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Pendulum question

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A pendulum consists of a mass m attached to a light rigid rod of length L. If there is friction in the hinge, then for small displacements theta can be shown to satisfy

    [tex]\ddot{\theta}+2\mu\dot{\theta}+\omega_{0}^{2}\theta = 0[/tex].

    Show that the solution is

    [tex]\theta(t)=\theta_{0}e^{-\mu*t}cos(\omega^{'}t+\phi)[/tex] .

    Take the case where [tex]\phi=-0.5*\pi[/tex]. If the pendulum is hit every other time it passes through the origin, deduce how much energy has to be given to the pendulum to ensure that, at this point, its energy is the same as at t=0.


    3. The attempt at a solution

    Showing that the solution is as they say is easy. The problem i have is with the energy question.

    Every swing you give the pendulum a push (like you would do with a child on a swing) to counter the loss in energy due to friction.

    It seems we must find the energy lost during one period from t=0 to t=T, and this is the energy that will need to be supplied by the push.

    In one period the amplitude will decrease by a factor [tex]e^{-\mu*T}[/tex]. Now the energy of the pendulum at t=0 is all kinetic and i think is equal to

    [tex]E_{0}=mgL(1-cos(\theta_{0}))[/tex].

    So would that then make the energy required by the push

    [tex]E_{p}=mgL(1-cos(\theta_{0})) - mgL(1-cos(e^{-\mu*T}\theta_{0})) = mgL(cos(e^{-\mu*T}\theta_{0})-cos(\theta_{0}))[/tex].

    I'm not sure if that's correct or not...?

    Thanks.
     
  2. jcsd
  3. Mar 22, 2009 #2
    Correct. But the thing that bothers me is:
    That formula is derived from the height the pendulum reaches. It has nothing to do with the kinetic energy at [tex]\theta=0[/tex]. Besides that in a frictionless environment the potential energy at maximum angle is the same as the kinetic energy at the lowest point. But since we have friction here we aren't justified in saying that the max kinetic energy is this and that. We want to use the fact that we always want that pendulum to get to same height, which leads to the formula above.
     
  4. Mar 23, 2009 #3
    I'm a bit confused. Do you mean my answer is correct, but you're not happy with my method of reasoning, or is the answer is wrong? :-)

    The way I was thinking about it was this:

    If there were no friction then the pendulum would oscillate with SHM back and forth between [tex]\pm\theta_{0}[/tex]. At t=0, it's at its lowest point, and since it hasn't moved yet no work has been done against friction. If it didn't have K.E. [tex]E_{0}=mgL(1-cos(\theta_{0}))[/tex] then were the friction not there, the pendulum wouldn't have amplitude [tex]\theta_{0}[/tex] - it would have some other amplitude.

    As it stands we do have friction, so how this K.E. actually ends up being divided between P.E. and work done against friction is determined by the value of [tex]\mu[/tex]. We know that [tex]\mu > 0[/tex] so that the pendulum will never actually reach [tex]\theta_{0}[/tex] but this doesn't seem to change the statement about the K.E. at t=0 in my view.

    Thanks for your help!
     
  5. Mar 23, 2009 #4
    Your answer is correct.

    I was just not completely happy with how you reached the K.E. formula. But it seems you understand what is going on so I will leave it at that.

    edit: [tex]E_{0}=mgL(1-cos(\theta_{0}))[/tex] is not a formula for kinetic energy as it comes from [tex]mg\Delta h[/tex], which is for potential energy. That's what I didn't like.
     
  6. Mar 23, 2009 #5
    Oh okay :-) Thanks again!!
     
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