Question bank — Gyroscope in spacecraft attitude control — preview

Ek torque-free spinning gyro $\vec{L}$ ki direction aur magnitude dono fixed rakhta hai. ::: True. $\vec{\tau }=d\vec{L}/dt=0$ ke saath, kuch bhi $\vec{L}$ ko change nahi karta — direction ki yahi constancy woh "rigidity" hai jo spacecraft reference ke roop mein use karta hai. Same torque ke under ek faster-spinning sensing gyro zyada dheere drift (precess) karta hai. ::: True. $\Omega =\tau /(I\omega )$ mein denominator mein spin $I\omega$ hai, isliye zyada spin ka matlab choti drift rate hai — ek stiffer, zyada stubborn reference. $\vec{L}$ ke parallel apply kiya gaya torque precession cause karta hai. ::: False. Ek parallel torque sirf $\vec{L}$ ki length badalta hai (rotor ko speed up ya down karta hai). Precession ke liye $\vec{L}$ ke perpendicular torque chahiye, jo iska direction turn karta hai. Ek reaction wheel spacecraft ko kuch nahi se net angular momentum gain karne deta hai. ::: False. Body + wheel isolated hai, isliye total $\vec{L}$ conserved hai; wheel sirf body ke saath momentum trade karta hai, kabhi create nahi karta. Body ko clockwise spin karne ke liye, wheel ko bhi clockwise spin karte hain. ::: False. Conservation force karta hai $I_b{\omega }_b=-I_w{\omega }_w$: body aur wheel opposite senses mein spin karte hain taaki starting total cancel ho jaaye. Reaction wheels bina help ke spacecraft ko hamesha ke liye point kar sakte hain. ::: False. Wheels ek maximum speed par saturate ho jaate hain; external torques unhe load karte rehte hain, isliye thrusters ya magnetorquers ko periodically unhe desaturate karna padta hai. Pure precession ke dauran, spin angular momentum ki length $L=I\omega$ constant rehti hai. ::: True. Torque $\vec{L}$ ke perpendicular hai, isliye $d\vec{L}$ sirf vector ko rotate karta hai — ek perpendicular addition first order par vector ki length nahi badal sakta. Torque apply karte hi precession turant hoti hai, bilkul bhi fall nahi hoti. ::: Fast gyro ke liye zyaadatar true. $\vec{L}$ ki tip sideways move karna turant shuru kar deti hai; koi bhi choti initial "fall" (nutation) ek chota wobble hai, negligible jab $I\omega$ bahut bada ho.

"Main axis ke top ko north push karta hoon, isliye axis north tip karta hai." Kahan galti hai? ::: Response push ke perpendicular hoti hai, push ke along nahi. Axes set karo: $\vec{L}$ $+x$ ke along (spin axis), push ek torque $\vec{\tau }$ $+y$ ke along create karta hai; toh $d\vec{L}=\vec{\tau }dt$ $+y$ ke along point karta hai, isliye tip $+y$ ki taraf (maano east) move karti hai, kabhi north nahi. East ya west kya milega yeh tumhare chosen frame mein right-hand rule se fix hai — figure s01/s03 dekho. "$\Omega =\tau /(I\omega )$ kyunki, zyada torque matlab axis hamesha zyada tezi se move karta hai, isliye mujhe steady reference ke liye slow gyro use karna chahiye." Kahan galti hai? ::: Tumne sirf $\tau$ par dhyaan diya. Steady reference ke liye tum $\Omega$ chhota chahte ho, jo $I\omega$ bada karke (tezi se spin karke) achieve hota hai. Dono variables matter karte hain — ratio padho, sirf numerator nahi. "$\vec{L}=I\vec{\omega }$ us direction mein point karta hai jis taraf wheel ka rim move kar raha hai." Kahan galti hai? ::: $\vec{L}$ right-hand rule se spin axis ke along point karta hai, rim ki velocity ke along nahi. Rim ki velocity tangential hai aur har jagah direction badal deti hai; axis woh ek fixed direction hai. "Wheel torque body ko turn karta hai, aur jab wheel accelerate karna band kar deta hai toh body speed up hoti rehti hai." Kahan galti hai? ::: Body par torque ${\tau }_{\text{body}}=-I_{w}d{\omega }_w/dt$ hai, jahan $I_w$ wheel ka moment of inertia hai aur ${\omega }_w$ uski speed. Jab wheel accelerate karna band kar deta hai ($d{\omega }_w/dt=0$), woh torque gaayab ho jaata hai, isliye body sirf apni current rate maintain karti hai — speed up nahi hoti rehti. "$\vec{\tau }=\vec{\Omega }\times \vec{L}$, isliye main cross product drop kar sakta hoon aur hamesha $\tau =\Omega L$ likh sakta hoon." Kahan galti hai? ::: $\tau =\Omega L$ sirf tab hold karta hai jab $\vec{\Omega }\perp \vec{L}$ ho (usual precession case). Cross product (dekho Cross Product) direction bhi encode karta hai aur $\tau =\Omega L\sin \theta$ ke through shrink karta hai jab vectors perpendicular nahi hote. "Slew rokne ke liye, bas wheel motor ki power kaat do." Kahan galti hai? ::: Torque kaatne se body ki angular acceleration rukti hai, lekin body apni current rate par ghoomti rehti hai. Ise rokne ke liye tumhe wheel ke momentum change ko opposite torque se reverse karna hoga. "Ek reaction wheel aur poora spacecraft dono same taraf spin kar sakte hain agar motor itna powerful ho." Kahan galti hai? ::: Motor strength kitni tezi se badle, kaunsi direction mein nahin. Angular momentum ka conservation signs opposite fix karta hai; koi bhi torque $I_b{\omega }_b=-I_w{\omega }_w$ ko override nahi kar sakta.

$|d\vec{L}|$ ko $L$ se divide karne par precession angle kyun milta hai, length nahi? ::: Ek chote angle se turn hote vector ke liye, arc-length = radius × angle, aur yahan swing ka "radius" $L$ khud hai. Isliye angle = sideways displacement / length = $|d\vec{L}|/L$. $\tau /(I\omega )$ angular rate ki tarah kyun aata hai? ::: Kyunki $\tau dt$ ek length hai jo sideways add hoti hai, $I\omega$ length se divide karne par angle $d\varphi$ milta hai, aur usse $dt$ se divide karne par radians per second milte hain — ek rate. Units confirm karte hain: (N·m)/(kg·m²/s) = 1/s. Precession ke liye sirf $\vec{\tau }=d\vec{L}/dt$ hi assumption kyun kaafi hai? ::: Yeh rotational Newton's second law hai — $\vec{F}=d\vec{p}/dt$ ka direct analogue. Jab torque angular momentum ki rate of change ke barabar ho jaata hai, perpendicular-torque geometry baaki kaam kar leti hai. Ek bade body ko dheere turn karne ke liye chhote reaction wheel ko itni tezi se kyun spin karna padta hai? ::: Conservation deta hai ${\omega }_w=-(I_b/I_w){\omega }_b$. Bada inertia ratio $I_b/I_w$ required wheel speed multiply karta hai, isliye ek light wheel bhaari body ki momentum match karne ke liye tezi se spin karta hai. Ek spacecraft rocket ki taraah kuch expel kiye bina kyun rotate kar sakta hai? ::: Yeh internal angular-momentum trading exploit karta hai (conservation), jaise mid-air mein ek billi khud ko seedha karti hai. Rocket iske bajaaye mass bahar phenkata hai linear momentum gain karne ke liye — ek alag conservation law. $I_b{\omega }_b=-I_w{\omega }_w$ mein sign "opposite directions" ka poora idea kyun carry karta hai? ::: Kyunki total $L$ zero se shuru hota hai aur zero rehna chahiye. Do nonzero terms ke sum ke zero hone ka ek hi tarika hai — opposite signs, jo physically opposite spin senses matlab hai. $\vec{L}$ ke perpendicular torque rotor ki speed kyun nahi badal sakta? ::: Speed $|\vec{L}|=I\omega$ ki length se related hai. Ek perpendicular push vector ki tip ko sideways move karta hai bina use lengthen ya shorten kiye, isliye $\omega$ fixed rehta hai jabki direction badal jaati hai. Gyroscopic rigidity specifically sensing orientation ke liye kyun useful hai? ::: Ek torque-free gyro ka axis space mein ek fixed direction hold karta hai, ek stable reference line deta hai. Spacecraft apni turning is unmoving axis ke against measure karta hai.

Jab spin $\omega \to 0$ ho toh precession ka kya hota hai? ::: $\Omega =\tau /(I\omega )$ blow up karta hai, matlab "gyro" ab resist nahi karta — bina spin ke koi rigidity nahi, aur object torque ke under ek ordinary body ki tarah tip ho jaata hai. Agar applied torque $\tau =0$ ho toh kya hoga? ::: Toh $\Omega =0$ aur $d\vec{L}=0$: na precession, na drift. Axis perfectly steady rehta hai — sensing ke liye use hone wala ideal rigidity limit. Agar torque $\vec{L}$ ke na parallel ho na perpendicular? ::: Ise split karo: parallel part $\omega$ change karta hai (rotor ko speed up/down karta hai), perpendicular part precession cause karta hai. Dono effects ek saath hote hain, har ek apne component se. Reaction wheel kitne time tak correct karta reh sakta hai iski kya limit hai? ::: Wheel ki maximum spin speed — saturation. Jab yeh ceiling hit karti hai toh yeh aur momentum absorb nahi kar sakta, isliye external torques ko pehle thrusters ya magnetorquers se dump karna padta hai. Jab wheel ek constant speed par pahunch jaaye aur accelerate karna band kar de toh body ki motion kya hogi? ::: Reaction torque $-I_{w}d{\omega }_w/dt$ zero ho jaata hai, isliye body jis rate par tha us par coast karta rehta hai — steady rotation, na speed up na stop. Agar do identical counter-rotating wheels load share karein, toh unka net spin angular momentum kya hoga? ::: Zero, kyunki unke equal aur opposite $\vec{L}$ cancel ho jaate hain. Yeh deliberately kiya jaata hai: yeh control ke liye capacity store karta hai jabki baseline system momentum neutral rehta hai. Jab $I\omega \to \infty$ ho (ek ideally stiff gyro), toh precession kaisi dikhti hai? ::: $\Omega \to 0$: axis kisi bhi finite torque ke under essentially immovable ho jaati hai — perfect fixed reference direction, jise ek real sensing gyro approach karta hai.