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Rolling

AP Physics 1· difficulty 3/5

For an object with moment of inertia I=kMR2I = k M R^2 rolling without slipping from rest down height hh, the translational speed at the bottom is

  • A

    gh\sqrt{gh}

  • B

    gh/(1+k)\sqrt{gh/(1+k)}

  • C

    2gh/(1+k)\sqrt{2gh/(1+k)}

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  • D

    2gh\sqrt{2gh}

Explanation

Mgh=12Mv2+12(kMR2)(v/R)2=12Mv2(1+k)M g h = \tfrac{1}{2}M v^2 + \tfrac{1}{2}(k M R^2)(v/R)^2 = \tfrac{1}{2}M v^2 (1+k). So v=2gh/(1+k)v = \sqrt{2gh/(1+k)}.

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