Inertia tensor — principal axes, principal moments: Worked examples

This page is the drill-ground for the parent topic. We build a scenario matrix — a checklist of every kind of situation this topic can throw at you — then work examples until every box is ticked. If a phrase like "eigenvalue" or "principal axis" feels shaky, keep Eigenvalues and eigenvectors and Symmetric and orthogonal matrices open in another tab.

Formula The component formulas we will reuse all page (stated once, up front)

For a set of point masses $m_{a}$ at positions $r_{a} = (x_{a}, y_{a}, z_{a})$ , the inertia tensor entries are: $I_{xx} = \sum_{a} m_{a} (y_{a}^{2} + z_{a}^{2}), I_{y y} = \sum_{a} m_{a} (x_{a}^{2} + z_{a}^{2}), I_{z z} = \sum_{a} m_{a} (x_{a}^{2} + y_{a}^{2}),$ $I_{x y} = I_{y x} = - \sum_{a} m_{a} x_{a} y_{a}, I_{x z} = - \sum_{a} m_{a} x_{a} z_{a}, I_{y z} = - \sum_{a} m_{a} y_{a} z_{a} .$ Read them as: each diagonal entry is total mass $\times$ squared-distance from that axis; each off-diagonal (a product of inertia) carries a minus sign. For a continuous body replace $\sum_{a} m_{a} \to \int d m$ . Everything below is just these formulas evaluated for a given layout.

One property we will lean on for checks: define $r_{a}^{2} = x_{a}^{2} + y_{a}^{2} + z_{a}^{2}$ (squared distance of mass $a$ from the origin). Then the trace — the sum of the three diagonal entries — is $Tr I = I_{xx} + I_{y y} + I_{z z} = 2 \sum_{a} m_{a} r_{a}^{2},$ and it equals $I_{1} + I_{2} + I_{3}$ (the sum of principal moments) in any frame. That gives us a one-line sanity check on every eigenvalue calculation.

The scenario matrix

Before any numbers, here is the full space of cases. Each worked example is tagged with the cell it covers.

Cell	Case class	What is tricky about it	Example
A	Masses on an axis (degenerate, $I = 0$ )	zero moment about the line they lie on	Ex 1
B	Off-diagonal negative, tilted $L$	product of inertia sign, non-parallel $L$	Ex 2
C	Off-diagonal positive (opposite quadrant layout)	sign of $- \sum m x y$ flips with quadrant	Ex 3
D	Full symmetry ( $I_{1} = I_{2} = I_{3}$ , spherical top)	every axis principal, isotropic	Ex 4
E	Continuous body + Parallel axis theorem shift	$\int d m$ and axis not through centre	Ex 5
F	Symmetric top ( $I_{1} = I_{2} \neq = I_{3}$ ) real word problem	two equal, one different; frisbee	Ex 6
G	Fully asymmetric $3 \times 3$ , exam twist	genuine cubic, trace/det checks	Ex 7
H	Continuous box, 3 distinct + thin-rod limit	all $I$ distinct, then a dimension $\to 0$	Ex 8

The axes of the matrix are: sign of off-diagonals (zero / plus / minus), degeneracy of eigenvalues (all distinct / two equal / all equal), and input type (point masses / continuous / limiting). The eight examples together hit every combination that appears in exams.

Ex 1 — Cell A: masses sitting on an axis (the degenerate $I = 0$ case)

Forecast: guess now — what is the moment of inertia about the $x$ -axis (the line the masses sit on)?

Figure — Inertia tensor — principal axes, principal moments

Read the figure: the two amber dots sit exactly on the cyan horizontal line (the $x$ -axis). Notice there is no perpendicular gap between either dot and that line — that visible zero gap is what will make $I_{xx} = 0$ below.

Compute $I_{xx} = \sum m (y^{2} + z^{2})$ . Both masses have $y = 0$ and $z = 0$ , so each term $y^{2} + z^{2} = 0$ and $I_{xx} = 0$ . Why this step? $I_{xx}$ measures mass squared-distance from the $x$ -axis; masses on that axis are at zero distance — exactly the zero gap you see in the figure.
Compute $I_{y y} = \sum m (x^{2} + z^{2}) = m d^{2} + m d^{2} = 2 m d^{2}$ . Same for $I_{z z} = 2 m d^{2}$ . Why this step? Distance from the $y$ -axis is $x^{2} + z^{2} = ∣ x ∣ = d$ ; spinning about $y$ swings the masses on radius $d$ .
Products of inertia: $I_{x y} = - \sum m x y = 0$ (all $y = 0$ ); likewise $I_{x z} = I_{y z} = 0$ . Why this step? Any product with a coordinate that is uniformly zero vanishes.
Assemble. $I = 000 0 2 m d^{2} 0 00 2 m d^{2} = 00003600036 kg\cdotm^{2} .$ It is already diagonal, so principal moments are $0, 36, 36 kg\cdotm^{2}$ and the principal axes are just $x, y, z$ .

Verify: using the trace property stated up front (with $r^{2} = x^{2} + y^{2} + z^{2} = d^{2}$ for each mass), trace $= 0 + 2 m d^{2} + 2 m d^{2} = 4 m d^{2} = 4 (2) (9) = 72$ , and this must equal $2 \sum m r^{2} = 2 (2 \cdot 9 + 2 \cdot 9) = 72$ . ✓

Ex 2 — Cell B: off-axis pair, tilted $L$ (negative product)

Forecast: if you spin this about the $x$ -axis, does $L$ point along $x$ ? Guess yes/no first.

Read the figure: the amber dots straddle the dashed cyan line $y = x$ . The white arrow is the spin $ω$ pointing straight along $x$ ; the amber arrow is the resulting $L = (8 ω, - 8 ω, 0)$ pointing down into the fourth quadrant. The visible angle between the white and amber arrows is the whole point of this example — $L$ is tilted away from $ω$ .

Diagonal entries. Start from the full definition $I_{xx} = \sum m (y^{2} + z^{2})$ . Every mass has $z = 0$ (both lie in the $x y$ -plane), so the $z^{2}$ term drops and $I_{xx} = \sum m y^{2} = m a^{2} + m a^{2} = 2 m a^{2}$ . Likewise $I_{y y} = \sum m (x^{2} + z^{2}) = \sum m x^{2} = 2 m a^{2}$ , and $I_{z z} = \sum m (x^{2} + y^{2}) = 4 m a^{2}$ . Why this step? Each diagonal is squared-distance from its axis; here $z = 0$ for all masses, so the general formula collapses to the in-plane terms, and both masses are at the same $∣ x ∣ = ∣ y ∣ = a$ .
Product $I_{x y} = - \sum m x y$ . First mass: $x y = (a) (a) = a^{2}$ . Second: $(- a) (- a) = a^{2}$ . Sum $= 2 a^{2}$ , so $I_{x y} = - 2 m a^{2}$ . Why this step? Both masses live in quadrants where $x$ and $y$ share a sign, so each $x y > 0$ — the product piles up, and the minus sign in the definition makes $I_{x y}$ negative.
Assemble with $m a^{2} = 1 \cdot 4 = 4$ : $I = 2 m a^{2} 1 - 1 0 - 1 10 002 = 8 - 8 0 - 8 80 0016 kg\cdotm^{2} .$
Spin about $x$ : $ω = (ω, 0, 0)$ , so $L = I ω = (8 ω, - 8 ω, 0)$ . Why this step? Matrix times a column vector; the $- 8$ off-diagonal drags an $L_{y}$ component into existence — that is the amber arrow's downward tilt in the figure.

$L$ has a $y$ -component even though we only spun about $x$ — not parallel. This is Rigid body rotation biting you.

Verify: eigenvalues of the $2 \times 2$ block $8 (1 - 1 - 1 1)$ are $8 (1 \mp 1) = 0$ and $16$ ; with $I_{z z} = 16$ the principal moments are $0, 16, 16 kg\cdotm^{2}$ . Trace $= 8 + 8 + 16 = 32 = 2 \sum m r^{2} = 2 (1 \cdot 8 + 1 \cdot 8) = 32$ . ✓

Ex 3 — Cell C: opposite-quadrant layout flips the product sign

Forecast: the masses moved to the other diagonal ( $y = - x$ ). Guess whether $I_{x y}$ keeps its sign or flips.

Read the figure: compare with the previous one — the amber dots now sit on the dashed cyan line $y = - x$ (the other diagonal), one dot in the second quadrant and one in the fourth. This mirror-image placement is exactly what flips the sign of the product of inertia, as the amber caption in the figure states.

Diagonal entries are unchanged. $I_{xx} = \sum m y^{2} = 2 m a^{2}$ , $I_{y y} = 2 m a^{2}$ , $I_{z z} = 4 m a^{2}$ . Why this step? Squaring kills the sign: $(- a)^{2} = a^{2}$ , so the masses are the same distance from every axis as in Ex 2.
Product $I_{x y} = - \sum m x y$ . First mass: $(a) (- a) = - a^{2}$ . Second: $(- a) (a) = - a^{2}$ . Sum $= - 2 a^{2}$ , so $I_{x y} = + 2 m a^{2}$ . Why this step? Now the masses sit where $x$ and $y$ have opposite signs, so $x y < 0$ ; the minus in the definition flips it to a positive $I_{x y}$ .
Assemble: $I = 8808800016 kg\cdotm^{2} .$
Eigenvectors switch diagonal. The $2 \times 2$ block $8 (1111)$ has eigenvalue $0$ for $\frac{1}{2} (1, - 1, 0)$ and $16$ for $\frac{1}{2} (1, 1, 0)$ . Why this step? The masses now line up along $y = - x$ ; that line is the zero-moment principal axis (they have no extent off it) — the dashed line in the figure.

Verify: eigenvalues $0, 16, 16$ ; trace $= 8 + 8 + 16 = 32 = 2 \sum m r^{2}$ . Compared to Ex 2, only the off-diagonal sign and hence the direction of the $I = 0$ axis changed — geometry mirrored, physics mirrored. ✓

Ex 4 — Cell D: full symmetry, the spherical top

Forecast: how many principal axes does this cube have — three, or infinitely many?

Off-diagonals vanish by symmetry. $I_{x y} = - \int x y d m = 0$ because for every mass element at $(x, y, z)$ there is a mirror one at $(- x, y, z)$ cancelling $x y$ . Why this step? Symmetric and orthogonal matrices: a reflection symmetry of the body forces the corresponding products to zero.
One diagonal moment. $I_{z z} = \int (x^{2} + y^{2}) d m = \frac{M}{12} (L^{2} + L^{2}) = \frac{1}{6} M L^{2}$ . Why this step? Standard uniform-cube integral; the $\frac{M}{12} L^{2}$ per axis is the box-inertia result.
By symmetry all three are equal: $I_{xx} = I_{y y} = I_{z z} = \frac{1}{6} M L^{2} = \frac{1}{6} (6) (1) = 1 kg\cdotm^{2} .$
So $I = I 1$ . Then $I ω = I ω$ for every $ω$ . Why this step? A scalar multiple of the identity commutes with all rotations, so every axis through the centre is principal.

Verify: principal moments $1, 1, 1 kg\cdotm^{2}$ ; trace $= 3 kg\cdotm^{2}$ . ✓

Ex 5 — Cell E: continuous body + parallel axis shift

Forecast: will the end-axis moment be bigger or smaller than the centre one, and by how much?

Centre moment by integration. With linear density $λ = M / ℓ$ , $I_{cm} = \int_{- ℓ /2}^{ℓ /2} x^{2} λ d x = λ \frac{x ^{3}}{3}_{- ℓ /2}^{ℓ /2} = \frac{M}{ℓ} \cdot \frac{ℓ ^{3}}{12} = \frac{1}{12} M ℓ^{2} .$ Why this step? Every element is at distance $∣ x ∣$ from the $z$ -axis; squared-distance is $x^{2}$ ; integrate over the rod.
Number it: $I_{cm} = \frac{1}{12} (3) (4) = 1 kg\cdotm^{2} .$
Shift to the end with the Parallel axis theorem. The end is $d = ℓ /2 = 1 m$ from the centre: $I_{end} = I_{cm} + M d^{2} = 1 + (3) (1)^{2} = 4 kg\cdotm^{2} .$ Why this step? The theorem $I = I_{cm} + M d^{2}$ lets us reuse the centre value instead of re-integrating. It only works from the centre of mass outward.

Verify: $I_{cm} = 1$ , $I_{end} = 4$ . Direct integration from an end $\int_{0}^{ℓ} x^{2} λ d x = \frac{1}{3} M ℓ^{2} = \frac{1}{3} (3) (4) = 4$ ✓ — matches the theorem.

Ex 6 — Cell F: symmetric top (word problem, a frisbee)

Forecast: a disc has one obvious axis (through the centre, perpendicular to the flat face). Guess whether the two in-plane moments are equal or different.

Confirm the setup. By our chosen axes the disc lies in the $x y$ -plane with $z = 0$ everywhere, so the perpendicular-axis theorem applies with the $z$ -axis as the perpendicular one. Why this step? The theorem is valid only for a lamina in the $x y$ -plane; stating the placement is what licenses using it.
Perpendicular (spin) axis $I_{3}$ . For a disc, $I_{3} = \frac{1}{2} M R^{2}$ . Why this step? Standard disc result; every mass ring at radius $r$ contributes $r^{2} d m$ .
In-plane axes $I_{1} = I_{2}$ by the perpendicular-axis theorem: $I_{3} = I_{1} + I_{2}$ for the flat lamina, and $I_{1} = I_{2}$ by circular symmetry, so $I_{1} = I_{2} = \frac{1}{2} I_{3} = \frac{1}{4} M R^{2}$ . Why this step? The disc is flat, so the perpendicular-axis theorem applies; symmetry then splits $I_{3}$ evenly between the two in-plane axes.
Numbers with $M R^{2} = 0.2 \cdot 0.0225 = 0.0045$ : $I_{3} = \frac{1}{2} (0.0045) = 0.00225, I_{1} = I_{2} = \frac{1}{4} (0.0045) = 0.001125 kg\cdotm^{2} .$
Interpretation. Two equal, one distinct: a symmetric top. The perpendicular axis is the distinguished one — spin it there and the flight is stable (this is why a thrown frisbee spins face-flat, per Euler's equations of rigid body motion).

Verify: $I_{3} = 0.00225$ , $I_{1} = I_{2} = 0.001125$ ; check $I_{1} + I_{2} = 0.00225 = I_{3}$ ✓ (perpendicular-axis theorem holds).

Ex 7 — Cell G: fully asymmetric tensor, exam twist

Forecast: one axis is already decoupled. Guess which, and whether the other two mix.

Read off the block structure. The first row/column is clean (only the $2$ on the diagonal), so the direction $\overset{n}{^}_{A} = (1, 0, 0)$ is already a principal axis with moment $2$ . Why this step? A zero row and column means $x$ does not mix — $I (1, 0, 0)^{T} = (2, 0, 0)^{T} = 2 (1, 0, 0)^{T}$ , which is the eigenvalue equation for $λ = 2$ .
Diagonalize the $2 \times 2$ block $(3113)$ . Characteristic equation $(3 - λ)^{2} - 1 = 0 \Rightarrow 3 - λ = \pm 1 \Rightarrow λ = 2$ or $λ = 4$ . Why this step? $det (I - λ 1) = 0$ is the eigenvalue condition; the block factorises the cubic into the isolated $λ = 2$ (from step 1) times this quadratic.
Eigenvectors of the block. For $λ = 4$ : $(3 - 4) v_{2} + v_{3} = 0 \Rightarrow v_{2} = v_{3}$ , direction $(0, 1, 1)$ . For $λ = 2$ : $(3 - 2) v_{2} + v_{3} = 0 \Rightarrow v_{3} = - v_{2}$ , direction $(0, 1, - 1)$ . Normalise: $\overset{n}{^}_{B} = \frac{1}{2} (0, 1, - 1)$ (moment $2$ ) and $\overset{n}{^}_{C} = \frac{1}{2} (0, 1, 1)$ (moment $4$ ). Why this step? Solve $(block - λ 1) v = 0$ for each root to get the axis direction, then normalise to unit length.
Collect and order. With the convention $I_{1} \leq I_{2} \leq I_{3}$ :
- $I_{1} = 2$ along $\overset{e}{^}_{1} = \overset{n}{^}_{A} = (1, 0, 0)$ ,
- $I_{2} = 2$ along $\overset{e}{^}_{2} = \overset{n}{^}_{B} = \frac{1}{2} (0, 1, - 1)$ ,
- $I_{3} = 4$ along $\overset{e}{^}_{3} = \overset{n}{^}_{C} = \frac{1}{2} (0, 1, 1)$ . Why this step? Pairing each eigenvalue with its normalised eigenvector completes the diagonalization; the three $\overset{e}{^}_{k}$ are the principal axes.

Verify: trace $= 2 + 3 + 3 = 8$ must equal $I_{1} + I_{2} + I_{3} = 2 + 2 + 4 = 8$ ✓. Determinant $= 2 \cdot (9 - 1) = 16$ must equal $I_{1} I_{2} I_{3} = 2 \cdot 2 \cdot 4 = 16$ ✓.

Ex 8 — Cell H: general box (3 distinct moments) and its thin-rod limit

Forecast: with three different side lengths, guess whether any two moments come out equal. Then guess which moment collapses when two sides shrink to a wire.

(a) The general box — Cell G-style but continuous, all distinct.

Off-diagonals are zero. Each coordinate plane is a symmetry plane, so $\int x y d m = \int y z d m = \int x z d m = 0$ ; the face-parallel axes are already principal. Why this step? Same reflection argument as the cube (Ex 4), but the unequal sides mean the diagonal entries will differ.
Plug numbers with $\frac{M}{12} = 1$ : $I_{x} = b^{2} + c^{2} = 4 + 9 = 13, I_{y} = a^{2} + c^{2} = 1 + 9 = 10, I_{z} = a^{2} + b^{2} = 1 + 4 = 5 kg\cdotm^{2} .$ Why this step? Direct substitution. All three are different — this is the "three distinct eigenvalues, zero off-diagonals" cell the matrix promised.
Order them: $I_{1} = 5 (z), I_{2} = 10 (y), I_{3} = 13 (x)$ — an asymmetric top (all moments distinct).

(b) The thin-rod limit.

Send $b \to 0, c \to 0$ (collapse to a wire along $x$ ). $I_{x} = \frac{M}{12} (b^{2} + c^{2}) \to 0$ . Why this step? $I_{x}$ is squared-distance from the $x$ -axis; a wire lying on the $x$ -axis has zero such distance — the same degeneracy we met in Ex 1.
The two transverse moments. $I_{y} = \frac{M}{12} (a^{2} + c^{2}) \to \frac{M}{12} a^{2}$ and $I_{z} = \frac{M}{12} (a^{2} + b^{2}) \to \frac{M}{12} a^{2}$ . Why this step? When $b, c \to 0$ only the length $a$ survives; both perpendicular axes now see an identical thin rod of length $a$ , so they must give the same moment — the top has become symmetric.
Conclusion. With $M = 12, a = 1$ the box's three-distinct moments $(13, 10, 5)$ collapse to ${0, 1, 1} kg\cdotm^{2}$ : one zero (spin about the wire's own line) and two equal — exactly the thin-rod result. Degeneracy is created by the limit.

Verify: box moments $I_{x} = 13, I_{y} = 10, I_{z} = 5$ (all distinct); limit gives $I_{x} = 0$ , $I_{y} = I_{z} = \frac{M}{12} a^{2}$ , which with $M = 12, a = 1$ is $1$ . ✓

Recap

Recall Which cell had the sign flip, and why?

Cells B vs C ::: same distances (so equal diagonals) but masses in same-sign vs opposite-sign quadrants, flipping $I_{x y} = - \sum m x y$ from $- 2 m a^{2}$ to $+ 2 m a^{2}$ and rotating the $I = 0$ principal axis from $y = x$ to $y = - x$ .

Recall What invariant did every "Verify" use?

The trace ::: $Tr I = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}$ is frame-independent, plus $det I = I_{1} I_{2} I_{3}$ — both are quick sanity checks that survive diagonalization.

Recall When is

every axis principal? When $I_{1} = I_{2} = I_{3}$ (Ex 4 cube) ::: then $I = I 1$ and $I ω = I ω$ for all $ω$ .

Every cell A–H of the scenario matrix is now worked. Next, connect these to Euler's equations of rigid body motion to see how these principal moments drive the time evolution of the spin.