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Moment of inertia tensor

Consider a rigid body rotating with fixed angular velocity $ \omega$ about an axis that passes through the origin. (See Figure 8.1.) 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 Section A.7 that

$\displaystyle \frac{d{\bf r}_i}{dt} =$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times {\bf r}_i.$ (8.4)

Thus, Equation (8.4) 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 8.1: A rigid rotating body.
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The total angular momentum of the body (about the origin) is written

$\displaystyle {\bf L} = \sum_{i=1,N} m_i\,{\bf r}_i\times\frac{d{\bf r}_i}{dt} = \sum_{i=1,N}m_i\,{\bf r}_i\times($$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \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],$ (8.5)

where use has been made of Equation (8.4), and some standard vector identities. (See Section A.4.) The preceding formula can be written as a matrix equation of the form

$\displaystyle \left(\begin{array}{c}L_x\\ L_y\\ L_z\end{array}\right)= \left(\b...
...}\right)\left(\begin{array}{c}\omega_x\\ \omega_y\\ \omega_z\end{array}\right),$ (8.6)

where

    $\displaystyle {\cal I}_{xx}$ $\displaystyle = \sum_{i=1,N}(y_i^{\,2}+z_i^{\,2}) \,m_i= \int(y^{\,2}+ z^{\,2})\,dm,$ (8.7)
    $\displaystyle {\cal I}_{yy}$ $\displaystyle = \sum_{i=1,N}(x_i^{\,2}+z_i^{\,2}) \,m_i= \int(x^{\,2}+ z^{\,2})\,dm,$ (8.8)
    $\displaystyle {\cal I}_{zz}$ $\displaystyle = \sum_{i=1,N}(x_i^{\,2}+y_i^{\,2}) \,m_i= \int(x^{\,2}+ y^{\,2})\,dm,$ (8.9)
    $\displaystyle {\cal I}_{xy}={\cal I}_{yx}$ $\displaystyle = - \sum_{i=1,N}x_i\,y_i \,m_i=- \int x\,y\,dm,$ (8.10)
    $\displaystyle {\cal I}_{yz}={\cal I}_{zy}$ $\displaystyle = - \sum_{i=1,N}y_i\,z_i \,m_i= -\int y\,z\,dm,$ (8.11)
and   $\displaystyle {\cal I}_{xz}={\cal I}_{zx}$ $\displaystyle = - \sum_{i=1,N}x_i\,z_i \,m_i= -\int x\,z\,dm.$ (8.12)

Here, $ {\cal I}_{xx}$ is called the moment of inertia about the $ x$ -axis, $ {\cal I}_{yy}$ the moment of inertia about the $ y$ -axis, $ {\cal I}_{xy}$ the $ xy$ product of inertia, $ {\cal I}_{yz}$ the $ yz$ product of inertia, and so on. The matrix of the $ {\cal I}_{ij}$ values is known as the moment of inertia tensor. 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\,d^{\,3}{\bf r}$ , where $ \rho $ is the mass density, and $ d^{\,3}{\bf r}$ a volume element. Equation (8.6) can be written more succinctly as

$\displaystyle {\bf L} = \textbf{\em I}\,$$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle .
$ (8.13)

Here, it is understood that $ {\bf L}$ and $ \omega$ are both column vectors, and $ \textbf{\em I}$ is the matrix of the $ {\cal I}_{ij}$ values. Note that $ \textbf{\em I}$ is a real symmetric matrix; that is, $ {\cal I}_{ij}^{\,\ast} = {\cal I}_{ij}$ and $ {\cal I}_{ji} = {\cal I}_{ij}$ .

In general, the angular momentum vector, $ {\bf L}$ , obtained from Equation (8.13), 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 preceding results were obtained assuming a fixed angular velocity they remain valid, at each instant in time, if the angular velocity varies.


next up previous
Next: Rotational kinetic energy Up: Rigid body rotation Previous: Fundamental equations
Richard Fitzpatrick 2016-03-31