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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

    Office_Shredder

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    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
    YES

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

    robphy

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    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

    tot

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    [​IMG]
    Here is the derivation.
     
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