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Kinetic Energy and Moment of Inertia

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A hoop with mass m= 10 kg, radius r=0.6 m, and moment of inertia I=mr^2 is rolling without slipping with a linear velocity of 8.5 m/s. What is the total kinetic energy (in J).


    2. Relevant equations
    K=1/2 m v^2
    Kr=1/2 I ω^2


    3. The attempt at a solution
    The kinetic energy of the object is
    K = 1/2 m v^2
    K = 1/2 (10) (8.5) = 42.5 J

    The rotational kinetic energy
    Kr = 1/2 (10 * 0.6^2) (8.5/0.6)^2 = 361.25 J

    Total kinetic energy
    42.5 + 361.25 = 403.75 J

    I am not sure about my solution since the answer I end up with is not included in the choices of the question.
     
  2. jcsd
  3. Jun 3, 2013 #2
    Have a look at this part again:

    Also, Kt = K + Kr = 1/2*m*v2 + 1/2*m*r2*(v/r)2 = ?
     
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