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Finding Angular Speed of Hoop Using Energy Methods

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data

    You hang a thin hoop using radius R over a nail at the rim of a the hoop. You displace it to the side (within the plane of the hoop) through angle β from its equilibrium position and let it go. Using U = M*g*y(center of mass), what is the angular speed when it returns to its equilibrium position

    2. Relevant equations

    ycm = R - R*cosβ

    K = U → 0.5*I*ω^2 = M*g*y(center of mass), where I = MR^2 for thin walled and hollow cylinders.

    3. The attempt at a solution

    0.5*I*ω^2 = M*g*ycm

    0.5*M*R^2*ω^2 = M*g*(R - R*cosβ)

    ω = √((2*g*(1 - cosβ))/R)

    But my book, which seems to never be wrong, has everything but the 2,
    ω = √((g*(1 - cosβ))/R)

    I just can't see how I could be wrong.
  2. jcsd
  3. Aug 4, 2013 #2
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