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Pendulum, conservation of energy theorem

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    A simple pendulum whose length is L=2 meters has a mass of m=2kg. When the angle between the pendulum and the vertical is 35 degrees, it has a speed of 1.2 m/s. Find the pendulum's speed when the pendulum is at its lowest height.


    2. Relevant equations
    K = 0.5mv2
    U = mgh
    E = K+U

    3. The attempt at a solution
    I arbitrarily set that h=0 when theta = 35 degrees
    [​IMG]
    NOTE: I have found the right answer by setting h=0 at the pendulum's lowest point, but I can't find the right answer when I set h=0 when theta = 35 degrees. Since h=0 can be arbitrarily set, I would like to know where is the mistake.

    Since E = K + U, and U = 0
    then E = K = 0.5mv2 = 0.5(2)(1.22)= 1.44 J
    Now, at any point E = K + U = mg*-[L-Lcos(theta)] + 0.5mv2 = mg[Lcos(theta) - L] + 0.5mv2

    Now, I am pretty sure the error is in what follows:
    At the pendulum's lowest point, theta = 0 degrees
    then mg[Lcos(theta) - L] + 0.5mv2 = 0.5mv2 = 1.44 J, solving for v gives back the 1.22 m/s, which is clearly not the answer. If i set theta = 35 degrees, I get v = 2.38 m/s, which is also not correct.

    The correct answer is 2.9 m/s
    Can anyone help?

    Thank you
     
  2. jcsd
  3. Nov 11, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The problem is in your PE term. You need the distance below the start point, which is where θ = 35 degrees. (Why not just calculate that distance for the bottom position? That's what your expression gives if you put θ=35.)
     
  4. Nov 11, 2008 #3
    hmm this is what I did: Since d = L cos theta and the pendulum's length = L, then the distance below the starting point is the -(distance between the blue sphere and the red horizontal line), which is = -(L - L*cos(theta)). Isn't that what we need?
    [​IMG]
     
  5. Nov 11, 2008 #4

    Doc Al

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    Staff: Mentor

    Sure. As long as θ = 35. (θ is your initial angle, not the final angle.)
     
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