# Finding Angular Speed

1. Nov 4, 2007

### Heat

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 cm and mass 0.180kg. The free end of the string is held in place and the hoop is released from rest (the figure View Figure ). After the hoop has descended 95.0cm, calculate

the angular speed of the rotating hoop and the speed of its center.

2. Relevant equations

3. The attempt at a solution

First attempt was like this

I thought that KE is 0 at initial ,and thought that I could get angular speed this way, but that would just give me that angular speed is 0. :O

Then I tried a formula given by the book for a yoyo, which is vcm = sqrt (4/3gh)
that did not result, and then is another formula that vcm = R(omega)

I don't have omega, and don't have vcm. So I can't use it at this instance. :(

2. Nov 4, 2007

### Astronuc

Staff Emeritus
The hoop is decending under the force of gravity by virtue of its weight, mg. At the same time, it is unraveling the string. It's vertical velocity is the same as the tangential velocity (speed) at it's radius. And then there is the rotational inertia to consider.

For moments of inertia and rotational motion equations, see -

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

3. Nov 13, 2007

### tmle04

i dont get it..so the angular speed of the rotating hoop w = ?

and

the angular speed of the speed of its center v = Square_Root_Of_((4/3)gh) .... ??