Worked examples — Forces on a rocket in flight — thrust, aerodynamic (normal, axial), gravity

3.4.3 · D3 · Physics › Rocket Flight Mechanics › Forces on a rocket in flight — thrust, aerodynamic (normal,

Yeh page parent topic ke liye drill floor hai. Parent ne teen force families (thrust, aerodynamic axial/normal, gravity) aur equations of motion build ki thi. Yahan hum har case exhaust karte hain: har sign, har degenerate input, har limit, plus ek real-world word problem aur ek exam twist.

Shuru karne se pehle, ek promise: neeche har symbol parent mein define kiya gaya tha, lekin hum tricky walon ko jab bhi aate hain re-anchor karte hain, taaki tumhe kabhi scroll back na karna pade.

Scenario matrix

Ise ek checklist ki tarah socho. Har row ek aisa tarika hai jisme physics tumhe surprise kar sakti hai; neeche har worked example us cell ke saath tagged hai jo woh cover karta hai.

Setup: symbols, re-anchored

Ex 1 — Teen pressures par Thrust · C1, C2 cover karta hai

1. Momentum thrust (woh part jo tumhe hamesha milta hai). $\dot{m}{v}_e=250\times 2900=725000\text{ N}=725\text{ kN}$ Yeh step kyun? Yeh woh momentum hai jo tum har second peeche phenkते ho — Newton ka 3rd law tumhe exactly isi se aage push karta hai. Yeh altitude par depend nahi karta.

2. Sea level par pressure term. Yahan $p_e<p_a$ hai (50 < 101), toh exhaust over-expanded hai (thick air se squeeze hota hai): $(p_e-p_a)A_e=(50000-101000)(0.6)=(-51000)(0.6)=-30600\text{ N}$ Yeh step kyun? Negative number ka matlab hai bahar ki air exit plane par exhaust se zyada push back karti hai — yeh ek penalty hai (case C1).

3. Sea-level thrust. $T_{\text{sea}}=725000-30600=694400\text{ N}=694.4\text{ kN}$ Yeh step kyun? Full thrust law hai $T=\dot{m}{v}_e+(p_e-p_a)A_e$, toh total thrust literally momentum term plus pressure term hai. Yahan pressure term negative hai, isliye yeh 725 kN se subtract karta hai — thick sea-level air quietly thrust chura rahi hai.

4. Vacuum: $p_a=0$, toh term ek pure bonus ban jaata hai (case C2, limiting case): $(50000-0)(0.6)=30000\text{ N},T_{\text{vac}}=725000+30000=755000\text{ N}=755\text{ kN}$ Yeh step kyun? Hum vacuum thrust banate hain same momentum term lekar (unchanged — engine flow identical hai) aur ab positive pressure term add karke. Koi bahar ki air push back nahi kar rahi, toh full exit pressure $p_e$ bonus ke roop mein $A_e$ par act karta hai, isliye vacuum thrust sea-level thrust se zyada hai.

Ex 2 — Perfect expansion, phir real altitude par under-expanded · C3, C2 cover karta hai

1. (2a) Pressure term. $p_e-p_a=19000-19000=0$: $(p_e-p_a)A_e=0$ Yeh step kyun? Case C3 — degenerate "clean" point. $A_e$ jo bhi ho, zero se multiply karne par poora pressure contribution khatam ho jaata hai.

2. (2a) Thrust = sirf momentum thrust. $T=\dot{m}{v}_e=250\times 2900=725000\text{ N}=725\text{ kN}$ Yeh step kyun? Kyunki Step 1 mein pressure term vanish ho gaya, full thrust law $T=\dot{m}{v}_e+(p_e-p_a)A_e$ sirf apne pehle piece par collapse ho jaata hai. Yeh single condition hai jahan thrust pure momentum flux ke barabar hoti hai.

3. (2b) 25 km par pressure term. Ab $0<p_a<p_e$, toh term ek positive lekin maximal nahi bonus hai (case C2, non-vacuum): $(p_e-p_a)A_e=(19000-2500)(0.9)=(16500)(0.9)=14850\text{ N}$ Yeh step kyun? Ex 1 ke vacuum ($p_a=0$) ke unlike, yahan kuch air abhi bhi push back kar rahi hai, isliye hum $A_e$ se multiply karne se pehle $p_e$ se $p_a$ subtract karte hain. Bonus real hai lekin partial — yeh everyday cruising case hai, "matched" aur "vacuum" ke beech.

4. (2b) Thrust. $T=\dot{m}{v}_e+(p_e-p_a)A_e=725000+14850=739850\text{ N}=739.85\text{ kN}$ Yeh step kyun? Hum Step 3 ka partial bonus momentum thrust mein add karte hain, ek aisi value milti hai jo matched 725 kN se upar hai kyunki ambient ab exit ko perfectly balance nahi karta.

Ex 3 — Lift/drag ko body axes mein resolve karna, positive $\alpha$ · C4 cover karta hai

Figure dekho: body axis (lavender) velocity (slate) se $\alpha$ se tilted hai. Drag velocity ke saath peeche point karta hai; lift uske perpendicular hai. Dono ko body axes par project karte hain.

1. Axial force — body ke saath component: $A=D\cos \alpha +L\sin \alpha =6\cos 15^{\circ }+14\sin 15^{\circ }$ $=6(0.9659)+14(0.2588)=5.795+3.623=9.42\text{ kN}$ Yeh step kyun? Kyunki body tilted hai, lift ka kuch hissa ($L\sin \alpha$) body ke forward-along press karta hai, axial load mein add hota hai. Toh $A=9.42>D=6$ — forecast trap.

2. Normal force — body ke across component: $N=L\cos \alpha -D\sin \alpha =14\cos 15^{\circ }-6\sin 15^{\circ }$ $=14(0.9659)-6(0.2588)=13.523-1.553=11.97\text{ kN}$ Yeh step kyun? Lift mostly body ke across act karta hai (+$N$), lekin tilted drag thoda isse chura leta hai (−$D\sin \alpha$).

Ex 4 — Negative aur zero angle of attack · C5, C6 cover karta hai

Figure mein body axis velocity ke neeche drop hoti dikhai deti hai — ek negative tilt.

1. Case (a), $\alpha =-15^{\circ }$. $\cos (-15^{\circ })=\cos 15^{\circ }=0.9659$ aur $\sin (-15^{\circ })=-0.2588$ use karo: $N=L\cos \alpha -D\sin \alpha =14(0.9659)-6(-0.2588)=13.523+1.553=15.08\text{ kN}$ $A=D\cos \alpha +L\sin \alpha =6(0.9659)+14(-0.2588)=5.795-3.623=2.17\text{ kN}$ Yeh step kyun? Yeh case C5 hai. Cosine even hai (unchanged), lekin sine odd hai (sign flip karta hai). Toh $A$ mein $L\sin \alpha$ term ab subtract karta hai — axial sharply drop karta hai; aur $N$ mein $-D\sin \alpha$ term ab add karta hai. Har sign track karna padega; yeh exactly "cover all quadrants" rule hai.

2. Case (b), $\alpha =0$. Tab $\cos 0=1$, $\sin 0=0$: $A=D(1)+L(0)=D=6\text{ kN},N=L(1)-D(0)=L=14\text{ kN}$ Yeh step kyun? Case C6, degenerate wala — body axes wind axes ke saath coincide karte hain. Yeh single angle hai jahan "axial = drag" aur "normal = lift". Koi bhi dusra $\alpha$ woh identity tod deta hai.

Ex 4b — Quadrant II mein angle of attack ($\cos \alpha <0$) · C6b cover karta hai

$\alpha =120^{\circ }$ par hum second quadrant mein hain, jahan $\cos \alpha <0$ lekin $\sin \alpha >0$: $\cos 120^{\circ }=-0.5$, $\sin 120^{\circ }=+0.8660$.

1. Axial force: $A=D\cos \alpha +L\sin \alpha =6(-0.5)+14(0.8660)=-3.0+12.124=9.12\text{ kN}$ Yeh step kyun? Drag term $D\cos \alpha$ negative ho gaya kyunki $\cos \alpha <0$ — drag ab body ke saath doosri taraf push karta hai. Lekin lift term $L\sin \alpha$ abhi bhi bada aur positive hai (quadrant II mein sine positive hota hai), isliye net axial force positive rehta hai. Yeh woh case hai jise pehle ke $\pm 15^{\circ }$ examples reach nahi kar sakte the: $\cos \alpha$ mein khud sign change.

2. Normal force: $N=L\cos \alpha -D\sin \alpha =14(-0.5)-6(0.8660)=-7.0-5.196=-12.20\text{ kN}$ Yeh step kyun? Yahan dono terms negative hain: $\cos \alpha <0$ lift contribution flip karta hai, aur $\sin \alpha >0$ $-D\sin \alpha$ term ko negative rakhta hai. Negative $N$ ka matlab hai normal force ab body ke opposite face par press karti hai — exactly woh loading jo ek tumbling stage ki fins snap kar deti hai.

Ex 5 — Tangential EOM: steep climb vs level flight · C7, C8 cover karta hai

Parent se tangential (along-velocity) equation: $m\frac{dv}{dt}=T\cos \alpha -D-mg\sin \gamma .$ $\sin \gamma$ factor isliye hai kyunki gravity ki grip depend karti hai ki tum kitne steeply climb kar rahe ho: sirf gravity ka flight path ke saath component hi tumhe slow karta hai.

1. $\gamma =75^{\circ }$ par gravity ka along-path pull. $mg\sin 75^{\circ }=5000(9.8)(0.9659)=47330\text{ N}=47.33\text{ kN}$ Yeh step kyun? Case C7 — steep. Drag ($10$ kN) se compare karo: gravity loss ~5× zyada hai. Forecast answer ho gaya: gravity dominate karta hai.

2. Case (a) mein acceleration: $a=\frac{130000-10000-47330}{5000}=\frac{72670}{5000}=14.53{\text{ m/s}}^2$ Yeh step kyun? Hum net along-path force (thrust minus drag minus gravity ka along-path pull) ko mass se divide karte hain, jo exactly $\vec{F}=m\vec{a}$ hai $a$ ke liye solve kiya gaya. Teen forces subtract hote hain kyunki drag aur gravity dono yahan forward motion ko oppose karte hain.

3. Level flight, $\gamma =0$ (case C8): $\sin 0=0$, toh gravity tangential equation mein kuch contribute nahi karta: $a=\frac{130000-10000-0}{5000}=\frac{120000}{5000}=24.0{\text{ m/s}}^2$ Yeh step kyun? Level hone par, gravity purely path ke across act karta hai — trajectory curve karta hai (agla example) lekin rocket ko slow nahi karta. $\sin \gamma$ ke liye yeh degenerate zero-input case hai.

Ex 6 — Normal EOM: nose kis taraf swing karta hai? · C9 cover karta hai

Parent se normal (turning) equation: $mv\frac{d\gamma }{dt}=T\sin \alpha +L-mg\cos \gamma .$ Right side ka sign $d\gamma /dt$ ka sign hai: positive → pitching up, negative → nosing over.

1. Thrust ka turning contribution: $T\sin \alpha =130000\sin 4^{\circ }=130000(0.06976)=9069\text{ N}$. Yeh step kyun? Ek tilted thrust mein ek small sideways component hota hai jo steer karne mein help karta hai.

2. Gravity ka cross-path pull: $mg\cos \gamma =5000(9.8)\cos 80^{\circ }=49000(0.1736)=8508\text{ N}$. Yeh step kyun? Steep $\gamma =80^{\circ }$ par, gravity almost along path hai, isliye uska cross component (∝ $\cos \gamma$) small hai.

3. Right side assemble karo: $T\sin \alpha +L-mg\cos \gamma =9069+20000-8508=20561\text{ N}>0$ Yeh step kyun? Positive ⇒ case C9 jisme $d\gamma /dt>0$: nose up pitch karta hai. Lift aur thrust-tilt modest gravity cross-pull ko beat karte hain.

4. Turn rate: $\frac{d\gamma }{dt}=\frac{20561}{mv}=\frac{20561}{5000\times 300}=\frac{20561}{1500000}=0.01371\text{ rad/s}$ Yeh step kyun? Normal equation hai $mvd\gamma /dt=(\text{net cross-path force})$, toh us net force ko $mv$ se divide karne par turn rate isolate hota hai. $mv$ factor isliye aata hai kyunki turning momentum $mv$ ko bend karta hai, sirf mass ko nahi — same force par ek faster rocket dheerey curve karta hai.

Ex 7 — Real-world word problem · C10 cover karta hai

Prose translate karo: thrust $T=\dot{m}c$ (refresher se effective exhaust velocity $c$ use karke), aur instantaneous mass $m(t)=m_0-\dot{m}t$ (Variable Mass Systems se). Vertical flight $\gamma =90^{\circ }$ ⇒ $\sin \gamma =1$, drag zero: $a(t)=\frac{T-m(t)g}{m(t)}=\frac{T}{m(t)}-g.$

1. Thrust (yahan constant hai): $T=\dot{m}c=500\times 3000=1,500,000\text{ N}=1.5\text{ MN}$. Yeh step kyun? Effective exhaust velocity $c$ pressure term pehle se fold kar chuka hai, isliye $T$ ek clean number hai aur humein $p_e,p_a,A_e$ separately touch nahi karna. Exactly isliye $c$ pehle define kiya gaya tha.

2. Liftoff ($t=0$): $m(0)=m_0=60000$ kg. $a_0=\frac{1500000}{60000}-9.8=25-9.8=15.2{\text{ m/s}}^2$ Yeh step kyun? Hum full launch mass $m_0$ $a=T/m-g$ mein plug karte hain kyunki $t=0$ par koi propellant abhi burn nahi hua hai. Subtract kiya gaya $g$ vertical flight mein gravity ka full along-path pull hai ($\sin 90^{\circ }=1$).

3. Burnout ($t=60$ s): tank exactly 60-second burn ke end par khali ho jaata hai, toh $m(60)=m_0-\dot{m}t=60000-500(60)=60000-30000=30000$ kg. $a_1=\frac{1500000}{30000}-9.8=50-9.8=40.2{\text{ m/s}}^2$ Yeh step kyun? Yahan hum same formula mein shrunken mass $m(60)$ dalte hain, kyunki 60 s ke burn ne rocket ka aadha hissa phenk diya. Same thrust aadhe mass se divide hone par thrust-to-weight double ho jaata hai — variable-mass system ka signature: acceleration burnout ke paas uchhal jaata hai.

Ex 8 — Exam twist: hidden variable ke liye solve karo · C11 cover karta hai

Full thrust law $T=\dot{m}{v}_e+(p_e-p_a)A_e$ se start karo aur $v_e$ isolate karo: $v_e=\frac{T-(p_e-p_a)A_e}{\dot{m}}.$

1. Pressure term (vacuum, $p_a=0$): $(p_e-p_a)A_e=(60000-0)(0.8)=48000\text{ N}=48\text{ kN}$ Yeh step kyun? Case C11 — exam chahta hai ki tum divide karne se pehle pressure bonus subtract karo, warna $v_e$ overstate ho jaayega. Vacuum mein full exit pressure act karta hai, toh yeh term measured thrust mein ek positive 48 kN slice hai.

2. $v_e$ ke liye solve karo: $v_e=\frac{600000-48000}{200}=\frac{552000}{200}=2760\text{ m/s}$ Yeh step kyun? Hum measured 600 kN se 48 kN pressure contribution hataate hain taaki sirf momentum thrust (552 kN) bache, phir $\dot{m}$ se divide karte hain kyunki momentum thrust $\dot{m}{v}_e$ hai. Kyunki reading ka kuch hissa pressure se aaya, $v_e=2760<3000$ m/s — forecast confirm: chhota.

Recall drill