Finding rate of change of moment of inertia tensor

[TadeusPrastowo]

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 $\sum_{i=1}^n \boldsymbol\Delta\mathbf{r}_i\times (m_i\mathbf{a}_i)$.

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.
  

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

$$\begin{align} \frac{\text{d}}{\text{d}t}\left(\begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\ \end{bmatrix}\right) &= \begin{bmatrix} \frac{\text{d}\,I_{1,1}}{\text{d}t} & \frac{\text{d}\,I_{1,2}}{\text{d}t} & \frac{\text{d}\,I_{1,3}}{\text{d}t} \\ \frac{\text{d}\,I_{2,1}}{\text{d}t} & \frac{\text{d}\,I_{2,2}}{\text{d}t} & \frac{\text{d}\,I_{2,3}}{\text{d}t} \\ \frac{\text{d}\,I_{3,1}}{\text{d}t} & \frac{\text{d}\,I_{3,2}}{\text{d}t} & \frac{\text{d}\,I_{3,3}}{\text{d}t} \\ \end{bmatrix} \end{align}$$

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

Then, I calculated the above one as follows:
$$\begin{align} \frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\ &= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}} \frac{\text{d} \,r_{p,k}} { \text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\omega_k - (\omega_i\,r_{p,j} + r_{p,i}\,\omega_j) \\ \frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & -(r_{p,1}\,\omega_2 + r_{p,2}\,\omega_1) & -(r_{p,1}\,\omega_3 + r_{p,3}\,\omega_1) \\ -(r_{p,2}\,\omega_1 + r_{p,1}\,\omega_2) & 2\,r_{p,1}\,\omega_1 + 2\,r_{p,3}\,\omega_3 & -(r_{p,2}\,\omega_3 + r_{p,3}\,\omega_2) \\ -(r_{p,3}\,\omega_1 + r_{p,1}\,\omega_3) & -(r_{p,3}\,\omega_2 + r_{p,2}\,\omega_3) & 2\,r_{p,1}\omega_1 + 2\,r_{p,2}\,\omega_2 \end{bmatrix} \end{align}$$

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

$$\begin{align} [\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times &= \sum_p m_p \left( \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}\begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix} + \begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix}\begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix} \right) \\ &= \sum_p m_p \left( \begin{bmatrix} r_{p,1}\,r_{p,2}\,\omega_3 - r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} + \begin{bmatrix} -r_{p,1}\,r_{p,2}\,\omega_3 + r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \right) \\ &= \sum_p m_p \begin{bmatrix} 0 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \ne \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \end{align}$$

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?

 

[mafagafo]

I think that you may have posted this in the wrong category. I may be wrong, though.

 

[BiGyElLoWhAt]

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 $\sum_{i=1}^n \boldsymbol\Delta\mathbf{r}_i\times (m_i\mathbf{a}_i)$.

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.
  

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

$$\begin{align} \frac{\text{d}}{\text{d}t}\left(\begin{bmatrix} I_{1,1} & I_{1,2} & I_{1,3} \\ I_{2,1} & I_{2,2} & I_{2,3} \\ I_{3,1} & I_{3,2} & I_{3,3} \\ \end{bmatrix}\right) &= \begin{bmatrix} \frac{\text{d}\,I_{1,1}}{\text{d}t} & \frac{\text{d}\,I_{1,2}}{\text{d}t} & \frac{\text{d}\,I_{1,3}}{\text{d}t} \\ \frac{\text{d}\,I_{2,1}}{\text{d}t} & \frac{\text{d}\,I_{2,2}}{\text{d}t} & \frac{\text{d}\,I_{2,3}}{\text{d}t} \\ \frac{\text{d}\,I_{3,1}}{\text{d}t} & \frac{\text{d}\,I_{3,2}}{\text{d}t} & \frac{\text{d}\,I_{3,3}}{\text{d}t} \\ \end{bmatrix} \end{align}$$

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

Then, I calculated the above one as follows:
$$\begin{align} \frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\ &= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}} \frac{\text{d} \,r_{p,k}} { \text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\omega_k - (\omega_i\,r_{p,j} + r_{p,i}\,\omega_j) \\ \frac{\text{d}\text{I}_{i,j}}{\text{d}t} &= \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & -(r_{p,1}\,\omega_2 + r_{p,2}\,\omega_1) & -(r_{p,1}\,\omega_3 + r_{p,3}\,\omega_1) \\ -(r_{p,2}\,\omega_1 + r_{p,1}\,\omega_2) & 2\,r_{p,1}\,\omega_1 + 2\,r_{p,3}\,\omega_3 & -(r_{p,2}\,\omega_3 + r_{p,3}\,\omega_2) \\ -(r_{p,3}\,\omega_1 + r_{p,1}\,\omega_3) & -(r_{p,3}\,\omega_2 + r_{p,2}\,\omega_3) & 2\,r_{p,1}\omega_1 + 2\,r_{p,2}\,\omega_2 \end{bmatrix} \end{align}$$

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

$$\begin{align} [\omega]_\times\,I_{\text{body}} + I_{\text{body}}\,[\omega]_\times &= \sum_p m_p \left( \begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix}\begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix} + \begin{bmatrix} r_{p,2}^2 + r_{p,3}^2 & -r_{p,1} r_{p,2} & -r_{p,1} r_{p,3} \\ -r_{p,2} r_{p,1} & r_{p,1}^2 + r_{p,3}^2 & -r_{p,2} r_{p,3} \\ -r_{p,3} r_{p,1} & -r_{p,3} r_{p,2} & r_{p,1}^2 + r_{p,2}^2 \end{bmatrix}\begin{bmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0 \end{bmatrix} \right) \\ &= \sum_p m_p \left( \begin{bmatrix} r_{p,1}\,r_{p,2}\,\omega_3 - r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} + \begin{bmatrix} -r_{p,1}\,r_{p,2}\,\omega_3 + r_{p,1}\,r_{p,3}\,\omega_2 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \right) \\ &= \sum_p m_p \begin{bmatrix} 0 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \ne \sum_p m_p \begin{bmatrix} 2\,r_{p,2}\,\omega_2 + 2\,r_{p,3}\,\omega_3 & \ldots \\ \vdots & \ddots \\ \end{bmatrix} \end{align}$$

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?

A couple things. Firstly, damn that's a lot of latex. Secondly, I'm not sure exactly what it is you're trying to do. $\omega$ is a vector, and there for what you have must be a cross product, but its $\omega \times \omega$ which is always 0, seeing as how omega is parallel to itself.

Next thing, I'm not following the notation very well. You're cutting you're object up into 9 chunks and approximating it? And $m_{p}$ is the mass of each chunk? What is $\delta_{i, j}$ ? Is thisrepresentative of the fact that you're trying to use a delta epsilon statement for your approximation?
when you sum over k, what are you actually summing over? The same with p.

 

[BiGyElLoWhAt]

I'm going to go to bed for tonight. I'm also going to blame my phone for the "there for".
XD
I'll check this again tomorrow.
Good night all

 

[TadeusPrastowo]

No, it is not $\boldsymbol\omega \times \boldsymbol\omega$ but [itex]\boldsymbol\omega \times \boldsymbol\omega[/itex] where [itex][/itex] is a second-order tensor, not a scalar. To be exact, we are dealing with the product of matrices [itex][\omega]_\times \left( [\omega]\right)[/itex]. See Wikipedia article on cross-product as a matrix.

I think mafagafo may be correct that my post is in the wrong category. But, I don't see any facility to move this post to another category. Let me see... I will re-post if I can't find one.

 

[bloby]

I would try to replace $\omega_i$ with $(\omega \times r)_i$ since dr/dt=$\omega \times r$

 

[bloby]

$$\begin{align} \frac{\text{d}I_{i,j}}{\text{d}t} &= \frac{\text{d}}{\text{d}t}\left(\sum_p m_p\,\left(\delta_{i,j} \sum_k r_k^2 - r_i\,r_j\right)\right) \\ &= \sum_p m_p \frac{\text{d}}{\text{d}t}\,\left(\delta_{i,j} \sum_k r_{p,k}^2 \right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \frac{\text{d}}{\text{d}t}\, \left(\sum_k r_{p,k}^2\right) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}}{\text{d}t}\,(r_{p,k}^2) - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k \frac{\text{d}\,r_{p,k}^2}{\text{d}r_{p,k}} \frac{\text{d} \,r_{p,k}} { \text{d}t} - \frac{\text{d}}{\text{d}t}\,(r_{p,i}\,r_{p,j}) \\ &= \sum_p m_p \delta_{i,j} \sum_k 2\,r_{p,k}\,\pmb{(\omega\,\times\,r_p)_k} - (\pmb{(\omega\,\times\,r_p)_i}\,r_{p,j} + r_{p,i}\,\pmb{(\omega\,\times\,r_p)_j}) \end{align}$$

 

[TadeusPrastowo]

I have tried to continue the derivation after correcting the dr/dt as you pointed out.

I did $(\boldsymbol\omega \times \mathbf{r}_p)_k$ by calculating $\boldsymbol\omega \times \mathbf{r}_p$ first as matrix multiplication $[\omega]_\times [r_p]$ and then taking the k-element.

Then, I continued the derivation up to the result of [itex]\frac{d}{dt}\boldsymbol\omega[/itex] to check if the result would equal to [itex][\omega]_\times (\boldsymbol\omega)[/itex].

Unfortunately, they differ by two or three terms based on my pencil & paper calculation.

So, it may be that I make another mistake on the paper. But, it might also be that the derivation cannot go that way (i.e., tensor cannot be derived in the way I suggested).

In the end, I decided to be done with tensor derivation and went another route by performing the calculation starting from the definition of angular momentum $\mathbf{L}_p = \mathbf{r}_p \times \mathbf{p}_p$ as follows:
$$\begin{align} \frac{\text{d}\mathbf{L}}{\text{d}t} &= \frac{\text{d}}{\text{d}t} \left(\sum_p \mathbf{r}_p \times \mathbf{p}_p \right) \\ &= \frac{\text{d}}{\text{d}t} \left(\sum_p \mathbf{r}_p \times (m_p \mathbf{v}_p) \right) \\ &= \frac{\text{d}}{\text{d}t} \left(\sum_p m_p \left( \mathbf{r}_p \times (\boldsymbol\omega \times \mathbf{r}_p) \right)\right) \end{align}$$

By that route, I can arrive at [itex]\boldsymbol\alpha + \boldsymbol\omega \times ( \boldsymbol\omega)[/itex] using the derivation property of cross product.

So, I guess tensor derivation cannot be performed element-by-element like what I suggested.