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Rotational Kinetic Energy and moment of inertia

  1. Apr 2, 2008 #1
    [SOLVED] Rotational Kinetic Energy

    1. The problem statement, all variables and given/known data

    A solid cylinder of mass 14.0 kg rolls without slipping on a horizontal surface.
    (a) At the instant its center of mass has a speed of 11.0 m/s, determine the translational kinetic energy of its center of mass.

    2. Relevant equations
    [tex] I_cm = \frac{1}{2}MR^2 [/tex]
    [tex]K_R = \frac{1}{2}I\omega^2 [/tex]

    3. The attempt at a solution
    I can't figure out how to find the moment of inertia without have a radius... any hints?
     
  2. jcsd
  3. Apr 2, 2008 #2

    Doc Al

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    Staff: Mentor

    You won't need the radius. Hint: What does "rolls without slipping" imply?
     
  4. Apr 2, 2008 #3
    That its moving forward and has translational kinetic energy? If thats not it, I do not know what it means.

    Also, I found the translational kinetic energy (if that has to deal with the problem) and I am still stuck...
     
  5. Apr 2, 2008 #4

    Doc Al

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    Staff: Mentor

    "Rolling without slipping" means that the translational and rotational speeds are matched so that the bottom surface doesn't slip with respect to the ground. That condition relates the translational speed to the rotational speed, such that [itex]v = \omega r[/itex].
     
  6. Apr 2, 2008 #5
    I understand

    But I don't have r...
     
  7. Apr 2, 2008 #6

    Doc Al

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    Staff: Mentor

    Calculate the rotational KE in terms of the translational speed. (Apply the condition for rolling without slipping.)
     
  8. Apr 2, 2008 #7
    Ah, I see how the Rs cancel now, thanks.
     
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