Equations of motion — 3DOF point mass (trajectory analysis): Question bank

3.4.4 · D5 · Physics › Rocket Flight Mechanics › Equations of motion — 3DOF point mass (trajectory analysis)

Shuru karne se pehle, ek picture jise hum baar baar refer karenge — velocity frame jahan har force ko velocity arrow ke saath aur aage-peeche split kiya jaata hai.

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

Arrow hai $V$ . "Along" ( $∥$ ) arrow ki length change karta hai (speed). "Across" ( $⊥$ ) uski direction change karta hai (use turn karta hai). Weight seedha neeche point karta hai aur $m g sin γ$ along aur $m g cos γ$ across mein split hota hai. Ise dhyan mein rakho.

True or false — justify karo

Answer cover karo, true/false decide karo, phir kyun bolo.

"3DOF" matlab rocket teen spatial dimensions

x, y, z

mein move karta hai.

False — "3" count karta hai translation ke degrees of freedom (planar model mein:

V

γ

, aur position). Ye is baare mein hai ki hum kitni independent motions track karte hain, spatial axes ki number ke baare mein nahi; yahan ye 2D vertical plane hai.

Tangential equation mein, weight ka contribution

m g sin γ

hai.

True —

γ

horizontal se measure kiya jaata hai, isliye "seedha neeche" ka woh piece jo velocity ke along lie karta hai woh

m g sin γ

hai; vertical flight par (

γ = 90°

) ye poora weight ban jaata hai, jo sahi hai.

Agar thrust bilkul velocity ke along hai aur koi lift nahi hai, toh gravity akeli trajectory ko turn kar sakti hai.

True —

\overset{γ}{˙} = (L - m g cos γ) / (mV)

;

L = 0

aur

T

V

ke across kuch contribute na karne par, sirf

- m g cos γ

bacha rehta hai path ko bend karne ke liye. Ye bilkul ek Gravity Turn Trajectory hai.

Ek rocket jo bilkul vertically fly kar raha hai (

γ = 90°

) uska flight-path angle phir bhi apne aap change ho sakta hai.

False —

\overset{γ}{˙} = - g cos 90°/ V = 0

. Weight ke zero across-component ke saath, vertical path vertical rehta hai; ye sirf tabhi turn karta hai jab koi sideways force (lift ya off-axis thrust) add kiya jaaye.

Speed

V

badhane se trajectory tezi se bend hoti hai.

False —

\overset{γ}{˙} = - g cos γ / V

mein denominator mein

V

hai, isliye tezi se fly karne par zyada dheere bend hota hai. Ek fast rocket gravity ke turning ko resist karta hai; ek slow rocket jaldi se palat jaata hai.

Coasting arc ke bilkul top par (

γ = 0

), gravity speed ko kuch nahi karta.

True — along-component hai

m g sin 0 = 0

, isliye gravity us instant par rocket ko na speed up karta hai na slow down; ye poori tarah path ke across kaam karta hai (maximum turning,

\overset{γ}{˙} = - g / V

Mass

m

charon trajectory equations mein se cancel ho jaata hai.

False — ye shape equations (

\overset{γ}{˙}

\overset{x}{˙}

\dot{h}

) se cancel hota hai lekin

\dot{V} = (T - D) / m - g sin γ

se nahi, jahan thrust-aur-drag term

m

par depend karta hai. Bhaari vehicle, chhoti acceleration.

Agar drag

D

thrust

T

se zyada ho jaata hai, toh rocket immediately girna shuru ho jaata hai.

False —

D > T

sirf

\dot{V}

ko possibly negative banata hai (slow down karna). "Girna" direction ke baare mein hai (

γ

), jo alag

\overset{γ}{˙}

equation se govern hoti hai; ek slow ho raha rocket phir bhi climb kar sakta hai.

Error dhundho

Har item mein ek galat statement ya step hai. Flaw ko naam do aur use correct karo.

\dot{V} = (T - D) / m - g cos γ

, kyunki cosine along-track direction handle karta hai jaise projectile motion mein."

Error hai

cos γ

: weight ka along-velocity piece

m g sin γ

hai,

m g cos γ

nahi.

γ = 90°

par sanity check — sara weight climb ko oppose karna chahiye, aur

sin 90° = 1

woh deta hai;

cos 90° = 0

galat tarike se kehta ki gravity irrelevant hai jab seedha upar ja rahe ho.

"Kyunki Earth rotate karta hai, main apne flat-Earth planar 3DOF mein Coriolis aur centrifugal terms add karunga."

Wo fictitious forces sirf rotating round-Earth model mein belong karti hain. Flat-Earth planar analysis mein frame (approximately) inertial aur non-rotating hai, isliye wo terms absent hain. Models mix karna effects ko double-count karta hai.

"Thrust hamesha seedha upar point karta hai, isliye main

+ T

\dot{h}

equation mein dalunga."

Kinematic equations

\overset{x}{˙} = V cos γ

\dot{h} = V sin γ

mein bilkul bhi forces nahi hain — ye sirf speed aur direction ko position rates mein convert karte hain. Thrust

\dot{V}

ke through enter karta hai (jo

V

change karta hai), directly

\dot{h}

mein kabhi nahi.

"Main

m \dot{V} = T - D - m g sin γ

m

ko constant treat karte hue integrate karunga kyunki ye 'rocket ka mass' hai."

Propellant jalta hai, isliye

\overset{m}{˙} = - T / (I_{s p} g_{0})

aur

m

continuously shrink karta hai. Use constant rakhne se late-flight acceleration underestimate hoti hai — same thrust ek lighter vehicle par near burnout bahut bada

\dot{V}

deta hai.

\overset{γ}{˙}

negative aaya, isliye zaroor sign mistake hai — rocket climb kar raha hai, angles badhni chahiye."

Negative

\overset{γ}{˙}

sahi aur expected hai: gravity ka across-component hamesha velocity arrow ko neeche curve karta hai, waqt ke saath

γ

kam karta hai. Climbing (positive

\dot{h}

) aur decreasing flight-path angle ek saath khooshi se exist karte hain.

"Gravity turn ke liye mujhe pura rasta pitch-over command karne ke liye control system chahiye."

Gravity Turn Trajectory ka poora point yahi hai ki ek thodi si initial pitch-over ke baad, gravity akele

V

ko horizontal ki taraf

\overset{γ}{˙} = - g cos γ / V

ke through rotate karti hai — koi continuous control effort nahi, isliye minimum steering losses.

"Drag hai

D = ρ V^{2} S C_{D}

\frac{1}{2}

factor missing hai:

D = \frac{1}{2} ρ V^{2} S C_{D}

. Half dynamic-pressure definition

q = \frac{1}{2} ρ V^{2}

se aata hai; Drag and Atmospheric Models dekho.

Why questions

Reasoning explain karo, sirf fact nahi.

Hum forces ko ground $x$ – $h$ frame ki jagah velocity frame mein kyun resolve karte hain?

Kyunki velocity vector mein exactly do independent cheezein change ho sakti hain — uski length aur uski direction — aur velocity frame force ko cleanly ek piece per job mein split karta hai (along = length, across = direction), ek vector equation ko do decoupled scalar mein convert karta hai. Flight-Path Angle and Velocity Frame dekho.

Speed equation mein

sin γ

kyun aata hai lekin turning equation mein

cos γ

γ

horizontal se measure hota hai, aur weight neeche point karta hai. "Neeche" aur "

V

ke along peeche" ke beech ka angle along-piece

sin γ

banata hai; perpendicular piece jo bacha hai woh

cos γ

hai. "S-along, C-across."

Perpendicular acceleration

mV \overset{γ}{˙}

kyun likha jaata hai?

Velocity vector ko rate

\overset{γ}{˙}

par turn karna jab uski length

V

ho, ye exactly circular-motion geometry hai: tip sideways

V \overset{γ}{˙}

speed par sweep karta hai, isliye across-acceleration

V \overset{γ}{˙}

hai aur across-force

mV \overset{γ}{˙}

hai.

Rotational (attitude) dynamics ignore karne ke baad bhi hum achhi trajectory kyun pa sakte hain?

Range, altitude, aur burnout speed sirf centre of mass ki motion (translation) par depend karte hain. Rotation pointing aur stability ke liye matter karta hai — tab tum 6DOF Rigid-Body Dynamics par upgrade karte ho.

Ye char/paanch ODEs ek neat formula se nahi balki numerically kyun solve karne padte hain?

Ye coupled aur nonlinear hain —

D

V

par depend karta hai (aur

ρ (h)

par bhi),

\overset{γ}{˙}

V

se divide karta hai,

m

waqt ke saath change hota hai — isliye koi closed form generally exist nahi karta; hum inhe Numerical Integration of Trajectories (RK4) jaisi scheme se aage step karte hain.

Vertical-flight case Tsiolkovsky equation se kyun connect karta hai?

γ = 90°

par bina drag ke,

\dot{V} = T / m - g

, aur

\overset{m}{˙} = - T / (I_{s p} g_{0})

ke saath integrate karne par ideal velocity gain minus gravity-loss term milta hai — ye Tsiolkovsky Rocket Equation ka regime hai.

Same thrust flight aage badhne par zyada badi acceleration kyun deta hai?

\dot{V} = (T - D) / m - \dots

mein, mass

m

propellant jalne ke saath shrink karta hai. Roughly steady thrust ko chhote mass se divide karne par growing acceleration milti hai — burnout acceleration liftoff se dramatically zyada hoti hai.

Edge cases

Wo scenarios jo lazy formulas tod dete hain — actually kya hota hai batao.

V \to 0

(rocket momentarily at rest, e.g. vertical hop ka apex):

\overset{γ}{˙} = - g cos γ / V

ka kya hoga?

Ye blow up ho jaata hai (division by zero) — "velocity arrow ko turn karne" ka model undefined hai jab koi arrow ho hi nahi. Physically direction momentarily ill-posed hai; near zero speed par hum directly velocity components track karna shuru karte hain,

γ

nahi.

γ = 0

(level flight): along/across weight pieces kya ban jaate hain?

Along-piece

m g sin 0 = 0

(gravity tumhe slow nahi karta), across-piece

m g cos 0 = m g

(poora weight path ko maximum rate

\overset{γ}{˙} = - g / V

par neeche bend karta hai). Level flight wahan hai jahan gravity tumhe sabse tezi se turn karti hai.

γ = - 90°

(seedha neeche dive):

\dot{h}

ka sign aur speed par gravity ka effect?

\dot{h} = V sin (- 90°) = - V < 0

(descending), aur

g sin (- 90°) = - g

isliye

\dot{V} = (T - D) / m + g

— gravity ab speed mein add karta hai.

γ

ke signs automatically direction carry karte hain; koi special-casing nahi chahiye.

T = 0

aur

D = 0

(vacuum mein pure coast): trajectory ka kya shape hoga?

Equations reduce ho jaate hain

\dot{V} = - g sin γ

\overset{γ}{˙} = - g cos γ / V

\overset{x}{˙} = V cos γ

\dot{h} = V sin γ

— exactly ek projectile parabola. 3DOF set mein free-fall ek limiting case ke roop mein contained hai.

L

itna bada ki

mV \overset{γ}{˙} > 0

ho jaaye: kya ye allowed hai?

Haan — agar lift

m g cos γ

se zyada ho, toh

\overset{γ}{˙} > 0

aur path upar curve karta hai. Model mein kuch bhi ise forbid nahi karta; aise hi ek winged vehicle pull up karta hai. Ballistic rockets mein sirf usually

L \approx 0

hota hai.

Drag

D

V \to 0

: kya drag vanish ho jaata hai, aur liftoff par ye kyun matter karta hai?

Haan —

D = \frac{1}{2} ρ V^{2} S C_{D} \to 0

V \to 0

, isliye liftoff ke instant par drag negligible hai aur

\dot{V} \approx T / m - g

. Drag tabhi important banta hai jab speed build up hoti hai; Drag and Atmospheric Models dekho.

High altitude jahan

ρ \to 0

\dot{V}

mein kaunsa term disappear hota hai aur kya bacha rehta hai?

Air density

ρ \to 0

D \to 0

hota hai, isliye

\dot{V} \to T / m - g sin γ

— vacuum tangential equation. Isliye upper-stage burns itne efficient hote hain: koi atmospheric drag loss nahi.

Recall Ek-line self-test

Sab cover karo aur zor se jawab do: kaunsa equation speed change karta hai, kaunsa direction, aur kaun se do position — aur $γ$ ka kaunsa single trig function ( $sin$ ya $cos$ ) har ek mein baitha hai? Answer ::: Speed: $\dot{V}$ with $g sin γ$ . Direction: $\overset{γ}{˙}$ with $g cos γ$ . Position: $\overset{x}{˙} = V cos γ$ , $\dot{h} = V sin γ$ .

Connections

Equations of motion — 3DOF point mass (trajectory analysis) (trajectory analysis) — woh parent jise ye bank drill karta hai
Flight-Path Angle and Velocity Frame — along/across split ke peeche ki geometry
Gravity Turn Trajectory — $\overset{γ}{˙}$ edge cases action mein
Tsiolkovsky Rocket Equation — vertical/no-drag limit
Drag and Atmospheric Models — jahan $ρ V^{2}$ traps rehte hain
6DOF Rigid-Body Dynamics — ignore ki gayi rotation kya add karti
Numerical Integration of Trajectories (RK4) — closed forms kyun fail karte hain