A solid uniform sphere is released from the top of an inclined plane .25 m tall. the sphere rolls down the plane without slipping and there is no energy lost from friction. What is the translational speed of the sphere at the bottom of the incline?

Answer 1

Ke = (1/2) m v^2 + (1/2) I omega^2

v = omega r
so omega = v/r

Ke = (1/2) m v^2 + (1/2) I v^2/r^2

= m g (.25)

(1/2) m v^2 + (1/2) (2/5) m v^2 = .25 m g

(7/10) v^2 = .25 * 9.81