1. ## cylindrical coordinates

Angular momentum is defined as L = r x mv , where v is contained in the xy plane. show that in cylindrical coordinates angular momentum takes the form $L = mr^{2}\dot{\phi}$

i know v = $\dot{r} \hat{r} + r \hat{\phi} \dot{\phi}$
$L =r \times m (\dot{r} \hat{r} + r \hat{\phi} \dot{\phi} )$

am not sure what i get when you take the corss product of

$r \times m\dot{r} \hat{r}$ ?

2. ## Re: cylindrical coordinates

$\vec{L} = \vec{r} \times m\vec{v}$

Setting $\vec{v} = \frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt} \hat{\phi}$, and $\vec{r} = r \hat{r}$ gives:

$\vec{L} = r \hat{r} \times m(\frac{dr}{dt} \hat{r} + r \frac{d\phi}{dt} \hat{\phi})$

$\vec{L} = r \hat{r} \times m\frac{dr}{dt} \hat{r} + r \hat{r} \times mr \frac{d\phi}{dt} \hat{\phi}$

and we have $\hat{r} \times \hat{r} = 0$ and $\hat{r} \times \hat{\phi} = \hat{z}$, so

$\vec{L} = mr^2 \frac{d\phi}{dt} \hat{z}$

The radial component of velocity doesn't contribute to the angular momentum.

- Hollywood