Moment of Inertia Tensor

Consider a rigid body rotating with fixed angular velocity $\omega$ about an axis that passes through the origin. See Figure 1.8. Let ${\bf r}_i$ be the displacement of the $i$th mass element, whose mass is $m_i$. We expect this displacement to precess about the axis of rotation (which is parallel to $\omega$) with angular velocity $\omega$. It, therefore, follows from Equation (A.52) that

$${\bf v}_i= \mbox{\boldmath$\omega$} \times {\bf r}_i.$$ (1.181)

Figure 1.8: A rigid rotating body.

The total angular momentum of the body (about the origin) is written

$${\bf L} = \sum_{i=1,N} m_i\,{\bf r}_i\times{\bf v}_i= \sum_{i=1,N}m_i\,{\bf r}_i\times( \mbox{\boldmath$\omega$} \times {\bf r}_i) = \sum_{i=1,N}m_i\left[r_i^{\,2}\,\mbox{\boldmath$\omega$}- ({\bf r}_i\cdot\mbox{\boldmath$\omega$})\,{\bf r}_i\right],$$ (1.182)

where use has been made of Equation (1.181), and some standard vector identities. (See Section A.11.) The previous formula can be written as a matrix equation of the form

$$\left(\begin{array}{c}L_x\\ L_y\\ L_z\end{array}\right)= \left(\b... ...}\right)\left(\begin{array}{c}\omega_x\\ \omega_y\\ \omega_z\end{array}\right),$$ (1.183)

where

$$I_{xx} = \sum_{i=1,N}\left(y_i^{\,2}+z_i^{\,2}\right) m_i,$$ (1.184)

$$I_{yy} =\sum_{i=1,N}\left(x_i^{\,2}+z_i^{\,2}\right) m_i,$$ (1.185)

$$I_{zz} =\sum_{i=1,N}\left(x_i^{\,2}+y_i^{\,2}\right) m_i,$$ (1.186)

$$I_{xy}=I_{yx} = - \sum_{i=1,N}x_i\,y_i \,m_i$$ (1.187)

$$I_{yz}=I_{zy} = - \sum_{i=1,N}y_i\,z_i \,m_i,$$ (1.188)

$$I_{xz}=I_{zx} = - \sum_{i=1,N}x_i\,z_i \,m_i.$$ (1.189)

Here, $I_{xx}$ is called the moment of inertia about the $x$-axis, $I_{yy}$ the moment of inertia about the $y$-axis, $I_{xy}$ the $x$-$y$ product of inertia, $I_{yz}$ the $y$-$z$ product of inertia, et cetera. The matrix of the $I_{ij}$ values is known as the moment of inertia tensor.

Suppose that our body is rotationally symmetric about the $z$-axis. In this case, it is easily seen that the products of inertia are all zero. Moreover, $I_{xx}=I_{yy} = I_\perp$. Let us write $I_{zz}=I_\parallel$. Note that, in general, $I_\parallel \neq I_\perp$ (unless the body is spherically symmetric). Thus, Equation (1.183) simplifies to give

$${\bf L} = I_\perp\,\omega_x\,{\bf e}_x + I_\perp\,\omega_y\,{\bf e}_y + I_\parallel\,\omega_z\,{\bf e}_z.$$ (1.190)

The angular momentum vector, ${\bf L}$, obtained from the previous equation, does not necessarily point in the same direction as the angular velocity vector, $\omega$ (because $I_\parallel \neq I_\perp$). In other words, ${\bf L}$ is generally not parallel to $\omega$. However, if the body rotates about ${\bf e}_z$ or any axis in the $x$-$y$ plane then ${\bf L}$ is parallel to $\omega$. These special axes of rotation are called principal axes of rotation, and the associated moments of inertia, $I_\parallel$ and $I_\perp$, respectively, are called principal moments of inertia.

It can be demonstrated that any rigid body (not just an axisymmetric one) has three mutually perpendicular principal axes of rotation. Furthermore, if a body is rotating about one of its principal axes of rotation then

$${\bf L} = I\, \mbox{\boldmath$\omega$} ,$$ (1.191)

where $I$ is the associated principal moment of inertia. More generally, assuming that the Cartesian axis are parallel to the principal moments of inertia, we can write

$${\bf L} = I_{xx}\,\omega_x\,{\bf e}_x+I_{yy}\,\omega_y\,{\bf e}_y+I_{zz}\,\omega_z\,{\bf e}_z,$$ (1.192)

where $I_{xx}$, $I_{yy}$, and $I_{zz}$ are the three principal moments of inertia.