Moment of inertia: vector derivation

[sparkle123]

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!

 

[I like Serena]

Hey sparkle!

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.)

 

[sparkle123]

Hi I like Serena!

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

Thanks again!

 

[I like Serena]

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?

 

[sparkle123]

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! :)

 

[I like Serena]

How does $$r^2 - matrix$$ work?
Thanks! :)

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

and
$$r^2 \mathbf{α} = r^2 \begin{pmatrix} 1 & 0 & 0 \\ 0 &1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \mathbf{α} = \begin{pmatrix} r^2 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \end{pmatrix} \mathbf{α}$$

 

[sparkle123]

Oh thank you! I should brush up on matrices hehe :)