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Calculate the mechanical energy

  1. Aug 20, 2010 #1
    Thanks in advance for the help!! I really appreciate it :D

    1. The problem statement, all variables and given/known data
    A pendulum was swung once and when the pendulum bob was at the lowest point of its swing, it broke a photogate light beam. The following information was collected:

    • Diameter of pendulum bob = 3.50cm = 0.035m
    • mass of pendulum bob= 240.3g = 0.2403kg
    • initial height of pendulum bob = 48.0cm = 0.48m
    • length of pendulum string = 2.14m
    • time interval of photogate light interruption = 11.8ms = 0.0118s

    Calculate the mechanical energy at the start position,S, and the lowest point of the pendulum's swing,L.

    2. Relevant equations

    vL=[tex]\frac{d}{t}[/tex] , where d is the diameter and t is the time
    Emechanical=[tex]\frac{1}{2}[/tex]mv2+mg[tex]\Delta[/tex]h

    3. The attempt at a solution

    yea i'm kind a positive this is not right but here's the only thing that i can come up with...

    vL=[tex]\frac{d}{t}[/tex]
    vL=[tex]\frac{0.035}{0.0118}[/tex]
    vL=2.97m/s

    at L,
    Emechanical=[tex]\frac{1}{2}[/tex](0.2403kg)(2.97m/s)2+(0.2403kg)(9.8m)(0)
    I'm guessing the height of the pendulum bob at it's lowest is 0m
    Emechanical= 1.06J

    and for S, I'm guessing it has something to do with the law of conservation of energy...

    yea i think i'm wrong since i didn't use some of the information given and later on it tells me to make a conclusion as to whether or not the pendulum demonstrated the law of conservation of energy ><
     
  2. jcsd
  3. Aug 21, 2010 #2

    kuruman

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    The lowest point of the motion is the equilibrium position when the pendulum just hangs at rest. Is the photobeam blocked or unblocked at that point?
     
  4. Aug 22, 2010 #3
    the photobeam should be blocked at that point and the question also give a diagram and the bob does continue on after reaching the lowest point, so it doesn't just hang at rest
     
  5. Aug 22, 2010 #4

    kuruman

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    I know that. So what is Δh from the point when the photogate is blocked for the first time until the pendulum reaches the lowest point?
     
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