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Elementary pendulum equation of motion

  1. Jan 16, 2017 #1
    1. The problem statement, all variables and given/known data
    To find the period of oscillation of the pendulum and the equation of motion.

    2. Relevant equations
    Conservation of energy.
    Potential energy in a constant field = mgh.
    Kinetic energy in polar coordinates with r constant = (1/2) m r2 (dΘ2/dt2)

    3. The attempt at a solution
    I won't use the newton laws because that means a second order differential equation. Instead I use the conservation of energy so I start from a first order equation.

    1.- I set the origin of coordinates at the anchor point of the pendulum.
    2.- I will use polar coordinates, the angle starts counting from the vertical counterclockwise.
    3.- The energy is conserved so at the point 1 (at the bottom when the pendulum is faster) we have:
    L is the lenght of the pendulum

    KE =(1/2) mL22/dt2

    Potential energy = -mgL ; I set the potential energy as zero at the origin of coordinates, the anchor point of the pendulum.

    At the point 2 when the speed is zero I have:

    KE = 0

    PE = -mg (lcosΘ)

    Now I equate the sum of the KE and PE at both points:

    (1/2)mL22/dt2 -mgL = 0 -mg (LcosΘ)

    (1/2) dΘ2/dt2 = g(1 - cos Θ ) /L

    If Θ is small cos Θ can be aproximated by 1 - Θ2/2, so:

    (1/2) dΘ2/dt2 = g ( 1- (1-Θ2/2))/L

    Finally:

    dΘ/dt = √(g/L) Θ2

    dΘ/√(g/L) Θ2 = dt

    Since Θ can get outside de square the integral is :

    Log Θ = √L/g t + constant

    And the basic expression is an exponential:

    So Θ= exp t

    And this is not the expected periodic movement and that does not make any sense. Does anybody know where is the mistake?

    THanks

    209r6mc.jpg
     
  2. jcsd
  3. Jan 16, 2017 #2

    TSny

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    You need an equation that holds at each point of the motion rather than an equation that just relates the lowest and highest points of the motion. Your equation relates dΘ/dt at one point to Θ at another point. Instead, try to get an equation that relates dΘ/dt and Θ at the same point.
     
  4. Jan 16, 2017 #3
    But shouldn't this be equivalent?

    I mean, if I differenciate my equation, SHouldn't I get the newton equations?
     
  5. Jan 16, 2017 #4

    TSny

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    If you were to add subscripts 1 and 2 to denote the two points that you selected, you have an equation that involves [dΘ/dt]2 and Θ1 . So, you aren't getting a differential equation that holds throughout the motion.
     
  6. Jan 16, 2017 #5
    Ok I understand, so basically what I did simply didn't make sense.

    THe equation could have been used to find only differences between energies, and nothing else.
     
  7. Jan 16, 2017 #6
    So it could be calculated this way:

    (1/2) m L2 (dΘ/dt)2 - mgL cos Θ= -mgL cos Θ0

    Θ0 is the maximum angle.
    Θ is a function of time

    It means, the energy at an arbitrary point equals the maximum energy at the top of the movement.

    From this it should be possible to obtain Θ(t) right?
     
  8. Jan 16, 2017 #7

    TSny

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    Yes. Good.
     
  9. Jan 16, 2017 #8
    Are you going to make a small angle approximation soon. Otherwise, you will end up with an elliptic integral.
     
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