Mass properties — CG location, inertia tensor changing with propellant depletion: Exercises

Recurring reference rocket:

Dry mass $m_{d} = 1000 kg$ at fixed station $x_{d} = 6 m$ .
Full propellant $m_{p 0} = 4000 kg$ , centroid $x_{p} = 2 m$ (treated fixed unless a problem says otherwise).

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

How to read Figure s01: the horizontal white arrow is the $x$ -axis (nose at $0$ , aft positive, labelled in metres). The cyan lens is the fuselage. The white dot at $x = 6$ is the dry mass; the cyan dot at $x = 2$ is the propellant centroid; the amber diamond at $x = 2.8$ is the full-load CG. The single message of the figure: the CG sits between the two bodies, pulled toward the heavier one (the 4000 kg fuel), so it starts near the tank and will later slide forward toward the dry mass.

Level 1 — Recognition

Goal: can you pick the right formula and read it correctly?

Exercise 1.1

State, in words and symbols, the definition of the center of gravity for two point-like sub-bodies, and identify which quantity in it depends on time for a burning rocket.

Recall Solution

The CG is the mass-weighted average position: $x_{cg} = \frac{m _{d} x _{d} + m _{p} x _{p}}{m _{d} + m _{p}} .$ What each symbol means: $m_{d}, x_{d}$ = dry mass and its station (both fixed); $m_{p}, x_{p}$ = propellant mass and its centroid. The time-dependent quantity is == $m_{p} (t)$ ==, the remaining propellant mass (and, in a refined model, its centroid $x_{p} (t)$ too). As $m_{p}$ falls, the weight given to the term $x_{p}$ shrinks, so the average slides toward $x_{d}$ .

Exercise 1.2

Write the scalar parallel-axis theorem and label every symbol. Which term is zero when you compute inertia about the CG itself?

Recall Solution

$I_{P} = I_{cg} + M d^{2} .$

$I_{cg}$ = moment of inertia about an axis through the CG.
$M$ = total mass of the body.
$d$ = perpendicular distance from the CG axis to the new parallel axis.
$I_{P}$ = moment of inertia about that shifted axis.

When the new axis is the CG axis, $d = 0$ , so the added term $M d^{2} = 0$ and $I_{P} = I_{cg}$ . The distance is what "costs" you inertia; standing on the CG costs nothing.

Level 2 — Application

Goal: plug real numbers into one tool and get a clean answer.

Exercise 2.1

Find $x_{cg}$ at ignition ( $m_{p} = 4000$ ) and at the 25% burnt point ( $m_{p} = 3000$ ).

Recall Solution

At ignition: $x_{cg} = \frac{1000 ( 6 ) + 4000 ( 2 )}{1000 + 4000} = \frac{6000 + 8000}{5000} = \frac{14000}{5000} = 2.8 m .$ At 25% burnt ( $m_{p} = 3000$ ): $x_{cg} = \frac{1000 ( 6 ) + 3000 ( 2 )}{1000 + 3000} = \frac{6000 + 6000}{4000} = \frac{12000}{4000} = 3.0 m .$ The CG has already crept $0.2 m$ forward (toward the dry mass at $6 m$ ) after burning just a quarter of the fuel.

Exercise 2.2

Using the point-mass model (ignore each body's own $I^{c g}$ ), find the pitch inertia $I_{y y}$ at the 25% burnt point ( $m_{p} = 3000$ , $x_{cg} = 3.0$ ).

Recall Solution

Offsets from the CG at $x_{cg} = 3.0$ : $D_{d} = ∣6 - 3.0∣ = 3.0 m, D_{p} = ∣2 - 3.0∣ = 1.0 m .$ $I_{y y} = m_{d} D_{d}^{2} + m_{p} D_{p}^{2} = 1000 (3.0)^{2} + 3000 (1.0)^{2} = 9000 + 3000 = 12000 kg\cdotm^{2} .$ Notice both bodies sit on opposite sides of the CG, so both contribute positively — inertia adds distances squared, sign-blind.

Level 3 — Analysis

Goal: reason about how a quantity changes, not just its value.

Exercise 3.1

The CG marched from $2.8 m$ (full) to $6.0 m$ (burnout). Show that most of that migration happens late in the burn, not early. Compare the CG shift over the first half of the propellant mass versus the second half.

Recall Solution

Compute the CG at three points ( $m_{p} = 4000, 2000, 0$ ):

$m_{p} = 4000$ : $x_{cg} = 2.8 m$ .
$m_{p} = 2000$ : $x_{cg} = \frac{6000 + 4000}{3000} = 3.333 m$ .
$m_{p} = 0$ : $x_{cg} = 6.0 m$ .

First half of fuel ( $4000 \to 2000$ ): shift $= 3.333 - 2.8 = 0.533 m$ . Second half ( $2000 \to 0$ ): shift $= 6.0 - 3.333 = 2.667 m$ .

The second half moves the CG 5× as far. Why: the denominator $m_{d} + m_{p}$ gets small near burnout, so each remaining kilogram of fuel has huge leverage over the average. The CG "runs away" toward the dry mass at the very end — precisely when the autopilot is most sensitive. See the amber curve in Figure s02.

How to read Figure s02: the horizontal axis is propellant remaining $m_{p}$ in kg — and it is reversed, so time flows left as fuel drains (full at the right, empty at the left). The vertical axis is CG position $x_{cg}$ in metres. The amber curve is $x_{cg} (m_{p})$ ; the marked dots at $m_{p} = 4000, 2000, 0$ read $2.8, 3.33, 6.0 m$ . Notice the curve is nearly flat on the right and steepens sharply toward the left — that steepening is the "CG runs away late in the burn" result of Exercise 3.1: equal chunks of fuel move the CG far more near burnout.

Exercise 3.2

At burnout the point-mass model gives $I_{y y} = 0$ (fuel gone, dry mass sits at the CG). Explain why the real rocket has nonzero pitch inertia at burnout, and identify the single quantity from the parent note that supplies it.

Recall Solution

In the point model each body is a dimensionless dot. At burnout the only body left is the dry structure, and the CG is the dry structure's location, so $D_{d} = 0$ and $m_{d} D_{d}^{2} = 0$ . But a real dry structure is an extended body — a long tube with mass spread out — so it has its own intrinsic == $I_{d}^{c g}$ == (its inertia about its own centroid). The composite formula from the parent keeps that term: $I_{y y}^{burnout} = I_{d}^{c g} + m_{d} = 0 D_{d}^{2} = I_{d}^{c g} .$ So the rocket bottoms out at $I_{d}^{c g}$ , never truly zero. The point model just drops the term that now dominates.

Level 4 — Synthesis

Goal: combine CG, Steiner, and extended-body inertia in one problem.

Exercise 4.1

Now give the dry body real size: model it as a uniform rod of length $L = 2 m$ centered at $x_{d} = 6$ , so its own pitch inertia is $I_{d}^{c g} = \frac{1}{12} m_{d} L^{2}$ . Keep propellant point-like at $x_{p} = 2$ . Find the full-load ( $m_{p} = 4000$ ) pitch inertia $I_{y y}$ about the vehicle CG.

Recall Solution

Step 1 — CG (unchanged from the point model, since we only changed the dry body's shape, not its centroid): $x_{cg} = 2.8 m$ . Step 2 — dry rod's own inertia: $I_{d}^{c g} = \frac{1}{12} (1000) (2)^{2} = \frac{1}{12} (4000) = 333.33 kg\cdotm^{2} .$ Step 3 — Steiner each body to the CG: $D_{d} = ∣6 - 2.8∣ = 3.2, D_{p} = ∣2 - 2.8∣ = 0.8.$ $I_{y y} = [I_{d}^{c g} + m_{d} D_{d}^{2}] + [0 + m_{p} D_{p}^{2}]$ $= [333.33 + 1000 (10.24)] + [4000 (0.64)] = (333.33 + 10240) + 2560 = 13133.33 kg\cdotm^{2} .$ Compared to the pure point model's $12800$ , the extended rod added exactly its $I_{d}^{c g} = 333.33$ . Why the extra step matters: the shape term is small at full load (the $m_{d} D_{d}^{2}$ Steiner term dwarfs it), but it becomes the only survivor at burnout — so you cannot drop it in a general model.

Exercise 4.2

Let propellant burn from the bottom of the tank upward, so its centroid drifts. Model it as: at half-burn ( $m_{p} = 2000$ ) the remaining fuel centroid has moved forward to $x_{p} = 1.5 m$ (not $2$ ). Recompute $x_{cg}$ and the point-mass $I_{y y}$ , and compare to the fixed-centroid answer.

Recall Solution

Moving-centroid CG: $x_{cg} = \frac{1000 ( 6 ) + 2000 ( 1.5 )}{3000} = \frac{6000 + 3000}{3000} = \frac{9000}{3000} = 3.0 m .$ (The fixed-centroid model gave $3.333 m$ ; the fuel drifting forward pulls the CG forward too.) Offsets: $D_{d} = ∣6 - 3.0∣ = 3.0$ , $D_{p} = ∣1.5 - 3.0∣ = 1.5$ . $I_{y y} = 1000 (3.0)^{2} + 2000 (1.5)^{2} = 9000 + 4500 = 13500 kg\cdotm^{2} .$ Why this differs: letting the centroid move changed both the CG location and the moment arms. A fixed-centroid assumption would have mis-placed the CG by $0.333 m$ — enough to mis-schedule TVC gains. Real vehicles track $x_{p} (t)$ from tank geometry and fill level.

Level 5 — Mastery

Goal: time-varying, rate, and off-diagonal reasoning — the real flight problem.

Exercise 5.1

Fuel burns at a constant rate $\overset{m}{˙} = - 50 kg/s$ from $m_{p 0} = 4000$ . Using the fixed-centroid point model, find the CG velocity $\overset{x}{˙}_{cg}$ at ignition ( $m_{p} = 4000$ ) and near burnout ( $m_{p} = 200$ ). Interpret the ratio.

Recall Solution

With $x_{d} = 6$ , $x_{p} = 2$ , $m_{d} = 1000$ , write $x_{cg} (m_{p}) = \frac{6000 + 2 m _{p}}{1000 + m _{p}}$ . Differentiate with respect to $m_{p}$ (quotient rule): $\frac{d x _{cg}}{d m _{p}} = \frac{2 ( 1000 + m _{p} ) - ( 6000 + 2 m _{p} ) ( 1 )}{( 1000 + m _{p} ) ^{2}} .$ Watch the numerator carefully: expand the top to $2000 + 2 m_{p} - 6000 - 2 m_{p}$ . The two $\pm 2 m_{p}$ terms are equal and opposite, so they cancel exactly, leaving only the constants $2000 - 6000 = - 4000$ : $\frac{d x _{cg}}{d m _{p}} = \frac{- 4000}{( 1000 + m _{p} ) ^{2}} .$ Why calculus here? We want the rate the CG moves, and a rate is exactly what a derivative measures — the slope of $x_{cg}$ versus fuel remaining. Then chain-rule to time: $\overset{x}{˙}_{cg} = \frac{d x _{cg}}{d m _{p}} \overset{m}{˙} = \frac{- 4000}{( 1000 + m _{p} ) ^{2}} \times (- 50) = \frac{200000}{( 1000 + m _{p} ) ^{2}} .$ At $m_{p} = 4000$ : $\overset{x}{˙}_{cg} = \frac{200000}{( 5000 ) ^{2}} = \frac{200000}{25 000 000} = 0.008 m/s .$ At $m_{p} = 200$ : $\overset{x}{˙}_{cg} = \frac{200000}{( 1200 ) ^{2}} = \frac{200000}{1 440 000} = 0.1389 m/s .$ Ratio $\approx 17.4$ : the CG races nearly 17× faster near burnout than at ignition, for the same fuel burn rate. This is why gain-scheduled autopilots update fastest at the end of the burn.

Exercise 5.2

A single canted vernier engine sits at $(x, y) = (5, 0.4) m$ in the page's coordinate system (so $y = + 0.4$ means $0.4 m$ off the centreline toward the positive- $y$ side), mass $m_{e} = 20 kg$ (point mass). Compute its contribution to the product of inertia $I_{x y} = - \int x y d m$ about the vehicle CG at $x_{cg} = 2.8$ (take $y_{cg} = 0$ ). Why does a nonzero $I_{x y}$ matter?

Recall Solution

Position relative to CG: $x^{'} = 5 - 2.8 = + 2.2 m$ , $y^{'} = 0.4 - 0 = + 0.4 m$ . Both coordinates are positive, so this mass sits in quadrant I of the $(x^{'}, y^{'})$ plane. For a point mass the product of inertia is $I_{x y} = - m_{e} x^{'} y^{'} = - (20) (2.2) (0.4) = - 17.6 kg\cdotm^{2} .$ Why it matters: off-diagonal $I_{x y} \neq = 0$ means the inertia tensor is not diagonal, so a pure pitch torque produces a bit of roll response and vice versa — the axes are cross-coupled. A controller that assumes decoupled axes will command the wrong corrections. Real vehicles either balance the layout to kill $I_{x y}$ or model the coupling explicitly.

Common mistake L5 trap — treating the product of inertia as always negative (ignoring the quadrant)

Why it feels right: you just computed $I_{x y} = - 17.6$ and the formula wears a leading minus, so the answer "obviously" is negative. Fix: the sign of $I_{x y} = - m x^{'} y^{'}$ is set by the signs of $x^{'}$ and $y^{'}$ together, i.e. by which quadrant the mass sits in relative to the CG:

Quadrant I ( $x^{'} > 0, y^{'} > 0$ ) or III ( $x^{'} < 0, y^{'} < 0$ ): the product $x^{'} y^{'} > 0$ , so $I_{x y} < 0$ .
Quadrant II ( $x^{'} < 0, y^{'} > 0$ ) or IV ( $x^{'} > 0, y^{'} < 0$ ): the product $x^{'} y^{'} < 0$ , so $I_{x y} > 0$ — it flips sign.

Concretely, if that same engine were mounted on the other side at $y = - 0.4$ (quadrant IV), you'd get $I_{x y} = - (20) (2.2) (- 0.4) = + 17.6 kg\cdotm^{2}$ : same magnitude, opposite cross-coupling direction. And a mass exactly on either axis ( $x^{'} = 0$ or $y^{'} = 0$ ) contributes $I_{x y} = 0$ . Never assume the sign — read the quadrant.

Related vault topics: Parallel-Axis Theorem · Rigid Body Rotational Dynamics · Thrust Vector Control · Gain Scheduling in Autopilots · Propellant Slosh Dynamics · Tsiolkovsky Rocket Equation.