posted by Nikki .
A 290-N sphere 0.20 m in radius rolls without slipping 6.0 m down a ramp that is inclined at 31° with the horizontal. What is the angular speed of the sphere at the bottom of the slope if it starts from rest?
Use conservation of energy and include the rotational kinetic energy, (1/2) I w^2. They should have told you whether the sphere has uniform density or is hollow inside. It makes a difference.
For a uniform-density sphere,
M g H = (1/2) M V^2 + (1/2) I w^2
= (1/2) M V^2 + (1/2)(2/5)M R^2*(V/R)^2
= (7/10)M V^2
M cancels out. H is the elevation loss, 6.0 sin 31 = 3.09 meters
Solve for V, the final velocity. For the final angular speed, use
w = V/R