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Rolling

AP Physics 1· difficulty 3/5

A solid cylinder (I=12MR2I = \tfrac{1}{2}M R^2) rolls without slipping from rest down a height hh. Its translational speed at the bottom is

  • A

    2gh\sqrt{2gh}

  • B

    gh\sqrt{gh}

  • C

    (10/7)gh\sqrt{(10/7)gh}

  • D

    (4/3)gh\sqrt{(4/3)gh}

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Explanation

Mgh=12Mv2+12Iω2=12Mv2+14Mv2=34Mv2M g h = \tfrac{1}{2}M v^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}M v^2 + \tfrac{1}{4}M v^2 = \tfrac{3}{4}M v^2. v=(4/3)ghv = \sqrt{(4/3)gh}.

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