Lesson 8 · 40 min

Principal Axes and Principal Moments

Through every point of every body there are three perpendicular axes about which all the products of inertia vanish. About these principal axes the inertia tensor is diagonal, a body spinning about one of them needs no bearing moment, and the equations of rigid-body motion take their simplest form.

Learning objectives

What makes an axis principal

An axis through \(O\) with unit vector \(\uvec\) is a principal axis if spinning the body about it gives an angular momentum along the same axis:

\[ \Imat_O\,\uvec = I\,\uvec. \]

In a frame made of three principal axes, every product of inertia is zero and

\[ \Imat = \begin{bmatrix} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{bmatrix}, \qquad I_1 \ge I_2 \ge I_3. \]

The principal moments of inertia \(I_1\) and \(I_3\) are the largest and smallest moments of inertia about any axis through \(O\). Because \(\Imat\) is real and symmetric, three perpendicular principal axes always exist, at every point of every body.

Principal axes of a plane body

When \(z\) is already principal (\(I_{xz} = I_{yz} = 0\), as for a thin plate in the \(xy\)-plane or a body symmetric about that plane), the other two principal axes lie in the \(xy\)-plane. Setting \(I_{x'y'} = 0\) in the plane transformation of Lesson 7 gives the angle:

Principal axes in the plane

\[ \tan 2\theta_p = \frac{2I_{xy}}{I_{yy} - I_{xx}} \] \[ I_{\max,\min} = \frac{I_{xx} + I_{yy}}{2} \pm \sqrt{\left(\frac{I_{xx} - I_{yy}}{2}\right)^2 + I_{xy}^2} \]

The two solutions for \(\theta_p\) are \(90^\circ\) apart. To tell which axis carries \(I_{\max}\), put one \(\theta_p\) back into the formula for \(I_{x'x'}\).

Mohr's circle for inertia

Lesson 7 showed that \(I_{x'x'}\) and \(I_{x'y'}\) are a centre plus a rotating vector in \(2\theta\). Plot the point \(X' = (I_{x'x'},\ I_{x'y'})\): as the axes turn through \(\theta\), it moves around a circle through \(2\theta\) in the same direction.

Figure 8.1 Mohr's circle for the L-plate of Example 7.3. Turn the axes with the slider, or drag the point \(\colX{X'}\) around the circle: it moves through \(2\theta\) while the axes \(\colX{x'}\), \(\colY{y'}\) on the plate turn through \(\theta\). Where the circle crosses the horizontal axis, the product is zero and \(x'\) is a principal axis (dashed on the plate). Moments in units of \(0.01\ \text{kg·m}^2\).

Example 8.1 — Principal axes of the L-plate

For the L-plate of Example 7.3 (\(I_{xx} = 0.020\), \(I_{yy} = 0.044\), \(I_{xy} = 0.012\ \text{kg·m}^2\) about the corner \(O\)), find the principal axes and principal moments about \(O\).

Show solution
\[ \tan 2\theta_p = \frac{2(0.012)}{0.044 - 0.020} = 1 \quad\Rightarrow\quad \theta_p = 22.5^\circ \ \text{or}\ 112.5^\circ \] \[ I_{\max,\min} = 0.032 \pm \sqrt{(-0.012)^2 + 0.012^2} = 0.032 \pm 0.01697 = 0.04897,\ 0.01503\ \text{kg·m}^2 \]

At \(\theta = 22.5^\circ\): \(I_{x'x'} = 0.032 + (-0.012)(0.7071) - 0.012(0.7071) = 0.01503\), so the axis at \(22.5^\circ\) from \(x\) carries \(I_{\min}\) and the axis at \(112.5^\circ\) carries \(I_{\max}\). The minimum axis runs roughly along the long leg: most of the mass lies close to it. \(z\) is the third principal axis, with \(I_{zz} = 0.064\ \text{kg·m}^2 = I_{\max} + I_{\min}\) for a thin plate.

Principal axes in three dimensions

In general all three products are nonzero, and the condition \(\Imat\uvec = I\uvec\) is an eigenvalue problem:

The principal-axis eigenproblem

\[ \begin{bmatrix} I_{xx} - I & -I_{xy} & -I_{xz} \\ -I_{xy} & I_{yy} - I & -I_{yz} \\ -I_{xz} & -I_{yz} & I_{zz} - I \end{bmatrix} \begin{Bmatrix} u_x \\ u_y \\ u_z \end{Bmatrix} = \begin{Bmatrix} 0 \\ 0 \\ 0 \end{Bmatrix} \]

A nonzero \(\uvec\) exists only if the determinant vanishes. Expanded, that is a cubic whose three real roots are the principal moments:

\[ I^3 - J_1 I^2 + J_2 I - J_3 = 0 \]

where \(J_1 = I_{xx} + I_{yy} + I_{zz}\) (the trace), \(J_2\) is the sum of the three \(2\times2\) principal minors of \(\Imat\), and \(J_3 = \det\Imat\). The three \(J\) values are the same in every frame.

For each root, substitute it back and solve two of the three equations for the ratios \(u_x : u_y : u_z\), then normalise. The three axes are perpendicular to each other.

Example 8.2 — Solving the cubic

About \(O\), a body has \(I_{xx} = 0.4\), \(I_{yy} = 0.6\), \(I_{zz} = 0.8\), \(I_{xy} = 0.1\), \(I_{yz} = 0\), \(I_{zx} = 0.2\ \text{kg·m}^2\). Find the principal moments and the axis of the largest one.

Show solution
\[ \Imat_O = \begin{bmatrix} 0.4 & -0.1 & -0.2 \\ -0.1 & 0.6 & 0 \\ -0.2 & 0 & 0.8 \end{bmatrix} \]

The invariants:

\[ J_1 = 1.8, \quad J_2 = (0.24 - 0.01) + (0.48 - 0) + (0.32 - 0.04) = 0.99, \quad J_3 = \det\Imat_O = 0.160 \] \[ I^3 - 1.8\,I^2 + 0.99\,I - 0.160 = 0 \quad\Rightarrow\quad I_1 = 0.8882, \quad I_2 = 0.6223, \quad I_3 = 0.2895\ \text{kg·m}^2 \]

(Solve the cubic numerically, or with a calculator's polynomial solver. Check: the roots add to \(J_1 = 1.8\) and multiply to \(J_3 = 0.160\).)

Axis of \(I_1\). The second and third rows of \((\Imat - I_1\mathbf{1})\uvec = \mathbf{0}\):

\[ -0.1\,u_x + (0.6 - 0.8882)\,u_y = 0 \ \Rightarrow\ u_y = -0.3470\,u_x, \qquad -0.2\,u_x + (0.8 - 0.8882)\,u_z = 0 \ \Rightarrow\ u_z = -2.268\,u_x \]

Normalising \((1,\ -0.3470,\ -2.268)\): \(\uvec_1 = \pm(0.3996,\ -0.1387,\ -0.9061)\).

Letting software do it

In practice, principal axes come from a numerical eigen-solver. Use the routine for symmetric matrices: it returns real values and perpendicular unit vectors.

Python (NumPy)

import numpy as np
I = np.array([[ 0.4, -0.1, -0.2],
              [-0.1,  0.6,  0.0],
              [-0.2,  0.0,  0.8]])
vals, vecs = np.linalg.eigh(I)   # ascending order
print(vals)        # principal moments
print(vecs[:, -1]) # axis of the largest

MATLAB

I = [ 0.4 -0.1 -0.2;
     -0.1  0.6  0.0;
     -0.2  0.0  0.8];
[V, L] = eig(I);   % columns of V are the axes
diag(L)            % principal moments, ascending
V(:, end)          % axis of the largest
Figure 8.2 Principal axes about \(O\) or about the center of mass \(G\), and the inertia ellipsoid: along any axis, its radius is proportional to \(1/\sqrt{I}\) for that axis, so it is longest along the minimum axis. Its three symmetry axes are the principal axes. Compare the axes about \(O\) and about \(G\): moving the point changes the directions as well as the values.

Example 8.3 — The bent rod about its center of mass

Find the principal moments and axes of the bent rod of Example 5.2 about its center of mass.

Show solution

From Example 5.2, \(m = 1.8\ \text{kg}\) and \(G = (0.3111,\ 0.1167,\ 0.0222)\ \text{m}\). The parallel-axis theorem in matrix form (Lesson 6) moves the tensor from \(O\) to \(G\):

\[ \Imat_G = \Imat_O - m\begin{bmatrix} \bar y^2 + \bar z^2 & -\bar x\bar y & -\bar x\bar z \\ -\bar x\bar y & \bar z^2 + \bar x^2 & -\bar y\bar z \\ -\bar x\bar z & -\bar y\bar z & \bar x^2 + \bar y^2 \end{bmatrix} = \begin{bmatrix} 0.03394 & -0.01867 & -0.00356 \\ -0.01867 & 0.03289 & -0.00733 \\ -0.00356 & -0.00733 & 0.05794 \end{bmatrix}\ \text{kg·m}^2 \]

A symmetric eigen-solver gives

\[ I_1 = 0.06009, \quad I_2 = 0.05130, \quad I_3 = 0.01339\ \text{kg·m}^2 \] \[ \uvec_1 = (0.104,\ -0.325,\ 0.940), \quad \uvec_2 = (0.727,\ -0.620,\ -0.295), \quad \uvec_3 = (0.678,\ 0.714,\ 0.172) \]

The minimum axis \(\uvec_3\) runs roughly from \(O\) toward \(B\), the direction along which most of the rod lies. Select the bent rod in Figure 8.2 to see these axes.

Principal axes from symmetry

Lesson 5 showed that a plane of symmetry makes two products vanish. That is enough to spot principal axes without any calculation:

Why principal axes matter

Check your understanding

Key takeaways