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
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.