next up previous contents
Next: Rotational Kinetic Energy Up: Rigid Body Motion Previous: Fundamental Equations   Contents


The Moment of Inertia Tensor

Consider a rigid body rotating with fixed angular velocity $\omega$ about an axis which passes through the origin (see Fig. 42). Let ${\bf r}_i$ be the position vector of the $i$th mass element, whose mass is $m_i$. We expect this position vector to precess about the axis of rotation (which is parallel to $\omega$) with angular velocity $\omega$. It, therefore, follows from Eq. (41) that
\begin{displaymath}
\frac{d{\bf r}_i}{dt} = \mbox{\boldmath$\omega$}\times {\bf r}_i.
\end{displaymath} (529)

Thus, the above equation specifies the velocity, ${\bf v}_i = d{\bf r}_i/dt$, of each mass element as the body rotates with fixed angular velocity $\omega$ about an axis passing through the origin.

Figure 42: A rigid rotating body.
\begin{figure}
\epsfysize =3in
\centerline{\epsffile{rig.eps}}
\end{figure}

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

\begin{displaymath}
{\bf L} = \sum_{i=1,N} m_i\,{\bf r}_i\times\frac{d{\bf r}_i}...
...}- ({\bf r}_i\cdot\mbox{\boldmath$\omega$})\,{\bf r}_i\right],
\end{displaymath} (530)

where use has been made of Eq. (529), and some standard vector identities. The above formula can be written as a matrix equation of the form
\begin{displaymath}
\left(\begin{array}{c}L_x\\ L_y\\ L_z\end{array}\right)=
\le...
...gin{array}{c}\omega_x\\ \omega_y\\ \omega_z\end{array}\right),
\end{displaymath} (531)

where
$\displaystyle I_{xx}$ $\textstyle =$ $\displaystyle \sum_{i=1,N}(y_i^{\,2}+z_i^{\,2}) \,m_i= \int(y^2+ z^2)\,dm,$ (532)
$\displaystyle I_{yy}$ $\textstyle =$ $\displaystyle \sum_{i=1,N}(x_i^{\,2}+z_i^{\,2}) \,m_i= \int(x^2+ z^2)\,dm,$ (533)
$\displaystyle I_{zz}$ $\textstyle =$ $\displaystyle \sum_{i=1,N}(x_i^{\,2}+y_i^{\,2}) \,m_i= \int(x^2+ y^2)\,dm,$ (534)
$\displaystyle I_{xy}=I_{yx}$ $\textstyle =$ $\displaystyle - \sum_{i=1,N}x_i\,y_i \,m_i=- \int x\,y\,dm,$ (535)
$\displaystyle I_{yz}=I_{zy}$ $\textstyle =$ $\displaystyle - \sum_{i=1,N}y_i\,z_i \,m_i= -\int y\,z\,dm,$ (536)
$\displaystyle I_{xz}=I_{zx}$ $\textstyle =$ $\displaystyle - \sum_{i=1,N}x_i\,z_i \,m_i= -\int x\,z\,dm.$ (537)

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 $xy$ product of inertia, $I_{yz}$ the $yz$ product of inertia, etc. The matrix of the $I_{ij}$ values is known as the moment of inertia tensor.1 Note that each component of the moment of inertia tensor can be written as either a sum over separate mass elements, or as an integral over infinitesimal mass elements. In the integrals, $dm = \rho\,dV$, where $\rho $ is the mass density, and $dV$ a volume element. Equation (531) can be written more succinctly as
\begin{displaymath}
{\bf L} = \tilde{\bf I}\,\mbox{\boldmath$\omega$}.
\end{displaymath} (538)

Here, it is understood that ${\bf L}$ and $\omega$ are both column vectors, and $\tilde{\bf I}$ is the matrix of the $I_{ij}$ values. Note that $\tilde{\bf I}$ is a real symmetric matrix: i.e., $I_{ij}^{\,\ast} = I_{ij}$ and $I_{ji} = I_{ij}$.

In general, the angular momentum vector, ${\bf L}$, obtained from Eq. (538), points in a different direction to the angular velocity vector, $\omega$. In other words, ${\bf L}$ is generally not parallel to $\omega$.

Finally, although the above results were obtained assuming a fixed angular velocity, they remain valid at each instant in time even if the angular velocity varies.


next up previous contents
Next: Rotational Kinetic Energy Up: Rigid Body Motion Previous: Fundamental Equations   Contents
Richard Fitzpatrick 2008-01-13