A solid, uniform disk of radius 0.250 m and mass 57.7 kg rolls down a ramp of length 3.70 m that makes an angle of 14° with the horizontal. The disk starts from rest from the top of the ramp. (a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp. (b) Find the angular speed of the disk at the bottom of the ramp. I've tried the first part. I set the equation up as... mgh=1/2mv2+1/2(1/2) I ω^2 then mgh=1/2mv2+1/4mv2 Might have been wrong there on the second part of the equation? Does the Iw^2 just turn into mv^2 since its not slipping? I then added together the fractions and got mgh=3/4mv2 Plugged in the values but didnt get the right answer. About 99% sure I went somewhere wrong in getting the right equation. I noticed too I never used the radius or angle, not sure if theyre needed. Any help would be great thanks.