Angular Velocity?

Question

An 7.30-cm-diameter, 370 g sphere is released from rest at the top of a 1.50-m-long, 19.0 degree incline. It rolls, without slipping, to the bottom. What is the sphere's angular velocity at the bottom of the incline? What fraction of its kinetic energy is rotational?

Answer 1

for a sphere I=(2/5)*m*R^2

The total energy gain is
0.370*9.81*1.5*sin(19)
1.77 J

The energies are
.5*m*v^2 and
.5*I*v^2/R^2

2*1.77/0.370=v^2*(1+0.4)

v=2.61 m/s
and
w=100*2.61/7.3
w=35.8 rad/sec

.5*0.370*6.83
0.5*0.4*0.370*6.83

Translational 1.26 J
Rotational 0.51 J

Rot/Tran=0.51/1.77
0.288

j