Principal Axes and Principal Moments of Inertia

(Topic)

Principal Axes and Principal Moments of Inertia

I

(1)

through

L = Iω.

(2)

In a general Cartesian basis, the tensor can contain off-diagonal terms:

⌊ ⌋ Ixx Ixy Ixz I = ⌈ Ixy Iyy Iyz⌉ . I I I xz yz zz

(3)

Because the inertia tensor is real and symmetric, there always exists an orthonormal basis in which it is diagonal.

The special directions of that basis are the principal axes of inertia.

A unit vector

ˆei

(4)

is a principal-axis direction if

|----------| -Iˆei =-Iiˆei.

(5)

Thus principal axes are eigenvectors of the inertia tensor.

The associated eigenvalues

|----------------| I1, I2, I3| ------------------

(6)

are the principal moments of inertia.

In the principal-axis basis,

|----------⌊----------⌋--| | I1 0 0 | |I = ⌈ 0 I 0 ⌉ .| | principal 2 | -------------0---0--I3----

(7)

This basis reveals the natural rotational geometry of a rigid body.

Figure 1. A principal axis is an eigenvector direction of the inertia tensor. Applying the inertia tensor to a principal-axis unit vector changes only its magnitude, not its direction.

1 Why principal axes matter

For arbitrary angular velocity,

L = Iω.

(8)

In a nonprincipal basis, an angular-velocity component along one coordinate direction can contribute to several components of angular momentum.

In a principal basis,

L₁	= I₁ω₁,	(9)
L₂	= I₂ω₂,	(10)
L₃	= I₃ω₃.	(11)

The components decouple algebraically.

This makes principal axes useful for:

angular momentum,
rotational kinetic energy,
free rigid body motion,
rotor balance,
gyroscope dynamics,
Euler’s equations.

2 The eigenvalue problem

Principal directions satisfy

Iˆe = Ipˆe.

(12)

Move all terms to one side:

(I − Ip1)ˆe = 0.

(13)

A nonzero eigenvector exists only when

|-----------------| det (I − Ip1) = 0.| -------------------

(14)

This is the characteristic equation of the inertia tensor.

For a three-dimensional body it is a cubic equation in

Ip.

(15)

Its three real eigenvalues are the principal moments.

3 Why the eigenvalues are real

The inertia tensor satisfies

IT = I.

(16)

A real symmetric matrix has real eigenvalues.

Therefore

|--------------| I1, I2, I3 ∈ ℝ.| ----------------

(17)

This agrees with their physical interpretation as moments of inertia.

Furthermore, because

I = ˆnTIˆn ≥ 0 nˆ

(18)

for every unit vector, the principal moments are nonnegative.

4 Orthogonality of principal axes

Suppose

Iˆe1 = I1ˆe1

(19)

and

Iˆe2 = I2ˆe2.

(20)

Take the scalar product of the first equation with

ˆe2 :

(21)

ˆeT2 Iˆe1 = I1ˆe2 ⋅ ˆe1.

(22)

Because I is symmetric,

ˆeT2Iˆe1 = ˆeT1Iˆe2 = I2ˆe1 ⋅ ˆe2.

(23)

Thus

(I1 − I2)eˆ1 ⋅ ˆe2 = 0.

(24)

I ⁄= I , 1 2

(25)

then

|----------| ˆe ⋅ ˆe = 0.| --1--2------

(26)

Eigenvectors belonging to distinct principal moments are orthogonal.

5 Diagonalization

Collect three orthonormal principal-axis unit vectors as columns of

⌊ | | | ⌋ ⌈ ⌉ Q = ˆe1 ˆe2 ˆe3 . | | |

(27)

Because the columns are orthonormal,

|----------| |QT Q = 1.| ------------

(28)

The principal-axis representation is

|--------⌊----------⌋--| | I1 0 0 | QT IQ = ⌈ 0 I 0⌉ .| | 2 | -----------0---0---I3----

(29)

Equivalently,

|------------------| |I = QIprincipalQT . | -------------------

(30)

Figure 2. An orthogonal matrix built from principal-axis eigenvectors rotates the coordinate representation into a basis where the inertia tensor is diagonal.

6 Products of inertia vanish in principal coordinates

In the principal basis,

⌊ I 0 0 ⌋ ⌈ 1 ⌉ Iprincipal = 0 I2 0 . 0 0 I3

(31)

Therefore all off-diagonal tensor components vanish:

|-------------------| I12 = I13 = I23 = 0. | ---------------------

(32)

Using the sign convention from M05-09, the corresponding geometric products of inertia also vanish:

P---=--P--=--P---=-0.| --12----13----23------

(33)

Thus principal axes can be described equivalently as coordinate axes for which the products of inertia are zero.

The eigenvector definition is more general and more systematic.

7 Principal axes through the center of mass

Principal axes can be defined about any chosen origin.

For rigid body dynamics, the most important set is usually the principal axes through the center of mass.

Using the center of mass as tensor origin:

separates translation from rotation cleanly,
minimizes parallel-axis complications,
gives the standard form used in Euler’s equations.

Unless otherwise stated, the principal axes in later rigid body dynamics articles will be taken through the center of mass.

8 Two-dimensional principal-axis problem

For a planar lamina in the xy plane,

[ ] Ixx Ixy I2D = Ixy Iyy .

(34)

The characteristic equation is

[ ] Ixx − Ip Ixy det Ixy Iyy − Ip = 0.

(35)

Thus

2 (Ixx − Ip)(Iyy − Ip) − Ixy = 0.

(36)

Expanding:

I2p − (Ixx + Iyy)Ip + IxxIyy − I2xy = 0.

(37)

Therefore the two in-plane principal moments are

|------------------∘-(----------)--------| | Ixx + Iyy Ixx − Iyy 2 | |I1,2 = ---------∓ --------- + I2xy.| -----------2---------------2-------------|

(38)

9 Principal-axis rotation angle in two dimensions

Rotate the coordinate axes through angle 𝜃.

The transformed off-diagonal tensor component is

Iyy − Ixx Ix′y′ = ---------sin 2𝜃 + Ixy cos 2𝜃. 2

(39)

A principal orientation satisfies

Ix′y′ = 0.

(40)

Therefore

|--------------------| | --2Ixy--- | |tan 2𝜃p = I − I .| ------------xx----yy--

(41)

Equivalent forms can appear with an overall sign change depending on the active or passive rotation convention.

The safest procedure is always to check the transformed off-diagonal entry or solve the eigenvector equation directly.

Figure 3. In a planar problem, rotating the coordinate axes to the principal orientation eliminates the off-diagonal inertia component. The principal directions are orthogonal.

10 Example 1: a two-dimensional inertia tensor

Consider

[ 5 − 2] I = kg m2. − 2 8

(42)

The characteristic equation is

0	= det	(43)
	= (5 − I_p)(8 − I_p) − 4	(44)
	= I_p² − 13I_p + 36.	(45)

Thus

(Ip − 4)(Ip − 9) = 0.

(46)

The principal moments are

|------------------------------| |I1 = 4kg m2, I2 = 9kg m2. | --------------------------------

(47)

11 Example 1 continued: principal directions

For

I1 = 4,

(48)

solve

[ 1 − 2][ex ] = 0. − 2 4 ey

(49)

Thus

ex = 2ey.

(50)

A normalized eigenvector is

|---------[-]--| |ˆe1 = √1-- 2 .| | 5 1 | ---------------

(51)

For

I2 = 9,

(52)

a normalized eigenvector is

|---------[---]--| | -1-- − 1 | |ˆe2 = √5-- 2 .| -----------------

(53)

Their dot product is zero:

ˆe1 ⋅ ˆe2 = 0.

(54)

12 Verification by diagonalization

Construct

[ ] -1-- 2 − 1 Q = √ -- 1 2 . 5

(55)

Then

|--------[-----]--| QT IQ = 4 0 . | | 0 9 | ------------------

(56)

This verifies both principal moments and principal directions.

13 Rotational kinetic energy in principal coordinates

In any coordinates,

1 K = --ωT Iω. 2

(57)

In principal coordinates,

|----------------------------| | 1( 2 2 2) | K = 2 I1ω 1 + I2ω 2 + I3ω 3 . ------------------------------

(58)

There are no cross terms.

This is one major practical advantage of the principal-axis basis.

14 Angular momentum in principal coordinates

Similarly,

------------------------------- |L = I ω ˆe + I ω ˆe + I ω ˆe .| ------1--1-1---2--2-2---3--3-3-|

(59)

Thus

|----------| |L = I ω .| --i----i-i-

(60)

The principal components of angular momentum are obtained by simple scalar multiplication.

15 When are angular momentum and angular velocity parallel?

Suppose

ω = ω ˆe1

(61)

lies exactly along principal axis 1.

Then

L = I(ω ˆe1) = I1ω ˆe1.

(62)

Therefore

|------| L ∥ ω.| --------

(63)

More generally, L is parallel to ω precisely when ω is an eigenvector of I, apart from degeneracies where any direction within an equal-eigenvalue eigenspace is also principal.

Figure 7. Rotation exactly about a principal axis makes angular momentum parallel to angular velocity. For a generic nonprincipal direction, the two vectors are generally not parallel.

16 Principal moments as extrema of axis moment

M05-09 showed that the scalar moment about a unit axis is

Iˆn = ˆnT Iˆn.

(64)

The unit vector must satisfy

ˆnT ˆn = 1.

(65)

Ask: for which directions is

Iˆn

(66)

stationary?

Use a Lagrange multiplier λ:

F = ˆnTIˆn − λ(ˆnT ˆn − 1) .

(67)

Differentiate with respect to the vector components:

2Iˆn − 2 λˆn = 0.

(68)

Therefore

|---------| Iˆn-=--λˆn.--

(69)

The stationary axis moments occur exactly along eigenvector directions.

Multiplying on the left by

ˆnT

(70)

gives

λ = nˆT Iˆn = Inˆ.

(71)

Thus

|-------| λ = Ip. | ---------

(72)

Principal moments are the stationary values of the scalar moment of inertia over all axis directions through the chosen origin.

17 Minimum, intermediate, and maximum principal moments

Order the principal moments as

|------------| I1 ≤ I2 ≤ I3.| --------------

(73)

Then for any unit direction,

|------------| I1 ≤ Iˆn ≤ I3.| --------------

(74)

The minimum scalar moment occurs along the minimum-principal-moment axis.

The maximum scalar moment occurs along the maximum-principal-moment axis.

The intermediate principal direction is also a stationary direction, but it is not the global minimum or maximum.

Figure 4. The principal moments are stationary values of the scalar moment of inertia over all axis directions through the chosen origin. They include the minimum and maximum axis moments.

18 Rayleigh quotient interpretation

The quantity

nT-In- R (n ) = nTn

(75)

is the Rayleigh quotient of the inertia tensor.

For a unit vector,

R (ˆn) = Iˆn.

(76)

For a symmetric tensor,

|----------------| -I1 ≤-R-(n-) ≤-I3.|

(77)

This linear-algebra result has a direct physical interpretation: no axis through the chosen origin can have a scalar moment smaller than the minimum principal moment or larger than the maximum principal moment.

19 Symmetry can reveal principal axes without solving a determinant

Geometry often identifies principal axes immediately.

If a body is symmetric about a plane through the tensor origin, then the Normal to that plane is a principal direction.

For example, symmetry under

x → − x

(78)

makes

I = I = 0. xy xz

(79)

Then

Iex = Ixxex.

(80)

Therefore

|--| -ex-

(81)

is a principal axis.

With three mutually orthogonal symmetry planes through the origin, all three coordinate axes are principal.

20 Axis of rotational symmetry

Suppose a body is invariant under rotations about the z axis.

Then no direction in the xy plane can be physically distinguished.

Therefore

|--------| -Ix =-Iy.|

(82)

The symmetry axis

ez

(83)

is principal, and every orthonormal pair in the transverse plane can serve as the other two principal axes.

Such a body has two equal principal moments.

21 Degenerate principal moments

I1 = I2,

(84)

the corresponding eigenspace is two-dimensional.

Every unit vector in that eigenspace is an eigenvector with the same eigenvalue.

Thus the individual principal axes within that subspace are not unique.

Only the subspace itself is physically distinguished.

Figure 5. When two principal moments are equal, every direction in the corresponding eigenspace is principal. An axisymmetric body therefore has a unique symmetry axis but no unique pair of transverse principal axes.

22 Spherical inertia tensor

I1 = I2 = I3 = I0,

(85)

then

|--------| |I = I01.| ----------

(86)

Every direction is a principal axis.

For every angular velocity,

L = I ω, 0

(87)

|------| |L ∥ ω | -------

(88)

for all directions.

A homogeneous sphere about its center is the standard example.

23 Example 2: uniform rectangular box

A uniform rectangular box centered at its center of mass has side lengths:

a along x,
b along y,
c along z.

M05-09 found

⌊ ⌋ b2 + c2 0 0 ICM = M--⌈ 0 a2 + c2 0 ⌉ . 12 0 0 a2 + b2

(89)

The tensor is already diagonal.

Therefore the box edges are principal directions and the principal moments are

I₁ = (b² + c²),		(90)
I₂ = (a² + c²),		(91)
I₃ = (a² + b²).		(92)

If two side lengths are equal, two principal moments become degenerate.

24 Example 3: thin disk

For a uniform thin disk of radius R centered at the origin,

1- 2 Iz = 2M R .

(93)

Any diameter has

I = I = 1-M R2. x y 4

(94)

Thus

|----------------------------------| | 1 1 | |I1 = I2 = -M R2, I3 = -M R2. | -----------4----------------2-------

(95)

The normal to the disk is a unique principal direction.

Any two perpendicular diameters can serve as the other principal axes.

25 Example 4: slender rod

For an ideal slender rod of length L along the x axis,

Ix = 0

(96)

in the zero-thickness approximation, while

I = I = 1-M L2. y z 12

(97)

Therefore

|------------------------------| | -1- 2 | |I1 = 0, I2 = I3 = 12 M L .| -------------------------------

(98)

The rod direction is the minimum-moment principal axis.

Every direction perpendicular to the rod is a principal direction with the same moment.

26 Inertia ellipsoid

The principal moments can be visualized geometrically.

One common inertia-ellipsoid convention is

|-T--------| -x-Ix-=--1.|

(99)

In principal coordinates,

I1x21 + I2x22 + I3x23 = 1.

(100)

This is an ellipsoid with semiaxes

|------------------------| |√1-- √-1- √1-- | | I1, I2, I3.| --------------------------

(101)

Its geometric axes are the principal axes of inertia.

The exact scale convention varies among texts, but the orientation information is the same.

Figure 6. In principal coordinates, the quadratic form of the inertia tensor defines an ellipsoid whose geometric axes align with the principal axes of inertia.

27 Why the inertia ellipsoid is useful

The inertia ellipsoid provides a geometric picture of anisotropy.

I1 = I2 = I3,

(102)

the ellipsoid is a sphere.

If two moments are equal, it is a spheroid.

If all three differ, it is triaxial.

Thus:

spherical inertia corresponds to complete rotational isotropy,
two equal principal moments correspond to axial symmetry,
three distinct moments correspond to a triaxial mass distribution.

28 Tensor invariants and principal moments

Coordinate rotation changes the matrix entries but not the eigenvalues.

Therefore several quantities are invariant:

|--------------------| -tr(I) =-I1-+-I2 +-I3,

(103)

|----------------| |det(I) = I1I2I3.| -----------------

(104)

The sum of pairwise products is also invariant:

|------------------| |I1I2 + I2I3 + I3I1. -------------------

(105)

These invariants provide strong checks on diagonalization calculations.

29 Example 5: invariant checks

For

[ ] I = 5 − 2 , − 2 8

(106)

the principal moments were

4, 9.

(107)

The trace is

5 + 8 = 13,

(108)

and

4 + 9 = 13.

(109)

The determinant is

(5)(8) − (− 2)2 = 36,

(110)

and

(4)(9) = 36.

(111)

Both invariants agree.

30 Rotation about a principal axis

Suppose a rigid body has

ω = ωeˆi.

(112)

Then

L = Iiωˆei.

(113)

The rotational kinetic energy is

|------------| |K = 1-Iiω2. | ------2------|

(114)

Thus rotation about a principal axis behaves instantaneously like the familiar scalar-inertia case.

This is why the scalar formula is so successful in fixed-axis problems when the rotation axis is a principal direction.

31 Rotation not aligned with a principal axis

Let

ω = ω1ˆe1 + ω2ˆe2.

(115)

Then

L = I1ω1ˆe1 + I2ω2ˆe2.

(116)

I1 ⁄= I2,

(117)

the relative weighting of the two components changes.

Therefore

|------| |L ⁄∥ ω | -------

(118)

in general.

This mismatch drives much of the interesting behavior of torque-free three-dimensional rotation.

32 Principal axes and rotor balance

For a rotor intended to spin steadily about a shaft, two geometric conditions are especially important:

the shaft should pass through the center of mass,
the shaft should align with a principal axis.

The first condition removes static mass eccentricity.

The second removes cross-coupling between the shaft direction and transverse angular-momentum components.

These ideas underlie static and dynamic balancing of rotating machinery.

33 Finding principal axes numerically

For a complicated body, the inertia tensor may be assembled numerically from:

a finite-element model,
a CAD mass-property calculation,
discrete measured masses,
a numerical volume integral.

Once I is known, compute:

eigenvalues for principal moments,
normalized eigenvectors for principal directions.

Because I is symmetric, numerical eigensolvers specialized for real symmetric matrices should be preferred.

They preserve orthogonality more reliably and are more efficient than a general nonsymmetric eigenvalue solver.

34 Example 6: numerical three-dimensional tensor

Consider

⌊ ⌋ 6 2 0 I = ⌈2 6 0⌉ kg m2. 0 0 10

(119)

The z direction is already uncoupled, so

|------------2-------------| I3-=-10-kg-m-,-----ˆe3-=-ez.-

(120)

The in-plane block is

[ ] 6 2 . 2 6

(121)

Its eigenvalues are

4, 8.

(122)

Normalized eigenvectors are

|------------------| |ˆe1 = √1-(ex − ey),| --------2-----------

(123)

and

|------------------| | -1-- | |ˆe2 = √ 2(ex + ey).| --------------------

(124)

Thus the principal moments are

|----------------------2-| -4,----8,-----10--kg-m--.-

(125)

35 A principal-axis workflow

For practical calculations:

choose the tensor origin, usually the center of mass,
write or compute the inertia tensor,
use symmetry to identify any obvious principal axes,
solve the remaining characteristic equation,
normalize the eigenvectors,
check orthogonality,
check trace and determinant invariants,
assemble the orthogonal principal-axis matrix,
verify diagonalization.

Figure 8. Principal-axis calculations are an eigenvalue problem with strong physical checks: symmetry, orthogonality, tensor invariants, and final diagonalization.

36 Bridge to Euler’s equations

In a principal-axis body frame,

I = diag (I1,I2,I3).

(126)

This greatly simplifies rotational dynamics.

However, the principal axes rotate with the rigid body.

Differentiating angular momentum in a rotating body frame therefore requires the rotating-frame derivative rule.

Combining that derivative with

L = Iω

(127)

leads to Euler’s rigid body equations:

τ₁	= I₁ω₁ + (I₃ − I₂)ω₂ω₃,	(128)
τ₂	= I₂ω₂ + (I₁ − I₃)ω₃ω₁,	(129)
τ₃	= I₃ω₃ + (I₂ − I₁)ω₁ω₂.	(130)

These equations are derived carefully in M05-11.

37 Common mistakes

Confusing a principal axis with any convenient coordinate axis.
Assuming a diagonal tensor is the same as an isotropic tensor.
Forgetting that principal axes depend on the chosen tensor origin.
Solving for eigenvalues but not computing or normalizing eigenvectors.
Assuming eigenvectors for repeated eigenvalues are unique.
Forgetting that distinct eigenvalues of a symmetric tensor have orthogonal eigenvectors.
Using the two-dimensional principal-angle formula without checking the rotation convention.
Forgetting that the principal moments are stationary axis moments.
Assuming every stationary principal moment is either the global minimum or maximum; the intermediate moment is also stationary.
Forgetting to use a unit vector in I_n = n^TIn.
Assuming L ∥ ω for arbitrary rotation in a principal basis.
Forgetting that L ∥ ω when ω itself lies along a principal direction.
Confusing symmetry axes with center of mass offsets.
Ignoring trace or determinant checks after diagonalization.
Treating body-frame principal axes as fixed in inertial space during dynamics.

38 Practice exercises

Starting from Ie = I_pe, derive the characteristic equation for principal moments.
Prove that eigenvectors belonging to distinct principal moments are orthogonal.
Show that Q^TIQ is diagonal when the columns of Q are orthonormal principal-axis eigenvectors.
Derive the two-dimensional principal moments for a symmetric 2 × 2 inertia matrix.
Derive the equation for the two-dimensional principal-axis angle by requiring the transformed product of inertia to vanish.
For I = kg m², reproduce the eigenvalues and normalized eigenvectors in this article.
Verify the trace and determinant invariants for the same tensor.
Use a Lagrange multiplier to prove that principal axes give stationary scalar moments of inertia.
Prove that I₁ ≤ I_n ≤ I₃ for any unit vector.
Explain why a plane of reflection symmetry produces a principal axis normal to the plane.
Find the principal moments of a thin disk, a homogeneous sphere, and an ideal slender rod.
For I = kg m², determine all principal axes and moments.
Show that a spherical inertia tensor makes every direction principal.
Explain physically why aligning a rotor shaft with a principal axis is desirable.
Starting from diagonal principal coordinates, explain why Euler’s equations still contain nonlinear coupling terms even though the inertia tensor is diagonal.