Gyroscope in spacecraft attitude control — preview: Worked examples

This page is the drill floor. The parent preview gave you two master formulas:

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

Here we hit every case those formulas can produce — every sign, every zero, every limit, every word-problem flavour. If you can work all cells below, nothing on an exam can surprise you.

The scenario matrix

Before any numbers, let us list every kind of situation these formulas can produce. Each worked example below is tagged with the cell it fills.

#	Cell (case class)	What makes it distinct	Example
A	Plain precession	positive torque, fast spin	Ex 1
B	Sign / direction of precession	which way does the tip swing? uses Cross Product	Ex 2 (+figure)
C	Degenerate: torque parallel to $L$	no precession — spin speeds up instead	Ex 3
D	Limit: $ω \to 0$ (spin dies)	$Ω \to \infty$ , gyro topples	Ex 4
E	Reaction wheel, sign bookkeeping	body one way, wheel the other	Ex 5 (+figure)
F	Wheel saturation limit	max wheel speed caps the turn	Ex 6
G	Real-world word problem	telescope slew through an angle in a time	Ex 7
H	Exam twist: torque + wheel combined	external torque loads a wheel over time	Ex 8

Every numeric answer here is machine-checked in the verify block — but always run the Verify line yourself first.

A — Plain precession

Worked example Ex 1 · Straightforward precession rate

A sensing gyro has moment of inertia $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ , spins at $ω = 1500 rad/s$ . A tilting torque $τ = 6 \times 1 0^{- 3} N\cdotm$ acts perpendicular to the spin axis. Find the precession rate $Ω$ .

Forecast: Will $Ω$ be larger or smaller than $τ$ itself? Guess before reading.

Step 1 — spin angular momentum $L = I ω$ . $L = (3 \times 1 0^{- 4}) (1500) = 0.45 kg\cdotm^{2} / s$ Why this step? $Ω$ measures how fast $L$ 's direction turns; we need the length of $L$ that is being turned, because a longer vector needs a bigger sideways push to swing the same angle.

Step 2 — precession rate $Ω = τ / L$ . $Ω = \frac{6 \times 1 0 ^{- 3}}{0.45} = 0.0133 rad/s$ Why this step? Directly from $Ω = τ / (I ω)$ . The tiny answer shows how stubborn a fast rotor is — the reference axis drifts only slowly.

Verify: Units: $\frac{N\cdotm}{kg\cdotm ^{2} / s} = \frac{kg\cdotm ^{2} / s ^{2}}{kg\cdotm ^{2} / s} = 1/ s = rad/s$ ✓. And $Ω ≪ ω$ , as expected for a fast gyro.

B — Which direction does the axis swing?

The magnitude was easy. The direction is where beginners fall — and it is pure Cross Product.

Figure — Gyroscope in spacecraft attitude control — preview

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

A gyro's spin angular momentum points along $+ \overset{x}{^}$ (axis pointing "East"). Gravity applies a torque along $+ \overset{y}{^}$ ("North"). Which way does the tip of $L$ move, and around which axis does it precess?

Forecast: Does the axis fall toward North, or swing somewhere else entirely?

Step 1 — write the vectors. $L = L \overset{x}{^}$ , and the torque $τ = τ \overset{y}{^}$ . Why this step? The Cross Product only makes sense once the two vectors have explicit directions. We put them on axes so the cross product is mechanical.

Step 2 — apply $d L = τ d t$ . $d L = τ d t \overset{y}{^}$ The tip of $L$ (which was on the $+ \overset{x}{^}$ axis) gains a small $+ \overset{y}{^}$ nudge — it moves North, not down. Why this step? Newton's rotational law says the change in $L$ points along $τ$ . The tip therefore moves in the torque's direction — look at the green arrow in the figure.

Step 3 — find the precession axis $Ω$ . We need $Ω$ such that $τ = Ω \times L$ . With $τ = τ \overset{y}{^}$ and $L = L \overset{x}{^}$ , try $Ω = Ω (- \overset{z}{^})$ : $Ω \times L = Ω (- \overset{z}{^}) \times L \overset{x}{^} = - Ω L (\overset{z}{^} \times \overset{x}{^}) = - Ω L \overset{y}{^}$ That gives $- \overset{y}{^}$ , wrong sign. So take $Ω = Ω (+ \overset{z}{^})$ : $Ω \overset{z}{^} \times L \overset{x}{^} = Ω L \overset{y}{^} = τ$ ✓. Why this step? We reverse-engineer the axis by testing candidates against the required $τ$ ; the Cross Product fixes both magnitude and sign uniquely.

Result: The axis precesses counter-clockwise about $+ \overset{z}{^}$ (vertical, viewed from above), and instantaneously the tip drifts North — perpendicular to the push. The "falls the way you push" intuition is wrong by exactly 90°.

Verify: $\overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ (standard right-handed triad) ✓, so $Ω \overset{z}{^} \times L \overset{x}{^}$ does equal $τ \overset{y}{^}$ when $Ω L = τ$ .

C — Degenerate case: torque along the spin

Worked example Ex 3 · Torque parallel to

$L$ — no precession at all Same rotor as Ex 1 ( $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ , $ω = 1500 rad/s$ ), but now the $6 \times 1 0^{- 3} N\cdotm$ torque is applied along the spin axis (parallel to $L$ ). What happens?

Forecast: Does it precess at $0.0133 rad/s$ like Ex 1, or do something completely different?

Step 1 — decompose. A torque parallel to $L$ has zero perpendicular component, so it cannot rotate $L$ 's direction. Why this step? Precession comes only from the perpendicular part of $τ$ (the part that pushes the tip sideways). Here that part is zero, so $Ω = 0$ .

Step 2 — it changes the length instead. $τ = I \frac{d ω}{d t}$ , so the spin speeds up: $\frac{d ω}{d t} = \frac{τ}{I} = \frac{6 \times 1 0 ^{- 3}}{3 \times 1 0 ^{- 4}} = 20 rad/s^{2}$ Why this step? A parallel torque is just angular acceleration about the same axis — pure Torque and Newton's Second Law for Rotation, no gyroscopic effect.

Result: $Ω = 0$ ; the rotor simply spins faster at $20 rad/s^{2}$ . This is the degenerate cell the master formula $Ω = τ / (I ω)$ silently assumes away (it presumes $τ ⊥ L$ ).

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

D — Limiting case: the spin dies

Worked example Ex 4 · What happens as

$ω \to 0$ ? Take the Ex 1 gyro and let its spin decay: compute $Ω$ at $ω = 1500$ , then $150$ , then $15 rad/s$ , with the same $τ = 6 \times 1 0^{- 3} N\cdotm$ and $I = 3 \times 1 0^{- 4} kg\cdotm^{2}$ . What is the trend?

Forecast: As the wheel slows, does the precession get gentler or wilder?

Step 1 — evaluate $Ω = τ / (I ω)$ at each spin. $Ω_{1500} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 1500 )} = 0.0133 rad/s$ $Ω_{150} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 150 )} = 0.133 rad/s$ $Ω_{15} = \frac{6 \times 1 0 ^{- 3}}{( 3 \times 1 0 ^{- 4} ) ( 15 )} = 1.33 rad/s$ Why this step? $ω$ sits in the denominator, so halving/tenthing the spin multiplies $Ω$ . The precession accelerates as the wheel slows.

Step 2 — take the limit. $lim_{ω \to 0^{+}} Ω = lim_{ω \to 0^{+}} \frac{τ}{I ω} = + \infty$ Why this step? This is the mathematical face of a real observation: a spinning top holds itself up while fast, but as friction bleeds its spin it wobbles faster and faster, then topples. Formally $Ω$ blows up — physically the "fast gyro" approximation breaks and it just falls.

Result: Slow gyro $\Rightarrow$ huge, unstable precession. This is exactly why sensing gyros are kept spinning hard.

Verify: The three answers are in ratio $1 : 10 : 100$ , matching the $1 : 1/10 : 1/100$ ratio of the $ω$ values (inverse proportion) ✓.

E — Reaction wheel with full sign bookkeeping

Worked example Ex 5 · Body one way, wheel the other

A satellite body ( $I_{b} = 800 kg\cdotm^{2}$ ) is at rest. We spin its reaction wheel ( $I_{w} = 0.4 kg\cdotm^{2}$ ) up to $ω_{w} = + 60 rad/s$ (counter-clockwise, seen from above). What is the body's resulting angular velocity, and which way does it turn?

Forecast: Will the body turn the same way as the wheel or the opposite way — and much slower or much faster?

Step 1 — conservation of total $L$ . Nothing external touches the craft, so: $I_{b} ω_{b} + I_{w} ω_{w} = 0$ Why this step? Conservation of Angular Momentum: the isolated system started at rest ( $L = 0$ ) and must stay at $L = 0$ forever.

Step 2 — solve for $ω_{b}$ . $ω_{b} = - \frac{I _{w}}{I _{b}} ω_{w} = - \frac{0.4}{800} (60) = - 0.03 rad/s$ Why this step? The minus sign is the whole point: to keep the total zero, the body must carry the opposite angular momentum. Wheel counter-clockwise ⇒ body clockwise.

Result: Body turns clockwise at $0.03 rad/s$ — the inertia ratio $I_{b} / I_{w} = 2000$ means the heavy body creeps while the light wheel races. See the figure: 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 · Maximum turn before the wheel maxes out

The Ex 5 wheel ( $I_{w} = 0.4 kg\cdotm^{2}$ , body $I_{b} = 800 kg\cdotm^{2}$ ) can spin at most $ω_{w, m a x} = 6000 rad/s$ before it saturates. Starting from rest, what is the fastest body rotation rate it can produce in a single push?

Forecast: Is the top body rate a big number or a frustratingly small one?

Step 1 — apply conservation at the saturation point. $∣ ω_{b, m a x} ∣ = \frac{I _{w}}{I _{b}} ω_{w, m a x} = \frac{0.4}{800} (6000) = 3 rad/s$ Why this step? The wheel cannot store more angular momentum past $ω_{w, m a x}$ ; that momentum ceiling caps how much the body can carry.

Step 2 — read the physical limit. Beyond this, the wheel is saturated: you must fire thrusters or magnetorquers to bleed off its momentum (desaturation) before you can push further. Why this step? This is the real engineering constraint hidden in the Conservation of Angular Momentum budget — reaction wheels store momentum, they don't create it endlessly.

Result: Max sustainable body rate from one spin-up is $3 rad/s$ ; to slew more, desaturate first.

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

G — Real-world word problem

Worked example Ex 7 · Slew a space telescope through 90° in 5 minutes

A telescope body has $I_{b} = 1200 kg\cdotm^{2}$ . To repoint, it must rotate through $θ = 9 0^{\circ} = \frac{π}{2} rad$ and it must start and stop at rest (so it accelerates for half the time, decelerates for the other half). Total time $T = 300 s$ . Its wheel has $I_{w} = 0.5 kg\cdotm^{2}$ . Find (a) the peak body rate, (b) the wheel speed at that moment.

Forecast: With a "spin up then spin down" profile, is the peak body rate bigger or smaller than the simple average $θ / T$ ?

Step 1 — geometry of a symmetric slew. Angle = area under the (triangular) angular-velocity vs time graph. A triangle of base $T$ and peak height $ω_{b, pk}$ has area $\frac{1}{2} T ω_{b, pk}$ . Set equal to $θ$ : $θ = \frac{1}{2} T ω_{b, pk} \Rightarrow ω_{b, pk} = \frac{2 θ}{T} = \frac{2 ( π /2 )}{300} = \frac{π}{300} \approx 0.01047 rad/s$ Why this step? Accelerate-then-decelerate is a triangular speed profile; the peak is twice the average because half the time is spent below it.

Step 2 — wheel speed at peak body rate. Conservation (from rest): $ω_{w, pk} = - \frac{I _{b}}{I _{w}} ω_{b, pk} = - \frac{1200}{0.5} (0.01047) = - 25.13 rad/s$ Why this step? At every instant Conservation of Angular Momentum holds; the wheel must carry the opposite momentum to whatever the body has right now.

Result: Peak body rate $\approx 0.0105 rad/s$ ; wheel must reach $\approx - 25.1 rad/s$ (opposite sense) at mid-slew, then reverse to brake the body.

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

H — Exam twist: external torque loads the wheel

Worked example Ex 8 · Combined — a disturbance torque saturates the wheel over time

A satellite body is held fixed in orientation (pointing at a star). A steady disturbance torque $τ_{d} = 2 \times 1 0^{- 4} N\cdotm$ (e.g. solar-radiation pressure) acts on the body. The reaction wheel ( $I_{w} = 0.5 kg\cdotm^{2}$ , saturates at $6000 rad/s$ ) must absorb all of it. Starting from a stationary wheel, how long until the wheel saturates?

Forecast: A tiny torque — will saturation take minutes, hours, or days?

Step 1 — the wheel absorbs the disturbance's angular momentum. To keep the body still, the wheel's angular momentum must change at exactly the rate the disturbance would otherwise change the body's: $τ_{d} = I_{w} \frac{d ω _{w}}{d t} \Rightarrow \frac{d ω _{w}}{d t} = \frac{τ _{d}}{I _{w}} = \frac{2 \times 1 0 ^{- 4}}{0.5} = 4 \times 1 0^{- 4} rad/s^{2}$ Why this step? By Torque and Newton's Second Law for Rotation the wheel's spin rises steadily to soak up the incoming momentum — this is the wheel "winding up."

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

Step 3 — put it in human units. $1.5 \times 1 0^{7} s \times \frac{1 day}{86400 s} \approx 173.6 days$ Why this step? An exam wants the "so what" — the answer is a real operations schedule: desaturate roughly every ~6 months for this disturbance.

Result: ~ $1.5 \times 1 0^{7} s \approx 174$ days until the wheel saturates; a thruster/magnetorquer desaturation must happen before then.

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 — did the matrix stick?

Recall Predict each answer, then reveal

Torque parallel to the spin axis causes what? ::: No precession; the spin just speeds/slows ( $τ = I d ω / d t$ ). As $ω \to 0$ , what happens to $Ω = τ / (I ω)$ ? ::: It blows up to $+ \infty$ — the gyro wobbles faster and topples. Wheel spun counter-clockwise, which way does the body turn? ::: Clockwise (opposite), by conservation. Why is peak slew rate twice the average? ::: A triangular accelerate-then-decelerate profile peaks at $2 θ / T$ . What limits how far a reaction wheel can slew a craft? ::: Saturation — the wheel's max speed caps stored momentum; then you desaturate.