Lesson 7 · 30 min

Moment of Inertia About Any Axis

A crankshaft turns about its bearing line, a satellite tumbles about whatever axis its thrusters leave it, and an aircraft's stability axes tilt with its angle of attack. The inertia tensor about one set of axes already contains the moment of inertia about every other axis through the same point.

Learning objectives

An axis in any direction

Let an axis \(Oa\) pass through \(O\) with unit vector \(\uvec = u_x\ihat + u_y\jhat + u_z\khat\). The perpendicular distance from the axis to a mass element at \(\rvec\) is \(|\uvec \times \rvec|\), so

\[ I_{Oa} = \int |\uvec \times \rvec|^2\,dm. \]
Expanding \(|\uvec \times \rvec|^2\)

Using \(|\uvec \times \rvec|^2 = |\rvec|^2 - (\uvec\cdot\rvec)^2\) and \(u_x^2 + u_y^2 + u_z^2 = 1\):

\[ |\uvec \times \rvec|^2 = (y^2 + z^2)u_x^2 + (z^2 + x^2)u_y^2 + (x^2 + y^2)u_z^2 - 2xy\,u_xu_y - 2yz\,u_yu_z - 2zx\,u_zu_x. \]

Integrating term by term gives the moments and products of inertia about \(x, y, z\).

Moment of inertia about an axis through \(O\)

\[ I_{Oa} = I_{xx}u_x^2 + I_{yy}u_y^2 + I_{zz}u_z^2 - 2I_{xy}\,u_xu_y - 2I_{yz}\,u_yu_z - 2I_{zx}\,u_zu_x \] \[ I_{Oa} = \uvec^\mathsf{T}\,\Imat_O\,\uvec = \uvec\cdot(\Imat_O\,\uvec) \]

The components of \(\uvec\) are its direction cosines. The axis must pass through the point \(O\) about which \(\Imat_O\) was found; for a parallel axis elsewhere, use the parallel-axis theorem afterwards.

Figure 7.1 An axis through \(O\), steered by its azimuth \(\phi\) (from \(x\), in the \(xy\)-plane) and elevation \(\lambda\) (above the \(xy\)-plane). The readouts split \(I_{Oa} = \uvec^\mathsf{T}\Imat_O\uvec\) into its moment terms and its product terms. Look for the directions that make \(I_{Oa}\) largest and smallest: Lesson 8 finds them directly.

Example 7.1 — A box about its diagonal

A \(12\ \text{kg}\) solid block measures \(a = 0.3\), \(b = 0.2\), \(c = 0.1\ \text{m}\) along \(x, y, z\). Find its moment of inertia about the body diagonal through its center \(G\).

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About its centroidal axes the block has no products, and

\[ \bar I_{xx} = \tfrac{12}{12}(0.2^2 + 0.1^2) = 0.050, \quad \bar I_{yy} = \tfrac{12}{12}(0.3^2 + 0.1^2) = 0.100, \quad \bar I_{zz} = \tfrac{12}{12}(0.3^2 + 0.2^2) = 0.130\ \text{kg·m}^2. \]

The diagonal runs along \((a, b, c)\), of length \(\sqrt{0.14} = 0.3742\ \text{m}\), so \(\uvec = (0.8018,\ 0.5345,\ 0.2673)\):

\[ I_{Ga} = 0.050(0.6429) + 0.100(0.2857) + 0.130(0.0714) = 0.0700\ \text{kg·m}^2 \]

In closed form, \(I_{Ga} = \dfrac{m}{6}\,\dfrac{a^2b^2 + b^2c^2 + c^2a^2}{a^2 + b^2 + c^2}\). The same line also passes through two opposite corners. Because it passes through \(G\), the moment of inertia about the corner-to-corner diagonal is the same \(0.0700\ \text{kg·m}^2\): try the box with a corner at \(O\) in Figure 7.1.

Example 7.2 — The bent rod about \(OC\)

For the bent rod of Example 5.2, find the moment of inertia about the line \(OC\) from the origin to the free end \(C(0.4,\ 0.3,\ 0.2)\ \text{m}\).

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\(|OC| = \sqrt{0.29} = 0.5385\ \text{m}\), so \(\uvec = (0.7428,\ 0.5571,\ 0.3714)\). With the tensor about \(O\) from Example 5.2:

\[ \begin{aligned} I_{OC} &= 0.05933(0.5517) + 0.2080(0.3103) + 0.2567(0.1379) \\ &\quad - 2(0.0840)(0.4138) - 2(0.0120)(0.2069) - 2(0.0160)(0.2759) \\ &= 0.1327 - 0.0833 = 0.0494\ \text{kg·m}^2 \end{aligned} \]

The products remove almost two thirds of the moment terms: most of the rod lies close to the line \(OC\).

Rotating the axes

The formula \(I_{Oa} = \uvec^\mathsf{T}\Imat\,\uvec\) gives one diagonal entry of the tensor in a new frame. The products in the new frame follow the same pattern: with new unit vectors \(\uvec_{x'}\) and \(\uvec_{y'}\),

\[ I_{x'x'} = \uvec_{x'}^\mathsf{T}\Imat\,\uvec_{x'}, \qquad I_{x'y'} = -\,\uvec_{x'}^\mathsf{T}\Imat\,\uvec_{y'}. \]

Stack the new unit vectors as the rows of a rotation matrix \(R\) and all nine entries come at once:

Inertia tensor in rotated axes

\[ \Imat' = R\,\Imat\,R^\mathsf{T}, \qquad R = \begin{bmatrix} \uvec_{x'}^\mathsf{T} \\ \uvec_{y'}^\mathsf{T} \\ \uvec_{z'}^\mathsf{T} \end{bmatrix} \]

Both frames share the same origin. The trace \(I_{xx} + I_{yy} + I_{zz}\) is the same in every frame (it equals \(2\int r^2\,dm\)).

Rotation about one axis

The most common case turns the \(x\)- and \(y\)-axes through an angle \(\theta\) about \(z\) (counterclockwise seen from \(+z\)). Then \(\uvec_{x'} = (\cos\theta,\ \sin\theta,\ 0)\), \(\uvec_{y'} = (-\sin\theta,\ \cos\theta,\ 0)\), and

Plane transformation (rotation \(\theta\) about \(z\))

\[ I_{x'x'} = \frac{I_{xx} + I_{yy}}{2} + \frac{I_{xx} - I_{yy}}{2}\cos 2\theta - I_{xy}\sin 2\theta \] \[ I_{y'y'} = \frac{I_{xx} + I_{yy}}{2} - \frac{I_{xx} - I_{yy}}{2}\cos 2\theta + I_{xy}\sin 2\theta \] \[ I_{x'y'} = \frac{I_{xx} - I_{yy}}{2}\sin 2\theta + I_{xy}\cos 2\theta \]

\(I_{z'z'} = I_{zz}\), and \(I_{x'x'} + I_{y'y'} = I_{xx} + I_{yy}\) for every \(\theta\). These are the same equations as for stress transformation and for second moments of area.

Figure 7.2 A thin L-shaped plate (\(20\ \text{kg/m}^2\)) in the \(xy\)-plane, with axes \(\colX{x'}\), \(\colY{y'}\) turned through \(\theta\) about \(z\) at the corner \(O\). The plot shows \(\colX{I_{x'x'}}\), \(\colY{I_{y'y'}}\) and \(I_{x'y'}\) over half a turn: every curve repeats after \(180^\circ\), and the two moments always add to the same total.

Example 7.3 — The L-plate, rotated

The thin L-shaped plate of Figure 7.2 is made of a \(0.8\ \text{kg}\) rectangle covering \(0 \le x \le 0.4\), \(0 \le y \le 0.1\ \text{m}\) and a \(0.4\ \text{kg}\) rectangle covering \(0 \le x \le 0.1\), \(0.1 \le y \le 0.3\ \text{m}\). Find \(I_{xx}\), \(I_{yy}\), \(I_{xy}\) about \(O\), then the moments and product about axes turned \(30^\circ\) about \(z\).

Show solution

For a thin plate in the \(xy\)-plane, \(I_{xx} = \int y^2\,dm\) and \(I_{yy} = \int x^2\,dm\). A rectangle whose sides run from \(0\) to \(a\) has \(\int x^2\,dm = \tfrac13 ma^2\), and \(\int xy\,dm = m\bar x\bar y\):

Parts about \(O\) (kg·m²)
Part\(\int y^2\,dm\)\(\int x^2\,dm\)\(\int xy\,dm\)
1: \(0.8\ \text{kg}\)\(\tfrac13(0.8)(0.1)^2 = 0.00267\)\(\tfrac13(0.8)(0.4)^2 = 0.04267\)\(0.8(0.2)(0.05) = 0.008\)
2: \(0.4\ \text{kg}\)\(0.4\,\tfrac{0.3^3 - 0.1^3}{3(0.2)} = 0.01733\)\(\tfrac13(0.4)(0.1)^2 = 0.00133\)\(0.4(0.05)(0.2) = 0.004\)
Total\(I_{xx} = 0.020\)\(I_{yy} = 0.044\)\(I_{xy} = 0.012\)

With \(\theta = 30^\circ\), \(\cos 2\theta = 0.5\) and \(\sin 2\theta = 0.8660\):

\[ I_{x'x'} = 0.032 + (-0.012)(0.5) - 0.012(0.8660) = 0.01561\ \text{kg·m}^2 \] \[ I_{y'y'} = 0.032 - (-0.012)(0.5) + 0.012(0.8660) = 0.04839\ \text{kg·m}^2 \] \[ I_{x'y'} = (-0.012)(0.8660) + 0.012(0.5) = -0.00439\ \text{kg·m}^2 \]

Check: \(I_{x'x'} + I_{y'y'} = 0.064 = I_{xx} + I_{yy}\). The product changed sign between \(0^\circ\) and \(30^\circ\), so somewhere in between it is zero: that angle is found in Lesson 8.

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Key takeaways