# Find the angular speed of the cylinder

1. Feb 22, 2014

### PKay

1. The problem statement, all variables and given/known data

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed ω0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=Mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

2. Relevant equations

$y = 1/2 M R^2$
ω=v/r

3. The attempt at a solution

I know there's a post for the same problem. But I'm still confused. If I was supposed to ask that person on their thread or something, please tell me so I can delete this thread. I'm new to the forum. Anyways, this is what I tried:
mgh +1/2Iω02 = 1/2mv2 + 1/2Iwf2
mgR+1/2(1/2MR2ω02 = 1/2mv21/2(1/2MR2wf2
ωf = √{(mGR+1/4MR2ω02)/(1/2mv2+1/4MR2}

Apparently, according to masteringphysics, the answer does not depend on mv but now I'm a little stuck. I'm not sure what I should be substituting mv with. Thank you in advance!

2. Feb 22, 2014

### SammyS

Staff Emeritus
Hello PKay. Welcome to PF !

If you are referring to this thread as being the previous post, then you are correct in starting this new thread. For one thing, that thread is rather old. Also, it has been closed.

There is a relationship between v and ω . That should enable you to eliminate one of them.

3. Feb 22, 2014

### PKay

Yes! I was referring to that thread. I didn't know it was closed. Glad I did it correctly.

I actually got it now. I actually still have a related question. C=2piR but why do we have to divide by 2 for the center of mass?

Last edited: Feb 23, 2014