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Moment of inertia: vector derivation

  1. May 6, 2012 #1
    We have the representation of torque attached. The components of r are (x,y,z).
    Where did the matrix come from and how did we get the stuff in the matrix? (Basically I understand all the steps except the step from the 3rd line to the 4th line.)
    Thank you very much!
     

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  2. jcsd
  3. May 6, 2012 #2

    I like Serena

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    Hey sparkle! :smile:

    Did you try to write out the 3rd line?

    $$mr^2 \mathbf{α} = (r^2 I) \mathbf{α}$$

    $$(\mathbf{r} \cdot \mathbf{α})\mathbf{r} = ([x\ y\ z] \cdot \mathbf{α}) \begin{bmatrix}x \\ y \\ z \end{bmatrix} = ([x\ y\ z] \cdot \begin{bmatrix}x \\ y \\ z \end{bmatrix})\mathbf{α}$$

    (You can check that last equality by writing it out in components.)
     
  4. May 6, 2012 #3
    Hi I like Serena!! :smile:

    Do we get:
    $$ (\mathbf{r} \cdot \mathbf{α})\mathbf{r} = ([x\ y\ z] \cdot \begin{bmatrix}x \\ y \\ z \end{bmatrix})\mathbf{α} = (x^2 + y^2 + z^2)\mathbf{α}$$
    $$∴ mr^2 \mathbf{α} - m(\mathbf{r} \cdot \mathbf{α})\mathbf{r} = m(r^2 - x^2 - y^2 - z^2)\mathbf{α}$$

    I still don't see how we get the 3X3 matrix :confused:

    Thanks again!
     
  5. May 6, 2012 #4

    I like Serena

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    I have to admit that my representation was not correct. Sorry.

    Let's write it out in its components.

    $$ (\mathbf{r} \cdot \mathbf{α})\mathbf{r} = (x α_x + y α_y + z α_z ) \begin{bmatrix}x \\ y \\ z \end{bmatrix}=...$$

    Can you turn this into a vector without a factor in front?
    And then split off ##\mathbf{α}## yielding a matrix?
     
  6. May 6, 2012 #5
    Okay, I got
    $$\left(
    \begin{array}{ccc}
    x^2 & xy & xz \\
    xy & y^2 & yz \\
    xz & yz & z^2 \end{array}
    \right)\mathbf{α}$$

    How does $$r^2 - matrix$$ work?
    Thanks! :)
     
  7. May 6, 2012 #6

    I like Serena

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    It's
    $$r^2 \mathbf{α} - matrix \mathbf{α}$$

    and
    $$r^2 \mathbf{α} = r^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \mathbf{α} = \begin{pmatrix} r^2 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \end{pmatrix} \mathbf{α}$$
     
  8. May 6, 2012 #7
    Oh thank you! I should brush up on matrices hehe :)
     
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