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.
2. Relevant equations
U = mgh
I = 1/2 mr2
3. The attempt at a solution
Ei = mgh = (0.180)(9.8)(0.6) = 1.0584
Ef = 1/2 mv2+1/2 Iω2 = 3/4(0.08)2ω2
Ei = Ef
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.