Inertia tensor — principal axes, principal moments: Question bank

True or false — justify

L

is always parallel to

ω

for a rigid body.

False. Only along a principal axis (or when all three principal moments are equal). Off-axis the angular momentum tilts because

I

has non-zero products of inertia.

The inertia tensor is always a symmetric matrix.

True. Each off-diagonal is

- \sum m x_{i} y_{j}

, unchanged by swapping the two indices, so

I_{ij} = I_{j i}

always — this is what guarantees real eigenvalues and orthogonal axes (see Symmetric and orthogonal matrices).

Diagonalizing

I

changes the physical object.

False. It only rotates your coordinate description. The body, its mass distribution, and invariants like

Tr I

and

det I

are untouched.

If a coordinate axis is an axis of symmetry of the body, it is automatically a principal axis.

True. Symmetry forces the products of inertia involving that axis to cancel in pairs, so no diagonalization is needed for it.

Every rigid body has exactly three distinct principal axes.

False. The directions always exist, but they need not be unique: a symmetric top has one special axis and a whole plane of equivalent ones; a spherical top makes every axis principal.

Principal moments of inertia can be negative.

False. Each is

\sum m (distance from axis)^{2} \geq 0

, a sum of non-negative terms, so eigenvalues of

I

are

\geq 0

A principal moment of inertia can equal exactly zero.

True. If all mass lies on the axis (e.g. point masses on a line), the perpendicular distance is zero, giving

I = 0

about that line.

Products of inertia carry a plus sign, matching the positive moments of inertia.

False. Products are

I_{x y} = - \sum m x_{a} y_{a}

— a minus sign from the

- r (r \cdot ω)

term in the derivation. Getting the sign wrong flips the eigenvectors.

The trace

I_{xx} + I_{y y} + I_{z z}

depends on which frame you compute it in.

False. Rotations preserve the trace, so

I_{xx} + I_{y y} + I_{z z} = I_{1} + I_{2} + I_{3} = 2 \sum m r^{2}

in every frame — a great sanity check on eigenvalues.

Two masses on the x-axis have

I_{xx} = 0

but

I_{y y} = I_{z z} \neq = 0

True. Their distance from the x-axis is zero (so

I_{xx} = 0

), but they sit at

x = \pm d

away from the y- and z-axes, giving

I_{y y} = I_{z z} = 2 m d^{2}

For a cube spun about a body diagonal,

L

tilts away from

ω

False. A uniform cube is a spherical top (

I_{1} = I_{2} = I_{3}

), so

I = I 1

and

L = I ω

is parallel for every axis, diagonal included.

I

is already diagonal in your chosen axes, those axes are principal.

True — provided the off-diagonals are exactly zero. Diagonal form is the definition of a principal frame.

Spot the error

I

is diagonal, so principal axes exist only because the body is symmetric."

Wrong. Principal axes exist for every rigid body because

I

is real and symmetric (see Eigenvalues and eigenvectors); symmetry only makes them easy to spot, it isn't required.

"The off-diagonal terms are tiny, so the axes are essentially principal."

Wrong. Principal means off-diagonals are exactly zero. "Small" products still tilt

L

and still give the wrong rotational KE decomposition.

T = \frac{1}{2} (I_{xx} ω_{x}^{2} + I_{y y} ω_{y}^{2} + I_{z z} ω_{z}^{2})

in any frame."

Wrong. The clean diagonal KE formula holds only in principal axes. In a general frame

T = \frac{1}{2} ω^{T} I ω

includes cross terms like

I_{x y} ω_{x} ω_{y}

"Since

L = I ω

, doubling

ω

leaves

L

's direction unchanged."

This is actually correct —

I

is linear, so scaling

ω

scales

L

identically. The trap is thinking direction ever depends on magnitude; it only depends on the direction of

ω

relative to the principal axes.

"Because

L = I_{k} ω

along a principal axis, kinetic energy is

\frac{1}{2} I_{k} ∣ ω ∣^{2}

for any spin."

Wrong in general. That formula holds only when

ω

points along that single principal axis; for a general

ω

you must sum

\frac{1}{2} \sum_{k} I_{k} ω_{k}^{2}

"The eigenvectors of

I

point in the direction of

L

Wrong. Eigenvectors are the special

ω

directions for which

L

happens to align with

ω

; they are directions of

ω

, not of a tilted

L

The spectral theorem: real symmetric matrices have real eigenvalues and mutually orthogonal eigenvectors, and an orthogonal (rotation) matrix built from them diagonalizes it — see Symmetric and orthogonal matrices.

Why does the minus sign appear in the products of inertia?

It descends from the

- r_{a} (r_{a} \cdot ω)

term in

L_{a} = m_{a} [ω r_{a}^{2} - r_{a} (r_{a} \cdot ω)]

; the off-diagonal contributions come only from that subtracted piece.

Why is a cube a "spherical top" despite not being a sphere?

Its three face-parallel principal moments come out equal by symmetry, so

I = I 1

; equal eigenvalues mean the inertia ellipsoid is a sphere and every central axis is principal.

Why is knowing the principal frame useful for Euler's equations?

In the principal frame

I

is diagonal, so the torque–angular-momentum equations decouple into the three clean Euler equations with just

I_{1}, I_{2}, I_{3}

— far simpler than a full matrix.

Why does the parallel axis theorem not, by itself, give principal axes?

It only shifts moments/products to a parallel offset origin; principal axes are about orientation, so you still must diagonalize (or use symmetry) after shifting.

Edge cases

What are the principal moments of a single point mass at the origin?

All three are zero: with

r = 0

every

m r^{2}

term vanishes, so

I

is the zero matrix and any axis is trivially principal.

For point masses lying entirely on one line, how many principal moments are zero?

Exactly one — the moment about that line (distance zero for all masses); the two perpendicular moments are equal and non-zero.

If all three principal moments are equal, which axes are principal?

All of them.

I = I 1

commutes with everything, so

I ω = I ω

for every direction — a degenerate (spherical) case.

If exactly two principal moments are equal (symmetric top), what is special about the eigenvectors?

The two equal eigenvalues share a whole plane of eigenvectors: any orthonormal pair in that plane works, so the axes there aren't unique — only the distinct third axis is fixed.

What happens to

L

when

ω = 0

L = I \cdot 0 = 0

. No rotation means no angular momentum, regardless of how lopsided the mass distribution is.

Does a flat lamina (all mass in the

z = 0

plane) satisfy any special relation among its moments?

Yes — the perpendicular axis relation

I_{z z} = I_{xx} + I_{y y}

, because

z = 0

makes

I_{xx} + I_{y y} = \sum m (y^{2} + x^{2}) = I_{z z}

Can a non-symmetric-looking body still have three equal principal moments?

Yes. Equal moments require only the right mass distribution (like a cube), not visual sphericity; "looks unsymmetric" does not forbid a spherical inertia tensor.

Recall One-line summary of every trap

$L = I ω$ is linear; $I$ is symmetric (real eigenvalues, orthogonal axes); products carry a minus; principal means off-diagonals exactly zero; the diagonal KE formula and $L ∥ ω$ hold only in the principal frame; and degeneracies ( $I_{1} = I_{2} = I_{3}$ or one zero moment) are genuine, common cases.

Question bank — Inertia tensor — principal axes, principal moments

True or false — justify

Spot the error

Why questions

Edge cases