# Conservation of Rotational Energy Question

by Whiterice
Tags: conservation, energy, hoop, rotational, string
 P: 1 1. The problem statement, all variables and given/known data A string is wrapped several times around the rim of a small hoop with radius 8.00 and mass 0.180 . The free end of the string is held in place and the hoop is released from rest. After the hoop has descended 95.0 , calculate the angular speed and speed of its center. 2. Relevant equations U = mgh K_trans = 1/2mv^2 K_rot = 1/2Iω^2 3. The attempt at a solution I set the starting point was at a height of .95m. K_i + U_i = K_f + U_f 0 + mgh = 1/2mv^2+1/2Iω^2 + 0 mgh = 1/2mv^2 + 1/2m(r^2)(ω^2) (.18)(9.8)(.95) = (.5)(.18)(v^2) + (.5)(.18)(.08^2)(ω^2) But then I have two variables in the same equation and I'm not sure where to go. I know that the speed at the center is going to be equal to v (the linear velocity) and angular speed is ω.
 HW Helper Thanks P: 10,754 It is as if the hoop rolled down on the string. There is a relation between the velocity of translation and angular speed of rotation for the case "rolling without slipping". You can figure it out if you answer the question: what distance does the hoop travel down on the string while it turns one? ehild

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