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Kinetic energy in parabolic coordinates

  1. Jul 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove that in parabolic coordinates [itex]\alpha,\beta[/itex] the kinetic energy is [itex]T=m/8(\alpha+\beta)(\dot\alpha^2/\alpha+\dot\beta^2/\beta)[/itex]

    2. Relevant equations

    Parabolic coordinates are defined as follows: [itex]\alpha=r+x, \beta=r-x[/itex] with [itex]r=\sqrt{x^2+y^2}[/itex]

    3. The attempt at a solution

    I don't know how to proceed in this situation: in simpler case (spherical or cylindrical coordinates) I write down the three components of velocity using geometrical intuition (e.g. [itex]v_\rho=\dot \rho, v_\phi=\rho \dot\phi,v_z=\dot z[/itex], because I see they are right...). What if I get only the definition of the new coordinates?
     
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  3. Jul 16, 2011 #2

    Redbelly98

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    Welcome to physics forums.

    I think the key here is to get x and y in terms of [itex]\alpha, \beta[/itex]. And you hopefully already know how to express the kinetic energy in x, y coordinates, right?
     
  4. Jul 16, 2011 #3
    That's right! I check explicitly and it works. Now that I have the kinetic energy, it will be doable to write down the equations of motion. Thank you.
     
  5. Jul 16, 2011 #4

    Redbelly98

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