A 89cmdiameter 406 g sphere is released from rest at the top

A 8.9-cm-diameter, 406 g sphere is released from rest at the top of a 2.4-m-long, 25° incline. It rolls, without slipping, to the bottom.

(a) What is the sphere\'s angular velocity at the bottom of the incline? _________ rad/s

(b) What fraction of its kinetic energy is translational? _________

h = 2.4 * sin 25 = 1.01 m

potential energy loss = gain of kinetic energy

m g h = (1/2) m v² + (1/2) I w²

m g h = (1/2) m v² + (1/2) * (2/5) m R² w²

g h = (1/2) v² + (1/5) v² = (7/10) v²

v = sqrt [ (10/7) g h] = sqrt [ (10/7) * 9.8 * 1.01]

v = 3.76 m/s

w = v / R = 3.76 / 0.0445 = 84.5 rad/s

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Total KE = (7/10) m v²

Rotational KE = (1/5) m v²

Rotational KE / Total KE

= (1/5) * (10/7) = 2 / 7 = 0.286