- #1

TadeusPrastowo

- 21

- 0

## Homework Statement

The Wikipedia article on spatial rigid body dynamics derives the equation of motion [itex]\boldsymbol\tau =

*\boldsymbol\alpha + \boldsymbol\omega\times*

*\boldsymbol\omega[/itex] from [itex]\sum_{i=1}^n \boldsymbol\Delta\mathbf{r}_i\times (m_i\mathbf{a}_i)[/itex].*

But, there is another way to derive the same result from [itex]\frac{\text{d}\mathbf{L}}{\text{d}t} = \frac{\text{d}}{\text{d}t}\left(But, there is another way to derive the same result from [itex]\frac{\text{d}\mathbf{L}}{\text{d}t} = \frac{\text{d}}{\text{d}t}\left(

*\boldsymbol\omega\right)[/itex]. Can it be derived by performing element-by-element derivation of the moment of inertia tensor?*

## Homework Equations

- [tex]\frac{\text{d}}{\text{d}t}\left(
*\boldsymbol\omega\right) = \frac{\text{d}**}{\text{d}t}\boldsymbol\omega +**\frac{\text{d}\boldsymbol\omega}{\text{d}t}[/tex]*

[*][tex]\frac{\text{d}*}{\text{d}t} = [\omega]_\times\,**+**\,[\omega]_\times[/tex] as shown in Point 5 of this article.*

[tex]\begin{align}

\frac{\text{d}}{\text{d}t}\left(## The Attempt at a Solution

[tex]\begin{align}

\frac{\text{d}}{\text{d}t}\left(

*\boldsymbol\omega\right) &= \frac{\text{d}**}{\text{d}t}\boldsymbol\omega +**\frac{\text{d}\boldsymbol\omega}{\text{d}t} \\*

&= \frac{\text{d}&= \frac{\text{d}

*}{\text{d}t}\boldsymbol\omega +**\boldsymbol\alpha \\*

\end{align}[/tex]

Now, I am focusing on solving [itex]\frac{\text{d}\end{align}[/tex]

Now, I am focusing on solving [itex]\frac{\text{d}

*}{\text{d}t}[/itex] with the goal of obtaining [itex][\omega]_\times\,**+**\,[\omega]_\times[/itex].*

I perform element-by-element derivation of the moment of inertia tensor as follows:

[tex]

\begin{align}

\frac{\text{d}}{\text{d}t}\left(\begin{bmatrix}

I_{1,1} & I_{1,2} & I_{1,3} \\

I_{2,1} & I_{2,2} & I_{2,3} \\

I_{3,1} & I_{3,2} & I_{3,3} \\

\end{bmatrix}\right) &= \begin{bmatrix}

\frac{\text{d}\,I_{1,1}}{\text{d}t} & \frac{\text{d}\,I_{1,2}}{\text{d}t} & \frac{\text{d}\,I_{1,3}}{\text{d}t} \\

\frac{\text{d}\,I_{2,1}}{\text{d}t} & \frac{\text{d}\,I_{2,2}}{\text{d}t} & \frac{\text{d}\,I_{2,3}}{\text{d}t} \\

\frac{\text{d}\,I_{3,1}}{\text{d}t} & \frac{\text{d}\,I_{3,2}}{\text{d}t} & \frac{\text{d}\,I_{3,3}}{\text{d}t} \\

\end{bmatrix}

\end{align}

[/tex]

where the moment of inertia tensor is as follows: [tex]I_{i,j} = \sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)[/tex]

Then, I calculated the above one as follows:

[tex]\begin{align}

\frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\

&= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}} \frac{\text{d} \,r_{p,k}} { \text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\omega_k - (\omega_i\,r_{p,j} + r_{p,i}\,\omega_j) \\

\frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \sum_p m_p \begin{bmatrix}

2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & -(r_{p,1}\,\omega_2 + r_{p,2}\,\omega_1) & -(r_{p,1}\,\omega_3 + r_{p,3}\,\omega_1) \\

-(r_{p,2}\,\omega_1 + r_{p,1}\,\omega_2) & 2\,r_{p,1}\,\omega_1 + 2\,r_{p,3}\,\omega_3 & -(r_{p,2}\,\omega_3 + r_{p,3}\,\omega_2) \\

-(r_{p,3}\,\omega_1 + r_{p,1}\,\omega_3) & -(r_{p,3}\,\omega_2 + r_{p,2}\,\omega_3) & 2\,r_{p,1}\omega_1 + 2\,r_{p,2}\,\omega_2

\end{bmatrix}

\end{align}

[/tex]

But, after trying some factorizations, I can't seem to get to the following:

[tex]\begin{align}

[\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times &= \sum_p m_p \left( \begin{bmatrix}

0 & -\omega_3 & \omega_2 \\

\omega_3 & 0 & -\omega_1 \\

-\omega_2 & \omega_1 & 0

\end{bmatrix}\begin{bmatrix}

r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\

-r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\

-r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2

\end{bmatrix} + \begin{bmatrix}

r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\

-r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\

-r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2

\end{bmatrix}\begin{bmatrix}

0 & -\omega_3 & \omega_2 \\

\omega_3 & 0 & -\omega_1 \\

-\omega_2 & \omega_1 & 0

\end{bmatrix} \right) \\

&= \sum_p m_p \left( \begin{bmatrix}

r_{p,1}\,r_{p,2}\,\omega_3 - r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} + \begin{bmatrix}

-r_{p,1}\,r_{p,2}\,\omega_3 + r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} \right) \\

&= \sum_p m_p \begin{bmatrix}

0 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} \ne \sum_p m_p \begin{bmatrix}

2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & \ldots \\

\vdots & \ddots \\

\end{bmatrix}

\end{align}

[/tex]

So, I should have made a mistake in the differentiation. But, where? And, how to proceed?

Or, is it because the definition of tensor cannot be used in performing the derivation? Why?I perform element-by-element derivation of the moment of inertia tensor as follows:

[tex]

\begin{align}

\frac{\text{d}}{\text{d}t}\left(\begin{bmatrix}

I_{1,1} & I_{1,2} & I_{1,3} \\

I_{2,1} & I_{2,2} & I_{2,3} \\

I_{3,1} & I_{3,2} & I_{3,3} \\

\end{bmatrix}\right) &= \begin{bmatrix}

\frac{\text{d}\,I_{1,1}}{\text{d}t} & \frac{\text{d}\,I_{1,2}}{\text{d}t} & \frac{\text{d}\,I_{1,3}}{\text{d}t} \\

\frac{\text{d}\,I_{2,1}}{\text{d}t} & \frac{\text{d}\,I_{2,2}}{\text{d}t} & \frac{\text{d}\,I_{2,3}}{\text{d}t} \\

\frac{\text{d}\,I_{3,1}}{\text{d}t} & \frac{\text{d}\,I_{3,2}}{\text{d}t} & \frac{\text{d}\,I_{3,3}}{\text{d}t} \\

\end{bmatrix}

\end{align}

[/tex]

where the moment of inertia tensor is as follows: [tex]I_{i,j} = \sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)[/tex]

Then, I calculated the above one as follows:

[tex]\begin{align}

\frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\

&= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}} \frac{\text{d} \,r_{p,k}} { \text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\

&= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\omega_k - (\omega_i\,r_{p,j} + r_{p,i}\,\omega_j) \\

\frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \sum_p m_p \begin{bmatrix}

2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & -(r_{p,1}\,\omega_2 + r_{p,2}\,\omega_1) & -(r_{p,1}\,\omega_3 + r_{p,3}\,\omega_1) \\

-(r_{p,2}\,\omega_1 + r_{p,1}\,\omega_2) & 2\,r_{p,1}\,\omega_1 + 2\,r_{p,3}\,\omega_3 & -(r_{p,2}\,\omega_3 + r_{p,3}\,\omega_2) \\

-(r_{p,3}\,\omega_1 + r_{p,1}\,\omega_3) & -(r_{p,3}\,\omega_2 + r_{p,2}\,\omega_3) & 2\,r_{p,1}\omega_1 + 2\,r_{p,2}\,\omega_2

\end{bmatrix}

\end{align}

[/tex]

But, after trying some factorizations, I can't seem to get to the following:

[tex]\begin{align}

[\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times &= \sum_p m_p \left( \begin{bmatrix}

0 & -\omega_3 & \omega_2 \\

\omega_3 & 0 & -\omega_1 \\

-\omega_2 & \omega_1 & 0

\end{bmatrix}\begin{bmatrix}

r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\

-r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\

-r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2

\end{bmatrix} + \begin{bmatrix}

r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\

-r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\

-r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2

\end{bmatrix}\begin{bmatrix}

0 & -\omega_3 & \omega_2 \\

\omega_3 & 0 & -\omega_1 \\

-\omega_2 & \omega_1 & 0

\end{bmatrix} \right) \\

&= \sum_p m_p \left( \begin{bmatrix}

r_{p,1}\,r_{p,2}\,\omega_3 - r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} + \begin{bmatrix}

-r_{p,1}\,r_{p,2}\,\omega_3 + r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} \right) \\

&= \sum_p m_p \begin{bmatrix}

0 & \ldots \\

\vdots & \ddots \\

\end{bmatrix} \ne \sum_p m_p \begin{bmatrix}

2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & \ldots \\

\vdots & \ddots \\

\end{bmatrix}

\end{align}

[/tex]

So, I should have made a mistake in the differentiation. But, where? And, how to proceed?

Or, is it because the definition of tensor cannot be used in performing the derivation? Why?