A bike tire (1.3 kg) tire with a radius of 0.33 m starts from rest and rolls down?

Question

from the top of a hill that is 26.0 m high. What is the translational speed of the tire when it reaches the bottom of the hill in units of m/s? Assume acceleration of gravity is 9.81 m/s^2 and that the tire is a hoop with I = mr^2.

Answer 1

from conservation of energy,
mgh = 0.5mv^2 + .5Iw^2

w = v/r

so 0.5Iw^2 = 0.5 * mr^2 * (v/r)^2 = 0.5mv^2

so mgh = 0.5mv^2+0.5mv^2 = mv^2
or, gh = v^2
or, v = sqrt(gh) = sqrt(9.81*26) = 15.97m/s