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Mathematical pendulum

  1. Jan 4, 2012 #1
    Can the ideal pendulum, i.e. massless string etc., be explained using torque?
     
  2. jcsd
  3. Jan 5, 2012 #2

    Simon Bridge

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    Yes - that's the usual way of doing it.
    Welcome to PF :)
     
  4. Jan 5, 2012 #3

    Philip Wood

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    Gravitational torque about top of string = Force x perpendicular distance = mgL sin θ in which L is string length and θ is angle of string to vertical.

    Using G = I [itex]\ddot{θ}[/itex],

    I [itex]\ddot{θ}[/itex] = mgL sin θ

    But I = mL[itex]^{2}[/itex] and provided θ<< 1 rad, sin θ = θ

    So mL[itex]^{2}[/itex][itex]\ddot{θ}[/itex]= mgL θ

    That is [itex]\ddot{θ}[/itex] = (g/L)θ
     
  5. Jan 6, 2012 #4

    Simon Bridge

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    ... solve for a function of time, or parameterize by position and momentum - yeah.
     
  6. Jan 6, 2012 #5

    BruceW

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    Even when the angle is not small, (and for simplicity, I'll assume it is a point mass on the end of a rigid rod of negligible mass), we have:
    [tex]\ddot{\theta} = - \frac{g}{L} sin(\theta) [/tex]
    Now, multiplying both sides by [itex]2 \dot{\theta}[/itex], we get:
    [tex]2 \dot{\theta} \ddot{\theta} = - 2 \frac{g}{L} \dot{\theta} sin(\theta) [/tex]
    And now, we can rewrite both sides to get:
    [tex] \frac{d \dot{\theta}^2}{dt} = 2 \frac{g}{L} \frac{d cos(\theta)}{dt} [/tex]
    And now rearranging:
    [tex]\frac{d ( \dot{\theta}^2 - 2 \frac{g}{L} cos(\theta))}{dt} = 0 [/tex]
    So we have a conserved quantity. And as it happens, this is conservation of energy. The reason energy is conserved for the pendulum is because the only force acting in the direction of its movement is gravity. So the energy is converted from GPE to KE and vice versa.
     
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