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Hoop kinetic energy

  1. Dec 7, 2013 #1
    Hey PF, this is just a quick question:

    If there's a hoop of mass M and radius a rotating around its vertical axis (see pic) and I want to write the kinetic energy for the Lagrangian is it just [itex] T = \frac{1}{2} M a^2 w^2 [/itex] ? Considering w the angular velocity
     

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  2. jcsd
  3. Dec 7, 2013 #2
    Not quite. What is the moment of inertia for this set up?
     
  4. Dec 7, 2013 #3
    [itex] T =\frac{1}{2} w^2 \frac{1}{2} M a^2 [/itex] ? is it more close to the answer now?
     
  5. Dec 7, 2013 #4
    Yeah it's right now
     
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