Exercises — Gyroscope in spacecraft attitude control — preview

What we use: the definition $\vec{L}=I\vec{\omega }$. This is a definition, not a derivation — angular momentum of a symmetric rotor is just "spin-inertia times spin-rate". We want the magnitude $L$, so we drop the arrows: $L=I\omega =(0.8)(300)=240{\text{kg⋅m}}^2/\text{s}.$ Direction: the vector $\vec{L}$ lies along the spin axis, sense given by the right-hand rule — curl the right-hand fingers the way the wheel turns, the thumb points along $\vec{L}$.

What we use: the magnitude form $\Omega =\frac{\tau }{L}$ — derived in the intuition box above and boxed in the parent note. Both quantities here are magnitudes, so no arrows are needed. $\Omega =\frac{0.002}{0.5}=0.004\text{rad/s}.$ The axis drifts around a cone at $0.004$ radians per second — very slow, which is exactly what a stable reference needs.

What we do: start from the magnitude form $\Omega =\frac{\tau }{I\omega }$ and solve for $\omega$. Why: the unknown is trapped in the denominator, so we rearrange to free it. $\omega =\frac{\tau }{I\Omega }=\frac{1\times 10^{-3}}{(1\times 10^{-4})(5\times 10^{-4})}=\frac{1\times 10^{-3}}{5\times 10^{-8}}=2\times 10^4\text{rad/s}.$ So $\omega =20,000\text{rad/s}$ (about $191,000$ rpm — real precision gyros do spin this fast). A slower rotor would drift too quickly to trust.

What we use: conservation of angular momentum (Conservation of Angular Momentum). Here we keep signs because direction is the whole question — a positive $\omega$ means one rotational sense, a negative $\omega$ the opposite sense (this is the 1-D signed shadow of the full vector statement $\vec{L}=0$). The craft is isolated in vacuum and started at rest, so total angular momentum is $0$ forever: $I_b{\omega }_b+I_w{\omega }_w=0\Rightarrow {\omega }_w=-\frac{I_b{\omega }_b}{I_w}.$ ${\omega }_w=-\frac{(800)(0.015)}{0.4}=-\frac{12}{0.4}=-30\text{rad/s}.$ The minus sign means opposite sense: to turn the body one way, spin the wheel the other way. The magnitude $30\text{rad/s}$ is large because the wheel is $2000\times$ smaller in inertia.

What we do: precession rate $\Omega =\tau /(I\omega )$, so with the same $\tau$, whoever has the larger spin magnitude $L=I\omega$ precesses less. Compute each $L$: $L_A=(2\times 10^{-4})(5000)=1.0{\text{kg⋅m}}^2/\text{s},$ $L_B=(5\times 10^{-4})(3000)=1.5{\text{kg⋅m}}^2/\text{s}.$ B has the larger $L$, so B precesses less. The 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 drifts 1.5 times faster than B, even though A spins faster — because B's larger inertia more than compensates. Read the product, not one variable.

(a) Wheel torque. Torque is the rate of change of angular momentum (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. By Newton's third law for rotation, the reaction on the body is equal and opposite: ${\tau }_{\text{body}}=-{\tau }_{\text{wheel}}=-2.0\text{N⋅m}.$ (c) Body angular acceleration. Using ${\tau }_{\text{body}}=I_b{\alpha }_b$: ${\alpha }_b=\frac{{\tau }_{\text{body}}}{I_b}=\frac{-2.0}{800}=-2.5\times 10^{-3}{\text{rad/s}}^2.$ The body accelerates the opposite way — tiny, because it is so massive. After $6\text{s}$ its speed is $|{\alpha }_b|t=2.5\times 10^{-3}\times 6=0.015\text{rad/s}$, matching the conservation answer from L2·Q2. Two roads, same destination — a good consistency check.

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

(b) Direction — read it off the figure. The figure below is a top-down map of the horizontal plane: East is the $+\hat{x}$ axis, North is $+\hat{y}$. The teal arrow is $\vec{L}$ (spin axis, pointing East); the orange arrow is the applied torque $\vec{\tau }$ (pointing North); the plum dashed arc shows the resulting sweep of the axis; and the small circled dot marks $\vec{\Omega }$ pointing straight out of the page (upward). Use it to see that the axis does not tip over toward the push — it swings sideways in the horizontal plane.

Now confirm with the Cross Product relation $\vec{\tau }=\vec{\Omega }\times \vec{L}$. In a right-handed frame $\hat{x}\times \hat{y}=\hat{z}$, $\hat{y}\times \hat{z}=\hat{x}$, $\hat{z}\times \hat{x}=\hat{y}$. We guess $\vec{\Omega }=\Omega \hat{z}$ (pointing up, out of the page) and test: $\vec{\Omega }\times \vec{L}=(\Omega \hat{z})\times (L\hat{x})=\Omega L(\hat{z}\times \hat{x})=\Omega L\hat{y}.$ That points along $+\hat{y}$ — exactly the direction of $\vec{\tau }$. ✓ So $\vec{\Omega }=+\Omega \hat{z}$: the axis sweeps horizontally, turning from East toward North, tracing a cone about the vertical — precisely the plum arc in the figure.

(a) Phase 1 body velocity. Total angular momentum is $0$ throughout (started at rest, isolated): ${\omega }_b=-\frac{I_w{\omega }_w}{I_b}=-\frac{(0.5)(+40)}{600}=-\frac{20}{600}=-0.0333\text{rad/s}.$ The minus sign means: since the wheel spins counter-clockwise ($+$), the body turns clockwise ($-$). (b) After Phase 2. When the wheel returns to ${\omega }_w=0$, conservation forces $I_b{\omega }_b=0$, so ${\omega }_b=0$. The body stops rotating. (Spinning the wheel back undoes the body's spin — this is how a craft stops slewing.) (c) Net angle and direction. The body reorients during Phase 1 (it rotates at $-0.0333\text{rad/s}$), then holds still when the wheel is braked. The net turn is the Phase-1 rotation. With constant wheel acceleration the body's speed ramps linearly from $0$ to $-0.0333\text{rad/s}$ over $8\text{s}$, so the average signed velocity is half the peak: ${\theta }_b=\frac{1}{2}(-0.0333)(8)=-0.1333\text{rad}\approx -7.6^{\circ }.$ The negative sign is the physics: the craft has turned $\approx 7.6^{\circ }$ clockwise (the sense opposite the wheel), and now sits still, pointed at a new target. Spin wheel to turn, brake wheel to hold.

(a) Time to saturate. The wheel must absorb momentum at rate ${\tau }_d$, and its capacity is $L_{\max }=I_w{\omega }_{\max }$: $L_{\max }=(0.6)(600)=360{\text{kg⋅m}}^2/\text{s}.$ Time = capacity ÷ fill-rate: $t_{\text{sat}}=\frac{L_{\max }}{{\tau }_d}=\frac{360}{3\times 10^{-4}}=1.2\times 10^6\text{s}\approx 13.9\text{days}.$ Why: the tiny disturbance drips angular momentum in; the wheel's finite capacity fills over about two weeks — then it can't help any more. (b) Desaturation time. The thruster must remove all $360{\text{kg⋅m}}^2/\text{s}$ at rate $0.05\text{N⋅m}$: $t_{\text{desat}}=\frac{L_{\max }}{{\tau }_{\text{thruster}}}=\frac{360}{0.05}=7200\text{s}=2\text{hours}.$ The lesson: the wheel handled steady disturbances silently for two weeks; only occasionally must a thruster dump the accumulated momentum externally. That's exactly the "desaturation" the parent note warns about.

Step 1 — link drift angle to precession rate. During the observation the axis precesses at a steady rate $\Omega$, so the accumulated angle is $\theta =\Omega t$. Why: angle = rate × time when the rate is constant. Step 2 — express $\Omega$ from the precession formula. From the intuition box, $\Omega =\frac{\tau }{I\omega }$. Substitute: $\theta =\Omega t=\frac{\tau t}{I\omega }.$ Step 3 — impose the budget $\theta \le {\theta }_{\max }$ and solve for $\omega$. Since $\omega$ sits in the denominator, a larger $\omega$ makes $\theta$ smaller, so we need $\omega$ at least the value that makes $\theta$ exactly ${\theta }_{\max }$: $\omega \ge \frac{\tau t}{I{\theta }_{\max }}=\frac{(2\times 10^{-6})(100)}{(3\times 10^{-5})(1\times 10^{-4})}=\frac{2\times 10^{-4}}{3\times 10^{-9}}\approx 6.67\times 10^4\text{rad/s}.$ Answer: the rotor must spin at at least $\approx 6.67\times 10^4\text{rad/s}$ (about $66,700\text{rad/s}$) to keep the star image steady enough for the whole $100\text{s}$ exposure. Every earlier idea appears at once: the precession rate $\Omega =\tau /(I\omega )$, drift-as-rate×time $\theta =\Omega t$, and design-by-inequality — this is exactly what an attitude-control engineer computes when sizing a gyro.

Spin momentum $\vec{L}=I\vec{\omega }$, magnitude $L=I\omega$ ::: along the axis, right-hand rule. Precession rate $\Omega =\tau /(I\omega )$ ::: bigger spin → slower drift. Vector precession $\vec{\tau }=\vec{\Omega }\times \vec{L}$ ::: axis sweeps perpendicular to the push. Reaction wheel $I_b{\omega }_b=-I_w{\omega }_w$ ::: spin wheel one way, body turns the other. Torque $\tau =I\Delta \omega /\Delta t$ ::: rate of change of angular momentum. Drift angle $\theta =\Omega t$ ::: steady-rate precession over an interval. Saturation $L_{\max }=I_w{\omega }_{\max }$ ::: finite momentum store.