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Rotational Kinetic Energy

The instantaneous rotational kinetic energy of a rotating rigid body is written
K = \frac{1}{2}\sum_{i=1,N} m_i\left(\frac{d{\bf r}_i}{dt}\right)^2.
\end{displaymath} (467)

Making use of Equation (457), and some vector identities (see Section A.9), the kinetic energy takes the form
K = \frac{1}{2}\sum_{i=1,N} m_i\,(\mbox{\boldmath$\omega$}\t...
...i\,{\bf r}_i\times (\mbox{\boldmath$\omega$}\times {\bf r}_i).
\end{displaymath} (468)

Hence, it follows from (458) that
K = \frac{1}{2}\,\, \mbox{\boldmath$\omega$} \cdot {\bf L}.
\end{displaymath} (469)

Making use of Equation (466), we can also write
K = \frac{1}{2}\,\mbox{\boldmath$\omega$}^T\,\tilde{\bf I}\,\mbox{\boldmath$\omega$}.
\end{displaymath} (470)

Here, $\mbox{\boldmath$\omega$}^T$ is the row vector of the Cartesian components $\omega_x$, $\omega_y$, $\omega_z$, which is, of course, the transpose (denoted $~^T$) of the column vector $\omega$. When written in component form, the above equation yields
K = \frac{1}{2}\left(I_{xx}\,\omega_x^{\,2}+ I_{yy}\,\omega_...
...z}\,\omega_y\,\omega_z + 2\,I_{xz}\,\omega_x\,\omega_z\right).
\end{displaymath} (471)

Richard Fitzpatrick 2011-03-31