Exercises — Equations of motion — 3DOF point mass (trajectory analysis)

WHAT: Hume speed ki rate of change chahiye, yaani $\dot{V}$. WHY: speed velocity arrow ki lambai hai, aur sirf tangential (along-$V$) rule lambai ko change karta hai. Isliye hum use karte hain $\dot{V}=\frac{T-D}{m}-g\sin \gamma .$ Climb ke against gravity term $-g\sin \gamma$ hai. Figure mein yeh "down" arrow ka shadow hai jo velocity arrow ke peeche ki taraf project hota hai.

• (a) neeche bend hona → $\dot{\gamma }$ rule (direction badlti hai): $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}$.
• (b) climb rate → $\dot{h}=V\sin \gamma$ (velocity arrow ka vertical component).
• (c) tezi se jaana → $\dot{V}$ rule (arrow ki lambai badhti hai). Parent se Mnemonic: S-along, C-across — sine gravity ko speed se jodta hai, cosine usse turning se.

WHAT: speed-rate compute karo. WHY: tangential rule. KAISA DIKHTA HAI: thrust arrow ko stretch karta hai, drag + gravity ka along-shadow usse shrink karta hai. $\dot{V}=\frac{350000-45000}{12000}-9.81\sin 50^{\circ }=25.4167-7.5147=17.90m/{\mathbf{s}}^2.$

WHY $\dot{\gamma }$ rule: hume chahiye ki arrow kaise mude, na ki uski lambai. Full turning rule se shuru karo aur $L=0$ rakho: $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}=\frac{0-mg\cos \gamma }{mV}=\frac{-mg\cos \gamma }{mV}.$ $m$ kyun cancel ho jaata hai? Numerator mein jo term bacha hai usme $m$ ka ek factor hai (weight $mg\cos \gamma$ hai), aur denominator mein bhi ek $m$ hai. Woh common factor upar-neeche cancel ho jaata hai — bhaari ho ya halka, gravity direction ko same rate se bend karta hai (sirf speed $V$ matter karti hai). Isliye hum lift-free turning rule bina mass ke likh sakte hain: $\dot{\gamma }=-\frac{g\cos \gamma }{V}=-\frac{9.81\cos 50^{\circ }}{420}=-\frac{6.3057}{420}=-0.015014\text{rad/s}.$ Convert karo: $-0.015014\times \frac{180}{\pi }=-0.860deg/s$. Negative kyunki gravity hamesha arrow ki tip ko neeche kheenchti hai.

$V=420$ lambai ka velocity arrow right-triangle trig se horizon aur vertical par split hota hai: $\dot{x}=420\cos 50^{\circ }=269.96m/s,\dot{h}=420\sin 50^{\circ }=321.75m/s.$

$\dot{\gamma }=-\frac{g\cos 0^{\circ }}{V}=-\frac{9.81(1)}{180}=-0.0545\text{rad/s}=-3.12^{\circ }/\text{s}.$ Isko padhna: $\gamma =0$ par cosine maximum hai, isliye gravity ki turning power yahaan sabse strong hai — arrow bilkul apex par sabse tezi se neeche tip karta hai. Isliye trajectories upar "round over" karti hain: flat part woh jagah hai jahan gravity sabse zyada bend karti hai.

$\dot{V}=\frac{0-8000}{5000}-9.81\sin (-30^{\circ })=-1.6-9.81(-0.5)=-1.6+4.905=+3.305m/{\mathbf{s}}^2.$ Isko padhna: $\gamma$ negative hai, isliye $\sin \gamma <0$, isliye gravity term ka sign flip hota hai aur woh motion ko help karta hai — dive mein speed badhti hai. Sirf drag use slow karta ($-1.6$), lekin gravity ka downhill pull ($+4.905$) jeet jaata hai. Yeh tezi ho raha hai.

Speed: $\dot{V}=\frac{900000-120000}{30000}-9.81\sin 75^{\circ }=26-9.4767=16.523m/{\mathbf{s}}^2.$ Turn: $\dot{\gamma }=-\frac{9.81\cos 75^{\circ }}{250}=-\frac{2.5389}{250}=-0.010156rad/s(=-0.582^{\circ }/\text{s}).$ Position: $\dot{x}=250\cos 75^{\circ }=64.70m/s,\dot{h}=250\sin 75^{\circ }=241.48m/s.$ Mass: $\dot{m}=-\frac{T}{I_{sp}{g}_0}=-\frac{900000}{290\times 9.81}=-316.4kg/s.$ WHY mass rule: yeh parent mass-flow rule $\dot{m}=-T/(I_{sp}{g}_0)$ hai — yeh kehta hai ki propellant us rate se nikalta hai jo thrust ko exhaust efficiency ($I_{sp}{g}_0$ effective exhaust speed hai) se divide karne par milti hai. Hum ise isliye include karte hain kyunki baaki chaar rules quietly $m$ ki ek value assume karte hain; jaise jaise $m$ shrink hota hai, wahi thrust zyada zyada acceleration kharidta hai, isliye trajectory ko honestly aage badhane ke liye hume $m$ track karna hoga. Story: abhi bhi steep climb kar raha hai aur zor se accelerate kar raha hai, barely turn ho raha hai (steep $\gamma$ ⇒ chota $\cos \gamma$), har second ~316 kg propellant jala raha hai.

Euler = "maano rate chhote step ke liye constant rahi, utni door chalo." $V\approx 250+16.523(2)=283.05m/s.$ $\gamma \approx 75^{\circ }+(-0.582^{\circ }/\text{s})(2)=73.836^{\circ }.$ $h\approx h_0+241.48(2)=h_0+482.96\mathbf{m}.$ Arrow lamba hua, thoda side jhuka, aur 2 s mein ~483 m upar gaya. Neeche wali figure exactly yahi step draw karti hai: magenta arrow start par velocity direction hai ($\gamma =75^{\circ }$), aur orange segment woh jagah hai jahan ek single Euler step un 2 seconds mein point mass ko le jaata hai — notice karo yeh along ($x$) se kahin zyada upar ($h$) jaata hai kyunki $\gamma$ steep hai, jo $\dot{h}\gg \dot{x}$ se match karta hai.

$\gamma =90^{\circ }$ daalo: $\cos 90^{\circ }=0$, isliye $\dot{\gamma }=-\frac{g\cos 90^{\circ }}{V}=-\frac{g\cdot 0}{V}=0.$ $\dot{\gamma }=0$ ke saath, $\gamma$ kabhi $90^{\circ }$ se nahi hatega — vertical, vertical hi rehta hai. Aur $\sin 90^{\circ }=1$ deta hai $\dot{V}=\frac{T-D}{m}-g\cdot 1=\frac{T-D}{m}-g,$ exactly ideal one-dimensional rocket. Kyun samajh aata hai: seedha upar, gravity ka turning shadow ($\cos$) vanish ho jaata hai aur uska braking shadow ($\sin$) total ho jaata hai.

$\dot{\gamma }=-\frac{g\cos \gamma }{V}⟶-\infty \text{as }V\to 0^{+}.$ Iska matlab: near-zero speed par zara si off-vertical tilt arrow ko wildly swing kar deti — model blow up ho jaata hai. WHY reality mein nahi hota: lift-off par rocket vertical hold kiya jaata hai ($\gamma =90^{\circ }$, isliye $\cos \gamma =0$ chhote $V$ ko cancel karta hai), aur sirf speed build hone ke baad ek small pitch-over introduce kiya jaata hai. Tab $\cos \gamma$ abhi bhi chota hai aur $V$ ab bada hai, isliye $\dot{\gamma }$ gentle hai. Woh controlled hand-off exactly Gravity Turn Trajectory hai: pitch over karo tabhi jab speed ho, aur gravity ko bending free mein karne do.

WHY full rule: yahaan direction sirf gravity se nahi balki aerodynamic force $L$ se bhi change hoti hai, isliye hume $L$ term rakhna hoga: $\dot{\gamma }=\frac{L-mg\cos \gamma }{mV}=\frac{80000-6000\cdot 9.81\cdot \cos 10^{\circ }}{6000\cdot 200}.$ Numerator: $80000-6000(9.81)(0.98481)=80000-57967.3=22032.7$. Denominator: $1,200,000$. $\dot{\gamma }=\frac{22032.7}{1200000}=+0.018361\text{rad/s}=+1.052^{\circ }/\text{s}.$ Isko padhna: $\dot{\gamma }$ ab positive hai — path upar bend ho raha hai. Lift ($80$ kN) ne gravity ke perpendicular shadow ($mg\cos \gamma \approx 58$ kN) ko beat kiya, isliye net sideways force velocity arrow ko neeche ki jagah upar curl karti hai. Yeh ek aerodynamic (lift-driven) turn hai, gravity turn ka ulta: wings, gravity nahi, arrow ko steer karti hain. $L=0$ set karo aur familiar $\dot{\gamma }=-g\cos \gamma /V<0$ wapas milta hai.

• (a) Speed loss $\propto \sin \gamma$. $\sin 70^{\circ }>\sin 20^{\circ }$, isliye steeper ($70^{\circ }$) rocket gravity ko zyada speed deta hai — climbing mein energy lagti hai.
• (b) Turn rate $\propto \cos \gamma$. $\cos 20^{\circ }>\cos 70^{\circ }$, isliye shallower ($20^{\circ }$) rocket tezi se bend karta hai — horizon ke paas gravity ka sideways pull dominant hai. Picture: jaise arrow vertical se horizontal ki taraf tip karta hai, "along" shadow shrink hota hai aur "across" shadow badhta hai — sine aur cosine roles trade karte hain. Yeh single trade-off har ascent-profile design ki jaan hai.

lambai change karta hai, aur kaun sa uski direction? Lambai (speed) → $\dot{V}=(T-D)/m-g\sin \gamma$. Direction (turning) → $\dot{\gamma }=(L-mg\cos \gamma )/(mV)$.

$\gamma <0$) mein drag hote hue bhi rocket speed kyun badhata hai? Kyunki $-g\sin \gamma$ positive ho jaata hai jab $\gamma <0$ (gravity ab motion ke along kheenchti hai); agar yeh drag ko beat kare, toh $\dot{V}>0$.