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?

Answer 1

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

Answer 2

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