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)?
1/2mv(int)^2 + 1/2Iw(int)^2 = 1/2mv(final)^2 + 1/2Iw(final)^2
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
from there I got that
I really have no idea where to go from here though. Any push in the right direction would be appreciated!