- #1

JulienB

- 408

- 12

## Homework Statement

Hi everybody! I would like to discuss with you a problem that I am wondering if I understand it correctly:

Find expressions for the cartesian components and for the magnitude of the angular momentum of a particle in cylindrical coordinates ##(r,\varphi,z)##.

## Homework Equations

##p_i = mx_i##, ##\vec{L} = \vec{r} \times \vec{p}##

## The Attempt at a Solution

The ##r## confuses me a lot, since we normally use ##\rho## in cylindrical coordinates. I've set up ##\vec{r}## laying in the ##x,y##-plane (i.e. ##z=0##) and ##z## as the rotation axis, but am I allowed to do that? If so, the rest seemed pretty simple:

##x = r \cos \varphi##

##y = r \sin \varphi##

##z = 0##

##\dot{x} = \dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi##

##\dot{y} = \dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi##

##\dot{z} = 0##

##p_x = m\dot{x} = m(\dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi)##

##p_y = m\dot{y} = m(\dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi)##

##p_z = m\dot{z} = 0##

##\vec{L} = \vec{r} \times \vec{p}##

##L_x = y p_z - z p_y = 0##

##L_y = z p_x - x p_z = 0##

##L_z = x p_y - y p_x = mx (\dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi) - my(\dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi)##

As expected from my initial conditions, the angular momentum vector points in the z-direction, and after simplification I get the well-known expression:

##L_z = mr^2\dot{\varphi}##

But I find it strange to set up the ##z## component to ##0##, and if it was not the case the angular momentum vector could be pointing basically any direction depending on what angle ##\theta## it is leaning (in spherical coordinates). I've tried to solve the problem that way, but I can't get my hands on a proper expression for ##x##,##y## and ##z## that does not involve ##\theta## or a very unpractical arctan or I don't know what. Any suggestion?Julien.