Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic energy in polar coordinates

  1. Nov 3, 2006 #1
    if i have a system that i'm describing using polar coordinates, do i need to have an additional term for rotational kinetic energy? it would seem like this is covered since my velocity is in terms of the r and theta basis vectors. (i.e. i will have a term that covers the rotational movement ala theta). i'm not really sure about this, though. thanks if anyone is able to help.
  2. jcsd
  3. Nov 3, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    More information would be helpful, because in a general way you're wrong (example, if your velocity just describes linear motion). What's it rotating about? Is it just spinning, etc.
  4. Nov 3, 2006 #3
    two particles constrained to each other, free to rotate. each one is described by the position vector a(rhat) + b(theta hat). my velocity would have two components, r and theta. the velocity for each one would be in terms of r and theta, which would (i would think) would cover translational and rotational movement. the last sentence sums up my question fairly well. thanks for your input!-
  5. Nov 4, 2006 #4
    I assume you are trying to write a lagrangian though I may be incorrect. You will have two kinetic energy terms in your lagrangian in this case, one proportional to the time derivative of your r position, and one proportional to the time derivative of the theta direction.
  6. Nov 5, 2006 #5

  7. Nov 5, 2006 #6
    could you elaborate, please?
  8. Nov 5, 2006 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Write the kinetic energy of each point-particle in rectangular coordinates, then convert to polar coordinates. If there are relations among your point-particles [e.g. extended rigid bodies], you may be able to group terms and reinterpret.
  9. Nov 5, 2006 #8
    yes, this is definitely the best way to see what is going on if you are confused. Start with just a 1/2mv^2 term for each direction and then sub in the substitutions into polar coordinates, yuo should much of it will cancel out and you will be left with a simple expression with a kinetic energy term for each of your coordinates, r and theta.
  10. Jun 8, 2010 #9


    User Avatar

    Here is the derivation.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook