Worked examples — Mass properties — CG location, inertia tensor changing with propellant depletion

This page is the drill floor for Mass properties — CG location, inertia tensor changing with propellant depletion. We do not learn new theory here — we make sure that no case can surprise you: every sign of a distance, every degenerate mass, every limit, a real word problem, and an exam-style twist all appear below, each solved from zero.

Before line one, reminders in plain words so no symbol is unearned:

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The scenario matrix

Every problem this topic can throw sits in one of these cells. The examples that follow each carry a tag [Cell N].

#	Cell (the case class)	What is special about it	Example
1	Baseline / mid-burn	ordinary two-body CG, both masses present	Ex 1
2	Degenerate mass → 0 (burnout)	propellant $m_p=0$: CG snaps to dry mass	Ex 2
3	Degenerate mass → 0 (liftoff heavy)	dry mass tiny vs. propellant: limit toward tank	Ex 3
4	Sign of the moment arm	dry mass forward vs. propellant aft of CG — opposite signs of $x-x_{\text{cg}}$, same squared contribution	Ex 4 (figure)
5	Moving propellant centroid	$x_p(t)$ itself drifts as the tank drains	Ex 5
6	Inertia with body's own $I^{cg}$ kept	not point masses — cylinders/rods that keep their own inertia	Ex 6
7	Limiting behaviour of $I(t)$	how fast inertia collapses; ratio at two times	Ex 7
8	Real-world word problem	TVC moment-arm / gain consequence	Ex 8
9	Exam twist (off-diagonal / product of inertia)	asymmetric drain makes $I_{xy}\ne 0$	Ex 9

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Ex 1 — Baseline mid-burn CG [Cell 1]

Forecast: guess before computing — will the balance point sit closer to $2\text{ m}$ or $6\text{ m}$? (Twice as much fuel as structure, so lean toward $2\text{ m}$.)

1. Multiply each mass by its station. $1000\times 6=6000$; $2000\times 2=4000$. Why this step? The formula is a mass-weighted average — heavier pieces pull the average toward them, so we weight positions by mass first.
2. Add the weighted positions and the masses. Top $=6000+4000=10000$; bottom $=1000+2000=3000$. Why this step? Numerator is total "moment," denominator is total mass; their ratio is the balance point.
3. Divide. $x_{\text{cg}}=10000/3000=3.333\dots \text{ m}$ (a repeating decimal $=\frac{10}{3}\text{ m}$; we round it to $3.33\text{ m}$ for reporting, but keep the exact $\frac{10}{3}$ whenever we square it later).

Verify: $3.33\text{ m}$ lies between $2$ and $6$ (a genuine average must), and it is closer to $2$ (the heavier side) — matching our forecast. Units: $(\text{kg}\cdot \text{m})/\text{kg}=\text{m}$. ✓

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Ex 2 — Degenerate: propellant gone (burnout) [Cell 2]

Forecast: with nothing left in the tank, the only mass is the dry structure — the CG must sit exactly on it.

1. Substitute $m_p=0$. $x_{\text{cg}}=\frac{1000(6)+0\cdot 2}{1000+0}=\frac{6000}{1000}$. Why this step? A zero mass contributes zero moment and zero to the total — it disappears cleanly from both parts of the fraction.
2. Simplify. $x_{\text{cg}}=6\text{ m}=x_d$. Why this step? Dividing the surviving moment $6000$ by the surviving mass $1000$ is the whole averaging formula with only one body left — so the "average position" can only be that one body's position.

Verify: the answer equals $x_d$ exactly, as it must when only one body remains. No division-by-zero occurs because $m_d\ne 0$. ✓ (Contrast the danger case: if BOTH masses were zero the formula is $0/0$ — undefined, meaning "no body, no balance point." Always check total mass $>0$.)

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Ex 3 — Degenerate limit: heavy-propellant liftoff [Cell 3]

Forecast: propellant now swamps the structure — the balance point should sit almost on top of the tank at $2\text{ m}$.

1. Weight and sum. Top $=1000(6)+99000(2)=6000+198000=204000$; bottom $=1000+99000=100000$. Why this step? Same mass-weighted average; we just chose an extreme mass ratio to probe the limit.
2. Divide. $x_{\text{cg}}=204000/100000=2.04\text{ m}$. Why this step? The ratio of total moment to total mass is the balance station; carrying out the division turns the two accumulated totals into a single position we can read on the axis.
3. Take the limit. As $m_p\to \infty$, the $m_p{x}_p$ term dominates top and $m_p$ dominates bottom, so $x_{\text{cg}}\to x_p=2\text{ m}$. Why this step? Dividing top and bottom by $m_p$: $x_{\text{cg}}=\frac{(m_d/m_p)x_d+x_p}{(m_d/m_p)+1}\to x_p$.

Verify: $2.04\text{ m}$ is barely past $2\text{ m}$ — the CG is pinned to the tank, confirming the limit. Between Ex 3 ($2.04$) → Ex 1 ($3.33$) → Ex 2 ($6.00$) you can watch the CG march toward the dry mass as fuel drains. ✓

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Ex 4 — Signs of the moment arm (the two-sided case) [Cell 4]

Forecast: dry mass at $6$ is behind the CG (positive offset); propellant at $2$ is ahead of the CG (negative offset). But since Steiner squares the distance, both add positively to inertia.

1. Dry-mass offset. $x_d-x_{\text{cg}}=6-\frac{10}{3}=\frac{8}{3}=+2.667\text{ m}$ (aft of CG → positive). Why this step? The sign tells us which side — vital for torque directions later — but look at the figure: the dry body is the violet dot to the right of the balance line.
2. Propellant offset. $x_p-x_{\text{cg}}=2-\frac{10}{3}=-\frac{4}{3}=-1.333\text{ m}$ (forward of CG → negative). Magenta dot, left of the line. Why this step? Opposite sign confirms the two bodies straddle the CG — that is why it balances.
3. Square and weight (Steiner, point model). $I_{yy}=1000{\left(\frac{8}{3}\right)}^2+2000{\left(\frac{4}{3}\right)}^2=1000(7.111)+2000(1.778)$. Why this step? $d=|x-x_{\text{cg}}|$, and squaring erases the sign — a body's twist contribution never cares which side it sits on, only how far.
4. Add. $I_{yy}=7111.1+3555.6=10666.7{\text{ kg⋅m}}^2\approx 10667{\text{ kg⋅m}}^2$. Why this step? Inertias measured about the same point (here the CG) simply sum — each body's resistance to pitching adds directly to the vehicle's total.

Verify: both terms positive despite opposite signs — exactly what squaring guarantees. Sanity: the closer, heavier propellant contributes less than the farther, lighter dry mass here, because distance is squared and beats the mass factor. Units ${\text{kg⋅m}}^2$. ✓

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Ex 5 — Moving propellant centroid [Cell 5]

Forecast: vs. Ex 1 (which fixed $x_p=2$), letting the fuel centroid slide back to $2.5$ should pull the CG slightly further aft than $3.33\text{ m}$.

1. Find the current propellant centroid. $x_p=2+1.0(0.5)=2.5\text{ m}$. Why this step? The centroid is no longer a constant; we must read it off at this instant before averaging.
2. Find the current propellant mass. $m_p=m_{p0}(1-f)=4000(0.5)=2000\text{ kg}$. Why this step? $f$ ties mass and centroid together — both move as one variable.
3. Weighted average. $x_{\text{cg}}=\frac{1000(6)+2000(2.5)}{3000}=\frac{6000+5000}{3000}=\frac{11000}{3000}=3.667\text{ m}$. Why this step? Same balance-station formula as always — but now fed the updated $x_p$ and $m_p$, so the answer reflects the mid-burn snapshot rather than the full-tank one.

Verify: $3.667\text{ m}>3.33\text{ m}$ from Ex 1 — the sliding centroid did push the CG aft, as forecast. Still between $2.5$ and $6$. ✓ (See Propellant Slosh Dynamics for why the fuel surface — and hence $x_p$ — is not perfectly steady.)

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Ex 6 — Inertia keeping each body's own $I^{cg}$ [Cell 6]

Forecast: we get the point-mass answer from Ex 4 ($10666.7$) plus two extra "own-inertia" bumps — so the total exceeds $10667$.

1. Dry rod's own inertia. $I_d^{cg}=\frac{1}{12}(1000)(4^2)=\frac{16000}{12}=1333.3{\text{ kg⋅m}}^2$. Why this step? A body spread out along the axis resists pitch even about its own centre — we cannot ignore it once it is not a point.
2. Propellant cylinder's own inertia. $3R^2=3(0.64)=1.92$; $L_p^2=9$; sum $=10.92$. $I_p^{cg}=\frac{1}{12}(2000)(10.92)=\frac{21840}{12}=1820{\text{ kg⋅m}}^2$. Why this step? Same reason — the fuel column has real length, so it has intrinsic pitch inertia.
3. Steiner-shift each to the vehicle CG and add the point terms (from Ex 4): $1000{\left(\frac{8}{3}\right)}^2=7111.1$, $2000{\left(\frac{4}{3}\right)}^2=3555.6$. Why this step? Inertias only add about a common point, so we drag each body's own inertia to $x_{\text{cg}}$ via $M{d}^2$. See Parallel-Axis Theorem.
4. Total. $I_{yy}=1333.3+7111.1+1820+3555.6=13820.0{\text{ kg⋅m}}^2$. Why this step? Once every piece — both own-inertias and both Steiner terms — is referenced to the same CG, the parallel-axis theorem lets us add them straight up into one vehicle inertia.

Verify: $13820>10667$ (Ex 4 point model), by exactly the two own-inertia bumps $1333.3+1820=3153.3$; indeed $10666.7+3153.3=13820.0$. ✓ Units ${\text{kg⋅m}}^2$.

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Ex 7 — Limiting behaviour: how fast does $I$ collapse? [Cell 7]

Forecast: half the fuel gone but inertia should drop by less than half, because the remaining dry mass sits far from the CG (big $d^2$).

1. Take the ratio. $I_{yy}(\text{half})/I_{yy}(0)=10667/12800=0.833$. Why this step? A ratio strips units and shows the proportional collapse the autopilot feels.
2. Interpret. $83.3\%$ of inertia remains though only $50\%$ of fuel is left — because the surviving dry mass has grown its lever arm (CG moved aft toward it). Why this step? $d^2$ dominates; losing near-CG fuel barely lowers $I$.
3. Burnout limit. As $m_p\to 0$, the point model gives $I_{yy}\to m_d(x_d-x_d{)}^2=0$; the real rocket keeps the dry body's own $I_d^{cg}=1333.3$ (Ex 6). Why this step? At burnout $x_{\text{cg}}=x_d$, so the dry lever arm vanishes and only intrinsic inertia survives — never truly zero.

Verify: $0.833\times 12800=10666.7\approx 10667$ ✓. And $0.833>0.5$ confirms inertia falls slower than fuel mass. ✓

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Ex 8 — Real-world word problem: TVC moment arm & gain [Cell 8]

Forecast: at burnout the moment arm shrinks but inertia shrinks far more, so $\dot{\omega }$ should be larger — the rocket becomes twitchy. This is exactly why gains must schedule (Gain Scheduling in Autopilots).

1. Side force. $T\sin \delta =60000\times 0.05234=3140.4\text{ N}$. Why this step? Only the sideways component of a canted thrust makes pitch torque; the axial part just pushes forward.
2. Mid-burn arm & torque. arm $=x_e-x_{\text{cg}}=7-\frac{10}{3}=\frac{11}{3}=3.667\text{ m}$; $\tau =3140.4\times 3.667=11515\text{ N⋅m}$. $\dot{\omega }=11515/10667=1.080{\text{ rad/s}}^2$. Why this step? Torque = force × lever arm about the current CG; divide by current inertia for the response.
3. Burnout arm & torque. arm $=x_e-x_{\text{cg}}=7-6=1\text{ m}$; $\tau =3140.4\times 1=3140.4\text{ N⋅m}$. $\dot{\omega }=3140.4/1333.3=2.355{\text{ rad/s}}^2$.

Verify: arm dropped $3.667\to 1$ (×0.27) but $\dot{\omega }$ rose $1.08\to 2.36$ (×2.18), because inertia dropped ×0.125 — inertia wins. A fixed-gain autopilot tuned mid-burn would over-react at burnout. ✓ (Links to Thrust Vector Control and Rigid Body Rotational Dynamics.)

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Ex 9 — Exam twist: asymmetric drain → nonzero product of inertia [Cell 9]

Forecast: if the tanks were equal, the $+y$ and $-y$ contributions would cancel and $I_{xy}=0$. Unequal masses break the cancellation → nonzero cross-coupling that leaks pitch commands into roll.

1. Left tank contribution. $m_{L}xy=600(-\frac{4}{3})(+0.9)=-720$. Why this step? Product of inertia measures mass distributed off both symmetry planes; each term is mass × two offsets. We keep the exact $-\frac{4}{3}$ so the numbers close cleanly.
2. Right tank contribution. $m_{R}xy=400(-\frac{4}{3})(-0.9)=+480$. Why this step? Opposite $y$ sign flips this term positive — the source of the would-be cancellation.
3. Sum inside, then apply the leading minus sign. $\sum mxy=-720+480=-240$, so $I_{xy}=-(-240)=+240{\text{ kg⋅m}}^2$. Why this step? The definition carries an explicit minus, so a negative bracket becomes a positive product of inertia.

Verify: set $m_R=m_L=600$: then $\sum mxy=600(-\frac{4}{3})(0.9)+600(-\frac{4}{3})(-0.9)=-720+720=0$, so $I_{xy}=0$ — the symmetric case has no coupling, as it must. The $200\text{ kg}$ imbalance leaves $I_{xy}=+240\ne 0$, so a pure pitch command bleeds into roll. ✓ This is the exam trap that punishes "products of inertia are always zero."

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Recall Which cell did each example fill?

Ex1 baseline ::: Cell 1 Ex2 burnout, $m_p=0$ ::: Cell 2 Ex3 heavy-fuel limit ::: Cell 3 Ex4 signed moment arms ::: Cell 4 Ex5 moving centroid ::: Cell 5 Ex6 own $I^{cg}$ kept ::: Cell 6 Ex7 inertia collapse ratio ::: Cell 7 Ex8 TVC gain consequence ::: Cell 8 Ex9 nonzero product of inertia ::: Cell 9