Question

An 8.80-cm-diameter, 380 g solid sphere is released from rest at the top of a 2.00-m-long, 16.0 ∘ incline. It rolls, without slipping, to the bottom.

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

What fraction of its kinetic energy is rotational?

Answer 1

translation kE

kEt = 0.5 m v^2

rotational kE

KEr = 0.5 I w^2 = 0.5 ( 2/5) m r^2 ( v^2/r^2)

kEr = 0.2 m v^2

using conservation of energy

mgh = kEr + kEt

m * 9.8* 2 sin 16 = 0.2* m v^2 + 0.5 m v^2

v = 2.778

w = 2.778 / (0.088 /2) = 63.14 ad/s

======

ratio = 0.2 m v^2 / ( 0.2 m v^2 + 0.5 m v^2) = 2/7

======

Comment before rate in case any doubt, will reply for sure.. goodluck