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Conservation of Energy

AP Physics 1· difficulty 3/5

A small block (frictionless track) is released from rest at height hh and enters a vertical loop of radius RR. The minimum height hh for it to complete the loop is

  • A

    RR

  • B

    32R\tfrac{3}{2}R

  • C

    2R2R

  • D

    52R\tfrac{5}{2}R

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Explanation

At loop top, v2=gRv^2 = gR. Energy conservation: gh=12v2+g(2R)=12gR+2gR=52gR⇒h=52Rg h = \tfrac{1}{2} v^2 + g(2R) = \tfrac{1}{2}gR + 2gR = \tfrac{5}{2}gR \Rightarrow h = \tfrac{5}{2}R.

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