Conservation of Rotational Energy Question

by Whiterice
Tags: conservation, energy, hoop, rotational, string
Whiterice is offline
Apr1-12, 11:58 PM
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.
Phys.Org News Partner Science news on
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
ehild is offline
Apr2-12, 12:52 AM
HW Helper
P: 9,834
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?


Register to reply

Related Discussions
Rotational Motion and Conservation of Energy Introductory Physics Homework 4
Rotational Energy Conservation Introductory Physics Homework 2
Conservation of energy in rotational motion Introductory Physics Homework 5
Rotational Energy conservation Introductory Physics Homework 4
Conservation of rotational energy Introductory Physics Homework 1