6DOF equations — translational (Newton), rotational (Euler's equations): Worked examples

3.4.5 · D3 · Physics › Rocket Flight Mechanics › 6DOF equations — translational (Newton), rotational (Euler's

Kuch bhi shuru karne se pehle: woh symbols jo hum baar baar use karenge, plain words mein.

Definition Symbols ki cast (sab ek baar, yahan define)

$m$ — rocket ki mass (kitna saamaan hai), kilograms mein.
$v = (u, v, w)$ — center of mass (balance point) ki velocity body axes mein likhi: $u$ = forward speed, $v$ = sideways speed, $w$ = up/down speed. Units m/s.
$ω = (p, q, r)$ — rocket kitni tezi se twist karta hai: $p$ = roll (ek log ki tarah barrel-roll), $q$ = pitch (nose up/down), $r$ = yaw (nose left/right). Units rad/s.
$F = (F_{x}, F_{y}, F_{z})$ — total external force (push), newtons mein.
$M = (M_{x}, M_{y}, M_{z})$ — total external moment/torque (twisting effort), N·m mein.
$I_{x}, I_{y}, I_{z}$ — teen body axes ke baare mein moments of inertia: "is axis ko spin karwana kitna mushkil hai". Units kg·m². Inertia tensor and principal axes mein build kiya gaya hai.
$I_{t}$ — ek axisymmetric body ka transverse moment of inertia: jab do "side" axes identical hote hain to hum $I_{y} = I_{z} = I_{t}$ likhte hain (do ki jagah ek shared number). Units kg·m².
$ℓ$ — ek lever arm: center of mass se kisi force ki line tak ki perpendicular distance. Force ko uske lever arm se multiply karne par moment milta hai, $M = F ℓ$ . Units metres.
Symbol $\times$ cross product hai — yeh do arrows leta hai aur ek teesra arrow return karta hai jo dono ke perpendicular ho, jiska length "yeh dono kitne parallel nahi hain" yeh batata hai.

Do master equations jo hum baar baar apply karenge (parent se):

The scenario matrix

Har row ek class of situation hai. Last column us worked example ka naam batata hai jo use cover karta hai.

#	Case class	Kya special hai	Covered by
A	Zero rotation ( $ω = 0$ )	sab cross terms vanish → plain 1D Newton	Ex 1
B	Pure turn, constant body speed	$\dot{v} = 0$ lekin real acceleration exist karta hai	Ex 2
C	Sign of a coupling term ( $q r > 0$ vs $< 0$ )	gyroscopic torque kis direction mein point karta hai	Ex 3
D	Symmetric spinner ( $I_{y} = I_{z}$ )	steady coning, oscillation frequency	Ex 4
E	Fully asymmetric, torque-free ( $I_{x} \neq = I_{y} \neq = I_{z}$ )	intermediate-axis instability	Ex 5
F	Degenerate inertia (sphere $I_{x} = I_{y} = I_{z}$ )	all coupling dies, $\dot{ω}$ from torque only	Ex 6
G	Decoupled single-axis maneuver	one moment, clean angular accel	Ex 7
H	Word problem (real thruster + gravity)	build $F, M$ from a story	Ex 8
I	Limiting / exam twist (mass → variable)	why naive $d (m v) / d t$ is wrong	Ex 9

Ex 1 — Case A: zero rotation (sanity floor)

Worked example Vertical climb, no spin

Ek rocket seedha upar jaata hai. Body- $x$ upar point karta hai. Thrust $T = 30, 000$ N upar, weight $m g$ with $m = 2000$ kg, $g = 9.8$ m/s². Koi rotation nahi: $p = q = r = 0$ , aur $v = w = 0$ . Nikalo forward acceleration $\overset{u}{˙}$ .

Forecast: padhne se pehle guess karo — kya answer sirf $(T - m g) / m$ hona chahiye? Kyun ya kyun nahi?

Newton- $x$ likho: $F_{x} = m (\overset{u}{˙} + q w - r v)$ . Yeh step kyun? Isme hi $\overset{u}{˙}$ hai, jo hamara unknown hai.
$q = r = 0$ set karo: terms $q w$ aur $r v$ khatam ho jaate hain, bacha $F_{x} = m \overset{u}{˙}$ . Yeh step kyun? Rotation na hone par koi transport correction nahi hoti — body frame aur inertial frame acceleration par agree karte hain.
Net force hai $F_{x} = T - m g = 30000 - 2000 (9.8) = 10400$ N. Yeh step kyun? Dono forces body- $x$ ke along act karte hain; upar positive hai, weight negative hai, isliye hum dono vertical forces ko unke signs ke saath add kar dete hain.
Solve karo: $\overset{u}{˙} = F_{x} / m = 10400/2000 = 5.2$ m/s². Yeh step kyun? Net force ko mass se divide karo — $F_{x} = m \overset{u}{˙}$ ka final rearrangement.

Verify karo: units $N / kg = m/s^{2}$ ✓. Agar $T = m g = 19600$ N to $\overset{u}{˙} = 0$ (hovering) — intuition se match karta hai. Yeh woh "floor" hai jis par har richer case reduce hona chahiye.

Ex 2 — Case B: pure turn, constant body speed

Worked example Coordinated turn — acceleration saamne chupi hui hai

Ek rocket constant body velocity $u = 200$ m/s, $v = w = 0$ se cruise karta hai, jabki steady $r = 0.1$ rad/s se yaw kar raha hai ( $p = q = 0$ ). Body speed feel hoti hai constant. Mass $m = 1500$ kg. Nikalo zaruri sideways force $F_{y}$ , aur true inertial acceleration.

Forecast: $\overset{v}{˙} = 0$ ke saath, kya $F_{y}$ zero hai? (Nahi hai — yahi toh pura lesson hai.)

Newton- $y$ : $F_{y} = m (\overset{v}{˙} + r u - pw)$ . Yeh step kyun? Sideways force yahan hai; term $r u$ transport correction hai.
Constant body velocity → $\overset{v}{˙} = 0$ . $p = 0$ ke saath: $F_{y} = m r u$ . Yeh step kyun? Bhale hi numbers $u, v, w$ na badlein, velocity arrow yaw ki wajah se swing ho raha hai — uski inertial direction badal rahi hai, isliye real acceleration exist karta hai.
Plug in karo: $F_{y} = 1500 \times 0.1 \times 200 = 30000$ N. Yeh step kyun? Required force nikaalane ke liye $F_{y} = m r u$ mein given numbers substitute karo.
Inertial acceleration ki magnitude hai $a = F_{y} / m = r u = 20$ m/s². Yeh step kyun? Yeh exactly centripetal acceleration $v^{2} / R$ hai jahan $R = u / r = 2000$ m: $u^{2} / R = 20 0^{2} /2000 = 20$ m/s² ✓.

Neeche figure dekho. Gray dashed curve $R = 2000$ m radius ki flight arc hai. Blue arrow velocity $v$ hai, arc ke tangent — uski length (200 m/s) kabhi nahi badlti, lekin green curved arrow (yaw $r$ ) uski direction ko swing karte rehta hai. Orange arrow force $F_{y} = 30000$ N hai jo turn center ki taraf point kar raha hai: wahi force seedhi flight ko circle mein bend karti hai. Toh "constant speed" turn ko bhi real sideways push chahiye.

Verify karo: $a$ tak pahunchne ke dono raaste (transport term $r u$ aur centripetal $u^{2} / R$ ) 20 m/s² par agree karte hain. $F_{y}$ ke units: kg·(rad/s)·(m/s) = kg·m/s² = N ✓ (radians dimensionless hote hain).

Ex 3 — Case C: coupling term ka sign

Worked example Gyroscopic torque kis direction mein push karta hai?

Ek rocket roll $p = 5$ rad/s aur yaw $r = 2$ rad/s ke saath spin karta hai, is instant pitch $q$ = 0 hai. Inertias: $I_{x} = 10$ , $I_{y} = 20$ , $I_{z} = 40$ kg·m². Koi applied moments nahi ( $M = 0$ ). Nikalo pitch acceleration $\overset{q}{˙}$ aur uska sign.

Forecast: zero pitch aur zero torque ke saath, kya nose usi jagah rahega? $\overset{q}{˙}$ ke sign ka guess karo.

Euler- $y$ : $M_{y} = I_{y} \overset{q}{˙} + (I_{x} - I_{z}) r p$ . $M_{y} = 0$ set karo. Yeh step kyun? Hume $\overset{q}{˙}$ chahiye; coupling term $(I_{x} - I_{z}) r p$ uska sign decide karta hai.
Solve karo: $\overset{q}{˙} = - \frac{( I _{x} - I _{z} )}{I _{y}} r p = - \frac{( 10 - 40 )}{20} (2) (5)$ . Yeh step kyun? Euler- $y$ ko rearrange karna; $(I_{x} - I_{z})$ ka sign times $r p$ jo hum padh rahe hain.
Sign pieces evaluate karo: $(I_{x} - I_{z}) = - 30$ (negative), $r p = 10$ (positive). Saamne wale minus aur negative bracket ka product ek positive $\overset{q}{˙}$ deta hai. Yeh step kyun? Number crunch karne se pehle signs track karo, taaki physics (nose kis taraf pitch hoga) arithmetic mein dabbi na rahe.
Number: $\overset{q}{˙} = - \frac{- 30}{20} (10) = + 15$ rad/s². Yeh step kyun? Do axes par ek saath spin karne wala body ek teesra twist develop karta hai khud hi — yahi gyroscopic coupling hai.

Verify karo: yaw sign ko $r = - 2$ kar do: to $r p = - 10$ aur $\overset{q}{˙} = - \frac{- 30}{20} (- 10) = - 15$ rad/s² — same size, opposite sign, jaisa formula ki symmetry demand karti hai ✓.

Ex 4 — Case D: symmetric spinner → steady coning

Worked example Sounding rocket coning frequency

Ek spin-stabilized rocket axisymmetric hai: $I_{y} = I_{z} = I_{t} = 30$ kg·m², spin inertia $I_{x} = 6$ kg·m². Yeh $p_{0} = 20$ rad/s par spin karta hai, torque-free. Ek chhoti si disturbance use transverse rates de deti hai. Nikalo coning (nutation) frequency.

Forecast: kya transverse wobble grow karega, decay karega, ya hamesha oscillate karta rahega?

Euler- $x$ : $I_{x} \overset{p}{˙} = 0 \Rightarrow p = p_{0}$ constant. Yeh step kyun? $I_{y} = I_{z}$ ke saath $x$ -coupling term $(I_{z} - I_{y}) q r = 0$ ; spin rate lock ho jaati hai.
Euler- $y, z$ : $I_{t} \overset{q}{˙} = (I_{x} - I_{t}) r p_{0}$ aur $I_{t} \overset{r}{˙} = - (I_{x} - I_{t}) q p_{0}$ . Yeh step kyun? Symmetric inertias substitute karna; note karo ki $q$ aur $r$ ek doosre ko feed karte hain.
Define karo $λ = \frac{I _{x} - I _{t}}{I _{t}} p_{0} = \frac{6 - 30}{30} (20) = - 16$ rad/s. Phir $\overset{q}{˙} = λ r$ , $\overset{r}{˙} = - λ q \Rightarrow \overset{q}{¨} = - λ^{2} q$ . Yeh step kyun? Do first-order coupled equations ek oscillator equation mein combine ho jaate hain — yeh vector $(q, r)$ ki circular motion ki pehchaan hai.
Coning frequency hai $∣ λ ∣ = 16$ rad/s (lagbhag 2.5 Hz). Yeh step kyun? $\overset{q}{¨} = - λ^{2} q$ ek spring ka equation hai; uski frequency $∣ λ ∣$ hai.

Neeche figure dekho. Blue aur orange curves do transverse rates $q (t)$ aur $r (t)$ hain. Notice karo ki yeh ek cosine aur ek (negative) sine hain — quarter-cycle apart — isliye jab ek badhta hai doosra girta hai: pair apne aap ko circle mein chase karta rahta hai. Khaas baat yeh hai ki curves na badhte hain na ghatte; yeh $\pm 1$ ke beech bounded rehte hain. Green dashed line ek full period $2 π /∣ λ ∣ \approx 0.39$ s mark karti hai. Woh bounded, repeating wobble exactly wahi hai jo spin stabilization ko safe banata hai.

Verify karo: wobble na grow karta hai na decay — yeh bounded oscillation hai, exactly isliye spin stabilization kaam karta hai. Units: (kg·m²/kg·m²)(rad/s) = rad/s ✓.

Ex 5 — Case E: fully asymmetric torque-free (intermediate axis)

Worked example Middle axis ke baare mein tumbling

Ek body ke teen alag inertias hain $I_{x} = 2 < I_{y} = 5 < I_{z} = 9$ kg·m² (sab distinct). Yeh almost purely intermediate axis $y$ ke baare mein spin karta hai: $q = 10$ rad/s, tiny $p = r = 0.01$ rad/s. Torque-free. Nikalo kya chhote $p, r$ grow karte hain ya chhote rehte hain.

Forecast: kaun sa axis unstable hai — sabse chhota, middle, ya sabse bada?

Euler- $x$ : $I_{x} \overset{p}{˙} = (I_{y} - I_{z}) q r$ , Euler- $z$ : $I_{z} \overset{r}{˙} = (I_{x} - I_{y}) pq$ ( $q \approx$ const ke saath). Yeh step kyun? Hum bade spin $q$ ke baare mein linearize karte hain; $p, r$ chhote deviations hain.
Euler- $x$ differentiate karo aur Euler- $z$ substitute karo: $\overset{p}{¨} = \frac{( I _{y} - I _{z} )}{I _{x}} q \overset{r}{˙} = \frac{( I _{y} - I _{z} ) ( I _{x} - I _{y} )}{I _{x} I _{z}} q^{2} p$ . Yeh step kyun? Hum $p$ mein ek single equation chahte hain taaki growth vs oscillation padh sakein.
Coefficient compute karo $k = \frac{( I _{y} - I _{z} ) ( I _{x} - I _{y} )}{I _{x} I _{z}} q^{2} = \frac{( 5 - 9 ) ( 2 - 5 )}{2 \cdot 9} (100) = \frac{( - 4 ) ( - 3 )}{18} (100) = \frac{12}{18} (100) = \frac{200}{3} \approx 66.7$ . Yeh step kyun? $k$ ka sign sab kuch decide karta hai: $k > 0$ matlab $\overset{p}{¨} = + k p$ → exponential blow-up.
Kyunki $k > 0$ , $p$ $e^{k t}$ ki tarah grow karta hai — spin unstable hai. Body flip ho jaata hai. Yeh step kyun? Middle axis ke baare mein spin karne par hamesha ek negative aur ek positive bracket milta hai, jinका product positive hota hai → instability.

Verify karo: largest axis $z$ ko spin axis lekar repeat karo ( $r$ large): do brackets $(I_{z} - I_{x}) (I_{y} - I_{z})$ -type ban jaate hain opposite signs ke saath, $k < 0$ dete hain → oscillation → stable. Yahi Dzhanibekov effect / intermediate axis theorem hai. Growth rate $k = 200/3 \approx 8.16$ /s ✓.

Ex 6 — Case F: degenerate inertia (perfect sphere)

Worked example Ball-shaped satellite

Ek satellite uniform sphere hai: $I_{x} = I_{y} = I_{z} = I = 12$ kg·m². Ek torque $M_{z} = 3$ N·m $z$ ke baare mein apply hota hai jabki woh pehle se $p = 4$ , $q = 5$ rad/s spin kar raha hai. Nikalo $\overset{r}{˙}$ .

Forecast: kya $p, q$ spins $z$ ke baare mein koi coupling torque produce karte hain? Yes/no guess karo.

Euler- $z$ : $M_{z} = I_{z} \overset{r}{˙} + (I_{y} - I_{x}) pq$ . Yeh step kyun? Teeno inertias aate hain; coupling bracket dekho.
Kyunki $I_{x} = I_{y}$ , bracket $(I_{y} - I_{x}) = 0$ ; coupling term bilkul khatam ho jaata hai. Yeh step kyun? Sphere ka koi "hard" ya "easy" spin axis nahi hota, isliye kabhi koi gyroscopic coupling nahi hoti.
Isliye $M_{z} = I \overset{r}{˙} \Rightarrow \overset{r}{˙} = M_{z} / I = 3/12 = 0.25$ rad/s². Yeh step kyun? Rotation ab bilkul translation ki tarah behave karta hai: torque = inertia × angular accel.

Verify karo: sphere ke liye har Euler equation $M_{i} = I \overset{ω}{˙}_{i}$ tak reduce ho jaata hai — teen independent scalar laws, koi cross-talk nahi. Units: N·m / kg·m² = 1/s² = rad/s² ✓.

Ex 7 — Case G: decoupled single-axis maneuver

Worked example Nose thruster se clean pitch-up

Ek rocket rotation mein rest par hai ( $p = q = r = 0$ ), ek side thruster $F = 800$ N fire karta hai CoM se $ℓ = 3$ m aage lever arm par, ek pure pitch moment produce karta hai. $I_{y} = 500$ kg·m². Nikalo pitch acceleration $\overset{q}{˙}$ aur $t = 2$ s burn ke baad pitch rate.

Forecast: kya yahan roll/yaw state matter karta hai? (Nahi karega — isliye humne $p = r = 0$ choose kiya.)

Pitch axis ke baare mein moment: $M_{y} = F ℓ = 800 \times 3 = 2400$ N·m. Yeh step kyun? Force times lever arm = torque; yeh drive term hai.
Euler- $y$ : $M_{y} = I_{y} \overset{q}{˙} + (I_{x} - I_{z}) r p$ . $r = p = 0$ ke saath coupling khatam ho jaati hai. Yeh step kyun? Non-spinning shuru karne ka matlab maneuver ek clean axis mein decouple ho jaata hai.
$\overset{q}{˙} = M_{y} / I_{y} = 2400/500 = 4.8$ rad/s². Yeh step kyun? Torque ko inertia se divide karo — $M_{y} = I_{y} \overset{q}{˙}$ ka final rearrangement.
Burn ke baad: $q = \overset{q}{˙} t = 4.8 \times 2 = 9.6$ rad/s. Yeh step kyun? Kyunki thruster force, lever arm aur inertia sab burn ke dauran constant hain, $\overset{q}{˙}$ constant hai, aur ek constant rate of change simply $q = \overset{q}{˙} t$ (rate × time) tak integrate ho jaata hai, exactly jaise distance = speed × time.

Verify karo: units N·m / kg·m² = rad/s² ✓ $\overset{q}{˙}$ ke liye, phir rad/s² · s = rad/s ✓ $q = 9.6$ rad/s ke liye. Sanity check: agar burn sirf $t = 0$ s chala to pitch rate $0$ hogi, aur yeh burn time ke saath linearly badhti hai — jaise constant acceleration mein expect hota hai. Agar rocket roll kar raha hota ( $p \neq = 0$ ), Case C dikhata hai ki ek cross term aa jaata — isliye "start from rest" trick kaam aati hai.

Ex 8 — Case H: full word problem (force + gravity + moment)

Worked example Off-axis thruster ke saath gravity turn

Ek rocket ( $m = 1000$ kg, $I_{z} = 400$ kg·m²) horizontally fly kar raha hai: body- $x$ forward $u = 150$ m/s, $v = w = 0$ , aur yeh yaw kar raha hai $r = 0.2$ rad/s par ( $p = q = 0$ ). Ek gimballed engine thrust $T = 12, 000$ N body- $x$ ke along deta hai, plus ek chhota yaw-control side force $F_{s} = 500$ N CoM se $ℓ = 4$ m peeche lever arm par. Gravity neeche act karti hai (body- $z$ yahan neeche point karta hai) $m g$ par, $g = 9.8$ . Nikalo (a) forward accel $\overset{u}{˙}$ , (b) sideways accel $\overset{v}{˙}$ , (c) yaw accel $\overset{r}{˙}$ .

Forecast: teen answers mein se kaun sa simply force/mass nahi hai? (Step 1 se pehle guess karo.)

Forward: $F_{x} = T = 12000$ N. Newton- $x$ : $F_{x} = m (\overset{u}{˙} + q w - r v)$ , aur $q = v = 0$ isliye $\overset{u}{˙} = F_{x} / m = 12000/1000 = 12$ m/s². Yeh step kyun? Yahan $x$ par koi transport term survive nahi karta (use $q$ ya $v$ chahiye).
Sideways: side force $F_{y} = F_{s} = 500$ N. Newton- $y$ : $F_{y} = m (\overset{v}{˙} + r u - pw)$ , $p = 0$ ke saath: $\overset{v}{˙} = F_{y} / m - r u = 500/1000 - (0.2) (150) = 0.5 - 30 = - 29.5$ m/s². Yeh step kyun? Yahi non-trivial wala hai — yaw velocity arrow ko swing karta hai, ek bada transport term $- r u$ add karta hai jo thruster ke direct contribution ko peeche chhod deta hai. Hum $\overset{v}{˙}$ isolate karte hain us transport term ko right-hand side par move karke.
Yaw: moment $M_{z} = F_{s} ℓ = 500 \times 4 = 2000$ N·m. Euler- $z$ : $M_{z} = I_{z} \overset{r}{˙} + (I_{y} - I_{x}) pq$ , $p = q = 0$ ke saath: $\overset{r}{˙} = M_{z} / I_{z} = 2000/400 = 5$ rad/s². Yeh step kyun? Rotation decouple ho jaata hai kyunki $p = q = 0$ ; sirf drive term $M_{z}$ survive karta hai, isliye promised yaw acceleration seedha moment over inertia se aata hai.

Verify karo: (a) $\overset{u}{˙} = 12$ m/s², (b) $\overset{v}{˙} = - 29.5$ m/s², (c) $\overset{r}{˙} = 5$ rad/s². $\overset{v}{˙}$ ka sign negative hai: bhale hi thruster $+ y$ push karta ho, turn ki centripetal demand body ko doosri taraf zyada tezi se sideways pull karti hai — ek real coordinated-turn effect. Units sab m/s² aur rad/s² ✓.

Ex 9 — Case I: limiting / exam twist (variable mass)

Worked example Naive

$\frac{d ( m v )}{d t}$ kyun fail ho jaata hai Ek rocket propellant burn karta hai: $\overset{m}{˙} = - 20$ kg/s, current $m = 800$ kg, rocket ke relative exhaust speed $u_{e} = 2500$ m/s. Yeh seedha fly karta hai, $ω = 0$ , no gravity, $u = 400$ m/s. Ek student likhta hai $F = \frac{d ( m u )}{d t} = m \overset{u}{˙} + \overset{m}{˙} u$ aur only force ko zero set karta hai. Nikalo sahi $\overset{u}{˙}$ aur error expose karo.

Forecast: student conclude karta hai $\overset{u}{˙} = - \overset{m}{˙} u / m$ . Kya woh real acceleration hai?

Sahi physics (Meshchersky): $F_{e x t} + \overset{m}{˙} u_{e, rel} = m \overset{u}{˙}$ . Thrust $T = - \overset{m}{˙} u_{e}$ ko external force treat karo. Yeh step kyun? Ejected gas momentum carry karta hai; $\overset{m}{˙} u$ ko $\frac{d ( m v )}{d t}$ mein lump karke Newton nahi keh sakte — reference momentum gas ka hai, rocket ka nahi.
Thrust: $T = - \overset{m}{˙} u_{e} = - (- 20) (2500) = 50, 000$ N (forward). Yeh step kyun? Mass peeche se jaata hai ( $\overset{m}{˙} < 0$ ), isliye thrust forward aur positive hai.
Koi aur external force nahi, isliye $m \overset{u}{˙} = T \Rightarrow \overset{u}{˙} = 50000/800 = 62.5$ m/s². Yeh step kyun? Thrust ko external force maante hue clean constant-mass Newton form.
Student ka galat answer: $\overset{u}{˙}_{wrong} = - \overset{m}{˙} u / m = - (- 20) (400) /800 = 10$ m/s² — six se zyada factor se galat, aur exhaust speed $u_{e}$ ki jagah flight speed $u$ use karta hai. Yeh step kyun? Trap vehicle momentum se exhaust momentum replace kar deta hai.

Verify karo: sahi $\overset{u}{˙} = 62.5$ m/s² vs naive $10$ m/s². Missing physics ka ratio $u_{e} / u = 2500/400 = 6.25$ hai, exactly discrepancy — proof hai ki error swapped velocity hai ✓. Units: N/kg = m/s² ✓.

Recall Har case ka ek-line summary

Case A ::: no spin → plain $F = ma$ . Case B ::: constant body speed lekin turning → $ω \times v$ term se real accel. Case C ::: $(I_{x} - I_{z}) r p$ ka sign sets karta hai ki gyroscopic torque kis direction mein point karega. Case D ::: symmetric spinner → bounded coning frequency $∣ λ ∣ = \frac{∣ I _{x} - I _{t} ∣}{I _{t}} p_{0}$ par. Case E ::: intermediate-axis spin → coefficient $k > 0$ → exponential flip. Case F ::: sphere → sab coupling zero → teen independent $M_{i} = I \overset{ω}{˙}_{i}$ . Case G ::: rest se shuru → single-axis maneuver cleanly decouple ho jaata hai. Case H ::: word problem → $F, M$ build karo, dekho transport term $\overset{v}{˙}$ par dominate karta hai. Case I ::: variable mass → thrust ek external force hai $- \overset{m}{˙} u_{e}$ , kabhi $\frac{d ( m v )}{d t}$ nahi.

Flashcards

In Ex 2,

F_{y}

nonzero kyun hai bhale hi

\overset{v}{˙} = 0

ho?

Yaw rate

r

velocity arrow ko swing karta hai, isliye transport term

r u

real inertial (centripetal) acceleration

r u = u^{2} / R

deta hai.

Fully asymmetric body ke liye kaun sa spin axis unstable hota hai?

Intermediate (middle) axis — coupling coefficient

k

positive ho jaata hai, giving exponential growth (Dzhanibekov effect).

Perfect sphere ke liye Euler ke coupling terms ka kya hota hai?

Sab vanish ho jaate hain kyunki

I_{x} = I_{y} = I_{z}

, teen independent scalar laws

M_{i} = I \overset{ω}{˙}_{i}

bacha jaate hain.

Burning rocket ke liye naive

\frac{d ( m v )}{d t}

kyun galat hai?

Yeh exhaust momentum ki jagah vehicle momentum use karta hai; sahi thrust external force

- \overset{m}{˙} u_{e}

hai (Meshchersky).