solution

Eqn. 1.52 can be used here. The center of mass of the hoop is in the middle. We're hanging it from the rim. So $L ~=~ R$. What's the moment of inertia? Be careful! We're not rotating it about the center. So $I ~=~ I_{cm}+ mR^2 ~=~ 2mR^2$. Putting this together we have

$$\omega^2 ~=~ {mgL\over I} ~=~ {mgR\over 2mR^2} ~=~ {g\over 2 R}$$ (1.54)