Exercises — Reaction wheels — momentum management, zero-crossing

Total angular momentum fixed hai. Yeh zero se shuru hua, toh zero hi rehna chahiye: $I_b{\omega }_b+I_w{\Omega }_w=0.$ Agar wheel negative sense mein spin kare (${\Omega }_w<0$), toh ${\omega }_b$ ko opposite sign carry karna hoga, isliye ${\omega }_b>0$ — body positive (right-hand) sense mein $+z$ ke baare mein turn karti hai. Same fixed axis ke baare mein opposite senses: yahi poora mechanism hai. Yeh kaisa dikhta hai: figure s01 dekho — orange wheel-arrow aur teal body-arrow hamesha plum $+z$ axis ke baare mein opposite ways point karte hain, jaise do gears jo cancel karni hi padti hain.

Wheel speed unbounded badhti rehti hai kyunki $H_w(t)=H_w(0)+{\int }_0^t{\tau }_{\text{ext}}d{t}^{\prime }$ ek constant-sign torque ko hamesha integrate karta rehta hai, isliye yeh maximum speed (saturation) hit karta hai aur aur kuch absorb nahi kar sakta. Tumhe ek external actuator (magnetorquer ya thruster) ke saath momentum dumping karni padti hai.

Shared axis ke baare mein conservation, total zero se shuru: ${\omega }_b=-\frac{I_w{\Omega }_w}{I_b}=-\frac{0.04\times 500}{400}=-0.05\text{rad/s}.$ Body $0.05\text{rad/s}$ par negative sense mein rotate karti hai (wheel ke opposite). Sign kyun hai: kisi target ki taraf point karne ke liye, wheel ko uससे door spin karo.

Torque = inertia times angular acceleration. Wheel ka acceleration ${\dot{\Omega }}_w=500/20=25{\text{rad/s}}^2$ hai: ${\tau }_m=I_w{\dot{\Omega }}_w=0.04\times 25=1.0\text{N⋅m}.$ Humne kya kiya aur kyun: $\Delta H=\tau \Delta t$ rearrange karne par constant push ke liye $\tau =I_w\Delta \Omega /\Delta t$ milta hai.

Constant torque ka matlab linear accumulation hai, $H_w(t)={\tau }_{\text{ext}}t$: $t_{\text{sat}}=\frac{H_{\max }}{{\tau }_{\text{ext}}}=\frac{20}{4\times 10^{-5}}=5\times 10^5\text{s}\approx 5.79\text{days}.$

Ek field mein magnetic dipole ka torque $\vec{\tau }=\vec{m}\times \vec{B}$ hai, magnitude $\tau =mB\sin \theta$, $\theta =90^{\circ }$ par maximum: ${\tau }_{\max }=40\times 3\times 10^{-5}=1.2\times 10^{-3}\text{N⋅m}.$ Is torque par momentum remove karne ka time: $t=\frac{H}{{\tau }_{\max }}=\frac{24}{1.2\times 10^{-3}}=2.0\times 10^4\text{s}\approx 5.56\text{h}.$ Kyun matter karta hai: real dumps ghanton lagte hain kyunki $\vec{B}$ weak hai aur sirf perpendicular component act karta hai (dekho Magnetorquers and Earth's Magnetic Field).

Dot product sahi test kyun hai: dot product $\vec{v}\cdot \hat{B}$ measure karta hai ki $\vec{v}$ ka kitna hissa $\hat{B}$ direction mein hai — yeh literally woh length hai jo $\vec{v}$ $\hat{B}$ axis par cast karta hai. Toh agar hum poochhna chahte hain "kya torque mein koi piece $\vec{B}$ ki taraf point karta hai?", toh honest check yeh hai ki use dot product se $\hat{B}$ par project karo. Agar woh projection zero hai, toh us direction mein bilkul koi component nahi hai. Ab calculate karo. Cross product ki definition se, $\vec{m}\times \vec{B}$ $\vec{B}$ ke perpendicular hai: ${\vec{\tau }}_{\text{mag}}\cdot \hat{B}=(\vec{m}\times \vec{B})\cdot \hat{B}=0,$ kyunki cross product apne dono factors ke orthogonal hota hai. $\hat{B}$ par projection vanish ho jaati hai, isliye desired momentum change ka component $\vec{B}$ ke parallel is instant mein touch nahi kiya ja sakta. Practice mein fix: tum $\vec{B}$ ke orbital rotation par rely karte ho — jaise spacecraft move karta hai, $\vec{B}$ alag-alag taraf point karta hai, aur ek orbit ke baad pehle blocked direction accessible ho jaati hai. Figure s02 dekho.

Ek stuck wheel tabhi free hota hai jab commanded torque static friction se zyada ho: kya ${\tau }_{\text{cmd}}>{\tau }_s$ hai? Yahan $5\times 10^{-4}<8\times 10^{-4}$, isliye nahi — wheel stick karta hai. Pointing error tab tak badhta rehta hai jab tak controller command ko $8\times 10^{-4}\text{N⋅m}$ se upar ramp nahi karta; phir wheel snap free ho jaata hai. Jerk "sirf stiction break karne" se bhi bada kyun hota hai: jis moment yeh slip karta hai, friction static ${\tau }_s$ se lower kinetic value ${\tau }_k$ par drop ho jaati hai. Resisting torque ka yeh sudden shortfall woh impulse hai jo spacecraft mein throw hota hai: $\Delta \tau ={\tau }_s-{\tau }_k=8\times 10^{-4}-5\times 10^{-4}=3\times 10^{-4}\text{N⋅m},$ suddenly jitter ke roop mein release hota hai. Kyunki kinetic friction usually static se kam hoti hai, wheel usi instant over-accelerate ho jaata hai jab yeh chhootta hai — Bearing Friction and Stiction ke zero-crossing ke peeche classic stick-slip snap.

(a) Bias momentum $H_{\text{bias}}=I_w{\Omega }_{\text{bias}}=0.05\times 200=10\text{N⋅m⋅s}$. (b) New momentum $H^{\prime }=10-6=4\text{N⋅m⋅s}$, isliye ${\Omega }^{\prime }=\frac{H^{\prime }}{I_w}=\frac{4}{0.05}=80\text{rad/s}.$ (c) $80\text{rad/s}>0$ — wheel zero ke same side par raha (still $+z$ ke baare mein positive sense), koi sign change nahi, isliye friction kabhi reverse nahi hui, koi jitter nahi. Yahi bias run karne ka point hai. Figure s03 dekho.

Abhi pair $9+(-3)=+6\text{N⋅m⋅s}$ hold kar raha hai. Humein $+2\text{N⋅m⋅s}$ tak pahunchna hai, $-4\text{N⋅m⋅s}$ ka change. Ise is tarah route karo ki koi bhi wheel sign na badle. Ek safe split: sara change A se lo.

• $H_A:9\to 5\text{N⋅m⋅s}$, giving ${\Omega }_A=5/0.05=100\text{rad/s}$.
• $H_B$ unchanged $-3\text{N⋅m⋅s}$ par, giving ${\Omega }_B=-3/0.05=-60\text{rad/s}$. Total check karo: $5+(-3)=2\text{N⋅m⋅s}$. ✓ "$-60\text{rad/s}$" ka asal matlab kya hai: minus sirf right-hand-rule direction hai — wheel B $+z$ ke baare mein negative sense mein perfectly happily spin kar raha hai. Negative speed dangerous nahi hai; sirf zero se guzarna (sign reversal) stiction trigger karta hai, kyunki tabhi friction torque khud direction flip karta hai. Wheel B kabhi zero ke paas nahi aata, isliye woh smooth rehta hai chahe uski speed negative ho. Redundant arrays exactly yahi exploit karte hain — Attitude Determination and Control System (ADCS) se array trick.

Number $25\text{N⋅m⋅s}$ bias se upar headroom hai: disturbance $H_{\text{bias}}$ se momentum add karta hai, aur saturation extra $25\text{N⋅m⋅s}$ accumulate hone ke baad reach hoti hai. Humein kabhi bias value ki zaroorat nahi — sirf woh margin. (a) Saturation time: usable margin over disturbance, $t_{\text{sat}}=\frac{25}{3\times 10^{-5}}=8.333\times 10^5\text{s}\approx 9.65\text{days}.$ (b) Dump time: momentum removed over dump torque, $t_{\text{dump}}=\frac{25}{6\times 10^{-4}}=4.167\times 10^4\text{s}\approx 11.6\text{h}.$ (c) Keep-up test dono torques compare karta hai: $\frac{{\tau }_{\text{dump}}}{{\tau }_{\text{ext}}}=\frac{6\times 10^{-4}}{3\times 10^{-5}}=20.$ Dump capability disturbance ka 20× hai, isliye momentum 20 guna faster remove hota hai jitna accumulate hota hai. System comfortably stable hai: har kuch din mein ek dump schedule karo aur woh aadhe din mein clear ho jaata hai.

Momentum destroy nahi hota — yeh closed spacecraft system se bahar Earth mein transfer hota hai, magnetic interaction ke through. Magnetorquer ka dipole Earth ke field ke against push karta hai, aur Newton's third law se equal-and-opposite reaction field ke source — yaani enormous Earth — par act karti hai. Earth ka rotation utterly negligible amount se change hota hai kyunki uska moment of inertia astronomically larger hai. Toh wheel momentum ka "sink" planet khud hai, exactly jaise Conservation of Angular Momentum demand karta hai: total (spacecraft + Earth) unchanged rehta hai; sirf spacecraft ka share drop hota hai. Thruster-based dump instead momentum ko ejected propellant mass ke saath door throw karta hai.

Neeche ka syntax Question ::: Answer hai — ek reveal line: question padho, phir teen colons ke baad wala part uncover karo. Wheels ko deliberately non-zero bias par kyun spin karte hain? ::: Unhe ${\Omega }_w=0$ se door rakhne ke liye, jahan sign reversal bearing stiction ko reverse karta hai aur impulsive jitter inject karta hai — zero-crossing disturbance.