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Rotational kinetic energy

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

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

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

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

Hence, it follows from Equation (8.5) that

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

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

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

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


next up previous
Next: Principal axes of rotation Up: Rigid body rotation Previous: Moment of inertia tensor
Richard Fitzpatrick 2016-03-31