- #1

PKay

- 10

- 0

## Homework Statement

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=Mgy

_{cm}.)

Express your answer in terms of the variables m, M, R, and appropriate constants.

## Homework Equations

[itex]y = 1/2 M R^2 [/itex]

ω=v/r

## 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 +1/2Iω

_{0}

^{2}= 1/2mv

^{2}+ 1/2Iw

_{f}

^{2}

mgR+1/2(1/2MR

^{2}ω

_{0}

^{2}= 1/2mv

^{2}1/2(1/2MR

^{2}w

_{f}

^{2}

ω

_{f}= √{(mGR+1/4MR

^{2}ω

_{0}

^{2})/(1/2mv

^{2}+1/4MR

^{2}}

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!