Moment of Inertia Tensor

Take a book, wrap a rubber band around it, and toss it spinning into the air three ways. About its longest axis it turns smoothly; about its shortest axis, smoothly again; but flipped about the intermediate axis it tumbles — half a twist appears from nowhere every toss. Astronauts see the same trick in free fall: a wing-nut spinning off a bolt aboard Salyut 7 flipped over, again and again, with clockwork regularity. Nothing touched it. The culprit is the moment of inertia tensor. For rotation about a fixed axis, one number $I = \sum mr^2$ suffices. But a rigid body free to rotate in three dimensions answers a harder question — how hard is it to spin about every possible axis at once? — and the answer is a $3\times3$ real symmetric matrix $\mathbf{I}$ . It maps angular velocity to angular momentum, $\mathbf{L} = \mathbf{I}\boldsymbol{\omega}$ , and because a matrix can rotate a vector as it rescales it, $\mathbf{L}$ generally does not point along $\boldsymbol{\omega}$ . That misalignment is the origin of wheel wobble, gyroscopic precession, and the tumbling book. The redemption is linear algebra: every real symmetric matrix has three perpendicular eigenvectors. Those are the principal axes — the directions along which $\mathbf{L}$ and $\boldsymbol{\omega}$ finally agree — and they reduce three-dimensional rotation to three decoupled numbers $I_1, I_2, I_3$ . Everything downstream — the inertia ellipsoid, Euler's equations, the classification of tops — is built on this one diagonalization.

Key takeaways

The moment of inertia tensor $\mathbf{I}$ is a $3\times3$ real symmetric matrix encoding how a rigid body's mass is distributed about a chosen origin — the 3D generalization of the scalar $I = \sum mr^2$ .
It maps angular velocity to angular momentum linearly, $\mathbf{L} = \mathbf{I}\,\boldsymbol{\omega}$ , so $\mathbf{L}$ and $\boldsymbol{\omega}$ are generally NOT parallel.
Diagonal entries are moments of inertia about the axes; off-diagonal entries are (negated) products of inertia — the body's lopsidedness.
The same tensor prices the energy: $T = \tfrac12\,\boldsymbol{\omega}^{\mathsf T}\mathbf{I}\,\boldsymbol{\omega} = \tfrac12\,\boldsymbol{\omega}\cdot\mathbf{L}$ , a positive quadratic form.
Real + symmetric ⇒ three real eigenvalues (principal moments) and orthogonal eigenvectors (principal axes) — the frame in which rotation decouples.
Key idea: $\mathbf{I}$ is the linear operator that turns the angular velocity vector into the angular momentum vector; its symmetry guarantees an orthonormal eigenbasis (the principal axes) in which rotation decouples.

Where this lives today

Every spacecraft carries its inertia tensor in its flight software. Attitude-control systems solve $\mathbf{L}=\mathbf{I}\boldsymbol{\omega}$ and Euler's equations in real time, and mission designers obsess over the eigenvalue ordering because the Explorer 1 lesson is permanent — an energy-dissipating body drifts toward spin about its maximum-moment axis, so satellites are spun about that axis deliberately or actively stabilized. The Dzhanibekov flip is now a standard validation case for attitude simulators, and ISS crews still film it with wing-nuts and T-handles. On the ground, the tensor is infrastructure. Every CAD package computes $\mathbf{I}$ for each designed part, and every robotics engine integrates rigid-body dynamics with a full inertia tensor per link. Tire shops diagonalize the tensor empirically every day: dynamic balancing measures the bearing wobble that nonzero products of inertia produce and prescribes the clip-on masses that make the spin axis principal again. And divers and figure skaters edit their own eigenvalues mid-air — tucking to slash a principal moment and spin faster, opening out to stop.

The physical picture

Think of $\mathbf{I}$ as directional mass. Just as force produces acceleration through the scalar $m$ , angular velocity produces angular momentum through $\mathbf{I}$ — but rotational inertia depends on which way you spin. A long rod twirls easily about its own axis (the mass hugs the axis) and reluctantly end-over-end (the mass sweeps wide circles). One number cannot hold both answers; a symmetric operator can. The picture to carry is the tilted dumbbell below. Spin it about the vertical axis and each mass runs in a horizontal circle — but each mass's angular momentum $\mathbf{r}\times m\mathbf{v}$ is perpendicular to its position vector, so $\mathbf{L}$ tilts away from $\boldsymbol{\omega}$ and, as the body turns, sweeps a cone. A rotating $\mathbf{L}$ means $\mathrm{d}\mathbf{L}/\mathrm{d}t \neq 0$ , demanding continuous torque from the bearings. That is what an unbalanced wheel is: a body spun about a non-principal axis, hammering its bearings once per revolution.

Figure

A tilted dumbbell — angular momentum refuses to follow the spin axis

When is $\mathbf{L}$ parallel to $\boldsymbol{\omega}$ ? Exactly when you spin the body about one of three special perpendicular directions — its principal axes, the eigenvectors of $\mathbf{I}$ , with eigenvalues $I_1 \le I_2 \le I_3$ the principal moments. Every rigid body has them, however lumpy. For a brick they point through the faces, and the ordering carries drama: spins about the largest- and smallest-moment axes are stable, while the intermediate axis is unstable — the tennis-racket theorem your tossed book keeps demonstrating.

Figure

The three principal axes of a brick — and which spins are stable

The last piece of intuition concerns the off-diagonal entries. The diagonal $I_{xx} = \sum m(y^2+z^2)$ weighs mass by its distance from the $x$ -axis; the off-diagonal $I_{xy} = -\sum m\,xy$ asks whether mass prefers the quadrants where $x$ and $y$ share a sign — a lopsidedness meter. Mirror symmetry kills the corresponding products of inertia, which is why symmetry axes of uniform bodies are always principal, and why balancing a car wheel (clipping lead weights on the rim until the wobble stops) is literally diagonalizing $\mathbf{I}$ by hand.

From angular momentum to the tensor

Everything starts from the kinematic fact defining a rigid body rotating about a fixed point: every particle rides the same angular velocity. Summing the particles' angular momenta forces the tensor into existence, and the relation is linear:

\mathbf{L}=\mathbf{I}\,\boldsymbol{\omega}\qquad L_i=\sum_{j}I_{ij}\,\omega_j \qquad (1)

where: $\mathbf{L}$ — angular momentum about the chosen origin; $\boldsymbol{\omega}$ — instantaneous angular velocity; $I_{ij}$ — components of the inertia tensor about that origin. Read it as a machine: feed in a spin, and $\mathbf{I}$ returns the angular momentum — rescaled and, in general, rotated. Only for special inputs (the eigenvectors) does the output come back parallel.

Derivation AWhy L and ω need not be parallel

The target is equation (1) made explicit: starting from nothing but rigidity, we derive $L_i=\sum_j I_{ij}\,\omega_j$ and read off the components $I_{ij}$ that will become equation (2). The route is direct assembly — insert the rigid-body velocity $\mathbf{v}=\boldsymbol{\omega}\times\mathbf{r}$ into the definition of angular momentum, expand the resulting triple product, and factor out the components of $\boldsymbol{\omega}$ . The expansion is where the physics lives: the BAC–CAB identity splits each particle's contribution into a piece along $\boldsymbol{\omega}$ and a piece along $\mathbf{r}$ , and that second piece is the whole reason $\mathbf{L}$ generically tilts away from the spin axis. Watch the index bookkeeping in the final step — the Kronecker delta that lets both terms share one sum over $j$ is exactly what makes the bracket a matrix rather than a scalar.

Step 1. Rigid-body velocity. Every point of a rigid body rotating about a fixed origin moves with $\mathbf{v}_\alpha=\boldsymbol{\omega}\times\mathbf{r}_\alpha$ . Why: rigidity fixes all distances, so each particle can only circle the instantaneous axis — one shared $\boldsymbol{\omega}$ describes the whole body.

Step 2. Sum the angular momenta. Insert this velocity into the definition of $\mathbf{L}$ :

\mathbf{L}=\sum_\alpha m_\alpha\,\mathbf{r}_\alpha\times\mathbf{v}_\alpha=\sum_\alpha m_\alpha\,\mathbf{r}_\alpha\times(\boldsymbol{\omega}\times\mathbf{r}_\alpha)

Why: angular momentum is additive, and every term is slaved to the same $\boldsymbol{\omega}$ — so $\mathbf{L}$ must be a linear function of it.

Step 3. Expand the triple product. The BAC–CAB identity gives

\mathbf{r}\times(\boldsymbol{\omega}\times\mathbf{r})=r^2\boldsymbol{\omega}-(\mathbf{r}\cdot\boldsymbol{\omega})\,\mathbf{r}

Why: this is the crux. The first term is parallel to $\boldsymbol{\omega}$ , but the second points along $\mathbf{r}$ — and unless the mass distribution conspires to cancel these pieces, $\mathbf{L}$ tilts away from $\boldsymbol{\omega}$ .

Step 4. Take components. Writing the $i$ -th component and pulling out $\omega_i=\sum_j\delta_{ij}\omega_j$ :

L_i=\sum_\alpha m_\alpha\Big(r_\alpha^2\,\omega_i-x_{\alpha,i}\sum_j x_{\alpha,j}\,\omega_j\Big)

Why: to expose the linear map, every term must read (coefficient) × $\omega_j$ ; the Kronecker delta is bookkeeping that lets both terms share one $\sum_j$ .

Step 5. Factor out $\omega_j$ . Collecting coefficients,

L_i=\sum_j\underbrace{\Big[\sum_\alpha m_\alpha(r_\alpha^2\delta_{ij}-x_{\alpha,i}x_{\alpha,j})\Big]}_{I_{ij}}\,\omega_j

Why: the bracket depends only on the mass distribution, not on $\boldsymbol{\omega}$ — a property of the body itself. Being generally non-diagonal, it returns $\mathbf{L}$ rotated away from $\boldsymbol{\omega}$ ; they coincide only when $\boldsymbol{\omega}$ is an eigenvector. The bracket deserves its own display — it is the definition you actually compute with. For point masses:

I_{ij}=\sum_\alpha m_\alpha\left(r_\alpha^2\,\delta_{ij}-x_{\alpha,i}\,x_{\alpha,j}\right) \qquad (2)

where: $m_\alpha$ — mass of particle $\alpha$ ; $x_{\alpha,i}$ — its $i$ -th coordinate relative to the origin; $r_\alpha^2 = x_{\alpha,1}^2+x_{\alpha,2}^2+x_{\alpha,3}^2$ ; $\delta_{ij}$ — Kronecker delta. On the diagonal, $I_{xx}=\sum m(y^2+z^2)$ is the familiar "mass times perpendicular distance squared" from the $x$ -axis; off it, $I_{xy}=-\sum m\,xy$ is negative when mass concentrates where $x$ and $y$ share a sign — the algebraic fingerprint of a lopsided body. For a continuous body the sum becomes a density integral — the version deployed on spheres, rods, and bricks, and the source of every tabulated moment of inertia:

I_{ij}=\int_V\rho(\mathbf{r})\left(r^2\,\delta_{ij}-x_i\,x_j\right)\mathrm{d}^3r \qquad (3)

where: $\rho(\mathbf{r})$ — mass density; $\mathrm{d}^3r$ — volume element. Symmetries of $\rho$ translate directly into zeros of the tensor: reflection through a coordinate plane kills the products of inertia involving that direction, which is why symmetry axes of uniform bodies are automatically principal. Assembled as a matrix, the structure is plain — six independent numbers, not nine, because it is symmetric:

\mathbf{I}=\begin{pmatrix}I_{xx}&I_{xy}&I_{xz}\\ I_{xy}&I_{yy}&I_{yz}\\ I_{xz}&I_{yz}&I_{zz}\end{pmatrix} \qquad (4)

where: $I_{xx}=\sum m(y^2+z^2)$ — moment of inertia about the $x$ -axis; $I_{xy}=-\sum m\,xy$ — a product of inertia (the minus sign is baked into the convention). The symmetry $I_{ij}=I_{ji}$ is manifest from equation (2), and it is the hook on which the entire principal-axis theory hangs.

One tensor, two jobs

A remarkable economy hides here: the same nine numbers that convert $\boldsymbol{\omega}$ into $\mathbf{L}$ also price the spin's kinetic energy, as a quadratic form:

T=\tfrac12\,\boldsymbol{\omega}^{\mathsf T}\mathbf{I}\,\boldsymbol{\omega}=\tfrac12\sum_{i,j}I_{ij}\,\omega_i\,\omega_j=\tfrac12\,\boldsymbol{\omega}\cdot\mathbf{L} \qquad (5)

where: $T$ — rotational kinetic energy about the origin. The last form is the rotational cousin of $T=\tfrac12\,\mathbf{v}\cdot\mathbf{p}$ , and it says something physical: only the component of $\mathbf{L}$ along the spin axis stores energy — the transverse part is pure wobble, shaking bearings without doing work.

Step 1. Start from particle kinetic energies. Substitute the rigid-body velocity into $T=\tfrac12\sum m_\alpha v_\alpha^2$ :

T=\tfrac12\sum_\alpha m_\alpha(\boldsymbol{\omega}\times\mathbf{r}_\alpha)\cdot(\boldsymbol{\omega}\times\mathbf{r}_\alpha)

Why: energy too is a sum over particles all locked to the same $\boldsymbol{\omega}$ , so it must be built from $\boldsymbol{\omega}$ and the mass geometry alone.

Step 2. Apply the Lagrange identity. $(\boldsymbol{\omega}\times\mathbf{r})^2=\omega^2 r^2-(\boldsymbol{\omega}\cdot\mathbf{r})^2$ . Why: this converts the cross product into dot products, exposing exactly the combination $r^2\delta_{ij}-x_ix_j$ that defined the tensor in (2).

Step 3. Recognize the tensor. Expanding both dot products in components,

T=\tfrac12\sum_\alpha m_\alpha\big[\omega^2 r_\alpha^2-(\boldsymbol{\omega}\cdot\mathbf{r}_\alpha)^2\big]=\tfrac12\sum_{i,j}I_{ij}\,\omega_i\,\omega_j

Why: the same $I_{ij}$ reappears with no freedom in the matter — momentum and energy share one tensor.

Changing frames and origins

Two bookkeeping rules make the tensor portable. First, its components depend on your axes' orientation, transforming exactly as a rank-2 tensor must — that is what "tensor" means:

I_{ij}=R_{ik}\,I'_{kl}\,R_{jl}\quad\Leftrightarrow\quad \mathbf{I}=\mathbf{R}\,\mathbf{I}'\,\mathbf{R}^{\mathsf T} \qquad (6)

where: $\mathbf{R}$ — orthogonal rotation matrix relating the two frames; $\mathbf{I}'$ — components in the rotated frame. A similarity transformation preserves eigenvalues, so the principal moments are frame-independent, as physical quantities must be. It also explains the central annoyance of rigid-body dynamics: lab-frame components of $\mathbf{I}$ change as the body turns, while body-frame components are constants. The body frame wins, and Euler's equations live there. Second, the tensor depends on the reference point. The clean statement relates any origin to the cm, and the correction is memorable — the inertia tensor of a point particle carrying the whole mass:

I_{ij}=I^{\text{cm}}_{ij}+M\left(a^2\delta_{ij}-a_i a_j\right) \qquad (7)

where: $I^{\text{cm}}_{ij}$ — inertia tensor about the center of mass; $M$ — total mass; $\mathbf{a}$ — displacement of the new origin from the cm. This is the parallel-axis (Steiner) theorem in tensor form. The correction is positive semi-definite, so the cm is the point about which the body is easiest to spin — moments only grow as the origin moves away.

Principal axes

The payoff of symmetry is the spectral theorem: $\mathbf{I}$ diagonalizes under a pure rotation, and the eigen-directions are physical, mutually perpendicular axes frozen into the body. Finding them is a standard eigenvalue problem:

\det(\mathbf{I}-\lambda\,\mathbb{1})=0\;\Rightarrow\;\lambda\in\{I_1,I_2,I_3\} \qquad (8)

where: $\lambda$ — eigenvalue = principal moment of inertia; the eigenvectors are the principal axes. Spin the body about one and $\mathbf{L}=I_k\boldsymbol{\omega}$ with an honest scalar — no wobble, no bearing forces: the one situation where freshman rotational mechanics is exactly right.

Go deeper · Derivation B · 2 minWhy principal axes must exist

Derivation BWhy principal axes must exist

The claim to establish is that every rigid body, however lopsided, owns three mutually perpendicular axes about which $\mathbf{L}$ and $\boldsymbol{\omega}$ align — that the eigenvalue problem of equation (8) always has a full set of real solutions. The strategy is to prove that $\mathbf{I}$ is real and symmetric directly from its definition (2), then let the spectral theorem do the heavy lifting: reality of the eigenvalues from the Rayleigh quotient, orthogonality of the eigenvectors from the symmetry. Along the way we harvest two physical bonuses — every principal moment is non-negative, and the three moments obey triangle inequalities — hard constraints on what tensors nature can serve up. The subtle case is degeneracy: when two eigenvalues coincide, orthogonality within their subspace is not forced but chosen, which is why any diameter of a sphere qualifies as a principal axis.

Step 1. Symmetry. From the definition, $I_{ij}=\sum_\alpha m_\alpha(r_\alpha^2\delta_{ij}-x_{\alpha,i}x_{\alpha,j})=I_{ji}$ . Why: the expression is manifestly unchanged under $i\leftrightarrow j$ — symmetry is not an assumption but a consequence of how the tensor was built.

Step 2. Reality of eigenvalues. For any eigenpair $\mathbf{I}\mathbf{u}=\lambda\mathbf{u}$ ,

\lambda=\frac{\mathbf{u}^\dagger\mathbf{I}\mathbf{u}}{\mathbf{u}^\dagger\mathbf{u}}\in\mathbb{R}

Why: for a real symmetric (hence Hermitian) matrix the Rayleigh quotient is real, so every principal moment is a genuine physical number — no complex moments of inertia.

Step 3. Orthogonality of eigenvectors. For two eigenpairs with $\lambda_1\neq\lambda_2$ ,

(\lambda_1-\lambda_2)\,\mathbf{u}_1\cdot\mathbf{u}_2=0

Why: symmetry gives $\mathbf u_2^{\mathsf T}\mathbf I\mathbf u_1=\mathbf u_1^{\mathsf T}\mathbf I\mathbf u_2$ ; distinct eigenvalues then force $\mathbf{u}_1\perp\mathbf{u}_2$ . Degenerate eigenvalues leave a free choice within their subspace — why any diameter of a sphere is principal.

Step 4. Positive semi-definiteness. For any axis direction $\hat{\mathbf n}$ ,

\hat{\mathbf n}^{\mathsf T}\mathbf{I}\,\hat{\mathbf n}=\sum_\alpha m_\alpha\,d_\alpha^2\ge0

Why: $d_\alpha$ is each particle's perpendicular distance from the axis — a sum of non-negative terms, so every $I_k\ge0$ and $T=\tfrac12\sum I_k\Omega_k^2\ge0$ .

Step 5. Triangle inequalities. $I_1\le I_2+I_3$ and cyclic permutations. Why: each principal moment is $I_i=\int\rho(x_j^2+x_k^2)$ ; adding two counts every coordinate at least as often as the third does. A pancake saturates the bound ( $I_3=I_1+I_2$ , the perpendicular-axis theorem); a rod pins one moment near zero. One last reading. The scalar moment about any axis $\hat{\mathbf n}$ through the origin — the number for $L=I\omega$ when rotation is locked to that axis — is the quadratic form evaluated on the direction:

I_a=\hat{\mathbf{n}}^{\mathsf T}\mathbf{I}\,\hat{\mathbf{n}}=\sum_{i,j}I_{ij}\,n_i\,n_j = I_1 n_1^2+I_2 n_2^2+I_3 n_3^2 \;\;\text{(principal frame)} \qquad (9)

where: $\hat{\mathbf{n}}$ — unit vector along the rotation axis; $I_a$ — scalar moment about that axis. The extremes of $I_a$ over all directions are $I_1$ and $I_3$ : the principal moments bracket every rotational inertia the body can show. Plotting $I_a(\hat{\mathbf n})^{-1/2}$ traces the inertia ellipsoid, the geometric portrait of the tensor you will meet next.

Limiting cases

Case	Result	Meaning
Spherical top	$\mathbf{I}=I\,\mathbb{1},\quad \mathbf{L}=I\boldsymbol{\omega}$	All moments equal (sphere, cube): $\mathbf{L}\parallel\boldsymbol{\omega}$ always; every axis principal.
Symmetric top	$I_1=I_2\neq I_3$	Two moments degenerate (cylinder, frisbee); any axis in that plane is principal; governs free precession.
Asymmetric top	$I_1<I_2<I_3$	All distinct (brick, racket, phone); the intermediate axis is unstable — the tennis-racket theorem.
Planar lamina	$I_z=I_x+I_y$	Perpendicular-axis theorem fixes the out-of-plane moment; the triangle inequality is saturated.
Point mass on the axis	contributes $0$	Only off-axis mass resists spin — why $I=0$ along a thin rod's own axis.

Worked examples

1 · The tilted dumbbell — L breaks away from ω (confidence builder)

Two point masses of $m = 2.0\ \mathrm{kg}$ sit at the ends of a light rod, at $\mathbf{r}_{1,2} = \pm(0.3,\,0,\,0.4)\ \mathrm{m}$ — half-length $0.5\ \mathrm{m}$ , tilted $36.9°$ from the $z$ -axis. The assembly spins about $z$ at $\boldsymbol{\omega} = (0,0,10)\ \mathrm{rad/s}$ . Find $\mathbf{I}$ , then $\mathbf{L}$ , and the angle between $\mathbf{L}$ and $\boldsymbol{\omega}$ .

Step 1. Assemble the tensor from (2). Each mass has $r^2 = 0.3^2+0.4^2 = 0.25\ \mathrm{m}^2$ , and both contribute identically (coordinates flip sign in pairs, products don't): $I_{xx} = 4.0\,(0.25-0.09) = 0.64$ ; $I_{yy} = 4.0\times0.25 = 1.00$ ; $I_{zz} = 4.0\,(0.25-0.16) = 0.36$ ; $I_{xz} = -4.0\,(0.3)(0.4) = -0.48$ ; $I_{xy}=I_{yz}=0$ (no mass off the $xz$ -plane). In $\mathrm{kg\,m^2}$ :

\mathbf{I}=\begin{pmatrix}0.64&0&-0.48\\ 0&1.00&0\\ -0.48&0&0.36\end{pmatrix}

Step 2. Apply the map (1): $\mathbf{L} = \mathbf{I}\boldsymbol{\omega} = (10\,I_{xz},\ 0,\ 10\,I_{zz}) = \boxed{(-4.8,\ 0,\ 3.6)\ \mathrm{kg\,m^2/s}}$ , magnitude $|\mathbf{L}| = \sqrt{4.8^2+3.6^2} = 6.0$ . The nonzero product of inertia $I_{xz}$ has fed the purely- $z$ spin into an $x$ -component of $\mathbf{L}$ .

Step 3. Angle between $\mathbf{L}$ and $\boldsymbol{\omega}$ : $\cos\theta = L_z/|\mathbf{L}| = 3.6/6.0 = 0.60$ , so $\boldsymbol{\theta = 53.1°}$ . The angular momentum points nowhere near the spin axis — indeed $\mathbf{L}\cdot\hat{\mathbf{r}}_{\text{rod}} = 0$ : exactly perpendicular to the rod, as it must be for point masses on a line.

Step 4. Energy check via (5): $T = \tfrac12\,\boldsymbol{\omega}\cdot\mathbf{L} = \tfrac12(10)(3.6) = \boldsymbol{18\ \mathrm{J}}$ — only $L_z$ , the component along the spin, stores energy; the $-4.8$ transverse piece is pure wobble, dragged around a cone once per turn by bearing torque.

Step 5. Principal moments (eigenvalues of $\mathbf{I}$ ): $\{0,\ 1.0,\ 1.0\}\ \mathrm{kg\,m^2}$ — zero along the rod (all mass on that axis), $2mr^2 = 1.0$ about any perpendicular. Spin the dumbbell about a rod-perpendicular and the wobble vanishes entirely.

2 · A brick spun about its body diagonal (exam level)

A uniform brick of mass $M = 12\ \mathrm{kg}$ has edges $a\times b\times c = 0.6\times0.4\times0.2\ \mathrm{m}$ along $x, y, z$ . It spins at $\omega = 20\ \mathrm{rad/s}$ about its main body diagonal through the center of mass. Find the principal moments, the moment about the diagonal, the kinetic energy, and the angle between $\mathbf{L}$ and $\boldsymbol{\omega}$ .

Step 1. Principal moments from the slab integrals $I = \tfrac{M}{12}(\text{sum of the two transverse edges squared})$ ; here $M/12 = 1.0$ exactly, so $I_1 = 0.4^2+0.2^2 = \boldsymbol{0.20}$ , $I_2 = 0.6^2+0.2^2 = \boldsymbol{0.40}$ , $I_3 = 0.6^2+0.4^2 = \boldsymbol{0.52\ \mathrm{kg\,m^2}}$ . The face axes are principal by the brick's three reflection symmetries — no products-of-inertia integrals needed.

Step 2. Diagonal unit vector: $\hat{\mathbf n} = (0.6,0.4,0.2)/\sqrt{0.56} = (0.802,\ 0.535,\ 0.267)$ , so $n_1^2 = 9/14$ , $n_2^2 = 4/14$ , $n_3^2 = 1/14$ .

Step 3. Moment about the diagonal from (9): $I_a = 0.20\cdot\tfrac{9}{14} + 0.40\cdot\tfrac{4}{14} + 0.52\cdot\tfrac{1}{14} = \tfrac{1.8+1.6+0.52}{14} = \boxed{0.28\ \mathrm{kg\,m^2}}$ — between $I_1$ and $I_3$ , as (9) guarantees.

Step 4. Kinetic energy: $T = \tfrac12 I_a\omega^2 = \tfrac12(0.28)(400) = \boldsymbol{56\ \mathrm{J}}$ .

Step 5. Angular momentum in the principal frame: $\boldsymbol{\omega} = 20\,\hat{\mathbf n} = (16.04,\ 10.69,\ 5.35)\ \mathrm{rad/s}$ , so $\mathbf{L} = (I_1\omega_1, I_2\omega_2, I_3\omega_3) = (3.21,\ 4.28,\ 2.78)\ \mathrm{kg\,m^2/s}$ , $|\mathbf{L}| = 6.02$ . Each component of $\boldsymbol{\omega}$ is stretched by a different factor, so the direction bends: $\cos\theta = 5.6/6.02$ , giving $\boldsymbol{\theta = 21.6°}$ . Released torque-free, the brick cannot keep spinning about the diagonal — $\mathbf{L}$ stays fixed in space while the body tumbles around it, the motion Euler's equations solve.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · St Petersburg and Berlin

Euler owns rigid-body mechanics the way Newton owns point particles. His 1765 treatise Theoria motus corporum solidorum seu rigidorum introduced the moments of inertia, proved that every body — however irregular — possesses three mutually perpendicular principal axes, and wrote down the equations of rotational motion that bear his name. Half a century before matrices existed, Euler had effectively diagonalized a symmetric operator and understood the eigen-directions as physically real, spin-stable axes — much of it done functionally blind, dictating calculations of legendary accuracy. The angles that parametrize a rotated body, the equations that govern its tumble, and the theorem that every rotation has an axis are all his.

History

1765

— Euler's Theoria motus corporum solidorum introduces principal axes and moments of inertia and derives the rigid-body equations of motion.

1834

— Poinsot's Théorie nouvelle de la rotation des corps gives the geometric picture: the inertia ellipsoid rolling without slipping on the invariable plane.

1852

— Foucault builds and names the gyroscope, turning principal-axis spin stability into a laboratory instrument for seeing Earth's rotation.

1958

— Explorer 1, spun about its long (minimum-I) axis, flips into a flat spin within hours as its flexible antennas dissipate energy — only the maximum-I axis is stable for a lossy body.

1985

— Cosmonaut Vladimir Dzhanibekov watches a wing-nut flip rhythmically aboard Salyut 7 — the intermediate-axis instability, filmed in free fall.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 5 · Landau & Lifshitz, Mechanics (Vol. 1), §32–33 · Morin, Introduction to Classical Mechanics, Ch. 9 · Taylor, Classical Mechanics, Ch. 10 Key results to keep in mind: the BAC–CAB expansion yields $I_{ij}=\sum m(r^2\delta_{ij}-x_ix_j)$ ; standard moments are sphere $\tfrac25MR^2$ , rod $\tfrac1{12}M\ell^2$ , cylinder axial $\tfrac12MR^2$ and transverse $\tfrac14MR^2+\tfrac1{12}ML^2$ ; the tensor parallel-axis correction is $M(a^2\delta_{ij}-a_ia_j)$ ; and the rotational kinetic energy is $T=\tfrac12\omega^{\mathsf T}\mathbf I\omega=\tfrac12\boldsymbol\omega\cdot\mathbf L$ . Worked-example values: dumbbell $\mathbf{L}=(-4.8,0,3.6)$ , $\theta=53.1°$ , $T=18$ J; brick $I_{1,2,3}=(0.20,0.40,0.52)$ , $I_a=0.28$ , $T=56$ J, $\theta=21.6°$ .

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

\mathbf L

and

\boldsymbol\omega

point in different directions, which one does the body actually spin around?

At any instant the body rotates about $\boldsymbol\omega$ — that is the definition of angular velocity: every particle circles the instantaneous axis. The real question is which vector gets to stay fixed, and that depends on the constraints. With bearings enforcing the axis, as for the tilted dumbbell, $\boldsymbol\omega$ is held constant and $\mathbf L$ is dragged around a cone — which is why the bearings must supply torque. Released torque-free, the roles swap: $\mathbf L$ is conserved and fixed in space, while $\boldsymbol\omega$ wanders both through space and through the body — the brick of worked example 2 cannot keep spinning about its diagonal and tumbles around the fixed $\mathbf L$ instead. So the instantaneous spin axis is $\boldsymbol\omega$ ; the dynamically protected direction is $\mathbf L$ .

The dumbbell's principal moments came out as

\{0, 1.0, 1.0\}

— how can a moment of inertia be exactly zero?

Because rotational inertia counts only mass off the axis: $I = \sum_\alpha m_\alpha d_\alpha^2$ with $d_\alpha$ the perpendicular distance. Both point masses of the dumbbell sit exactly on the rod's line, so spinning about the rod moves nothing — zero kinetic energy, zero angular momentum, zero resistance. This is the degenerate edge of positive semi-definiteness: every $I_k \ge 0$ , with equality only when all the mass lies on the axis itself. It is an idealization, of course; a real rod has thickness, giving a small but nonzero moment. The limiting-cases table says the same thing: a point mass on the axis contributes nothing, which is why a thin rod's moment about its own axis is (nearly) zero while its transverse moment is $\tfrac{1}{12}M\ell^2$ .

What does a negative product of inertia like

I_{xz} = -0.48

physically mean when nothing about the body is negative?

Diagonal entries must be non-negative — they are sums of $m\,d^2$ — but off-diagonal entries are signed by construction: $I_{xz} = -\sum m\,xz$ . The dumbbell's masses sit at $\pm(0.3, 0, 0.4)$ , so $xz = +0.12$ for both, making $I_{xz}$ negative. The sign encodes which quadrants hold the mass: negative $I_{xz}$ means mass favors regions where $x$ and $z$ share a sign. And the sign has teeth — it fixes which way $\mathbf L$ tilts. Feeding in $\boldsymbol\omega = (0,0,10)$ produces $L_x = 10\,I_{xz} = -4.8$ , tipping $\mathbf L$ toward $-x$ ; a mirror-image dumbbell would flip both signs. The overall minus itself is pure convention, baked in by the BAC–CAB expansion in the derivation.

Why is spinning about the intermediate axis unstable when both the largest and smallest axes are fine?

Perturb a spin about a principal axis and watch the small transverse components of $\boldsymbol\omega$ . Euler's equations couple them in pairs, and the character of the feedback depends on where the spin axis sits in the ordering $I_1 < I_2 < I_3$ . About the largest- or smallest-moment axis the coupled pair feeds back with a restoring sign — the wobble just circulates and stays small. About the intermediate axis the signs conspire the other way and the perturbation grows exponentially: half a flip later the body spins nearly the opposite way, and the cycle repeats — Dzhanibekov's wing-nut. A conserved-quantity picture says the same: on the sphere of constant $|\mathbf L|$ , the constant-energy curves circle the $I_1$ and $I_3$ poles but cross at saddle points on the $I_2$ axis.

Explorer 1 was spun about its minimum-moment axis, which is supposed to be stable. Why did it flip?

The stability analysis assumes a perfectly rigid body, for which both extreme axes are stable. Explorer 1 was not rigid: its flexible antennas flexed as it spun, and flexing dissipates energy as heat. In torque-free flight $\mathbf L$ is strictly conserved, but kinetic energy is not — for spin about a principal axis $T = L^2/2I$ , so at fixed $L$ the energy can only drift downward, toward the largest moment. Spin about the minimum- $I$ axis is the energy maximum at given $\mathbf L$ : any internal dissipation, however slight, destabilizes it, and within hours the satellite was in a flat spin. That is why spacecraft are spun about their maximum-moment axis deliberately or actively stabilized.

If the lab-frame components of

\mathbf I

change as the body rotates, what did I actually compute when I found the tensor?

You computed its components in one frame, at one instant. The tensor transforms as $\mathbf I = \mathbf R\,\mathbf I'\,\mathbf R^{\mathsf T}$ — equation (6) — so lab-frame components are genuinely time-dependent for a spinning body: the mass distribution seen from fixed axes keeps changing. What is permanent are the body-frame components, nine constants riding along with the material, and above all the eigenvalues $I_1, I_2, I_3$ , which no rotation can touch. That is why the useful deliverable is the principal moments plus the principal axes: a CAD package reports exactly that, and flight software stores the constant body-frame tensor and rotates it as needed. It is also why Euler's equations are written in the body frame, where $\mathbf I$ sits still.

Can I use the parallel-axis theorem to jump directly between two pivots, neither of which is the center of mass?

No — equation (7) is anchored to the cm, and only to the cm. The derivation relies on $\sum m_\alpha \mathbf r'_\alpha = 0$ about the center of mass, which kills the cross terms; about any other reference point they survive, so a "direct hop" $M(d^2\delta_{ij} - d_i d_j)$ between two arbitrary pivots is simply wrong. The correct route is two steps: subtract the Steiner term to get back to $\mathbf I^{\text{cm}}$ , then add the new one for the second pivot. The one-way structure is vivid in the positive semi-definiteness of the correction: every moment is smallest about the cm and only grows as the origin moves away, whereas between two off-cm points the change can go either direction.

The transverse part of

\mathbf L

needs bearing torque to drag it around the cone — doesn't that torque pump in energy?

No, and the geometry says why. For the steadily spinning dumbbell, $\mathbf L = (-4.8, 0, 3.6)$ rotates about the $z$ -axis, so $\mathrm d\mathbf L/\mathrm dt = \boldsymbol\omega \times \mathbf L$ is horizontal — and the required torque $\boldsymbol\tau = \mathrm d\mathbf L/\mathrm dt$ is therefore perpendicular to $\boldsymbol\omega$ . Rotational power is $\boldsymbol\tau \cdot \boldsymbol\omega$ , which vanishes: the bearings redirect $\mathbf L$ without doing net work, just as a string turning a whirling stone does none. That is consistent with $T = \tfrac12\,\boldsymbol\omega\cdot\mathbf L = 18\ \mathrm J$ staying constant, fed only by $L_z$ . The transverse $-4.8$ piece costs no energy — it costs force: a sideways hammering once per revolution, which is what wears out the bearings of an unbalanced wheel.