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## Homework Statement

A uniform disk is set into rotation with an initial angular speed about its axis through its center. While still rotating at this speed, the disk is placed in contact with a horizontal surface and released. What is the angular speed of the disk once pure rolling takes place (in terms of omega final and initial)?

## Homework Equations

1/2mv(int)^2 + 1/2Iw(int)^2 = 1/2mv(final)^2 + 1/2Iw(final)^2

Torque=d(L)/d(t)

Normal Force=mg

I=1/2mr^2

L=Iw

## The Attempt at a Solution

I knew that the only once a disk rolls without slipping does w=v/r

I simplified the first equation as follows:

v(int)^2 + 1/2r^2w(int)^2 = v(final)^2 + 1/2v(final)^2

v(int)^2 + 1/2r^2w(int)^2 = 3/2v(final)^2

I also knew that the only force acting on the disk was the surface. So I knew that

Fr= d(L)/d(t)

mgr=[w(final)I-w(int)I]/d(t)

from there I got that

g=[r(w(final)-w(int))]/2d(t)

I really have no idea where to go from here though. Any push in the right direction would be appreciated!

Thanks!