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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asked by ML
  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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