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Finding rate of change of moment of inertia tensor

  1. Jun 2, 2014 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.


    3. 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:

    [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?
     
  2. jcsd
  3. Jun 2, 2014 #2
    I think that you may have posted this in the wrong category. I may be wrong, though.
     
  4. Jun 3, 2014 #3

    BiGyElLoWhAt

    User Avatar
    Gold Member




    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.
     
  5. Jun 3, 2014 #4

    BiGyElLoWhAt

    User Avatar
    Gold Member

    I'm gonna go to bed for tonight. I'm also gonna blame my phone for the "there for".
    XD
    I'll check this again tomorrow.
    Good night all
     
  6. Jun 3, 2014 #5
    No, it is not [itex]\boldsymbol\omega \times \boldsymbol\omega[/itex] 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.
     
  7. Jun 5, 2014 #6
    I would try to replace ##\omega_i## with ##(\omega \times r)_i## since dr/dt=##\omega \times r##
     
  8. Jun 7, 2014 #7
    [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}\,\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}
    [/tex]
     
  9. Jun 7, 2014 #8
    I have tried to continue the derivation after correcting the dr/dt as you pointed out.

    I did [itex](\boldsymbol\omega \times \mathbf{r}_p)_k[/itex] by calculating [itex]\boldsymbol\omega \times \mathbf{r}_p[/itex] first as matrix multiplication [itex][\omega]_\times [r_p][/itex] 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 [itex]\mathbf{L}_p = \mathbf{r}_p \times \mathbf{p}_p[/itex] as follows:
    [tex]\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}[/tex]

    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.
     
    Last edited: Jun 7, 2014
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