PX155 - E8 - angular momentum of a particle moving in a circle

\vec{L} = \vec{r} \times \vec{p} = r m v \vec{k} = m r^{2} ω \vec{k}

L = | \vec{L} | = m r^{2} ω

\vec{L} = I \vec{ω} $ $ a n d $ $ T = \frac{1}{2} I ω^{2}

$I$ could be taken as the angular counter part of mass $m$ , but unlike $m$ , $I$ depends on the axis of rotation
note: $\vec{L} = I \vec{ω}$ is not a general result
- $I \vec{ω} \neq \vec{r} \times \vec{p}$ if:
  - plane of rotation doesn't contain the origin of $\vec{r}$
  - origin isn't the centre of rotation