Question bank — Reaction wheels — momentum management, zero-crossing

False. It holds pointing for a while, but a constant-sign external torque integrates without bound ($H_w=\int {\tau }_{\text{ext}}dt$), so the wheel saturates in finite time and then loses control. The wheel is a buffer, not a sink.

False. The motor torque ${\tau }_m$ is internal, so it only moves momentum between the wheel and the body; ${\vec{H}}_{\text{tot}}$ stays fixed. Only an external actuator (magnetorquers or thrusters) can change the total.

True, when no external torque acts: $I_b{\dot{\omega }}_b=-I_w{\dot{\Omega }}_w$, so the two rates always change with opposite signs — the sum is pinned, so a gain by the wheel is exactly a loss by the body.

False. Zero speed is the worst place: bearing friction reverses sign there — around ${\Omega }_w=0$ the sticking (static) friction flips direction, so the wheel briefly locks then jerks free (the friction curve in figure s02 shows the discontinuity — see Bearing Friction and Stiction), and motor cogging is worst with no rotational averaging. The resulting jitter ruins precise pointing, so wheels are biased away from zero.

False. Even a perfectly still, non-maneuvering satellite accumulates wheel momentum from external secular torques (gravity gradient, solar pressure, drag). Dumping is driven by the environment, not by your maneuvers.

False. The torque is $\vec{\tau }=\vec{m}\times \vec{B}$, which is always perpendicular to Earth's field $\vec{B}$; the component of momentum along $\vec{B}$ is untouchable at that instant. You rely on $\vec{B}$ rotating over the orbit to eventually reach all directions.

True. Redundancy gives freedom in how total momentum is distributed among wheels; you can route the required momentum so every wheel stays on one side of zero (the pyramid schematic in figure s03 shows the four tilted axes that give this freedom), avoiding stiction jitter entirely.

Half true. Both apply an external torque and so both can dump, but thrusters spend propellant and can act any direction, while magnetorquers are propellant-free yet constrained perpendicular to $\vec{B}$ and only useful where a field exists (LEO). The trade-off, not equivalence, is the point.

False in general. When wheel and body axes are not perfectly aligned (or the body is already rotating), a spinning wheel acts like a gyroscope: its momentum ${\vec{H}}_w$ crossed with the body rate gives a cross-coupling torque ${\vec{\omega }}_b\times {\vec{H}}_w$ about a third axis (figure s04 shows the two input arrows and the perpendicular output). Real 3-axis controllers must compensate this coupling.

Wrong sign. With our right-hand convention ($+$ = "left"), ${\dot{\omega }}_b=-(I_w/I_b){\dot{\Omega }}_w$: spinning the wheel in the $+$ ("left") sense turns the body in the $-$ ("right") sense. To point the body left you must spin the wheel right (away from the target).

Only valid when ${\tau }_{\text{ext}}$ is constant. In general $H_w(t)=H_w(0)+{\int }_0^t{\tau }_{\text{ext}}d{t}^{\prime }$; the linear-in-$t$ form is the special case of a constant integrand.

Wrong: near zero speed static (stick) friction can exceed the tiny torque needed for fine pointing, so the wheel sticks, pointing error builds, then the motor breaks free and snaps — an impulsive disturbance. Small on average, catastrophic at the instant it matters.

No — it sets the dipole $\vec{m}$ (the coil's magnetic moment), which produces a torque $\vec{m}\times \vec{B}$ that only drives ${\vec{H}}_w$ toward the bias ${\vec{H}}_{\text{target}}$ over time, and only in directions perpendicular to $\vec{B}$. It is a slow feedback nudge, not an instantaneous reset.

Error: for a given motor torque ${\tau }_m$, ${\dot{\Omega }}_w={\tau }_m/I_w$, so a bigger $I_w$ spins up slower. Larger $I_w$ stores more momentum but does not by itself grant faster body acceleration — that's set by the reaction ${\tau }_m/I_b$.

Wrong. Holding the body still against an external torque forces the wheel to absorb all of it: $I_w{\dot{\Omega }}_w={\tau }_{\text{ext}}$, so the wheel's stored momentum grows precisely because the body is kept still.

The motor torque ${\tau }_m$ that spins the wheel reacts back on the body (Newton's third law / momentum conservation), and it is this reaction on the body that produces the desired rotation — the body moves in reaction to the wheel.

Wheels only redistribute ${\vec{H}}_{\text{tot}}$ internally; no internal cleverness changes the total. Removing accumulated momentum requires changing the total, which only an external torque ${\tau }_{\text{ext}}$ can do.

To avoid the zero-crossing region where bearing friction reverses and cogging is worst; regulating $H_w$ toward a bias value ${\vec{H}}_{\text{target}}$ keeps friction unidirectional, eliminating stick-slip jitter during fine pointing.

Because it is constant-sign: its integral grows linearly forever, so however small the torque, $H_w$ inevitably reaches $H_{\max }=I_w{\Omega }_{\max }$ given enough time. Saturation is about accumulation, not instantaneous magnitude.

The available torque $mB\sin \theta$ (with $m$ the dipole magnitude, $\theta$ the angle between $\vec{m}$ and $\vec{B}$) is very small because Earth's field is weak in LEO, and it acts only perpendicular to $\vec{B}$; you must wait for $\vec{B}$ to rotate along the orbit to reach the momentum component you couldn't touch earlier.

With ${\dot{\omega }}_b\approx 0$, the momentum balance $\frac{d}{dt}(I_b{\omega }_b+I_w{\Omega }_w)={\tau }_{\text{ext}}$ reduces to $I_w{\dot{\Omega }}_w={\tau }_{\text{ext}}$ — every bit of injected external momentum goes into the wheel.

They re-point using electric power only (solar → motor), consuming no propellant for routine attitude changes, whereas thrusters burn finite fuel and would run dry. Momentum is borrowed internally rather than thrown overboard.

The wheel can spin no faster, so it can no longer absorb momentum; the excess now torques the body, and attitude control is lost until momentum is dumped. Saturation = end of buffering capacity.

No — with ${\tau }_{\text{ext}}=0$ the total is constant and the wheel only trades momentum with the body during maneuvers, returning to bias afterward. Dumping is required only because real environments inject net momentum.

Zero, because $\vec{m}\times \vec{B}=0$ when they are parallel ($\sin \theta =0$). No torque about any axis is produced in that geometry — you must wait for the geometry to change.

The magnetorquer cannot touch it right then (it only acts perpendicular to $\vec{B}$). You must wait for $\vec{B}$ to reorient over the orbit so that component becomes perpendicular and dumpable.

It enters the stick-slip regime: static friction can exceed the fine-pointing torque command, the wheel hangs, error grows, then breaks free impulsively — injecting jitter exactly during the finest pointing.

A tiny one: ${\omega }_b=-(I_w/I_b){\Omega }_w\to 0$ as the ratio shrinks. Small wheels give fine, slow control authority — good resolution, limited slew rate and limited momentum storage.

A spinning wheel carries momentum ${\vec{H}}_w$; combined with body rotation it produces a gyroscopic cross-coupling torque ${\vec{\omega }}_b\times {\vec{H}}_w$ pointing along a third axis (the same physics exploited by Control Moment Gyroscopes (CMG)). A good Attitude Determination and Control System (ADCS) / controller must model and cancel it.