Lesson 5 · 35 min
Products of Inertia
A moment of inertia says how far the mass is from one axis. A product of inertia says whether the mass leans toward one diagonal of a coordinate plane, and in 3D that lean is what makes a spinning body wobble on its bearings.
Learning objectives
- Define the products of inertia \(I_{xy}\), \(I_{yz}\), \(I_{zx}\) and compute them for particles and simple bodies.
- Explain why a product of inertia can be positive, negative or zero.
- Use planes of symmetry to recognize zero products without calculating.
- Apply the parallel-axis theorem for products, \(I_{xy} = \bar I_{x'y'} + m\,\bar x\,\bar y\).
Definition
For a body described in axes \(x, y, z\) through a point \(O\), the three products of inertia are
Products of inertia
\[ I_{xy} = \int xy\,dm, \qquad I_{yz} = \int yz\,dm, \qquad I_{zx} = \int zx\,dm \]For particles, \(I_{xy} = \sum m\,x\,y\), and so on. The order of the letters does not matter: \(I_{yx} = I_{xy}\). Units: kg·m².
A moment of inertia adds up squares, so it is always positive. A product multiplies two signed coordinates, so each element's contribution takes the sign of its quadrant. In the \(xy\)-plane, mass in the first and third quadrants makes \(I_{xy}\) positive; mass in the second and fourth makes it negative. A product of inertia therefore measures an imbalance: whether more mass leans toward one diagonal of the \(xy\)-plane than toward the other.
Symmetry makes products vanish
Suppose a body is symmetric with respect to the \(yz\)-plane: for every element at \((x, y, z)\) there is an equal one at \((-x, y, z)\). In \(\int xy\,dm\) and \(\int zx\,dm\) the pair contributes \(xy - xy = 0\) and \(zx - zx = 0\). So
Symmetry rule
If a body has a plane of symmetry, the two products of inertia that involve the coordinate perpendicular to that plane are zero.
\[ \text{symmetric about the } yz\text{-plane} \ \Rightarrow\ I_{xy} = I_{zx} = 0, \qquad \text{about the } xz\text{-plane} \ \Rightarrow\ I_{xy} = I_{yz} = 0 \]A body of revolution about the \(z\)-axis (cylinder, disk, cone, sphere) has all three products zero for any \(x\) and \(y\) through the axis.
The rule needs the symmetry plane to be a coordinate plane of the axes you are using. The same rectangular plate has zero \(I_{xy}\) about axes through its center, and a nonzero \(I_{xy}\) about axes through a corner.
The parallel-axis theorem for products
Let \(G\) be at \((\bar x, \bar y, \bar z)\) in the \(xyz\) axes, and let \(x', y', z'\) be parallel axes through \(G\). With \(x = \bar x + x'\) and \(y = \bar y + y'\),
\[ I_{xy} = \int (\bar x + x')(\bar y + y')\,dm = \int x'y'\,dm + \bar x \int y'\,dm + \bar y \int x'\,dm + \bar x\,\bar y \int dm, \]and the two middle integrals vanish because \(G\) is the center of mass. The same happens for the other two products:
Parallel-axis theorem for products of inertia
\[ I_{xy} = \bar I_{x'y'} + m\,\bar x\,\bar y, \qquad I_{yz} = \bar I_{y'z'} + m\,\bar y\,\bar z, \qquad I_{zx} = \bar I_{z'x'} + m\,\bar z\,\bar x \]\(\bar x, \bar y, \bar z\) are the signed coordinates of \(G\): the transfer term can be negative.
Example 5.1 — A plate about its corner
A thin \(3\ \text{kg}\) rectangular plate lies in the \(xy\)-plane with one corner at \(O\), sides \(a = 0.4\ \text{m}\) along \(+x\) and \(b = 0.3\ \text{m}\) along \(+y\). Find \(I_{xy}\), \(I_{yz}\) and \(I_{zx}\) about \(O\). What changes if the plate lies along \(-x\) instead?
Show solution
About its own center the plate is symmetric in \(x'\) and in \(y'\), so \(\bar I_{x'y'} = 0\). Its center is at \((0.2,\ 0.15,\ 0)\):
\[ I_{xy} = 0 + m\,\bar x\,\bar y = 3(0.2)(0.15) = 0.0900\ \text{kg·m}^2 \qquad \left(= \tfrac14 m a b\right) \]Every element has \(z = 0\), so \(I_{yz} = I_{zx} = 0\): a flat body in the \(xy\)-plane has only \(I_{xy}\).
Along \(-x\), the center is at \((-0.2,\ 0.15,\ 0)\) and \(I_{xy} = -0.0900\ \text{kg·m}^2\): same size, opposite sign, because the plate is now in the second quadrant.
Example 5.2 — A rod bent in three dimensions
A slender rod with a mass of \(2\ \text{kg}\) per metre is bent into three straight segments along the coordinate directions: \(OA\) from \(O\) to \(A(0.4,\ 0,\ 0)\), \(AB\) to \(B(0.4,\ 0.3,\ 0)\) and \(BC\) to \(C(0.4,\ 0.3,\ 0.2)\) (metres). Find its three products of inertia about the \(x, y, z\) axes.
Show solution
Each segment is a slender rod parallel to a coordinate axis, so its points differ from its own center in one coordinate only. Its centroidal products \(\bar I_{x'y'}, \bar I_{y'z'}, \bar I_{z'x'}\) are therefore zero, and only the transfer terms remain:
| Segment | \(m\) | \((\bar x, \bar y, \bar z)\) | \(m\bar x\bar y\) | \(m\bar y\bar z\) | \(m\bar z\bar x\) |
|---|---|---|---|---|---|
| \(OA\) | 0.8 | \((0.2,\ 0,\ 0)\) | 0 | 0 | 0 |
| \(AB\) | 0.6 | \((0.4,\ 0.15,\ 0)\) | 0.036 | 0 | 0 |
| \(BC\) | 0.4 | \((0.4,\ 0.3,\ 0.1)\) | 0.048 | 0.012 | 0.016 |
| Rod | 1.8 | 0.084 | 0.012 | 0.016 |
So \(I_{xy} = 0.0840\), \(I_{yz} = 0.0120\) and \(I_{zx} = 0.0160\ \text{kg·m}^2\). The rod reappears in Lesson 6, spinning.
Check your understanding
Key takeaways
- \(I_{xy} = \int xy\,dm\), \(I_{yz} = \int yz\,dm\), \(I_{zx} = \int zx\,dm\): positive, negative or zero.
- A plane of symmetry makes the two products involving its normal coordinate zero; a body of revolution has all products zero about axes through its axis.
- \(I_{xy} = \bar I_{x'y'} + m\,\bar x\,\bar y\) (and cyclic), with the signed coordinates of \(G\).
- A flat body in the \(xy\)-plane has \(I_{yz} = I_{zx} = 0\); a segment parallel to an axis has zero centroidal products.
- Next: Lesson 6 assembles moments and products into the inertia tensor and uses it for angular momentum.