How Do You Calculate the Angular Speed and Center Speed of a Descending Hoop?

[PKay]

Homework Statement

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180kg. The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 60.0cm , calculate the angular speed of the rotating hoop and the speed of its center.

Homework Equations

K=Iω²
U = mgh
ω=v/r
I = 1/2 mr²

The Attempt at a Solution

E_i = mgh = (0.180)(9.8)(0.6) = 1.0584
E_f = 1/2 mv²+1/2 Iω² = 3/4(0.08)²ω²
E_i = E_f
I got ω = 14.9 rad/s (I've also tried 35 rad/s) Is my math wrong or something?

For the second part, I saw somewhere the equation was v=√gh and but I was thinking I could just use ω=v/r but since I didn't have ω I couldn't solve that one.

 

[TSny]

Hello, PKay.

I = 1/2 mr²

Is this the correct formula for a hoop?

 

[PKay]

I think so. The problem is the axis is in the middle for that equation but since it's on the side, I'm not too sure anymore.

 

[TSny]

The equation E_f = 1/2 mv²+1/2 Iω² is correct if you take v to be the speed of the center of mass and I as the moment of inertia about the center of mass.

For a hoop, see http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html