Rotational Kinetic Energy

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

$$K = \frac{1}{2}\sum_{i=1,N} m_i\left(\frac{d{\bf r}_i}{dt}\right)^2.$$ (539)

Making use of Eq. (529), and some vector identities, the kinetic energy takes the form

$$K = \frac{1}{2}\sum_{i=1,N} m_i\,(\mbox{\boldmath$\omega$}\t... ...i\,{\bf r}_i\times (\mbox{\boldmath$\omega$}\times {\bf r}_i).$$ (540)

Hence, it follows from (530) that

$$K = \frac{1}{2}\,\, \mbox{\boldmath$\omega$} \cdot {\bf L}.$$ (541)

Making use of Eq. (538), we can also write

$$K = \frac{1}{2}\,\mbox{\boldmath$\omega$}^T\,\tilde{\bf I}\,\mbox{\boldmath$\omega$}.$$ (542)

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

$$K = \frac{1}{2}\left(I_{xx}\,\omega_x^{\,2}+ I_{yy}\,\omega_... ...z}\,\omega_y\,\omega_z + 2\,I_{xz}\,\omega_x\,\omega_z\right).$$ (543)