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1/2 m v^2 in a circle?

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    in circular motion (e.g. a pendulum) is the kinetic energy still 1/2 m v ^2 or is it a different equation?

    2. Relevant equations
    1/2 m v ^2

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 12, 2017 #2

    gneill

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    Staff: Mentor

    It depends on what the object is. For a point mass at the end of a massless string you might use 1/2 m v^2. For a more complicated object that can't be viewed as a point mass you'll want to look at the moment of inertia and rotational kinetic energy. Investigate: "Physical Pendulum".
     
  4. Feb 17, 2017 #3

    James R

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    Science Advisor
    Homework Helper
    Gold Member

    Just to add: for a point mass, the kinetic energy depends on the speed (##|\vec{v}|^2##) rather than the (vector) velocity, so the direction of travel (e.g. circular motion) doesn't matter.
     
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