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Conservation of Rotational Energy Question

  1. Apr 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A string is wrapped several times around the rim of a small hoop with radius 8.00 and mass 0.180 . The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 95.0 , calculate the angular speed and speed of its center.


    2. Relevant equations
    U = mgh
    K_trans = 1/2mv^2
    K_rot = 1/2Iω^2


    3. The attempt at a solution
    I set the starting point was at a height of .95m.
    K_i + U_i = K_f + U_f
    0 + mgh = 1/2mv^2+1/2Iω^2 + 0
    mgh = 1/2mv^2 + 1/2m(r^2)(ω^2)
    (.18)(9.8)(.95) = (.5)(.18)(v^2) + (.5)(.18)(.08^2)(ω^2)

    But then I have two variables in the same equation and I'm not sure where to go. I know that the speed at the center is going to be equal to v (the linear velocity) and angular speed is ω.
     
  2. jcsd
  3. Apr 2, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    It is as if the hoop rolled down on the string. There is a relation between the velocity of translation and angular speed of rotation for the case "rolling without slipping". You can figure it out if you answer the question: what distance does the hoop travel down on the string while it turns one?

    ehild
     
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