Gyroscope in spacecraft attitude control — preview: Worked examples

1.5.17 · D3 · Physics › Rotational Mechanics › Gyroscope in spacecraft attitude control — preview

Yeh page drill floor hai. Parent preview ne tumhe do master formulas diye the:

$Ω = \frac{τ}{I ω} (precession rate)$ $I_{b} ω_{b} + I_{w} ω_{w} = 0 (reaction-wheel conservation)$

Yahan hum har woh case cover karenge jo yeh formulas produce kar sakte hain — har sign, har zero, har limit, har word-problem flavor. Agar tum neeche ke saare cells solve kar sako, toh exam mein koi cheez surprise nahi kar sakti.

The scenario matrix

Koi bhi number se pehle, chalte hain har tarah ki situation list karte hain jo yeh formulas produce kar sakti hain. Neeche ka har worked example us cell ko fill karta hai jisme woh fit hota hai.

#	Cell (case class)	Kya cheez ise alag banati hai	Example
A	Plain precession	positive torque, fast spin	Ex 1
B	Sign / direction of precession	kis taraf tip swing karta hai? Cross Product use karta hai	Ex 2 (+figure)
C	Degenerate: torque parallel to $L$	no precession — spin speed up ho jaata hai	Ex 3
D	Limit: $ω \to 0$ (spin dies)	$Ω \to \infty$ , gyro topple ho jaata hai	Ex 4
E	Reaction wheel, sign bookkeeping	body ek taraf, wheel doosri taraf	Ex 5 (+figure)
F	Wheel saturation limit	max wheel speed turn ko cap kar deta hai	Ex 6
G	Real-world word problem	telescope ko ek angle se doosre angle par ek time mein slew karna	Ex 7
H	Exam twist: torque + wheel combined	external torque ek wheel ko time ke saath load karta hai	Ex 8

Yahan har numeric answer machine-checked hai verify block mein — lekin Verify line khud bhi zaroor run karo.

A — Plain precession

Worked example Ex 1 · Seedha precession rate

Ek sensing gyro ka moment of inertia $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ hai, $ω = 1500 rad/s$ par spin karta hai. Ek tilting torque $τ = 6 \times 1 0^{- 3} N\cdotm$ spin axis ke perpendicular act karta hai. Precession rate $Ω$ nikalo.

Forecast: Kya $Ω$ , $τ$ se bada hoga ya chota? Padhne se pehle guess karo.

Step 1 — spin angular momentum $L = I ω$ . $L = (3 \times 1 0^{- 4}) (1500) = 0.45 kg\cdotm^{2} / s$ Yeh step kyun? $Ω$ measure karta hai ki $L$ ki direction kitni tez turn hoti hai; hume $L$ ki length chahiye jo turn ho rahi hai, kyunki ek lamba vector same angle swing karne ke liye zyada sideways push maangta hai.

Step 2 — precession rate $Ω = τ / L$ . $Ω = \frac{6 \times 1 0 ^{- 3}}{0.45} = 0.0133 rad/s$ Yeh step kyun? Seedha $Ω = τ / (I ω)$ se. Yeh chota answer dikhata hai ki ek fast rotor kitna stubborn hota hai — reference axis bahut dheere drift karta hai.

Verify: Units: $\frac{N\cdotm}{kg\cdotm ^{2} / s} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = 1/ s = rad/s$ ✓. Aur $Ω ≪ ω$ , jaisa ek fast gyro ke liye expected hai.

B — Axis kis direction mein swing karta hai?

Magnitude toh easy tha. Direction woh jagah hai jahan beginners gir jaate hain — aur yeh pure Cross Product hai.

Figure — Gyroscope in spacecraft attitude control — preview

Worked example Ex 2 · Direction of precession (swing ka sign)

Ek gyro ka spin angular momentum $+ \overset{x}{^}$ ke along point karta hai (axis "East" ki taraf pointing). Gravity $+ \overset{y}{^}$ ("North") ke along torque apply karta hai. $L$ ki tip kis taraf move karti hai, aur kis axis ke around yeh precess karta hai?

Forecast: Kya axis North ki taraf girta hai, ya bilkul alag jagah swing karta hai?

Step 1 — vectors likho. $L = L \overset{x}{^}$ , aur torque $τ = τ \overset{y}{^}$ . Yeh step kyun? Cross Product tabhi kaam karta hai jab dono vectors ki explicit directions ho. Hum unhe axes par rakhte hain taaki cross product mechanical ho jaaye.

Step 2 — $d L = τ d t$ apply karo. $d L = τ d t \overset{y}{^}$ $L$ ki tip (jo $+ \overset{x}{^}$ axis par thi) ko ek chota $+ \overset{y}{^}$ nudge milta hai — woh North ki taraf move karti hai, neeche nahi. Yeh step kyun? Newton ka rotational law kehta hai $L$ mein change $τ$ ke along point karta hai. Tip isliye torque ki direction mein move karti hai — figure mein green arrow dekho.

Step 3 — precession axis $Ω$ nikalo. Hume $Ω$ chahiye jaise ki $τ = Ω \times L$ . $τ = τ \overset{y}{^}$ aur $L = L \overset{x}{^}$ ke saath, $Ω = Ω (- \overset{z}{^})$ try karo: $Ω \times L = Ω (- \overset{z}{^}) \times L \overset{x}{^} = - Ω L (\overset{z}{^} \times \overset{x}{^}) = - Ω L \overset{y}{^}$ Yeh $- \overset{y}{^}$ deta hai, sign galat hai. Toh $Ω = Ω (+ \overset{z}{^})$ lo: $Ω \overset{z}{^} \times L \overset{x}{^} = Ω L \overset{y}{^} = τ$ ✓. Yeh step kyun? Hum axis ko reverse-engineer karte hain required $τ$ ke against candidates test karke; Cross Product magnitude aur sign dono uniquely fix karta hai.

Result: Axis $+ \overset{z}{^}$ (vertical, upar se dekha) ke around counter-clockwise precess karta hai, aur instantaneously tip North ki taraf drift karti hai — push se perpendicular. "Jis taraf push karo usi taraf girta hai" wala intuition exactly 90° se galat hai.

Verify: $\overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ (standard right-handed triad) ✓, isliye $Ω \overset{z}{^} \times L \overset{x}{^}$ wakai $τ \overset{y}{^}$ ke equal hota hai jab $Ω L = τ$ .

C — Degenerate case: torque spin ke along

Worked example Ex 3 ·

$L$ ke parallel torque — bilkul bhi precession nahi Ex 1 wala hi rotor ( $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ , $ω = 1500 rad/s$ ), lekin ab $6 \times 1 0^{- 3} N\cdotm$ torque spin axis ke along apply hota hai ( $L$ ke parallel). Kya hota hai?

Forecast: Kya yeh Ex 1 ki tarah $0.0133 rad/s$ par precess karta hai, ya kuch bilkul alag karta hai?

Step 1 — decompose karo. $L$ ke parallel torque ka zero perpendicular component hota hai, isliye yeh $L$ ki direction rotate nahi kar sakta. Yeh step kyun? Precession sirf $τ$ ke perpendicular part se aati hai (jo tip ko sideways push karta hai). Yahan woh part zero hai, isliye $Ω = 0$ .

Step 2 — yeh length badalta hai instead. $τ = I \frac{d ω}{d t}$ , isliye spin speed up hoti hai: $\frac{d ω}{d t} = \frac{τ}{I} = \frac{6 \times 1 0 ^{- 3}}{3 \times 1 0 ^{- 4}} = 20 rad/s^{2}$ Yeh step kyun? Ek parallel torque same axis ke around sirf angular acceleration hai — pure Torque and Newton's Second Law for Rotation, koi gyroscopic effect nahi.

Result: $Ω = 0$ ; rotor simply $20 rad/s^{2}$ par tez spin karta hai. Yeh woh degenerate cell hai jise master formula $Ω = τ / (I ω)$ silently assume away karta hai (yeh maanta hai ki $τ ⊥ L$ ).

Verify: $d ω / d t$ ke units: $\frac{N\cdotm}{kg\cdotm ^{2}} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2}} = s^{- 2} = rad/s^{2}$ ✓.

D — Limiting case: spin khatam ho jaata hai

Worked example Ex 4 · Jab

$ω \to 0$ tab kya hota hai? Ex 1 gyro lo aur uski spin decay karne do: $Ω$ calculate karo $ω = 1500$ , phir $150$ , phir $15 rad/s$ par, same $τ = 6 \times 1 0^{- 3} N\cdotm$ aur $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ ke saath. Trend kya hai?

Forecast: Jaise wheel slow hota hai, precession gentle hoti hai ya wild?

Step 1 — $Ω = τ / (I ω)$ har spin par evaluate karo. $Ω_{1500} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 1500 )} = 0.0133 rad/s$ $Ω_{150} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 150 )} = 0.133 rad/s$ $Ω_{15} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 15 )} = 1.33 rad/s$ Yeh step kyun? $ω$ denominator mein hai, isliye spin half/tenth karne par $Ω$ multiply hoti hai. Precession accelerate hoti hai jaise wheel slow hota hai.

Step 2 — limit lo. $lim_{ω \to 0^{+}} Ω = lim_{ω \to 0^{+}} \frac{τ}{I ω} = + \infty$ Yeh step kyun? Yeh ek real observation ka mathematical face hai: ek spinning top khud ko tab tak hold karta hai jab tak fast ho, lekin jaise friction uski spin bleed karta hai woh pehle se tez wobble karta hai, phir topple ho jaata hai. Formally $Ω$ blow up karta hai — physically "fast gyro" approximation break ho jaati hai aur woh simply gir jaata hai.

Result: Slow gyro $\Rightarrow$ huge, unstable precession. Yahi reason hai ki sensing gyros hard spin kiye jaate hain.

Verify: Teeno answers $1 : 10 : 100$ ke ratio mein hain, $ω$ values ke $1 : 1/10 : 1/100$ ratio se match karta hai (inverse proportion) ✓.

E — Reaction wheel with full sign bookkeeping

Worked example Ex 5 · Body ek taraf, wheel doosri taraf

Ek satellite body ( $I_{b} = 800 kg\cdotm^{2}$ ) rest par hai. Hum iska reaction wheel ( $I_{w} = 0.4 kg\cdotm^{2}$ ) spin up karte hain $ω_{w} = + 60 rad/s$ tak (counter-clockwise, upar se dekha). Body ki resulting angular velocity kya hai, aur woh kis taraf turn karti hai?

Forecast: Kya body wheel ki same direction mein turn karegi ya opposite direction mein — aur kaafi slow ya kaafi fast?

Step 1 — total $L$ ka conservation. Koi bhi external cheez craft ko touch nahi karti, isliye: $I_{b} ω_{b} + I_{w} ω_{w} = 0$ Yeh step kyun? Conservation of Angular Momentum: isolated system rest par start hua tha ( $L = 0$ ) aur hamesha $L = 0$ par rehna chahiye.

Step 2 — $ω_{b}$ solve karo. $ω_{b} = - \frac{I _{w}}{I _{b}} ω_{w} = - \frac{0.4}{800} (60) = - 0.03 rad/s$ Yeh step kyun? Minus sign hi poori baat hai: total zero rakhne ke liye body ko opposite angular momentum carry karni chahiye. Wheel counter-clockwise ⇒ body clockwise.

Result: Body clockwise $0.03 rad/s$ par turn karti hai — inertia ratio $I_{b} / I_{w} = 2000$ ka matlab hai ki heavy body creep karti hai jabki light wheel race karta hai. Figure dekho: opposite curved arrows, wildly different lengths.

Verify: Total: $I_{b} ω_{b} + I_{w} ω_{w} = 800 (- 0.03) + 0.4 (60) = - 24 + 24 = 0$ ✓.

F — The saturation limit

Worked example Ex 6 · Wheel max out hone se pehle maximum turn

Ex 5 wala wheel ( $I_{w} = 0.4 kg\cdotm^{2}$ , body $I_{b} = 800 kg\cdotm^{2}$ ) maximum $ω_{w, m a x} = 6000 rad/s$ tak spin kar sakta hai saturate hone se pehle. Rest se start karke, ek single push mein yeh body rotation rate kitni fast produce kar sakta hai?

Forecast: Kya top body rate ek bada number hai ya frustratingly chota?

Step 1 — saturation point par conservation apply karo. $∣ ω_{b, m a x} ∣ = \frac{I _{w}}{I _{b}} ω_{w, m a x} = \frac{0.4}{800} (6000) = 3 rad/s$ Yeh step kyun? Wheel $ω_{w, m a x}$ ke baad zyada angular momentum store nahi kar sakta; woh momentum ceiling cap karta hai ki body kitna carry kar sakti hai.

Step 2 — physical limit padho. Iske aage, wheel saturated hai: aage push karne se pehle tum thrusters ya magnetorquers fire karke uska momentum bleed karo (desaturation). Yeh step kyun? Yeh Conservation of Angular Momentum budget mein chupi real engineering constraint hai — reaction wheels momentum store karte hain, endlessly create nahi karte.

Result: Ek single spin-up se max sustainable body rate $3 rad/s$ hai; aur zyada slew karne ke liye pehle desaturate karo.

Verify: $\frac{0.4}{800} \times 6000 = 0.0005 \times 6000 = 3$ ✓.

G — Real-world word problem

Worked example Ex 7 · Space telescope ko 90° mein 5 minutes mein slew karo

Ek telescope body ka $I_{b} = 1200 kg\cdotm^{2}$ hai. Repoint karne ke liye, use $θ = 9 0^{\circ} = \frac{π}{2} rad$ rotate karna hai aur use rest par start aur stop karna hai (isliye yeh aadhe time accelerate karta hai, aadhe time decelerate). Total time $T = 300 s$ . Iska wheel $I_{w} = 0.5 kg\cdotm^{2}$ ka hai. Nikalo (a) peak body rate, (b) us moment par wheel speed.

Forecast: "Spin up then spin down" profile ke saath, kya peak body rate simple average $θ / T$ se badi hai ya choti?

Step 1 — symmetric slew ki geometry. Angle = angular-velocity vs time graph ke neeche ka area (triangular). $T$ base aur $ω_{b, pk}$ peak height wale triangle ka area $\frac{1}{2} T ω_{b, pk}$ hai. $θ$ ke barabar set karo: $θ = \frac{1}{2} T ω_{b, pk} \Rightarrow ω_{b, pk} = \frac{2 θ}{T} = \frac{2 ( π /2 )}{300} = \frac{π}{300} \approx 0.01047 rad/s$ Yeh step kyun? Accelerate-then-decelerate ek triangular speed profile hai; peak average se double hoti hai kyunki aadha time usse neeche spend hota hai.

Step 2 — peak body rate par wheel speed. Conservation (rest se): $ω_{w, pk} = - \frac{I _{b}}{I _{w}} ω_{b, pk} = - \frac{1200}{0.5} (0.01047) = - 25.13 rad/s$ Yeh step kyun? Har instant par Conservation of Angular Momentum hold karta hai; wheel ko body ke abhi jo momentum hai uska opposite carry karna chahiye.

Result: Peak body rate $\approx 0.0105 rad/s$ ; wheel ko mid-slew par $\approx - 25.1 rad/s$ (opposite sense) tak pahunchna chahiye, phir body ko brake karne ke liye reverse karna chahiye.

Verify: Average rate $θ / T = (π /2) /300 \approx 0.00524 rad/s$ ; peak $= 2 \times$ woh $\approx 0.01047$ ✓. Aur $\frac{1200}{0.5} = 2400$ , times $0.01047 \approx 25.13$ ✓.

H — Exam twist: external torque wheel ko load karta hai

Worked example Ex 8 · Combined — ek disturbance torque wheel ko time ke saath saturate karta hai

Ek satellite body orientation mein fixed hai (ek star ko point kar rahi hai). Ek steady disturbance torque $τ_{d} = 2 \times 1 0^{- 4} N\cdotm$ (e.g. solar-radiation pressure) body par act karta hai. Reaction wheel ( $I_{w} = 0.5 kg\cdotm^{2}$ , $6000 rad/s$ par saturate hota hai) ko sab absorb karna chahiye. Stationary wheel se start karke, wheel saturate hone mein kitna time lagta hai?

Forecast: Ek tiny torque — kya saturation minutes, hours, ya days mein hogi?

Step 1 — wheel disturbance ka angular momentum absorb karta hai. Body ko still rakhne ke liye, wheel ki angular momentum exactly utni rate se change honi chahiye jis rate par disturbance otherwise body ki change karta: $τ_{d} = I_{w} \frac{d ω _{w}}{d t} \Rightarrow \frac{d ω _{w}}{d t} = \frac{τ _{d}}{I _{w}} = \frac{2 \times 1 0 ^{- 4}}{0.5} = 4 \times 1 0^{- 4} rad/s^{2}$ Yeh step kyun? Torque and Newton's Second Law for Rotation se wheel ki spin steadily rise karti hai incoming momentum soak up karne ke liye — yeh wheel ka "winding up" hai.

Step 2 — saturation tak time. $t_{sat} = \frac{ω _{w, m a x}}{d ω _{w} / d t} = \frac{6000}{4 \times 1 0 ^{- 4}} = 1.5 \times 1 0^{7} s$ Yeh step kyun? Constant angular acceleration ⇒ time = final speed ÷ acceleration.

Step 3 — human units mein rakho. $1.5 \times 1 0^{7} s \times \frac{1 day}{86400 s} \approx 173.6 days$ Yeh step kyun? Ek exam "so what" chahta hai — answer ek real operations schedule hai: is disturbance ke liye roughly har ~6 months mein desaturate karo.

Result: ~ $1.5 \times 1 0^{7} s \approx 174$ days tak wheel saturate hota hai; thruster/magnetorquer desaturation usse pehle honi chahiye.

Verify: $\frac{6000}{4 \times 1 0 ^{- 4}} = 6000 \times 2500 = 1.5 \times 1 0^{7}$ s; $\frac{1.5 \times 1 0 ^{7}}{86400} \approx 173.6$ days ✓.

Recall — kya matrix yaad raha?

Recall Har answer predict karo, phir reveal karo

Spin axis ke parallel torque kya cause karta hai? ::: Koi precession nahi; spin sirf speed/slow hoti hai ( $τ = I d ω / d t$ ). Jab $ω \to 0$ , $Ω = τ / (I ω)$ ka kya hota hai? ::: Yeh $+ \infty$ tak blow up karta hai — gyro tez wobble karta hai aur topple ho jaata hai. Wheel counter-clockwise spun, body kis taraf turn karti hai? ::: Clockwise (opposite), conservation se. Peak slew rate average se double kyun hoti hai? ::: Triangular accelerate-then-decelerate profile $2 θ / T$ par peak karti hai. Reaction wheel ek craft ko kitna slew kar sakta hai, yeh kya limit karta hai? ::: Saturation — wheel ki max speed stored momentum cap karti hai; phir desaturate karo.