Question bank — Mass properties — CG location, inertia tensor changing with propellant depletion

False — the CG drifts toward the mass that is left behind, which for most rockets is the dry structure (engine + payload) sitting forward of the tanks. Fuel leaving the back means less mass pulling the average rearward, so the CG moves forward.

False — inertia is not proportional to mass here. Removing fuel also moves the CG, which changes every Steiner arm $d$, so each surviving body's $M{d}^2$ contribution changes too. You must recompute CG and re-run the parallel-axis shift, not just scale.

True — parallel-axis inertia is minimized at the CG, so any parallel offset $d$ can only add the positive quantity $M{d}^2$. Going the other way (toward the CG) you subtract it, provided both axes are parallel and $d$ is measured between them.

Roughly true while symmetry holds — but real rockets lose it through asymmetric tank drain, sloshing propellant, and canted engines. Those break symmetry and populate the off-diagonals, cross-coupling roll and pitch. See Propellant Slosh Dynamics.

True — with all propellant gone, only the dry body remains, and the vehicle CG coincides with the dry-body CG (no other mass to average with). So $I_{veh}=I_d^{cg}$ with a zero Steiner arm. A point-mass model wrongly gives $I=0$ because it ignores the dry body's own spatial extent.

False — that equation is the definition. We demand a point about which the mass-weighted position offsets sum to zero (the balance point); the familiar formula ${\mathbf{r}}_{cg}=\frac{1}{M}\sum m_i{\mathbf{r}}_i$ is just algebra solving that condition.

False — the opposite risk appears. Lower inertia means the same control torque produces a larger angular acceleration, so a gain tuned for liftoff can overshoot or destabilize at burnout. This is exactly why gains are scheduled; see Gain Scheduling in Autopilots.

It's a mass-weighted average, not a plain average. $x_{cg}=\frac{m_d{x}_d+m_p{x}_p}{m_d+m_p}$; a simple midpoint would only be right if $m_d=m_p$.

The Steiner arm is the distance between the two axes, i.e. $d=|x_p-x_{cg}|$, not the propellant's absolute station $x_p$. Using the raw coordinate confuses position with offset.

Inertias are additive only about a common point. You must Steiner-shift each body from its own CG to the shared vehicle CG before summing; adding own-CG values directly ignores every $M{d}^2$ term.

Only the linear cross-terms ($2d_y\int y^{\prime }dm$, etc.) vanish because $\int {\mathbf{r}}^{\prime }dm=0$ measured from the CG. The quadratic term $M{d}^2$ survives — that is the entire content of the parallel-axis theorem.

Even at constant thrust, the moment arm changes as the CG migrates, so the control/misalignment torque about the CG changes. TVC (see Thrust Vector Control) must schedule its deflection against the moving CG.

Only when $\mathbf{\omega }$ lies along a principal axis (an eigenvector of $\mathbf{I}$). With nonzero products of inertia, $\mathbf{L}$ and $\mathbf{\omega }$ point in different directions, producing gyroscopic cross-coupling — the whole reason products of inertia matter. See Rigid Body Rotational Dynamics.

Because CG location and inertia change continuously as fuel drains, and both set the control response ($\tau =I\dot{\omega }$, moment arms about CG). A controller frozen at liftoff values grows mistuned — sluggish or unstable — by burnout.

Because $\int {\mathbf{r}}^{\prime }dm=0$ by definition of the CG — the primed coordinates are measured from it. Shift from any non-CG reference and that integral is nonzero, so the cross-term does not die.

Because each body's inertia integral is taken about a specific axis; the $r^2$ weighting depends on distance to that axis. Only after Steiner-shifting every body to the same axis do the integrals share a common measuring stick and become legitimately summable.

We demand a linear map sending $\mathbf{\omega }$ to $\mathbf{L}=\int \mathbf{r}\times (\mathbf{\omega }\times \mathbf{r})dm$. Expanding the triple product and factoring $\mathbf{\omega }$ out forces exactly that bracketed operator — it is the map, not an arbitrary definition.

A cant breaks the mirror symmetry that made $\int xydm$-type integrals cancel. The off-axis mass distribution (and the off-axis thrust line) introduces nonzero off-diagonal $I$ terms, coupling roll and pitch responses.

With $\dot{\omega }=\tau /I$, a smaller $I$ makes the vehicle react more violently to the same commanded torque. To keep the loop from overshooting, the controller must command less torque per unit error — i.e. lower gain.

The formula $\frac{m_d{x}_d+m_p{x}_p}{m_d+m_p}$ tends to $x_p$ as $m_d\to 0$ — the CG sits entirely at the propellant centroid, since there is nothing else to average with.

It collapses to $x_{cg}=x_d$: the CG lands exactly on the dry structure's station, because the only remaining mass is the dry body. This is the endpoint of the whole CG migration.

Yes — the formula is instantaneous. At each moment you plug in the current $m_p(t)$ and current $x_p(t)$. The subtlety is that $x_p$ is now time-dependent (and slosh can move it dynamically), so it must be re-evaluated each step, not held fixed.

Then both Steiner arms $D_d=D_p=0$, so the vehicle inertia is just the sum of each body's own-CG inertia with no parallel-axis contribution — a degenerate case where the CG doesn't move as fuel burns.

Yes, if the remaining propellant's centroid drifts such that its mass-weighted pull outruns the shrinking $m_p$ — e.g. a tank draining from the top so its centroid moves rearward. The rule "toward the leftover mass" still holds; the leftover just isn't where you assumed.

They contribute nothing to $\mathbf{L}$ for that particular spin — $\mathbf{L}$ stays parallel to $\mathbf{\omega }$. But they still exist in the tensor and will bite the moment the spin axis tilts off the principal direction (e.g. after a slosh disturbance).

These are the four instincts that fail: "CG follows the exhaust", "$I\propto m$", "add own-CG inertias directly", and "lighter is easier to fly".