Rolling

AP Physics 1· difficulty 4/5

A solid sphere (I=25MR2I = \frac{2}{5}MR^2) rolls without slipping down an incline from height hh. Its speed at the bottom is

  • A

    gh/2\sqrt{gh/2}

  • B

    gh\sqrt{gh}

  • C

    10gh/7\sqrt{10gh/7} (KE split between translation and rotation)

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  • D

    2gh\sqrt{2gh}

Explanation

Energy conservation: mgh=12mv2+12Iω2mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 with ω=v/R\omega = v/R. For solid sphere: mgh=12mv2+15mv2=710mv2mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2, so v=10gh/7v = \sqrt{10gh/7}.

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