Finding Angular Speed of Hoop Using Energy Methods

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Homework Statement



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

Homework 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.

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.
 

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