Exercises — Gyroscope in spacecraft attitude control — preview

Hum kya use karte hain: definition $\vec{L}=I\vec{\omega }$. Yeh ek definition hai, derivation nahi — ek symmetric rotor ka angular momentum bas "spin-inertia times spin-rate" hai. Hum magnitude $L$ chahte hain, toh arrows hatate hain: $L=I\omega =(0.8)(300)=240{\text{kg⋅m}}^2/\text{s}.$ Direction: vector $\vec{L}$ spin axis ke saath lie karta hai, sense right-hand rule se milta hai — right-hand fingers ko usi taraf curl karo jis tarah wheel ghoomta hai, thumb $\vec{L}$ ki taraf point karega.

Hum kya use karte hain: magnitude form $\Omega =\frac{\tau }{L}$ — upar intuition box mein derive kiya gaya aur parent note mein box kiya gaya hai. Yahan dono quantities magnitudes hain, toh koi arrows ki zarurat nahi. $\Omega =\frac{0.002}{0.5}=0.004\text{rad/s}.$ Axis ek cone ke around $0.004$ radians per second ki speed se drift karta hai — bahut slow, jo exactly woh hai jo ek stable reference ko chahiye.

Hum kya karte hain: magnitude form $\Omega =\frac{\tau }{I\omega }$ se shuru karo aur $\omega$ ke liye solve karo. Kyun: unknown denominator mein trap hai, toh isse free karne ke liye rearrange karte hain. $\omega =\frac{\tau }{I\Omega }=\frac{1\times 10^{-3}}{(1\times 10^{-4})(5\times 10^{-4})}=\frac{1\times 10^{-3}}{5\times 10^{-8}}=2\times 10^4\text{rad/s}.$ Toh $\omega =20,000\text{rad/s}$ (lagbhag $191,000$ rpm — real precision gyros itni tezi se hi spin karte hain). Slower rotor bahut jaldi drift kar deta aur trustworthy nahi hota.

Hum kya use karte hain: conservation of angular momentum (Conservation of Angular Momentum). Yahan hum signs rakhte hain kyunki direction hi poora sawaal hai — positive $\omega$ matlab ek rotational sense, negative $\omega$ matlab ulta sense (yeh poore vector statement $\vec{L}=0$ ka 1-D signed shadow hai). Craft vacuum mein isolated hai aur rest par shuru hua, toh total angular momentum $0$ hamesha ke liye hai: $I_b{\omega }_b+I_w{\omega }_w=0\Rightarrow {\omega }_w=-\frac{I_b{\omega }_b}{I_w}.$ ${\omega }_w=-\frac{(800)(0.015)}{0.4}=-\frac{12}{0.4}=-30\text{rad/s}.$ Minus sign ka matlab ulta sense: body ko ek taraf turn karne ke liye, wheel ko doosri taraf spin karo. Magnitude $30\text{rad/s}$ badi hai kyunki wheel inertia mein $2000\times$ chhota hai.

Hum kya karte hain: precession rate $\Omega =\tau /(I\omega )$, toh same $\tau$ ke saath, jiske paas larger spin magnitude $L=I\omega$ hai woh kam precesses karega. Har ek $L$ calculate karo: $L_A=(2\times 10^{-4})(5000)=1.0{\text{kg⋅m}}^2/\text{s},$ $L_B=(5\times 10^{-4})(3000)=1.5{\text{kg⋅m}}^2/\text{s}.$ B ke paas larger $L$ hai, toh B kam precesses karta hai. Factor: $\frac{{\Omega }_A}{{\Omega }_B}=\frac{\tau /L_A}{\tau /L_B}=\frac{L_B}{L_A}=\frac{1.5}{1.0}=\mathrm{1.5.}$ Gyro A, B se 1.5 times tezi se drift karta hai, chahe A tezi se spin kare — kyunki B ki badi inertia zyada compensate kar deti hai. Product padho, ek variable nahi.

(a) Wheel torque. Torque angular momentum ka rate of change hai (Torque and Newton's Second Law for Rotation): $\tau =I_w\frac{\Delta {\omega }_w}{\Delta t}$. ${\tau }_{\text{wheel}}=0.4\times \frac{30}{6}=0.4\times 5=2.0\text{N⋅m}.$ (b) Body torque. Rotation ke liye Newton's third law se, body par reaction equal aur opposite hai: ${\tau }_{\text{body}}=-{\tau }_{\text{wheel}}=-2.0\text{N⋅m}.$ (c) Body angular acceleration. ${\tau }_{\text{body}}=I_b{\alpha }_b$ use karke: ${\alpha }_b=\frac{{\tau }_{\text{body}}}{I_b}=\frac{-2.0}{800}=-2.5\times 10^{-3}{\text{rad/s}}^2.$ Body ulti taraf accelerate karti hai — chhoti, kyunki yeh itni massive hai. $6\text{s}$ ke baad uski speed hai $|{\alpha }_b|t=2.5\times 10^{-3}\times 6=0.015\text{rad/s}$, jo L2·Q2 ke conservation answer se match karta hai. Do raaste, same destination — yeh ek achha consistency check hai.

(a) Rate. Sirf magnitudes: $\Omega =\tau /L=0.003/0.6=5\times 10^{-3}\text{rad/s}$.

(b) Direction — figure se read karo. Neeche ka figure horizontal plane ka ek top-down map hai: East $+\hat{x}$ axis hai, North $+\hat{y}$ hai. Teal arrow $\vec{L}$ hai (spin axis, East ki taraf point karta hua); orange arrow applied torque $\vec{\tau }$ hai (North ki taraf point karta hua); plum dashed arc axis ki resulting sweep dikhata hai; aur chhota circled dot $\vec{\Omega }$ ko page ke bahar (upar ki taraf) straight point karte hua mark karta hai. Isse dekho ki axis push ki taraf tip nahi hoti — yeh horizontal plane mein sideways swing karti hai.

Ab Cross Product relation $\vec{\tau }=\vec{\Omega }\times \vec{L}$ se confirm karo. Right-handed frame mein $\hat{x}\times \hat{y}=\hat{z}$, $\hat{y}\times \hat{z}=\hat{x}$, $\hat{z}\times \hat{x}=\hat{y}$. Hum guess karte hain $\vec{\Omega }=\Omega \hat{z}$ (upar, page ke bahar point karta hua) aur test karte hain: $\vec{\Omega }\times \vec{L}=(\Omega \hat{z})\times (L\hat{x})=\Omega L(\hat{z}\times \hat{x})=\Omega L\hat{y}.$ Yeh $+\hat{y}$ ki taraf point karta hai — exactly $\vec{\tau }$ ki direction. ✓ Toh $\vec{\Omega }=+\Omega \hat{z}$: axis horizontally sweep karta hai, East se North ki taraf ghoomta hua, vertical ke baare mein ek cone trace karta hua — exactly figure mein plum arc.

(a) Phase 1 body velocity. Total angular momentum poore waqt $0$ hai (rest par shuru hua, isolated): ${\omega }_b=-\frac{I_w{\omega }_w}{I_b}=-\frac{(0.5)(+40)}{600}=-\frac{20}{600}=-0.0333\text{rad/s}.$ Minus sign ka matlab: kyunki wheel counter-clockwise spin karta hai ($+$), body clockwise ($-$) ghoomti hai. (b) Phase 2 ke baad. Jab wheel ${\omega }_w=0$ par wapas aata hai, conservation $I_b{\omega }_b=0$ force karta hai, toh ${\omega }_b=0$. Body rotate karna band kar deti hai. (Wheel ko wapas spin karna body ki spin ko undo karta hai — isi tarah ek craft slewing band karta hai.) (c) Net angle aur direction. Body Phase 1 ke dauran reorient karti hai (yeh $-0.0333\text{rad/s}$ par rotate karti hai), phir wheel brake hone par still rehti hai. Net turn Phase-1 rotation hai. Constant wheel acceleration ke saath body ki speed linearly $0$ se $-0.0333\text{rad/s}$ tak $8\text{s}$ mein ramp karti hai, toh average signed velocity peak ki half hai: ${\theta }_b=\frac{1}{2}(-0.0333)(8)=-0.1333\text{rad}\approx -7.6^{\circ }.$ Negative sign physics hai: craft $\approx 7.6^{\circ }$ clockwise (wheel ke ulte sense mein) ghoom chuka hai, aur ab still hai, naye target ki taraf point karta hua. Wheel spin karo turn karne ke liye, wheel brake karo hold karne ke liye.

(a) Saturate hone ka time. Wheel ko ${\tau }_d$ ki rate par momentum absorb karna hai, aur uski capacity $L_{\max }=I_w{\omega }_{\max }$ hai: $L_{\max }=(0.6)(600)=360{\text{kg⋅m}}^2/\text{s}.$ Time = capacity ÷ fill-rate: $t_{\text{sat}}=\frac{L_{\max }}{{\tau }_d}=\frac{360}{3\times 10^{-4}}=1.2\times 10^6\text{s}\approx 13.9\text{days}.$ Kyun: chhota disturbance angular momentum drip karta rehta hai; wheel ki finite capacity karib do hafte mein bhar jaati hai — phir woh aur madad nahi kar sakta. (b) Desaturation time. Thruster ko saare $360{\text{kg⋅m}}^2/\text{s}$ $0.05\text{N⋅m}$ ki rate par hatane hain: $t_{\text{desat}}=\frac{L_{\max }}{{\tau }_{\text{thruster}}}=\frac{360}{0.05}=7200\text{s}=2\text{hours}.$ Sabak: wheel ne do hafte steady disturbances ko silently handle kiya; kabhi kabhi hi thruster ko accumulated momentum bahar dump karna padta hai. Yahi "desaturation" hai jiske baare mein parent note warn karta hai.

Step 1 — drift angle ko precession rate se link karo. Observation ke dauran axis steady rate $\Omega$ par precesses karta hai, toh accumulated angle hai $\theta =\Omega t$. Kyun: angle = rate × time jab rate constant ho. Step 2 — precession formula se $\Omega$ express karo. Intuition box se, $\Omega =\frac{\tau }{I\omega }$. Substitute karo: $\theta =\Omega t=\frac{\tau t}{I\omega }.$ Step 3 — budget impose karo $\theta \le {\theta }_{\max }$ aur $\omega$ ke liye solve karo. Kyunki $\omega$ denominator mein hai, zyada $\omega$ $\theta$ ko chhota karta hai, toh hame $\omega$ kam se kam woh value chahiye jo $\theta$ ko exactly ${\theta }_{\max }$ banaye: $\omega \ge \frac{\tau t}{I{\theta }_{\max }}=\frac{(2\times 10^{-6})(100)}{(3\times 10^{-5})(1\times 10^{-4})}=\frac{2\times 10^{-4}}{3\times 10^{-9}}\approx 6.67\times 10^4\text{rad/s}.$ Answer: rotor kam se kam $\approx 6.67\times 10^4\text{rad/s}$ (lagbhag $66,700\text{rad/s}$) par spin kare taaki poore $100\text{s}$ exposure ke liye star image kaafi steady rahe. Pehle ke har idea ek saath aata hai: precession rate $\Omega =\tau /(I\omega )$, drift-as-rate×time $\theta =\Omega t$, aur design-by-inequality — yahi ek attitude-control engineer gyro size karte waqt compute karta hai.

Spin momentum $\vec{L}=I\vec{\omega }$, magnitude $L=I\omega$ ::: axis ke saath, right-hand rule. Precession rate $\Omega =\tau /(I\omega )$ ::: zyada spin → slower drift. Vector precession $\vec{\tau }=\vec{\Omega }\times \vec{L}$ ::: axis push ke perpendicular sweep karta hai. Reaction wheel $I_b{\omega }_b=-I_w{\omega }_w$ ::: wheel ek taraf spin karo, body doosri taraf ghoomti hai. Torque $\tau =I\Delta \omega /\Delta t$ ::: angular momentum ka rate of change. Drift angle $\theta =\Omega t$ ::: ek interval mein steady-rate precession. Saturation $L_{\max }=I_w{\omega }_{\max }$ ::: finite momentum store.