nahi, jaisi honi chahiye. $200\text{ kg}$ ka imbalance $I_{xy}=+240\ne 0$ chhodta hai, isliye ek pure pitch command roll mein bleed karta hai. ✓ Yeh woh exam trap hai jo "products of inertia hamesha zero hote hain" ko punish karta hai.

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Recall Har example ne kaunsa cell fill kiya?

Ex1 baseline ::: Cell 1 Ex2 burnout, $m_p=0$ ::: Cell 2 Ex3 heavy-fuel limit ::: Cell 3 Ex4 signed moment arms ::: Cell 4 Ex5 moving centroid ::: Cell 5 Ex6 own $I^{cg}$ kept ::: Cell 6 Ex7 inertia collapse ratio ::: Cell 7 Ex8 TVC gain consequence ::: Cell 8 Ex9 nonzero product of inertia ::: Cell 9