Conservation of Energy and Moment of Inertia

[mit_hacker]

Homework Statement

(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?

Homework 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.

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!

 

[Astronuc]

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

Don't forget the parallel axis theorem.

http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html#pax

 

[mit_hacker]

Thanks a ton! That was really a wake-up call. I can't believe how careless I've been. Thanks again!