A solid uniform sphere is released from the top of an inclined plane .25 m tall. the sphere rolls down the plane without slipping and there is no energy lost from friction. What is the translational speed of the sphere at the bottom of the incline?

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

    v = omega r
    so omega = v/r

    Ke = (1/2) m v^2 + (1/2) I v^2/r^2

    = m g (.25)

    (1/2) m v^2 + (1/2) (2/5) m v^2 = .25 m g

    (7/10) v^2 = .25 * 9.81

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