Physics help please?

Question

A 270 N sphere 0.20 m in radius rolls, without slipping 6.0 m down a ramp that is inclined at 37° with the horizontal. What is the angular speed of the sphere at the bottom of the hill if it starts from rest?

Answer 1

assuming no loss, the energy of the system is
270*6*sin(37)
975 J

the KE of the system is
.5*m*v^2+.5*I*w^2
I will assume a solid sphere
I=2*m*r^2/5

w=v/r
so
.5*m*w^2*r^2*(7/5)

using g=9.81
w=sqrt(975*981/(7*270*.2^2))

w=112 rad/s
j