pleaseeeee help mee?

Question

A 190 N sphere 0.20 m in radius rolls, without slipping 6.0 m down a ramp that is inclined at 28° with the horizontal. What is the angular speed of the sphere at the bottom of the hill if it starts from rest?

that answer doesn't work :'( I am so confused

Answer 1

I = 2mr^2/5
α = T/I = 5Wgsinφ/(2Wr^2)
θ = s/(2πr)
ω^2 = 2αθ
ω^2 = 5sgsinφ/(2πr^3)
ω^2 = (5)(6.0)(9.80665)(sin(28°))/(2π(0.2)^3)
ω ≈ 52.41924 rad./s ≈ 501 rpm