1. The problem statement, all variables and given/known data (Q) You hang a thin hoop with Radius R over a nail at the rim of the hoop. You displace it to the side through an angle B (Beta) from its equilibrium position and let it go. What is its angular speed when it returns through its equilibrium position? 2. Relevant equations Change in Potential energy = Mghcm where hcm is the distance moved by the center of mass. Rotational Kinetic Energy = 1/2 (IW^2) where W is the angular speed. 3. The attempt at a solution The distance upward moved by the center of mass = R-RCosB. Thus, change in potential energy = MgR(1-CosB). This should be equal to the rotational Kinetic energy which is 1/2(IW^2). Since this is a hoop(not sure what that means), I = MR^2 (Hope I am correct!!) Thus, MgR(1-CosB) = 1/2 (MR^2)(W^2) Thus, W = sqrt [(2g/R)(1-CosB)]. Unfortunately, according to my book, the 2 isn't supposed to be there. Can someone please help me figure out why this is so? Thank-you very much!!