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Energy of translation and energy of rotation

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data
    How would you go about figuring out the ratio of components (rotational and translational energy) of total kinetic energy?

    2. Relevant equations

    3. The attempt at a solution
    i started with KEtotal=KEr+KEtr. KEr=1/2Iw^2; KEtr=1/2mv^2
    use I for a disk-I=1/2mr^2
    I put KEtr over KEr KEtr/KEr for a ratio

    (1/2 mv^2)/1/2Iw^2
    substitute in 1/mr^2 for I and v/r for w

    (1/2 mv^2)/(1/2)(1/2mr^2)(v/r)^2

    cancel m, v and r and 1/2

    left with (1)/(1/2)

    ratio is 2:1 KEtr:KEr ??
  2. jcsd
  3. Jun 24, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's right. For a uniform disk that is rolling without slipping, the translational KE is twice the rotational KE. Good!
  4. Jun 24, 2009 #3
    YAY! Thanks for confirming. I am struggling with physics.
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