Need Some Assistance With Angular Motion Problem?

Question

An 8.30-cm-diameter, 360 g sphere is released from rest at the top of a 2.00-m-long, 19.0 degree incline. It rolls, without slipping, to the bottom.

1) What is the sphere's angular velocity at the bottom of the incline?

2) What fraction of its kinetic energy is rotational?

Answer 1

Initial potential energy:

PE = mgh = 0.360 * 9.81 * 2 * sin 19 = 2.30 J

Initial kinetic energy = 0

Final potential energy = 0

Final kinetic energy = 1/2 m v^2 + 1/2 I  omega ^2

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

For a solid sphere: I = 2mr^2/5

KE = 1/2 mv^2 (1 + 2/5) = 0.7 mv^2

The final kinetic energy = initial potential energy = 2.30 J

solving for v: v = 3.02 m/s

omega = v/r = 72.8 rad/s

fraction of KE that is rotational = (2/5) / (1 + 2/5) = 2/7 = 28.6%