Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Advanced Physics Homework Help
Angular momentum in cartesian coordinates (Lagrangian)
Reply to thread
Message
[QUOTE="JulienB, post: 5485794, member: 574060"] [h2]Homework Statement [/h2] 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)##. [h2]Homework Equations[/h2] ##p_i = mx_i##, ##\vec{L} = \vec{r} \times \vec{p}## [h2]The Attempt at a Solution[/h2] 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Advanced Physics Homework Help
Angular momentum in cartesian coordinates (Lagrangian)
Back
Top