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How to integrate partial differential eqn

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data

    im not sure the elegant way to put it, i need to integrate this somehow (ie, throw over the dt and integrate each side) but that squared is really tripping me up. is there a trick i should be using?
    -g(cos(x))=r(dx/dt)^2

    2. Relevant equations



    3. The attempt at a solution

    my attempt was initially to integrate 1/(cos(x))^(1/2), that is not pretty

    i heard classmates talking about multiplying though by (dx/dt), not sure if that applies

    any help appreciated, thanks!
     
  2. jcsd
  3. Feb 1, 2012 #2

    vela

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    Where did this equation come from?
     
  4. Feb 1, 2012 #3
    Im solving for the tension in a pendulum, right now im equating the radial accel with the force of gravity to try and integrate to something with a theta-not (representing the max amplitude of the pendulum) and theta (t). Hope that makes some sense
     
  5. Feb 1, 2012 #4

    vela

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    Your equation looks wrong since it suggests the only force in the radial direction is gravity, but the tension also acts radially.

    You'd be better off integrating Newton's second law to find ##\theta(t)##, and then use F=ma to solve for the tension.
     
  6. Feb 1, 2012 #5
    You're right, the full radial equation as I have it is this,

    T=mgcos(theta)+mL(dtheta/dt)^2

    My hope was that in integrating the right half independently of the tension then I could make a substitution (from the result of that integration, into the full equation) that would allow me to have an answer with theta-not component(s)
     
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