Exercises — 6DOF equations — translational (Newton), rotational (Euler's equations)

• $\dot{v}$ = sideways velocity component ka rate of change body frame mein measure kiya gaya — yeh $\dot{\vec{v}}{|}_B$ ka $y$-component hai (upar define kiya gaya "seen in the body" derivative).
• $ru$ aur $-pw$ = $\vec{\omega }\times \vec{v}$ ka $y$-component. Explicitly $\vec{\omega }\times \vec{v}$ ka $y$-component $ru-pw$ hai (from $(\vec{\omega }\times \vec{v}{)}_y=ru-pw$). Kaisa dikhta hai: chahe body mein $v$ kabhi change na ho ($\dot{v}=0$), ek rolling ($p$) ya yawing ($r$) rocket inertial frame mein phir bhi sideways accelerate karta hai — yahi $ru-pw$ encode karta hai.

Har woh term jo do $\omega$-components ka product hai, ya ek $\omega$ times velocity/momentum hai, vanish ho jaati hai.

• Newton → $F_x=m\dot{u},F_y=m\dot{v},F_z=m\dot{w}$ (plain $\vec{F}=m\dot{\vec{v}}$).
• Euler → $M_x=I_x\dot{p},M_y=I_y\dot{q},M_z=I_z\dot{r}$ (plain torque = inertia × angular acceleration). Yeh "textbook" non-spinning limit hai.

$F_y=m(\dot{v}+ru-pw)=0$. $u=200,r=0.5,p=0$ ke saath: $0=\dot{v}+(0.5)(200)-0\Rightarrow \dot{v}=-100{\text{m/s}}^2.$ Iska matlab: koi side force na hone par bhi, sideways body-velocity component negative grow karta hai kyunki poora velocity vector yaw se sweep ho raha hai. Yeh ek bookkeeping acceleration hai, real inertial-frame wala nahin.

Lever arm $\ell$ ek force ko moment mein badalta hai: $M_y=F\ell =800\times 3=2400\text{N⋅m}$ (force times perpendicular distance). Euler-$y$: $M_y=I_y\dot{q}+(I_x-I_z)rp$. $r=p=0$ ke saath coupling term $0$ hai: $\dot{q}=\frac{M_y}{I_y}=\frac{2400}{1200}=2{\text{rad/s}}^2.$ Kaisa dikhta hai: ek clean, decoupled angular acceleration — pitch axis abhi kisi roll/yaw se nahin lad raha.

Step 0 — figure ke axes padhna. Figure mein horizontal axis $q$ (pitch rate) hai aur vertical axis $r$ (yaw rate) hai. Ek single point $(q,r)$ angular velocity ka current transverse hissa hai — woh hissa jo spin axis ke perpendicular hai. Hamara poora kaam yeh pata karna hai ki yeh point kaise move karta hai. Yeh yellow dot $(q_0,0)$ par start karta hai (given: $q(0)=q_0$, $r(0)=0$).

Step 1 — spin rate constant hai. Euler-$x$: $M_x=I_x\dot{p}=0\Rightarrow \dot{p}=0\Rightarrow p=p_0$. ✓ Toh roll rate kabhi nahin badlata; sirf transverse point $(q,r)$ move kar sakta hai.

Step 2 — do transverse Euler equations likhna, sign carefully. Parent ke boxed forms ko exactly jaisa likha hai lo aur symmetric values $I_y=I_z=I_t$, $p=p_0$, aur $M_y=M_z=0$ substitute karo: $\text{Euler-}y:0=I_y\dot{q}+(I_x-I_z)rp\Rightarrow 0=I_t\dot{q}+(I_x-I_t)r{p}_0.$ $\dot{q}$ ke liye coupling term ko doosri side le jaao (yahi sirf algebra hai — koi cyclic re-indexing nahin chahiye): $\boxed{I_t\dot{q}=-(I_x-I_t)r{p}_0}.$ Euler-$z$ mein identical substitution karo, jiska parent coefficient $(I_y-I_x)$ hai aur $I_y=I_t$ hai: $\text{Euler-}z:0=I_z\dot{r}+(I_y-I_x)pq\Rightarrow 0=I_t\dot{r}+(I_t-I_x)p_{0}q,$ $\boxed{I_t\dot{r}=-(I_t-I_x)p_{0}q=(I_x-I_t)p_{0}q}.$ Do boxed equations mein $(I_x-I_t)$ par opposite signs kyun hain — asli "kyun": $\dot{q}$ equation ne coefficient $(I_x-I_z)=(I_x-I_t)$ Euler-$y$ se inherit kiya, jabki $\dot{r}$ equation ne $(I_y-I_x)=(I_t-I_x)$ Euler-$z$ se inherit kiya. Yeh do coefficients ek doosre ke negatives hain — yeh parent ke cyclic pattern mein baked in hai ($x:I_z-I_y$, $y:I_x-I_z$, $z:I_y-I_x$). Yahi $q$- aur $r$-equations ke beech sign flip hai jo straight-line growth ko point $(q,r)$ ke rotation mein badal deta hai, jaisa aagla step dikhata hai.

Step 3 — frequency name karna aur simple-harmonic motion lena. Dono boxed equations ko $I_t$ se divide karo aur define karo $\lambda =\frac{I_x-I_t}{I_t}{p}_0.$ Phir, signs tidying karke, $\dot{q}=-\lambda r$ ... aao instead woh convention choose karein jo initial data ke saath cleanly match ho. Boxed pair ko rewrite karo: $\dot{q}=-\frac{(I_x-I_t)}{I_t}r{p}_0=-\lambda r,\dot{r}=\frac{(I_x-I_t)}{I_t}q{p}_0=\lambda q.$ Pehle ko differentiate karo aur doosre se $\dot{r}$ substitute karo: $\ddot{q}=-\lambda \dot{r}=-\lambda (\lambda q)=-{\lambda }^{2}q.$ Yeh simple harmonic motion hai: $\ddot{q}=-{\lambda }^{2}q$. $q(0)=q_0$, $r(0)=0$ ke saath (toh $\dot{q}(0)=-\lambda r(0)=0$), $q(t)=q_0\cos (\lambda t),r(t)=q_0\sin (\lambda t),$ jo aap verify kar sakte ho ki $\dot{q}=-\lambda r$ aur $\dot{r}=\lambda q$ aur dono initial conditions satisfy karta hai.

Step 4 — figure se answer wapas padhna. Kyunki $q^2+r^2=q_0^2{\cos }^2(\lambda t)+q_0^2{\sin }^2(\lambda t)=q_0^2$, point $(q,r)$ origin se exactly $q_0$ door rehta hai — yeh figure mein radius $q_0$ ki pink circle par ride karta hai. Blue arrow travel direction dikhata hai; yeh angular rate $|\lambda |$ par circle complete karta hai. Woh closed, non-growing circle hi steady coning hai: spin axis ek cone mein nod karta hai lekin kabhi tumble nahin karta. SHM aur blow-up kyun nahin? Step 2 ka sign flip matlab hai ki $q$ aur $r$ ek doosre mein $\dot{q}=-\lambda r$, $\dot{r}=+\lambda q$ ki tarah feed karte hain — vector $(q,r)$ ka pure rotation, toh iska length conserved hai aur energy sirf pitch aur yaw ke beech slosh karti hai. Yahi poori wajah hai ki spinning stabilize karta hai: koi bhi transverse rate grow nahin kar sakta.

• Lamba patla rocket (prolate): $I_x<I_t\Rightarrow \lambda <0$. Coning doosri taraf circulate karta hai lekin phir bhi SHM hai, phir bhi bounded: $|\lambda |$ real.
• Flat disk (oblate): $I_x>I_t\Rightarrow \lambda >0$, opposite circulation direction, phir bhi bounded. Dono cases mein $\lambda$ real hai, toh $\ddot{q}=-{\lambda }^{2}q$ oscillation deta hai, kabhi exponential growth nahin. Ek symmetric body dono signs ke liye apne spin axis ke baare mein coning-stable hai; sirf cone ka sense (direction) flip hota hai. Dangerous case $I_x\to I_t$ ($\lambda \to 0$) hai: coning infinitely slow ho jaata hai — axis freely wander karta hai, koi restoring "stiffness" nahin. Full instability ke liye teen alag moments chahiye (L4 trap aur Dzhanibekov effect / intermediate axis theorem dekho).

Euler-$y$ aur Euler-$z$ $M=0$ ke saath, $p=p_0$ constant rakhte hue aur $q,r$ ko small treat karte hue: $I_y\dot{q}=(I_z-I_x)r{p}_0,I_z\dot{r}=(I_x-I_y)q{p}_0.$ $\ddot{q}$ ke liye solve karo: pehle ko differentiate karo, doosre se $\dot{r}$ substitute karo: $\ddot{q}=\frac{(I_z-I_x)p_0}{I_y}\dot{r}=\frac{(I_z-I_x)(I_x-I_y)}{I_y{I}_z}{p}_0^{2}q.$ Bracketed coefficient ko $C=\frac{(I_z-I_x)(I_x-I_y)}{I_y{I}_z}{p}_0^2$ naam do. Numbers: $(I_z-I_x)=420-90=330$, $(I_x-I_y)=90-400=-310$, toh numerator $=330\times (-310)=-102300$, denominator $=400\times 420=168000$. $C=\frac{-102300}{168000}\times 10^2=-60.89{\text{(rad/s)}}^2.$ $C<0\Rightarrow \ddot{q}=Cq=-|C|q$: oscillation frequency $\sqrt{|C|}\approx 7.80\text{ rad/s}$ par. Smallest axis ke baare mein spinning stable hai (bounded coning), major/minor-axis stability rule se consistent. Sign test kyun kaam karta hai: stability $\iff C<0\iff (I_z-I_x)$ aur $(I_x-I_y)$ ke opposite signs hain $\iff I_x$ teeno mein largest YA smallest hai. Intermediate axis $\Rightarrow$ same signs $\Rightarrow C>0\Rightarrow$ blow-up.

$T$ ko thrust denote karne do — exhaust rocket par jo forward force lagata hai. Iska magnitude $T=-\dot{m}{u}_e$ hai (minus $T>0$ banata hai kyunki $\dot{m}<0$ hai): $T=-\dot{m}{u}_e=-(-4)(2500)=10000\text{N},\text{along }+x.$ Total body-$x$ external force: $F_x=T+F_x^{\text{grav}}=10000-4000=6000\text{N}$. Newton-$x$: $F_x=m(\dot{u}+qw-rv)$. $q=0.2,w=10,r=0$ ke saath: $qw-rv=0.2\times 10=2$. $6000=500(\dot{u}+2)\Rightarrow \dot{u}+2=12\Rightarrow \dot{u}=10{\text{m/s}}^2.$ Yeh sahi route kyun hai: humne kabhi $\frac{d(mv)}{dt}$ nahin likha; exhaust se le jaaya gaya momentum already external thrust force $T$ mein baked in hai, toh clean $\vec{F}=m\dot{\vec{v}}$ form valid hai. Mass loss $\dot{m}$ ko agle time-step ke liye alag track kiya jaata hai. Meshchersky / Tsiolkovsky variable-mass dynamics dekho.

(a) Coning frequency. L3-1 se $\lambda =\frac{I_x-I_t}{I_t}{p}_0$ use karo: $\lambda =\frac{120-500}{500}\times 8=\frac{-380}{500}\times 8=-6.08\text{rad/s}\Rightarrow |\lambda |=6.08\text{rad/s}.$ Sign negative hai (yahan prolate-like, $I_x<I_t$), matlab cone ek particular taraf circulate karta hai; sirf magnitude $|\lambda |$ rate set karta hai.

(b) Ek full circle ke liye time. L3-1 se transverse point $(q,r)$ $\cos (\lambda t),\sin (\lambda t)$ ki tarah jaata hai, toh yeh apne start par wapas aata hai jab angle $\lambda t$ $2\pi$ advance kare: $T_{\text{cone}}=\frac{2\pi }{|\lambda |}=\frac{2\pi }{6.08}\approx 1.0334\text{s}.$

(c) Constant transverse magnitude. L3-1 se, $q(t)=q_0\cos (\lambda t)$ aur $r(t)=q_0\sin (\lambda t)$, toh $\sqrt{q^2+r^2}=\sqrt{q_0^2{\cos }^2(\lambda t)+q_0^2{\sin }^2(\lambda t)}=|q_0|=0.05\text{rad/s},$ ek constant — tip radius $0.05$ rad/s ki circle par ride karti hai, steady coning with no growth confirm karta hai. Yeh exactly L3-1 ki figure ki pink circle picture hai, ab concrete numbers ke saath.

Ek axis ke baare mein spin ki stability require karti hai ki woh axis largest ya smallest moment of inertia ho (do transverse differences ke opposite signs hone chahiye). Yahan sorted: $I_x=90$ (smallest), $I_y=95$ (intermediate), $I_z=400$ (largest).

• $x$ (smallest) ke baare mein spin: safe — coefficient $\propto (I_z-I_x)(I_x-I_y)=(310)(-5)<0\Rightarrow$ oscillatory (stable).
• $z$ (largest) ke baare mein spin: safe — $\propto (I_x-I_z)(I_z-I_y)=(-310)(305)<0\Rightarrow$ stable.
• $y$ (intermediate) ke baare mein spin: forbidden — $\propto (I_x-I_y)(I_y-I_z)=(-5)(-305)>0\Rightarrow$ exponential growth, Dzhanibekov flip. Design answer: $x$ ya $z$ ke baare mein spin karo; kabhi $y$ ke baare mein nahin. Near-equal $I_x\approx I_y$ (90 vs 95) $x$-spin ko sirf weakly stable banata hai, toh $z$ (large margin) robust choice hai.

L1: terms padhna. ::: $\omega \times$ pieces cycle karte hain $x\to y\to z\to x$ aur sirf instantaneously $\vec{\omega }=0$ par vanish hote hain. L2: plug in karo. ::: Signs cyclic template se rakho; ek decoupled axis ($p=r=0$) clean $\dot{q}=M_y/I_y$ deta hai. L3–L4: coupling analyze karo. ::: Symmetric body cones (real $\lambda$); stability sign hai $(I_a-I_b)(I_b-I_c)$. L5: design karo. ::: Sirf largest ya smallest principal axis ke baare mein spin karo; kabhi intermediate ke baare mein nahin.