Conservation of Energy

AP Physics 1· difficulty 4/5

R

A small bead enters a frictionless vertical circular loop of radius RR at the bottom. What is the minimum entry speed v0v_0 at the bottom that allows it to maintain contact at the top of the loop?

  • A

    v0=3gRv_0=\sqrt{3gR}

  • B

    v0=gRv_0=\sqrt{gR}

  • C

    v0=5gRv_0=\sqrt{5gR}

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  • D

    v0=2gRv_0=\sqrt{2gR}

Explanation

At the top, minimum condition gives vtop2=gRv_{top}^2=gR. Energy conservation between bottom and top: 12v02=12vtop2+g(2R)\tfrac{1}{2}v_0^2 = \tfrac{1}{2}v_{top}^2 + g(2R), so v02=gR+4gR=5gRv_0^2 = gR + 4gR = 5gR.

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