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Conservation of Energy and Moment of Inertia

  1. Oct 13, 2007 #1
    1. The problem statement, all variables and given/known data

    (Q) You hang a thin hoop with Radius R over a nail at the rim of the hoop. You displace it to the side through an angle B (Beta) from its equilibrium position and let it go. What is its angular speed when it returns through its equilibrium position?

    2. Relevant equations

    Change in Potential energy = Mghcm where hcm is the distance moved by the center of mass.

    Rotational Kinetic Energy = 1/2 (IW^2) where W is the angular speed.

    3. The attempt at a solution

    The distance upward moved by the center of mass = R-RCosB.

    Thus, change in potential energy = MgR(1-CosB).

    This should be equal to the rotational Kinetic energy which is 1/2(IW^2).

    Since this is a hoop(not sure what that means), I = MR^2 (Hope I am correct!!)

    Thus, MgR(1-CosB) = 1/2 (MR^2)(W^2)
    Thus, W = sqrt [(2g/R)(1-CosB)].

    Unfortunately, according to my book, the 2 isn't supposed to be there. Can someone please help me figure out why this is so? Thank-you very much!!
  2. jcsd
  3. Oct 13, 2007 #2


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    Staff: Mentor

  4. Oct 13, 2007 #3
    Thanks a ton! That was really a wake-up call. I can't believe how careless I've been. Thanks again!
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