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Pendulum Speed and total ME

  1. Oct 16, 2007 #1
    [SOLVED] Pendulum Speed and total ME

    1. The problem statement, all variables and given/known data

    The length of a simple pendulum is 0.84 m and the mass of the particle (the "bob") at the end of the cable is 0.68 kg. The pendulum is pulled away from its equilibrium position by an angle of 7.8 ° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

    2. Relevant equations

    E = .5mv^2+.5Iw^2+mgh+.5kx^2
    w = sqrt(g/L)
    L-Lcos@ = max height of pendulum bob
    I of pendulum = (1/3)ML^2

    3. The attempt at a solution

    so w = 3.416 rad/s
    The total energy:
    .5(.68)(v^2) + .5(1/3 ML^2)(w^2) + mgh + .5kx^2
    kx^2 is potential energy, which is 0 at the bottom of the swing
    .34v^2 + (1/6)(.68*.84^2)(3.416^2) + .68*-9.8*.00777
    .34v^2 + .88137
    v^2 = 2.5922
    v = 1.61 m/s

    Plugging this in gives a total ME of 1.7627 J

    The v and total E are wrong, but I'm not sure how to remedy this. Ideas, please? Thanks!
  2. jcsd
  3. Oct 16, 2007 #2


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    Whao, slow down there, your using way too many equations! When the plumb bob is at it's highest point, what type of energy does it have?
  4. Oct 16, 2007 #3
    It has potential energy, mgh. But the question asks for total mechanical energy at the bob's LOWEST point.
  5. Oct 17, 2007 #4


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    Yes, but if you know that at it's highest point it only has potential energy and you know the energy is conserved, then what is the value of the total energy at the lowest point?
  6. Oct 17, 2007 #5
    So I totally overanalyzed that one... oh well.

    Thanks for the help!
  7. Oct 17, 2007 #6


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    No worries :approve:
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