# Physics

A solid ball of mass 3 kg, rolls down a hill that is 7 meters high.  What is the rotational KE at the bottom of the hill?

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2. 0
1. loss of potential energy = m g h = 3*9.81* 7 = 206 Joules

= translational Ke + rotational Ke

= (1/2) m v^2 + (1/2)I omega^2

Note
I = (2/5) m r^2
v = omega r
so
206=(1/2)m v^2 + (1/2)(2/5)m r^2 v^2

412 = 1.4 *3 v^2
v = 9.9 m/s

then calculate (1/2) (2/5)(3) v^2

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posted by Damon

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