# Question #a1395

Oct 26, 2016

see below

#### Explanation:

For point mass $m$ at generalised position $\vec{r}$ with respect to origin, its angular momentum is:
$\vec{L} = \vec{r} \times m \vec{v} = m \vec{r} \times \left(\vec{\omega} \times \vec{r}\right)$

From the bac-cab rule

$\frac{\vec{L}}{m} = \vec{\omega} \left({r}^{2}\right) - \vec{r} \left(\vec{r} \cdot \vec{\omega}\right)$

In cartesian
$\frac{\vec{L}}{m} = \left(\begin{matrix}{\omega}_{x} \\ {\omega}_{y} \\ {\omega}_{z}\end{matrix}\right) \left({x}^{2} + {y}^{2} + {z}^{2}\right) - \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \left(\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \cdot \left(\begin{matrix}{\omega}_{x} \\ {\omega}_{y} \\ {\omega}_{z}\end{matrix}\right)\right)$

$\frac{\vec{L}}{m} = \left(\begin{matrix}{\omega}_{x} \\ {\omega}_{y} \\ {\omega}_{z}\end{matrix}\right) \left({x}^{2} + {y}^{2} + {z}^{2}\right) - \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) \left(x {\omega}_{x} + y {\omega}_{y} + z {\omega}_{z}\right)$

If we pick out the x-component of angular momentum, we have

${L}_{x} / m = {\omega}_{x} \left({x}^{2} + {y}^{2} + {z}^{2}\right) - x \left(x {\omega}_{x} + y {\omega}_{y} + z {\omega}_{z}\right)$

$= {\omega}_{x} \left({y}^{2} + {z}^{2}\right) - x y {\omega}_{y} - x z {\omega}_{z}$

${L}_{x} = \textcolor{red}{{\omega}_{x}} m \left({y}^{2} + {z}^{2}\right) - \textcolor{g r e e n}{{\omega}_{y}} m x y - \textcolor{b l u e}{{\omega}_{z}} m x z$

$= {I}_{x x} {\omega}_{x} + {I}_{x y} {\omega}_{y} + {I}_{x z} {\omega}_{z}$

We can pattern match to find a 3x3 matrix:

$\vec{L} = m \left(\begin{matrix}{y}^{2} + {z}^{2} & - x y & - x z \\ - y x & {x}^{2} + {z}^{2} & - y z \\ - z x & - z y & {x}^{2} + {y}^{2}\end{matrix}\right) \left(\begin{matrix}{\omega}_{x} \\ {\omega}_{y} \\ {\omega}_{z}\end{matrix}\right)$

$= \left(\begin{matrix}{I}_{x x} & {I}_{x y} & {I}_{x z} \\ {I}_{y x} & {I}_{y y} & {I}_{y z} \\ {I}_{z x} & {I}_{z y} & {I}_{z z}\end{matrix}\right) \left(\begin{matrix}{\omega}_{x} \\ {\omega}_{y} \\ {\omega}_{z}\end{matrix}\right)$

So that

$I = m \left(\begin{matrix}{y}^{2} + {z}^{2} & - x y & - x z \\ - y x & {x}^{2} + {z}^{2} & - y z \\ - z x & - z y & {x}^{2} + {y}^{2}\end{matrix}\right)$

In this generalised form, inertia is a 3x3 tensor (ie of rank 2).

If we consider simple circular rotation of the mass in a plane about the origin, we can set the co-ordinate system so that rotation is about the z-axis at z = 0, ie with ${\omega}_{x} = {\omega}_{y} = 0$ and ${\omega}_{z} = \omega$.

Angular momentum then is

$L = m \left(\begin{matrix}{y}^{2} & - x y & 0 \\ - y x & {x}^{2} & 0 \\ 0 & 0 & {x}^{2} + {y}^{2}\end{matrix}\right) \left(\begin{matrix}0 \\ 0 \\ {\omega}_{z}\end{matrix}\right)$

$= m \left({x}^{2} + {y}^{2}\right) {\omega}_{z} = m {r}^{2} {\omega}_{z} = I \omega$

In this simplified form, $I = m {r}^{2}$ which is a scalar (but still also a tensor of rank zero).