MHB Inertia matrix from orbital angular momentum of the ith element (please check)

[ognik]

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 ?

 

[topsquark]

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 ?

Two comments. When you use I as the moment of inertia please exclusively (if you can) use i to indicate the ith particle. [math]L_I[/math] is somewhat unclear. Also, you have a typo: The last expression should be [math]\vec{L}_i = \vec{ \omega } ~ m_i ( \vec{r}_i \cdot \vec{r} _i )[/math].

-Dan

 

[ognik]

Hi - that I subscript is an annoying auto-correct in browsers (unless it is in this editor?) , which I haven't figured out how to turn off - normally I catch them all...

I am not sure my conclusion that $r_i$ and $\omega$ are orthog. is valid?

Please confirm, in that last expression, because $m_i$ is scalar it can be moved anywhere in the term?

This is a section on matrices & eigenvalues, so I'd like a matrix solution for I, something like $ \begin{bmatrix}L_1\\L_2\\L_3\end{bmatrix} = \begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&...\\...&&I_{33}\end{bmatrix} \begin{bmatrix}\omega_1\\ \omega_2\\ \omega_3\end{bmatrix} $ but I'm struggling to figure out the I elements from the 2 term eqtn I got (assuming r & $\omega$ orthog. that is)

 

[topsquark]

I am not sure my conclusion that $r_i$ and $\omega$ are orthog. is valid?

Yeah, autocorrect features are a pain in the pahtootie.

[math]\vec{r}[/math] and [math]\vec{\omega}[/math] are defined to be perpendicular. Note though, that both can be functions of position or time.

Please confirm, in that last expression, because $m_i$ is scalar it can be moved anywhere in the term?

Correct.

This is a section on matrices & eigenvalues, so I'd like a matrix solution for I, something like $ \begin{bmatrix}L_1\\L_2\\L_3\end{bmatrix} = \begin{bmatrix}I_{11}&I_{12}&I_{13}\\I_{21}&...\\...&&I_{33}\end{bmatrix} \begin{bmatrix}\omega_1\\ \omega_2\\ \omega_3\end{bmatrix} $ but I'm struggling to figure out the I elements from the 2 term eqtn I got (assuming r & $\omega$ orthog. that is)

You have one. Rewrite the expression as [math]L_i = \left ( m_i \vec{r}_i \cdot \vec{r}_i \right ) \omega _i[/math]. This is now of the form: [math]\vec{L} = I \vec{\omega}[/math] where [math]I_{ii} = m_i~ \vec{r}_i \cdot \vec{r}_i[/math], to be explicit. Since the matrix elements [math]I_{ij}[/math], where i and j aren't equal, don't appear in the expressions then I is diagonal.

-Dan