- #1
ognik
- 643
- 2
Starting with the orbital angular momentum of the ith element of mass, $ \vec{L}_I = \vec{r}_I \times \vec{p}_I = m_i \vec{r}_i \times \left( \omega \times \vec{r}_i\right) $, derive the inertia matrix such that
$\vec{L} =I\omega, |\vec{L} \rangle = I |\vec{\omega} \rangle $
I used a X b X c = -c X a X b:
$ \vec{L}_i = -m_i \left( \vec{r}_i \times \vec{r}_i \times \omega \right) $
Using bac-cab, $ \vec{L}_i = -m_i \left( \vec{r}_i \left( \vec{r}_i \cdot \omega \right) -\vec{\omega}\left( \vec{r}_i \cdot\vec{r}_i \right) \right)$ and $\vec{r}_i \cdot \vec{\omega} = 0$ (orthogonal), so $ \vec{L}_I = m_i \left( \vec{\omega}\left( \vec{r}_i \cdot\vec{r}_i \right) \right)$ and $ I =m_i \vec{r}^2_i $ ...
That look right ?
$\vec{L} =I\omega, |\vec{L} \rangle = I |\vec{\omega} \rangle $
I used a X b X c = -c X a X b:
$ \vec{L}_i = -m_i \left( \vec{r}_i \times \vec{r}_i \times \omega \right) $
Using bac-cab, $ \vec{L}_i = -m_i \left( \vec{r}_i \left( \vec{r}_i \cdot \omega \right) -\vec{\omega}\left( \vec{r}_i \cdot\vec{r}_i \right) \right)$ and $\vec{r}_i \cdot \vec{\omega} = 0$ (orthogonal), so $ \vec{L}_I = m_i \left( \vec{\omega}\left( \vec{r}_i \cdot\vec{r}_i \right) \right)$ and $ I =m_i \vec{r}^2_i $ ...
That look right ?