The Inertia Tensor and Products of Inertia

The Inertia Tensor and Products of Inertia

For rotation about one fixed axis, a scalar moment of inertia is sufficient:

(1)

In general three-dimensional rotation, however, one scalar is not enough.

The angular velocity

(2)

can point in any direction, and the angular momentum

(3)

need not be parallel to it.

The mass distribution must therefore be described by a second-order tensor:

(4)

The matrix representation of the inertia tensor in Cartesian coordinates is

(5)

For a mass distribution,

(6)

This article derives that expression, explains the products of inertia, and connects the tensor back to the familiar scalar moment of inertia.

1 Why a tensor is needed

For a point mass,

(7)

For rigid body rotation about an origin,

(8)

Therefore

(9)

Use the vector triple-product identity

(10)

Then

(11)

Thus a particle contributes

(12)

This dependence on the direction of ω is more complicated than multiplication by one scalar.

2 Derivation for a system of particles

For particles indexed by a,

(13)

Write the Cartesian components:

(14)

Using the summation convention,

(15)

where

(16)

For a continuous body,

(17)

3 Matrix form

In Cartesian coordinates,

(18)

The diagonal entries are familiar scalar moments:

Ixx = ∫ (y2 + z2) dm,		(19)
Iyy = ∫ (x2 + z2) dm,		(20)
Izz = ∫ (x2 + y2) dm.		(21)

4 Products of inertia and sign convention

The off-diagonal integrals involve coordinate products.

Define the geometric products of inertia

Pxy = ∫ xy dm,		(22)
Pxz = ∫ xz dm,		(23)
Pyz = ∫ yz dm.		(24)

With this convention, the inertia tensor is

(25)

Physics texts also commonly denote the actual tensor component itself by

(26)

with analogous definitions for I_xz and I_yz.

Therefore:

(27)

The sign convention must always be checked when comparing references.

In this article:

• P_xy means the unsigned coordinate-product integral ∫ xy dm,
• I_xy means the off-diagonal tensor component −∫ xy dm.

5 The inertia tensor is symmetric

Because

(28)

(29)

Thus

(30)

Only six independent components exist:

(31)

This symmetry will be crucial when principal axes are introduced in M05-10.

6 Angular momentum components

From

(32)

Lx	= Ixxωx + Ixyωy + Ixzωz,	(33)
Ly	= Ixyωx + Iyyωy + Iyzωz,	(34)
Lz	= Ixzωx + Iyzωy + Izzωz.	(35)

Therefore an angular velocity along one coordinate direction can generate angular momentum components along other directions if the products of inertia are nonzero.

7 When angular momentum is not parallel to angular velocity

Suppose the inertia tensor is diagonal:

(36)

Then

(37)

If

(38)

and both

(39)

the direction of L generally differs from the direction of ω.

Thus even zero products of inertia do not guarantee

(40)

for every angular velocity.

Parallelism occurs when ω lies along a principal axis or within a degenerate equal-inertia subspace.

8 Symmetry and vanishing products of inertia

Products of inertia are especially sensitive to symmetry.

Consider

(41)

If the mass distribution is symmetric under

(42)

then every mass element at (x,y,z) is paired with an equal mass element at (−x,y,z).

Their contributions to xy cancel.

Therefore

(43)

for symmetry about the yz plane.

Similarly:

• symmetry under y →−y also makes P_xy = 0,
• symmetry under x →−x or z →−z makes P_xz = 0,
• symmetry under y →−y or z →−z makes P_yz = 0.

9 Example 1: four point masses in a rectangle

Four equal point masses m lie at

(44)

with all sign combinations present.

The diagonal moments are

Ixx	= ∑ my2 = 4mb2,	(45)
Iyy	= ∑ mx2 = 4ma2,	(46)
Izz	= ∑ m(x2 + y2) = 4m(a2 + b2).	(47)

The product

(48)

contains equal positive and negative terms:

(49)

Also,

(50)

because

(51)

Thus

(52)

10 Example 2: two asymmetric point masses

Let two point masses have:

m1	= 2 kg,	r1	= (1, 2, 0) m,	(53)
m2	= 1 kg,	r2	= (−1, 1, 0) m.	(54)

Then

Ixx	= 2(22) + 1(12) = 9 kg m2,	(55)
Iyy	= 2(12) + 1((−1)2) = 3 kg m2,	(56)
Izz	= Ixx + Iyy = 12 kg m2.	(57)

The geometric product is

Pxy	= 2(1)(2) + 1(−1)(1)	(58)
	= 3 kg m2.	(59)

Therefore

(60)

The tensor is

(61)

11 Planar lamina

For a thin lamina in the xy plane,

(62)

Therefore

(63)

The tensor becomes

(64)

Also,

(65)

Thus the perpendicular-axis theorem appears naturally as a component identity of the inertia tensor:

(66)

for a planar lamina.

12 Uniform rectangular plate

A rectangular plate of mass M has side lengths a along x and b along y, centered at the origin.

By reflection symmetry,

(67)

The moments are

Ixx	= Mb2,	(68)
Iyy	= Ma2,	(69)
Izz	= M(a2 + b2).	(70)

Therefore

(71)

13 Uniform rectangular box

For a uniform rectangular box centered at the origin with side lengths:

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

symmetry makes all products of inertia vanish.

The inertia tensor is

(72)

These coordinate directions are principal axes.

The full meaning of that statement is developed in M05-10.

14 Rotated slender rod and nonzero product of inertia

Consider a slender uniform rod of length L and mass M, centered at the origin in the xy plane.

Let the rod make angle ϕ with the x axis.

Define

(73)

A point on the rod can be written

(74)

The tensor components become

Ixx	= I⊥ sin 2ϕ,	(75)
Iyy	= I⊥ cos 2ϕ,	(76)
Ixy	= −I⊥ sin ϕ cos ϕ,	(77)
Izz	= I⊥.	(78)

Thus

(79)

Unless

(80)

or another symmetry-aligned orientation is chosen, the product of inertia is nonzero.

15 Example 3: rod at forty-five degrees

Let

(81)

Then

(82)

and

(83)

Thus

(84)

The off-diagonal terms are not a defect.

They simply indicate that the chosen coordinate axes are not aligned with the rod’s principal directions.

16 Moment of inertia about an arbitrary axis

Let

(85)

be a unit vector defining an axis through the tensor origin.

If

(86)

then the rotational kinetic energy is

(87)

Therefore

(88)

Define the scalar moment about the axis:

(89)

This is the tensor form of

(90)

17 Direct proof of the arbitrary-axis formula

The squared perpendicular distance from r to the axis

(91)

is

(92)

Therefore

In

= ∫ dm.

(93)

In component form,

(94)

Thus

In	= ninj	(95)
	= niIijnj.	(96)

Hence

(97)

18 Example 4: arbitrary axis for a diagonal tensor

Suppose

(98)

Let the unit axis lie in the xy plane:

(99)

Then

In

= 2 cos 2𝜃 + 5 sin 2𝜃.

(100)

Therefore

(101)

At

(102)

the result is

(103)

At

(104)

it is

(105)

19 Rotational kinetic energy

For a system of particles,

(106)

The tensor form is

(107)

Equivalently,

(108)

This relation remains valid even when

(109)

20 Positive character of the inertia tensor

For any angular velocity,

(110)

Since kinetic energy cannot be negative,

(111)

Thus the inertia tensor is positive semidefinite.

For a genuinely three-dimensional body with nonzero mass away from every possible axis through the origin, it is positive definite.

For idealized lower-dimensional objects, one eigenvalue can be zero.

For example, an ideal slender rod has zero moment of inertia about its own longitudinal axis if its thickness is neglected.

21 Trace of the inertia tensor

Add the diagonal components:

tr(I)	= Ixx + Iyy + Izz	(112)
	= ∫ dm.	(113)

Therefore

(114)

The trace is unchanged by a rotation of coordinate axes.

It equals the sum of the principal moments and provides a useful consistency check.

22 Changing coordinate axes

Suppose coordinate components transform under an orthogonal matrix Q:

(115)

Then

(116)

and

(117)

Because

(118)

the tensor components in the rotated coordinates satisfy

(119)

The physical mass distribution has not changed.

Only its matrix representation has changed.

23 Products of inertia depend on the coordinate axes

A body can have

(120)

in one coordinate system and a nonzero product in another rotated system.

Therefore a nonzero product of inertia does not mean anything is physically wrong or unbalanced.

It indicates that the coordinate axes are not aligned with directions that diagonalize the inertia tensor.

M05-10 finds those directions systematically.

24 Tensor parallel-axis theorem

M05-05 introduced the scalar parallel-axis theorem.

The three-dimensional tensor generalization is

(121)

where:

• I_CM is the inertia tensor about the center of mass,
• O is the new origin,
• d is the vector from the new origin to the center of mass,
• 1 is the identity matrix.

In components,

(122)

25 Recovering the scalar parallel-axis theorem

Let the desired axis have unit direction

(123)

Use

(124)

Then

In,O

= In,CM + M.

(125)

The bracket is the squared perpendicular separation of the two parallel axes:

(126)

Therefore

(127)

This exactly reproduces the scalar parallel-axis theorem.

26 Example 5: tensor shift along the x direction

Suppose

(128)

and the new origin is displaced from the center of mass by distance d along the x direction.

Then

(129)

The shift matrix is

(130)

Thus

(131)

The moment about the x axis is unchanged because shifting the origin along the axis does not change perpendicular distances to that axis.

27 Example 6: tensor shift in two coordinate directions

Let

(132)

Then

(133)

The shift term is

(134)

Notice that shifting the reference origin can create nonzero off-diagonal tensor components even when the center of mass tensor is diagonal.

28 Products of inertia and balance

A nonzero product of inertia should not be confused automatically with static or dynamic imbalance.

Balance depends on both:

• the center of mass location,
• the orientation of the inertia tensor relative to the intended rotation axis.

For a rotor intended to spin about a fixed shaft, desirable design often places:

• the center of mass on the shaft,
• a principal axis along the shaft.

The second requirement concerns products of inertia and principal directions.

29 Connection to principal axes

Because the inertia tensor is real and symmetric, it has orthogonal eigenvectors.

Those eigenvectors are the principal axes.

In coordinates aligned with them,

(135)

The eigenvalues

(136)

are the principal moments of inertia.

For rotation exactly about a principal axis,

(137)

M05-10 develops this eigenvalue problem in detail.

30 Common mistakes

1. Treating moment of inertia as one scalar in arbitrary three-dimensional rotation.
2. Assuming L is always parallel to ω.
3. Forgetting the minus sign between P_xy = ∫ xy dm and the tensor component I_xy = −P_xy.
4. Comparing formulas from different references without checking their product-of-inertia sign convention.
5. Assuming all products of inertia are automatically zero.
6. Failing to use symmetry to eliminate odd coordinate-product integrals.
7. Assuming a diagonal tensor guarantees L ∥ ω for every ω.
8. Confusing a diagonal coordinate representation with a spherical inertia tensor having equal diagonal entries.
9. Using I_n without normalizing the axis vector.
10. Forgetting that inertia-tensor components depend on both the origin and orientation of the coordinate axes.
11. Applying the scalar parallel-axis theorem independently to matrix entries without using the tensor shift formula.
12. Forgetting that an origin shift can create off-diagonal terms.
13. Treating nonzero products of inertia as proof of center of mass offset.
14. Confusing the inertia tensor with the stress tensor or area moment tensor.
15. Forgetting that the inertia tensor is symmetric.

31 Practice exercises

1. Starting from L = r × m(ω × r), derive the particle inertia tensor.
2. Extend the particle result to a discrete system and then to a continuous body.
3. Write the inertia tensor explicitly in Cartesian coordinates and identify all three products of inertia.
4. Explain the distinction between P_xy = ∫ xy dm and the tensor component I_xy used in this article.
5. Prove that the inertia tensor is symmetric.
6. Show that symmetry under x →−x forces P_xy = P_xz = 0.
7. Compute the inertia tensor of four equal masses at (±a,±b, 0).
8. Compute the tensor of the two-mass asymmetric example in this article and verify I_zz = I_xx + I_yy.
9. Derive the inertia tensor of a slender rod lying at angle ϕ in the xy plane.
10. Evaluate the rotated-rod tensor at ϕ = 45^∘.
11. Prove that the scalar moment about an arbitrary unit axis is I_n = n^T In.
12. Derive K = ω^T Iω.
13. Show that tr(I) = 2 ∫ r² dm.
14. Derive the tensor parallel-axis theorem and recover the scalar theorem from it.
15. Explain why principal axes correspond to eigenvectors of the inertia tensor.

32 Summary

The angular momentum of a rigid body rotating about a chosen origin is

(138)

The inertia tensor is

(139)

Its Cartesian form is

(140)

The tensor is symmetric:

(141)

The scalar moment about an arbitrary unit axis is

(142)

Rotational kinetic energy is

(143)

Under a coordinate rotation,

(144)

Under an origin shift from the center of mass,

(145)

M05-10 uses the symmetry of I to identify principal axes and principal moments of inertia.

References

References