Finding Angular Speed of Hoop Using Energy Methods

[student34]

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.

 

[milesyoung]

Your moment of inertia is correct if the axis of rotation of the hoop is through its center of mass, but that's not true for your problem. Easy fix:
https://en.wikipedia.org/wiki/Parallel_axis_theorem