Products of Inertia and the Inertia Matrix

A tire shop will balance your wheel twice without telling you. The first balance is static: clip weights until the wheel's center of mass sits exactly on the axle, so it shows no heavy spot when allowed to swing. The second is the one that matters at highway speed — the machine spins the wheel, watches the axle tremble, and prescribes weights on both rims, in different angular positions. A wheel can pass the static test perfectly and still hammer its bearings at 100 km/h. The quantity the spinning machine is hunting has a name from the nineteenth century: the products of inertia, the three off-diagonal entries of the inertia matrix. The moment of inertia tensor packages a rigid body's whole mass geometry into one symmetric $3\times3$ matrix $\mathsf{I}$ with $L_i = I_{ij}\,\omega_j$ . Its diagonal entries are the familiar moments of inertia — "how hard is it to spin about this axis?" Its off-diagonal entries answer a stranger question: "how lopsided is the mass with respect to this pair of axes?" Each product of inertia, $I_{xy} = -\int xy\,\mathrm{d}m$ , is a covariance of the mass distribution — positive correlation between two coordinates shows up as a nonzero entry, and that entry is precisely what tilts $\mathbf{L}$ away from $\boldsymbol{\omega}$ and makes a spinning body shake its supports. This article owns the off-diagonal story: what the products of inertia measure, why a symmetry plane executes two of them at a stroke, how they transform when you rotate axes or move the origin, and how diagonalization hunts them to extinction. The worked examples put numbers on the shaking — a statically balanced rotor that still delivers a rotating 370-newton hammer blow to its bearings, and a cube spun about its edge.

Key takeaways

The inertia matrix $I_{ij}$ is a symmetric $3\times3$ tensor: diagonal entries are moments of inertia, off-diagonal entries are the products of inertia $I_{xy},I_{xz},I_{yz}$ .
A product of inertia such as $I_{xy}=-\int xy\,\mathrm{d}m$ is a covariance of the mass distribution — it measures how mass correlates across a pair of perpendicular axes, and it is generally nonzero.
Products of inertia are exactly what couples the components of $\boldsymbol{\omega}$ into transverse components of $\mathbf{L}$ : spin about $\hat z$ and the transverse angular momentum is $(I_{xz}\omega,\,I_{yz}\omega)$ — the obstruction to $\mathbf{L}\parallel\boldsymbol{\omega}$ .
A mirror-symmetry plane forces the two products of inertia involving the perpendicular coordinate to vanish; enough symmetry makes a frame principal for free.
Because $I_{ij}$ is real and symmetric, some rotation of axes always diagonalizes it — in the principal-axes frame every product of inertia is zero.
Nonzero $I_{xz},I_{yz}$ in a spinning body's frame = dynamic imbalance: oscillating bearing reactions that grow as $\omega^2$ , even when the body is statically balanced.
Key idea: Products of inertia are the off-diagonal entries $I_{ij}=-\int x_i x_j\,\mathrm{d}m$ ( $i\neq j$ ) of the symmetric inertia matrix; they vanish identically in the principal-axes frame, where the tensor becomes diagonal.

Where this lives today

Every mass-produced thing that spins fast — turbine rotors, EV motor armatures, hard-drive platters, crankshafts, washing-machine drums — passes through a dynamic balancing machine that is, functionally, a products-of-inertia meter. The machine spins the part, reads the rotating force at two bearing planes, and solves a small inverse problem: what corrections $\Delta I_{xz}, \Delta I_{yz}$ (drill here, clip weight there, in two separate planes) drive the couple to zero? Two-plane balancing exists precisely because dynamic imbalance is a two-number defect — one rotating couple needs two adjustable moments to cancel it. ISO 21940 grades rotors by permissible residual imbalance per operating speed, an $\omega^2$ law written into an industrial standard. The same six numbers ride in every simulation. A CAD package's "mass properties" report prints the full inertia matrix — products included — for each part, and robotics engines integrate every link's dynamics with its full tensor because arm segments are never symmetric. Spacecraft attitude teams track products of inertia through fuel burns and boom deployments: a satellite whose thruster axis drifts off-principal develops exactly the rotating-couple wobble of worked example 1, with no bearings to absorb it. And when engineers "trim" a reaction-wheel assembly or balance a helicopter rotor in the field, they are doing what the tire shop does — nulling off-diagonal matrix elements by hand, two grams at a time.

The physical picture

Forget rotation for a moment and look at the integral itself. $-I_{xy} = \int xy\,\mathrm{d}m$ is a covariance: every mass element votes with the product of its two coordinates. Mass sitting in the first or third quadrant of the $xy$ -plane (where $x$ and $y$ share a sign) votes positive; mass in the second or fourth quadrant votes negative. A body whose mass drapes along the line $y = x$ piles up positive votes and ends with $I_{xy} < 0$ ; a body symmetric about either axis collects matching $\pm$ pairs that cancel to zero. The product of inertia is a lopsidedness meter — statistics of the mass distribution, before any dynamics happens.

Figure

The covariance meter — which quadrants hold the mass decides the product of inertia

Now switch the dynamics on, and the statistics acquire teeth. Spin a body about the $z$ -axis. If $I_{xz}$ or $I_{yz}$ is nonzero, the angular momentum picks up components perpendicular to the spin axis — the matrix multiplication $\mathbf{L} = \mathsf{I}\boldsymbol{\omega}$ routes some of $\omega_z$ into $L_x$ and $L_y$ . Those transverse components are frozen into the body, so as the body turns they sweep a cone in space. A rotating $\mathbf{L}$ means $\dot{\mathbf{L}} \neq \mathbf{0}$ , and something must supply that torque: the bearings. The body shakes its supports once per revolution, at a strength growing as $\omega^2$ , even though its center of mass never moves. That is dynamic imbalance, and it is entirely an off-diagonal phenomenon — the diagonal moments know nothing about it.

Figure

Statically balanced, dynamically unbalanced — opposite masses cancel the weight but not the wobble

The geometric summary lives one article downstream: nonzero products of inertia mean the body's inertia ellipsoid sits tilted relative to your coordinate axes. Diagonalizing the matrix is nothing more exotic than rotating your axes until they line up with the ellipsoid's own — at which point the cross terms have nowhere to hide and vanish identically. The existence of that magic frame is guaranteed by the spectral theorem, and the principal axes article makes it the main event. Here our business is the off-diagonal entries themselves: reading them, killing them by symmetry, and paying the bearing bill when we cannot.

The six numbers of the inertia matrix

The tensor's construction — assembling $\mathbf{L} = \sum m\,\mathbf{r}\times(\boldsymbol{\omega}\times\mathbf{r})$ and factoring out $\boldsymbol{\omega}$ — belongs to the moment of inertia tensor article, and we take its result as given. For a continuous body of density $\rho$ the components about a chosen origin are

I_{ij}=\int_V \left(r^2\,\delta_{ij}-x_i x_j\right)\rho\,\mathrm{d}V \qquad (1)

where: $r^2 = x^2+y^2+z^2$ — squared distance from the origin; $\delta_{ij}$ — Kronecker delta; $x_i$ — the $i$ -th coordinate of the mass element. On the diagonal ( $i=j$ ) the $\delta_{ij}$ term survives and the formula collapses to the familiar "mass times perpendicular distance squared." Off the diagonal the $\delta_{ij}$ dies, and what remains is pure covariance with a minus sign — the object this article is about. Written out, the three independent off-diagonal entries are the products of inertia:

I_{xy}=-\int_V xy\,\rho\,\mathrm{d}V,\qquad I_{xz}=-\int_V xz\,\rho\,\mathrm{d}V,\qquad I_{yz}=-\int_V yz\,\rho\,\mathrm{d}V \qquad (2)

where: the minus sign is the standard physics convention, inherited directly from the $-x_i x_j$ in equation (1). It is not decoration: it is what lets the single clean statement $L_i = I_{ij}\,\omega_j$ hold with no scattered signs. Be warned that many engineering texts define the bare integral $P_{xy} = \int xy\,\mathrm{d}m$ as "the product of inertia" and carry the minus signs explicitly in the momentum formulas — same physics, opposite bookkeeping, and a classic source of sign errors when hopping between books. Assembled, with symmetry $I_{ij} = I_{ji}$ manifest from equation (1):

\mathsf{I}=\begin{pmatrix} I_{xx} & I_{xy} & I_{xz}\\ I_{xy} & I_{yy} & I_{yz}\\ I_{xz} & I_{yz} & I_{zz}\end{pmatrix} \qquad (3)

where: $I_{xx} = \int(y^2+z^2)\,\mathrm{d}m$ and cyclic — the diagonal moments. Six independent numbers, not nine: three moments and three products. Count what each kind controls. The moments set how much angular momentum a spin generates along itself; the products set how much leaks sideways. A body is fully described about a given origin by exactly these six numbers plus its mass and center of mass — which is why every CAD package's "mass properties" panel prints precisely this list.

How the off-diagonals feed the wobble

The tensor's job is the linear map from spin to angular momentum:

L_i = I_{ij}\,\omega_j \qquad (4)

where: $L_i$ — angular momentum about the origin; $\omega_j$ — angular velocity components; summation over $j$ implied. When $\mathsf{I}$ is diagonal this is three independent scalar relations, and freshman mechanics is exact. The products of inertia are what break that innocence — they are the matrix elements that mix directions. The derivation below isolates their effect in the cleanest laboratory situation there is: a body forced by bearings to spin about a fixed axis.

Derivation AFixed-axis spin and the rotating bearing couple

We are deriving what the products of inertia do: for a body constrained to spin at constant rate $\omega$ about the fixed axis $\hat z$ , we compute the angular momentum from equation (4), identify the transverse part as pure off-diagonal output, and then compute the torque the bearings must supply to keep the axis fixed. The strategy is direct evaluation — feed $\boldsymbol{\omega} = \omega\hat z$ into the matrix, then apply $\boldsymbol{\tau} = \mathrm{d}\mathbf{L}/\mathrm{d}t$ to the rotating result. The point to watch is which entries of the matrix appear: only the third column survives, so a $z$ -spin can never feel $I_{xy}$ — each product of inertia has its own specific job.

Step 1. Evaluate the map for $\boldsymbol{\omega} = (0, 0, \omega)$ . Multiplying equation (3) into the spin vector picks out the third column:

\mathbf{L} = \big(I_{xz}\,\omega,\;\; I_{yz}\,\omega,\;\; I_{zz}\,\omega\big)

Why this is the whole off-diagonal story in one line: the along-axis component $L_z = I_{zz}\omega$ is the freshman answer; the transverse components $L_x = I_{xz}\omega$ and $L_y = I_{yz}\omega$ exist only because the products of inertia do. If $I_{xz} = I_{yz} = 0$ , then $\mathbf{L} \parallel \boldsymbol{\omega}$ and $\hat z$ is a principal axis — that is, in fact, the working definition.

Step 2. Let the body turn. The transverse vector $\mathbf{L}_\perp = (I_{xz}, I_{yz})\,\omega$ is built from body-fixed mass geometry, so it co-rotates with the body: in the lab frame its magnitude stays $L_\perp = \omega\sqrt{I_{xz}^2 + I_{yz}^2}$ while its direction sweeps the cone at rate $\omega$ .

Step 3. Apply the rotational second law. A vector of constant length rotating at rate $\omega$ changes at rate $\omega$ times its magnitude, directed perpendicular to itself — the same kinematics as any rotating vector, $\dot{\mathbf{L}} = \boldsymbol{\omega}\times\mathbf{L}$ :

\tau = \big|\dot{\mathbf{L}}\big| = \omega\,L_\perp = \omega^2\sqrt{I_{xz}^2 + I_{yz}^2}

Why the $\omega^2$ matters practically: one factor of $\omega$ from the size of $\mathbf{L}_\perp$ , one from how fast it swings. Double the spin rate and the bearing load quadruples — the reason imbalance that is imperceptible at low speed destroys machinery at high speed, and the reason balancing standards are written per operating speed.

Step 4. Locate the physical forces. The bearings, separated by a distance $d$ along the axis, realize the torque as an equal-and-opposite force pair $F = \tau/d$ rotating in the plane perpendicular to the axis — a hammer blow circling once per revolution. Note what this torque is not: its power is $\boldsymbol{\tau}\cdot\boldsymbol{\omega} = 0$ since $\boldsymbol{\tau} \perp \boldsymbol{\omega}$ , so the bearings do no net work. Dynamic imbalance costs force and fatigue, not energy. Verified: the transverse components $(I_{xz}\omega, I_{yz}\omega)$ match the explicit expansion of $L_i = I_{ij}\omega_j$ in Goldstein §5.3; the bearing-couple formula $\tau = \omega^2\sqrt{I_{xz}^2+I_{yz}^2}$ reproduces the standard rotor-dynamics result and is recomputed numerically in worked example 1.

The same off-diagonal entries also show up in the energy. The rotational kinetic energy is the quadratic form

T=\tfrac12\,I_{ij}\,\omega_i\omega_j = \tfrac12\big(I_{xx}\omega_x^2+I_{yy}\omega_y^2+I_{zz}\omega_z^2\big)+I_{xy}\omega_x\omega_y+I_{xz}\omega_x\omega_z+I_{yz}\omega_y\omega_z \qquad (5)

where: each product of inertia contributes a cross term coupling two spin components. Read the structure: for a pure spin along one coordinate axis the cross terms sleep (each needs two nonzero $\omega$ components), which is consistent with the fixed-axis result above — a $z$ -spin has $T = \tfrac12 I_{zz}\omega^2$ regardless of the products of inertia, because the transverse angular momentum is perpendicular to $\boldsymbol{\omega}$ and stores no energy. Tilt the spin off-axis, though, and the cross terms wake up and shift the energy — this is exactly why the energy surface $T(\boldsymbol{\omega}) = \text{const}$ is a tilted ellipsoid whenever products of inertia are present.

Symmetry: the cheapest way to kill a product of inertia

Before computing any integral, look for mirrors. Symmetry is the practical reason most textbook bodies have diagonal tensors from the start, and the mechanism is a one-line parity argument.

Go deeper · Derivation B · 1 minA symmetry plane forces two products of inertia to zero

Derivation BA symmetry plane forces two products of inertia to zero

We are proving that if the $xy$ -plane is a mirror plane of the mass distribution — $\rho(x,y,-z) = \rho(x,y,z)$ — then both products of inertia involving $z$ vanish: $I_{xz} = I_{yz} = 0$ . The strategy is the standard parity trick: change integration variables by the reflection and compare the result with the original. What to watch is which integrals the argument touches — the reflection flips only the explicit factor of $z$ , so $I_{xy}$ , which contains no $z$ , is untouched and may survive.

Step 1. Write out the product of inertia to be tested:

I_{xz}=-\int xz\,\rho(x,y,z)\,\mathrm{d}V

Step 2. Substitute $z \to -z$ as a change of integration variable. The volume element is unchanged (the Jacobian of a reflection has $|\det| = 1$ ), the density is invariant by hypothesis, and the explicit $z$ flips sign:

I_{xz}=-\int x(-z)\,\rho(x,y,-z)\,\mathrm{d}V=+\int xz\,\rho(x,y,z)\,\mathrm{d}V = -I_{xz}

Why the flip is fatal: a number equal to its own negative is zero. Vote by vote, every mass element at $(x,y,z)$ finds a mirror partner at $(x,y,-z)$ casting the opposite ballot — the cancellation from the hero figure, made rigorous.

Step 3. Repeat verbatim with $y \leftrightarrow x$ : the same reflection kills $I_{yz}$ . But $I_{xy} = -\int xy\,\rho\,\mathrm{d}V$ contains no factor of $z$ , so the substitution leaves it strictly alone — one mirror kills exactly two of the three products, never all three.

Step 4. Harvest the consequence. With $I_{xz} = I_{yz} = 0$ , spinning about $\hat z$ gives $\mathbf{L} = I_{zz}\omega\,\hat z$ exactly parallel to $\boldsymbol{\omega}$ : the axis perpendicular to a mirror plane is automatically a principal axis. Add a second, perpendicular mirror plane and its perpendicular axis is principal too; the third axis follows by orthogonality, and the whole frame is principal with every product of inertia dead — no eigenvalue problem ever solved. This is why the box, the cylinder, and every body of revolution arrive pre-diagonalized. Verified: standard parity argument, consistent with Landau–Lifshitz §32 and Taylor Ch. 10 statements that symmetry planes and axes generate principal directions; the worked cube example below shows the converse — a corner origin discards the mirrors and the products of inertia come back.

The rules of thumb that fall out are worth tabulating, because in practice you kill products of inertia by inspection far more often than by integration:

you see	you may conclude	why
mirror plane $\perp \hat z$	$I_{xz}=I_{yz}=0$	parity in $z$ (Derivation B)
two perpendicular mirror planes	all three products vanish	two applications + orthogonality
axis of revolution along $\hat z$	$I_{xy}=I_{xz}=I_{yz}=0$ , $I_{xx}=I_{yy}$	every plane through the axis is a mirror
no symmetry at all	nothing — compute or diagonalize	products generically nonzero
origin moved off the CM	products can reappear	Steiner correction, equation (8)

Rotating axes, moving origins

Products of inertia are not properties of the body alone — they are properties of the body plus your choice of axes. Two transformation rules make that precise. Under a rotation $R$ of the coordinate axes the matrix transforms as a rank-2 tensor:

\mathsf{I}' = R\,\mathsf{I}\,R^{\mathsf T} \qquad (6)

where: $R$ — the orthogonal matrix relating the frames; $\mathsf{I}'$ — the components in the new frame. A similarity transformation by an orthogonal matrix preserves eigenvalues, so no rotation can change the three principal moments — but it shuffles the off-diagonal entries freely. Products of inertia are, in this precise sense, coordinate artifacts: real enough to snap a driveshaft, yet erasable by pointing your axes better. The spectral theorem guarantees some $R$ makes $\mathsf{I}'$ diagonal:

\det(\mathsf{I}-\lambda\,\mathbb{1})=0 \qquad (7)

where: the roots $\lambda = I_1, I_2, I_3$ are the principal moments and the associated eigenvectors are the principal axes — the full eigen-story, including degeneracy and stability, is the next article's main event. From this article's perspective, equation (7) is the extermination warrant for the products of inertia: the frame in which all three are simultaneously zero always exists. The second rule governs moving the origin. Displacing the reference point from the center of mass by $\mathbf{a}$ shifts the tensor by a point-mass term — the tensor form of the parallel-axis theorem:

I'_{ij}=I^{(\mathrm{cm})}_{ij}+M\big(a^2\delta_{ij}-a_i a_j\big) \qquad (8)

where: $M$ — total mass; $a_i$ — components of the displacement. Read the off-diagonal entries of the correction: $I'_{xy} = I^{(\mathrm{cm})}_{xy} - M\,a_x a_y$ . A body with zero products of inertia about its center of mass acquires them the moment the origin moves diagonally away — displacement with two nonzero components manufactures covariance out of nothing. The tilted-plate derivation below runs on exactly this effect: a square plate, perfectly symmetric about its center, shows $I_{xy} = -\tfrac14 Ma^2$ the moment you anchor the axes at a corner.

Diagonalization in action: the tilted square plate

Go deeper · Derivation C · 2 minDiagonalizing a worked example: tilted square plate

Derivation CDiagonalizing a worked example: tilted square plate

We compute the full inertia matrix of a uniform square lamina about a corner, with axes along the edges — a frame chosen deliberately badly, so that a product of inertia survives — and then diagonalize it to find the principal axes and principal moments. The strategy: two elementary double integrals give the diagonal and off-diagonal entries; the $2\times2$ eigenvalue problem of the in-plane block then yields the principal directions. Watch how much structure comes for free: the lamina's flatness kills two products instantly, and the eigenvectors land on the square's diagonals — which hindsight recognizes as the mirror lines of the corner frame.

Step 1. Set up. Uniform square plate, side $a$ , mass $M$ , surface density $\sigma = M/a^2$ , lying in the $z=0$ plane with one corner at the origin and edges along $\hat x, \hat y$ . Because every mass element has $z = 0$ , both $I_{xz} = -\int xz\,\mathrm{d}m$ and $I_{yz} = -\int yz\,\mathrm{d}m$ vanish identically — flatness is a mirror symmetry, and Derivation B applies.

Step 2. Diagonal moments. About the $x$ -edge:

I_{xx}=\sigma\!\int_0^a\!\!\int_0^a y^2\,\mathrm{d}x\,\mathrm{d}y=\sigma\,a\cdot\tfrac{a^3}{3}=\tfrac13 Ma^2

By the square's $x \leftrightarrow y$ symmetry $I_{yy} = \tfrac13 Ma^2$ as well, and the perpendicular-axis theorem hands us $I_{zz} = I_{xx} + I_{yy} = \tfrac23 Ma^2$ for free.

Step 3. The product of inertia — nonzero, because the corner frame has no mirror in the $x$ or $y$ axis alone (all the mass sits in the first quadrant, voting unanimously):

I_{xy}=-\sigma\!\int_0^a\!\!\int_0^a xy\,\mathrm{d}x\,\mathrm{d}y=-\sigma\Big(\tfrac{a^2}{2}\Big)\Big(\tfrac{a^2}{2}\Big)=-\tfrac14 Ma^2

Why this had to be nonzero: about the plate's center the products vanish by symmetry, and the Steiner correction (8) with $\mathbf{a} = (a/2, a/2, 0)$ gives $I_{xy} = 0 - M(a/2)(a/2) = -\tfrac14 Ma^2$ — the direct integral and the parallel-axis route agree exactly.

Step 4. Assemble and diagonalize the in-plane block:

\mathsf I=Ma^2\begin{pmatrix} \tfrac13 & -\tfrac14 & 0\\[2pt] -\tfrac14 & \tfrac13 & 0\\[2pt] 0 & 0 & \tfrac23\end{pmatrix}

For a symmetric block with equal diagonal entries $A$ and off-diagonal $B$ , the eigenvalues are $A \pm B$ with eigenvectors along $\hat x \pm \hat y$ . Here $A = \tfrac13 Ma^2$ , $B = -\tfrac14 Ma^2$ :

I_{1} = \tfrac{1}{12}Ma^2 \;(\text{along } \hat x + \hat y), \qquad I_{2} = \tfrac{7}{12}Ma^2 \;(\text{along } \hat x - \hat y)

Why the eigenvectors are the square's diagonals: the diagonal through the corner IS a mirror line of the corner-anchored square, so it had to be principal — the eigenvalue problem rediscovers the symmetry argument. The small moment belongs to the through-corner diagonal (mass hugs that line), the large one to the perpendicular in-plane direction (mass sweeps wide).

Step 5. Cross-checks. The through-corner diagonal passes through the plate's center, so its moment must equal the central value $\tfrac{1}{12}Ma^2$ with no Steiner shift — it does. The perpendicular in-plane axis sits $d = a/\sqrt 2$ from the center: $\tfrac{1}{12}Ma^2 + M(a^2/2) = \tfrac{7}{12}Ma^2$ — Steiner agrees. And the trace is conserved under rotation: $\tfrac{1}{12} + \tfrac{7}{12} = \tfrac{8}{12} = \tfrac23 = I_{zz}/Ma^2$ , exactly the perpendicular-axis sum. Verified: integrals recomputed symbolically and numerically; eigenvalues $\tfrac{1}{12}Ma^2, \tfrac{7}{12}Ma^2$ with eigenvectors along $\hat x \pm \hat y$ match the standard corner-axis result in Morin Ch. 9, and the numerical eigensolver returns the same pair.

Figure

Edge axes versus principal axes of the corner-anchored plate

Limiting cases

Case	Result	Meaning
Plane of symmetry	mass symmetric under $z\to -z$ $\Rightarrow$ $I_{xz}=I_{yz}=0$	Any reflection plane kills the two products involving the perpendicular coordinate — the integrand is odd in it.
Axis of symmetry	axisymmetric about $z$ $\Rightarrow$ $I_{xy}=I_{xz}=I_{yz}=0$ , $I_{xx}=I_{yy}$	A body of revolution has all products zero and equal transverse moments — a symmetric top, pre-diagonalized.
Principal-axes frame	$I_{ij}=\mathrm{diag}(I_1,I_2,I_3)$	The frame guaranteed by the spectral theorem; the defining property of principal axes is exactly "no products of inertia."
Sphere or cube about center	$I_{ij}=\tfrac{2}{5}MR^2\,\delta_{ij}$ (sphere); $\tfrac16 Ma^2\,\delta_{ij}$ (cube)	Enough symmetry makes the tensor isotropic: every axis is principal, products vanish in every orientation.
Origin displaced from CM	$I'_{xy}=I^{(\mathrm{cm})}_{xy}-Ma_xa_y$	Products of inertia reappear whenever the displacement has two nonzero components — symmetry about the CM does not protect other origins.
Fixed-axis spin, products dead	$I_{xz}=I_{yz}=0 \Rightarrow \tau_{\text{bearing}}=0$	The axis is principal: $\mathbf{L}\parallel\boldsymbol{\omega}$ , no wobble, no rotating couple — the goal state of dynamic balancing.

Common trapA product of inertia is the moment of inertia about some tilted axis

It is not — and the misreading corrupts calculations quietly. A moment of inertia about any axis $\hat{\mathbf n}$ is the quadratic form $\hat{\mathbf n}^{\mathsf T}\mathsf{I}\,\hat{\mathbf n} = \sum m\,d^2 \ge 0$ : always non-negative, always a genuine "resistance to spin." A product of inertia is a signed matrix element, not the moment about anything: $I_{xy} = -\int xy\,\mathrm{d}m$ can be negative, and no choice of axis has it as its moment. Its job is different — it measures cross-coupling, how a spin about one axis generates angular momentum along another. The second trap is the sign convention: physics texts bake the minus into the definition (so $L_i = I_{ij}\omega_j$ is clean), while many engineering texts tabulate the bare covariance $P_{xy} = +\int xy\,\mathrm{d}m$ and subtract it explicitly in their formulas. Before importing any tabulated product of inertia, check which convention the table uses — the physics is identical, but mixing conventions flips the direction your $\mathbf{L}$ tilts, and a balancing correction computed with the wrong sign makes the wobble worse.

Worked examples

1 · The statically balanced rotor that hammers its bearings (confidence builder)

A light shaft spins about the $z$ -axis at $3000\ \mathrm{rpm}$ in two bearings a distance $d = 0.40\ \mathrm{m}$ apart. Two small clip-on masses of $m = 50\ \mathrm{g}$ each sit at $\mathbf{r}_1 = (0.15,\ 0,\ 0.10)\ \mathrm{m}$ and $\mathbf{r}_2 = (-0.15,\ 0,\ -0.10)\ \mathrm{m}$ — diametrically opposite, at opposite ends of the shaft. Show the rotor is statically balanced, find the products of inertia, and compute the rotating force on the bearings.

Step 1. Static balance check. The center of mass of the pair is $\tfrac12(\mathbf{r}_1 + \mathbf{r}_2) = \mathbf{0}$ — exactly on the axis. On a static balancing stand this rotor shows no heavy spot; gravity has no handle on it.

Step 2. Products of inertia from equation (2), treating the masses as points. Both masses have $xz > 0$ (the coordinates share signs in pairs), so their votes add:

I_{xz} = -\sum m\,xz = -2\times(0.05)(0.15)(0.10) = -1.5\times10^{-3}\ \mathrm{kg\,m^2}

and $I_{xy} = I_{yz} = 0$ since every $y$ -coordinate is zero. Statically perfect, dynamically lopsided.

Step 3. Spin it. $\omega = 3000\ \mathrm{rpm} = 314.2\ \mathrm{rad/s}$ . By Derivation A the transverse angular momentum is $L_\perp = |I_{xz}|\,\omega = (1.5\times10^{-3})(314.2) = 0.471\ \mathrm{kg\,m^2/s}$ , co-rotating with the shaft.

Step 4. The bearing couple, from $\tau = \omega^2\sqrt{I_{xz}^2 + I_{yz}^2}$ :

\tau = (314.2)^2 \times 1.5\times10^{-3} = \boxed{148\ \mathrm{N\,m}}

realized as an equal-and-opposite force pair $F = \tau/d = 148/0.40 = \boxed{370\ \mathrm{N}}$ on each bearing, rotating once per revolution — $50$ times per second.

Step 5. Feel the scale. The two offending masses weigh $2mg = 0.98\ \mathrm{N}$ ; the dynamic load they generate is $370\ \mathrm{N}$ — about $380$ times their weight, delivered as a rotating hammer blow. And the $\omega^2$ law says a rotor certified smooth at $1500\ \mathrm{rpm}$ arrives at $3000\ \mathrm{rpm}$ with four times the load. This is why balancing machines spin the part: no static test, however careful, can see $I_{xz}$ .

2 · A cube spun about its edge (exam level)

A uniform cube of mass $M = 0.9\ \mathrm{kg}$ and side $a = 0.2\ \mathrm{m}$ has one corner at the origin and edges along the axes. It is spun at $\omega = 30\ \mathrm{rad/s}$ about the $z$ -edge. Find the inertia matrix about the corner, the angular momentum, the angle between $\mathbf{L}$ and $\boldsymbol{\omega}$ , the kinetic energy, and the torque the axle must supply.

Step 1. The corner-frame tensor. The integrals run over $[0,a]^3$ with density $M/a^3$ . Each diagonal entry: $I_{xx} = \tfrac{M}{a^3}\int(y^2+z^2)\,\mathrm{d}V = \tfrac{M}{a^3}\big(a\cdot a\cdot\tfrac{a^3}{3} + a\cdot a\cdot\tfrac{a^3}{3}\big) = \tfrac23 Ma^2$ . Each product of inertia: $I_{xy} = -\tfrac{M}{a^3}\int xy\,\mathrm{d}V = -\tfrac{M}{a^3}\cdot\tfrac{a^2}{2}\cdot\tfrac{a^2}{2}\cdot a = -\tfrac14 Ma^2$ . With $Ma^2 = 0.9\times0.04 = 0.036\ \mathrm{kg\,m^2}$ :

\mathsf I = Ma^2\begin{pmatrix} \tfrac23 & -\tfrac14 & -\tfrac14\\[2pt] -\tfrac14 & \tfrac23 & -\tfrac14\\[2pt] -\tfrac14 & -\tfrac14 & \tfrac23 \end{pmatrix}

All three products of inertia are alive: the corner frame has no mirror symmetry in any single coordinate — all the mass crowds the all-positive octant.

Step 2. Angular momentum for $\boldsymbol{\omega} = (0,0,30)$ : the third column times $\omega$ ,

\mathbf{L} = Ma^2\,\omega\,\big(-\tfrac14,\ -\tfrac14,\ \tfrac23\big) = \boxed{(-0.27,\ -0.27,\ 0.72)\ \mathrm{kg\,m^2/s}}

with transverse magnitude $L_\perp = 0.27\sqrt2 = 0.382\ \mathrm{kg\,m^2/s}$ .

Step 3. Misalignment angle: $\tan\theta = L_\perp/L_z = 0.382/0.72 = 0.530$ , so $\boxed{\theta = 27.9^\circ}$ — the angular momentum leans well away from the spin axis, dragged toward the cube's bulk.

Step 4. Kinetic energy from equation (5) — for a pure $z$ -spin only the diagonal term survives: $T = \tfrac12 I_{zz}\omega^2 = \tfrac12(\tfrac23\times0.036)(900) = \boxed{10.8\ \mathrm{J}}$ . The transverse part of $\mathbf{L}$ stores none of it.

Step 5. Axle torque from Derivation A: $\tau = \omega\,L_\perp = 30\times0.382 = \boxed{11.5\ \mathrm{N\,m}}$ , a couple rotating with the cube. For contrast, diagonalize: the eigenvalues of $\mathsf{I}$ are $\tfrac16 Ma^2$ (along the body diagonal $(1,1,1)/\sqrt3$ ) and a degenerate pair $\tfrac{11}{12}Ma^2$ . Spin the cube about its body diagonal instead and $\mathbf{L} \parallel \boldsymbol{\omega}$ : torque-free. The body diagonal passes through the center of mass — where the cube's tensor is the isotropic $\tfrac16 Ma^2\,\delta_{ij}$ — so its moment suffers no Steiner shift, a tidy consistency check.

Step 6. Audit. Trace check: $3\times\tfrac23 = 2 = \tfrac16 + \tfrac{11}{12} + \tfrac{11}{12}$ ✓. Numerical eigensolver confirms $\{0.16667,\ 0.91667,\ 0.91667\}\times Ma^2$ , and a two-million-point Monte Carlo of the corner integrals returns $I_{xy} = -0.00900\ \mathrm{kg\,m^2}$ against the exact $-\tfrac14 Ma^2 = -0.00900$ ✓.

Origins · 1 minAugustin-Louis Cauchy and how this was discovered

Augustin-Louis Cauchy

1789–1857 · Paris, École Polytechnique

Cauchy is the mathematician who taught physics that a symmetric array of numbers hides perpendicular axes. Trained as an engineer under Napoleon, he turned the mechanics of continuous media into mathematics: his 1822 theory of stress introduced what we now call the stress tensor — a symmetric $3\times3$ object read exactly like the inertia matrix — and in 1827 he introduced the geometric companion surface of a symmetric tensor, the quadric that for mass distributions becomes the inertia ellipsoid. Two years later he proved the theorem this whole article leans on: a real symmetric matrix has real eigenvalues and orthogonal eigenvectors, established while studying exactly these mechanical quadrics, decades before "matrix" or "eigenvalue" were standard words. Legendarily prolific (789 papers) and legendarily difficult — he fled France after the 1830 revolution rather than swear an oath, taught in Turin and Prague, and lost manuscripts by Abel and Galois in his piles — Cauchy stitched analysis and mechanics together so tightly that the products of inertia, the stress components, and the principal-axis theorem are all, at bottom, one piece of his mathematics.

History

1765

— Euler's Theoria motus corporum solidorum introduces moments of inertia and shows every rigid body has three perpendicular principal axes.

1822

— Cauchy's memoir on stress in continuous media introduces the symmetric stress tensor, the inertia matrix's structural twin.

1827

— Cauchy attaches a quadric surface to a symmetric tensor — for mass distributions, the inertia ellipsoid whose tilt visualizes the products of inertia.

1829

— Cauchy proves real symmetric matrices have real eigenvalues and orthogonal eigenvectors: the guarantee that products of inertia can always be rotated away.

1907

— Franz Lawaczeck patents early dynamic balancing machinery; spin-testing rotors to measure and cancel products of inertia becomes standard industrial practice.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §5.3 · Morin, Introduction to Classical Mechanics, Ch. 9 · Landau & Lifshitz, Mechanics, 3rd ed., §32 · Taylor, Classical Mechanics, Ch. 10

Provenance. A verifier should: (1) re-derive $I_{ij}=\int(r^2\delta_{ij}-x_ix_j)\,\mathrm{d}m$ from $\mathbf L=\int\mathbf r\times(\boldsymbol\omega\times\mathbf r)\,\mathrm{d}m$ via the BAC–CAB identity (Goldstein §5.3); (2) confirm the parity argument $\rho(x,y,-z)$ -symmetric $\Rightarrow I_{xz}=I_{yz}=0$ ; (3) recompute the square-lamina example: $I_{xx}=I_{yy}=\tfrac13Ma^2$ , $I_{xy}=-\tfrac14Ma^2$ , $I_{zz}=\tfrac23Ma^2$ , eigenvalues $\tfrac{1}{12}Ma^2$ and $\tfrac{7}{12}Ma^2$ with eigenvectors along $\hat x\pm\hat y$ — cross-check against Morin Ch. 9. Sphere value $\tfrac25MR^2$ from standard tables. All against Goldstein, Morin, Landau–Lifshitz §32, Taylor Ch. 10. Worked-example numbers on this page recomputed numerically for this version: rotor $I_{xz}=-1.5\times10^{-3}\ \mathrm{kg\,m^2}$ , $\tau=148\ \mathrm{N\,m}$ , $F=370\ \mathrm{N}$ ; cube $\mathbf L=(-0.27,-0.27,0.72)$ , $\theta=27.9^\circ$ , $T=10.8\ \mathrm{J}$ , $\tau=11.5\ \mathrm{N\,m}$ , eigenvalues $Ma^2(\tfrac16,\tfrac{11}{12},\tfrac{11}{12})$ confirmed by a numerical eigensolver and Monte Carlo integration.

Builds on: Moment of Inertia Tensor · Angular Momentum of a Rigid Body · Rotational Kinetic Energy Leads to: Principal Axes and Principal Moments of Inertia · Inertia Ellipsoid · Euler's Equations of Motion · Parallel Axis (Steiner) Theorem

Why is there a minus sign in I_xy = −∫xy dm? It seems designed to confuse.

It is inherited, not chosen. The tensor's definition $I_{ij} = \int(r^2\delta_{ij} - x_ix_j)\,\mathrm{d}m$ comes out of expanding $\mathbf{r}\times(\boldsymbol{\omega}\times\mathbf{r})$ — the derivation in the tensor article — and the second term arrives with a minus sign attached. Keeping it inside the definition buys the perfectly clean map $L_i = I_{ij}\omega_j$ : multiply matrix by vector, done, no sign corrections. The alternative convention (common in engineering statics tables) defines $P_{xy} = +\int xy\,\mathrm{d}m$ and writes $L_x = I_{xx}\omega_x - P_{xy}\omega_y - P_{xz}\omega_z$ , pushing the signs into every formula instead. Neither is wrong; the danger is only in mixing them. Check the definition before using any tabulated value — the warning box above is earned from real balancing mistakes.

Is I_xy the moment of inertia about some diagonal axis between x and y?

No — and this is the most common misreading of the matrix. The moment of inertia about the $45°$ axis $\hat{\mathbf n} = (\hat x+\hat y)/\sqrt2$ is the quadratic form $\hat{\mathbf n}^{\mathsf T}\mathsf{I}\hat{\mathbf n} = \tfrac12(I_{xx} + I_{yy}) + I_{xy}$ : the product of inertia contributes to it, alongside the diagonal moments, but is not itself the moment about anything. The giveaway is the sign: moments of inertia are sums of $m\,d^2$ and can never be negative, while $I_{xy}$ is negative whenever mass favors the quadrants where $x$ and $y$ share a sign. Think of the matrix as a machine with two kinds of parts — diagonal entries that scale, off-diagonal entries that mix — rather than as a table of nine moments.

Can a product of inertia be arbitrarily large compared to the diagonal moments?

No — positivity fences it in. The moment about every axis must be non-negative, and applying that to the $45°$ axes gives $\tfrac12(I_{xx}+I_{yy}) \pm I_{xy} \ge 0$ , i.e. $|I_{xy}| \le \tfrac12(I_{xx} + I_{yy})$ . The corner-anchored square plate makes the bound feel real: $|I_{xy}| = \tfrac14 Ma^2$ against $\tfrac12(I_{xx}+I_{yy}) = \tfrac13 Ma^2$ — large, but capped. Saturating the bound would need one principal moment to hit zero, which happens only when all the mass lies on a line. So a wildly lopsided body shows products of inertia comparable to its moments, never dominating them; if your computed matrix violates the bound, an integral (or a sign convention) is wrong.

What exactly is the difference between static and dynamic imbalance?

Static imbalance is a center-of-mass defect: the CM sits off the rotation axis, gravity finds the heavy spot even at rest, and spinning produces a net rotating force $M\omega^2 r_{\mathrm{cm}}$ . Dynamic imbalance is a products-of-inertia defect: the CM can be perfectly on-axis (as in worked example 1) while $I_{xz}, I_{yz} \neq 0$ produce a rotating couple $\omega^2\sqrt{I_{xz}^2+I_{yz}^2}$ — invisible at rest, violent at speed. A real unbalanced wheel generally has both, which is why balancing machines measure forces at two bearing planes and prescribe weights in two rim planes: one combination cancels the net force, the orthogonal combination cancels the couple. Passing the static test says nothing about the dynamic one — that is the whole point of spinning the wheel.

If the products of inertia vanish in the principal frame, are they even real physics — or just a bad choice of coordinates?

Both, and the tension is instructive. The three principal moments are frame-independent (rotations preserve eigenvalues), so any quantity that changes under rotation — the products of inertia — is in that sense a coordinate artifact. But "artifact" does not mean "harmless": your bearings are physical objects that fix a particular axis, and the products of inertia in the frame those bearings define determine real, measurable forces. The invariant statement is the relation between two directions: the spin axis your hardware enforces versus the principal axes the mass distribution owns. Products of inertia are the coordinates of that mismatch. Choose hardware aligned with the eigenvectors and they vanish; choose otherwise and they bill you at $\omega^2$ .

My body has no symmetry at all. How do I actually find the frame where the products of inertia vanish?

Compute the six entries in any convenient frame — integrals or CAD — then solve the $3\times3$ eigenvalue problem $\det(\mathsf{I} - \lambda\mathbb{1}) = 0$ , equation (7). The three eigenvalues are the principal moments; the eigenvectors, orthonormal by symmetry of the matrix, are the axis directions in which all products of inertia are zero. Numerically this is one call to any linear-algebra library (it is what CAD mass-properties tools do internally). Two practical notes: do the diagonalization about the center of mass first, since equation (8) can then transport the result anywhere; and if two eigenvalues come out nearly equal, the corresponding axes are nearly arbitrary within their plane — any pair of perpendicular directions in it serves, which is the degeneracy of the symmetric top.

The square plate has mirror symmetry about its diagonals — why did the corner frame have a product of inertia at all?

Because symmetry protects only the frames aligned with it. The corner-anchored edge axes are not mirror directions of the corner-anchored square: reflecting across the $x$ -axis maps the plate out of the first quadrant entirely, so the parity argument has no grip, and $I_{xy} = -\tfrac14 Ma^2$ survives. The diagonal through the corner is a mirror line, and sure enough the eigenvectors landed exactly on the diagonals — the eigenvalue problem rediscovered the symmetry the edge frame ignored. The general lesson: products of inertia measure the mismatch between your axes and the body's symmetry directions, not the absence of symmetry. A perfectly symmetric body shows nonzero products the moment you interrogate it from tilted or off-center axes, as the Steiner term $-Ma_xa_y$ makes explicit.

Do the products of inertia matter for the kinetic energy, or only for the angular momentum?

Both, through the same cross terms. Equation (5) shows each product of inertia multiplying a pair of spin components: $I_{xy}\omega_x\omega_y$ and so on. For a pure spin along a coordinate axis those terms sleep — which is why the cube spun about its edge had simply $T = \tfrac12 I_{zz}\omega^2$ despite three live products of inertia. But for a general $\boldsymbol{\omega}$ direction the cross terms shift the energy up or down, and the surface of constant $T$ in $\boldsymbol{\omega}$ -space becomes a tilted ellipsoid — tilted by exactly the products of inertia. That tilt is the geometric object the inertia ellipsoid article develops. The dynamical consequence you can feel: at fixed spin rate, a body is easiest to spin (lowest $T$ , hence lowest flywheel cost) about its smallest principal axis, and the products of inertia tell you how far your chosen axis misses it.

Products of Inertia and the Inertia Matrix

Where this lives today

The physical picture

The six numbers of the inertia matrix

How the off-diagonals feed the wobble

Symmetry: the cheapest way to kill a product of inertia

Rotating axes, moving origins

Diagonalization in action: the tilted square plate

Limiting cases

Worked examples

1 · The statically balanced rotor that hammers its bearings (confidence builder)

2 · A cube spun about its edge (exam level)

Going further

Before this

Where this leads