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Pendulum speed with only height from equilibrium

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A pendulum is dropped from 0.50 metres above the equilibrium position. What is the speed of the pendulum as it passes through the equilibrium position? (The pendulum is in a isolated system.)

    h1= 0.05m
    h2= 0.00m
    g= 9.81 m/s^2
    Length of string (L)= 1.0m.

    2. Relevant equations
    Em=Ek+Ep
    Em=[(1/2)mv^2]+[mgh]
    v=√(2Ek/m)


    3. The attempt at a solution

    I'm not entirely sure what to try. I'm not given mass, or any value of any form of Energy, be it Kinetic or Gravitational Potential. I've tried to find the value of mass, but I don't have the energy or velocity factor to figure it out;

    v=√(2Ek/m)
    m=(2Ek/v^2), or
    m=(Ep/gh)

    Just don't work because I don't have that information. I'm totally lost... And that's all the formulas I can extrapolate from what information they've given me. All I need is a formula to follow, and I can get it from there, thanks muchly.
     
  2. jcsd
  3. Apr 25, 2009 #2

    rock.freak667

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    At 0.05 metres what is the potential energy?

    This same energy is converted entirely to ke as it passes through the equilibrium position.

    1/2 mv2=mgh
     
  4. Apr 25, 2009 #3
    I'm not supplied potential, kinetic, or mechanical energy.
     
  5. Apr 25, 2009 #4

    rock.freak667

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    Using the formula, m cancels out. you have g and h. find v.
     
  6. Apr 25, 2009 #5
    Oh, doy, thanks so much. Got it now.
     
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