$$\require{cancel}$$

# 13.4: Inertia Tensor

The square bracket term in $$(13.3.9)$$ is called the moment of inertia tensor, $$\mathbf{I}$$, which is usually referred to as the inertia tensor

$I_{ij} \equiv \sum^{N}_{\alpha} m_{\alpha} \left[ \delta_{ij} \left( \sum^3_k x^2_{\alpha , k} \right) − x_{\alpha , i} x_{\alpha , j} \right] \label{13.12}$

In most cases it is more useful to express the components of the inertia tensor in an integral form over the mass distribution rather than a summation for $$N$$ discrete bodies. That is,

$I_{ij} = \int\rho (\mathbf{r}^{\prime} ) \left( \delta_{ij} \left( \sum^3_k x^2_{k} \right) − x_{i} x_{ j} \right) dV$

The inertia tensor is easier to understand when written in cartesian coordinates $$\mathbf{r}^{\prime}_{\alpha} = (x_{\alpha}, y_{\alpha}, z_{\alpha})$$ rather than in the form $$\mathbf{r}^{\prime}_{\alpha} = (x_{\alpha ,1}, x_{\alpha ,2}, x_{\alpha ,3})$$. Then, the diagonal moments of inertia of the inertia tensor are

\begin{align} I_{x x} & \equiv \sum_{\alpha}^{N} m_{\alpha}\left[x_{\alpha}^{2}+y_{\alpha}^{2}+z_{\alpha}^{2}-x_{\alpha}^{2}\right]=\sum_{\alpha}^{N} m_{\alpha}\left[y_{\alpha}^{2}+z_{\alpha}^{2}\right] \\[4pt] \notag I_{y y} & \equiv \sum_{\alpha}^{N} m_{\alpha}\left[x_{\alpha}^{2}+y_{\alpha}^{2}+z_{\alpha}^{2}-y_{\alpha}^{2}\right]=\sum_{\alpha}^{N} m_{\alpha}\left[x_{\alpha}^{2}+z_{\alpha}^{2}\right] \\[4pt] I_{z z} & \equiv \sum_{\alpha}^{N} m_{\alpha}\left[x_{\alpha}^{2}+y_{\alpha}^{2}+z_{\alpha}^{2}-z_{\alpha}^{2}\right]=\sum_{\alpha}^{N} m_{\alpha}\left[x_{\alpha}^{2}+y_{\alpha}^{2}\right] \notag \end{align}

while the off-diagonal products of inertia are

\begin{align} I_{yx} & = I_{xy} \equiv - \sum^N_{\alpha} m_{\alpha} [x_{\alpha} y_{\alpha}] \\[4pt] \notag I_{zx} & = I_{xz} \equiv - \sum^N_{\alpha} m_{\alpha} [x_{\alpha} z_{\alpha}] \\[4pt] \notag I_{zy} & = I_{yz} \equiv - \sum^N_{\alpha} m_{\alpha} [y_{\alpha} z_{\alpha}] \end{align}

Note that the products of inertia are symmetric in that

$I_{ij} = I_{ji}$

The above notation for the inertia tensor allows the angular momentum \ref{13.12} to be written as

$L_i = \sum^3_j I_{ij} \omega_j$

Expanded in cartesian coordinates

\begin{align} L_x & = I_{xx} \omega_x + I_{xy} \omega_y + I_{xz} \omega_z \\[4pt] \notag L_y & = I_{yx} \omega_x + I_{yy} \omega_y + I_{yz} \omega_z \\[4pt] \notag L_z & = I_{zx} \omega_x + I_{zy} \omega_y + I_{zz} \omega_z \end{align}

Note that every fixed point in a body has a specific inertia tensor. The components of the inertia tensor at a specified point depend on the orientation of the coordinate frame whose origin is located at the specified fixed point. For example, the inertia tensor for a cube is very different when the fixed point is at the center of mass compared with when the fixed point is at a corner of the cube.