Rotational kinetic energy

The instantaneous rotational kinetic energy of a rotating rigid body is written

$$K = \frac{1}{2}\sum_{i=1,N} m_i\left(\frac{d{\bf r}_i}{dt}\right)^2.$$ (8.14)

Making use of Equation (8.4), and some vector identities (see Section A.4), the kinetic energy takes the form

$$K = \frac{1}{2}\sum_{i=1,N} m_i\,( \mbox{\boldmath$\omega$} \times {\bf r}_i)\cdot( \mbox{\boldmath$\omega$} \times {\bf r}_i) = \frac{1}{2}\, \mbox{\boldmath$\omega$} \cdot\!\! \sum_{i=1,N} m_i\,{\bf r}_i\times ( \mbox{\boldmath$\omega$} \times {\bf r}_i).$$ (8.15)

Hence, it follows from Equation (8.5) that

$$K = \frac{1}{2}\,\, \mbox{\boldmath$\omega$} \cdot {\bf L}.$$ (8.16)

Making use of Equation (8.13), we can also write

$$K = \frac{1}{2}\, \mbox{\boldmath$\omega$} ^T\,\textbf{\em I}\, \mbox{\boldmath$\omega$} .$$ (8.17)

Here, $\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, Equation (8.17) yields

$$K = \frac{1}{2}\left({\cal I}_{xx}\,\omega_x^{\,2}+ {\cal I}_{yy}... ...\cal I}_{yz}\,\omega_y\,\omega_z + 2\,{\cal I}_{xz}\,\omega_x\,\omega_z\right).$$ (8.18)