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Homework Help: Pendulums, total energy?, and Mathematica

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A)Show that for a non-frictional, simple linear pendulum (Sin(theta) ~ theta) the total energy of the pendulum (K + U) or kinetic plus potential is given by

    E = (1/2) m l^2 (d(theta)/dt)^2 + (1/2) mgl (theta)^2

    and therefor E = (1/2) mgl(theta0)^2

    theta0 = theta(t=0)

    2. Relevant equations
    F=ma , delta K = 1/2mv^2 , delta U = mgh

    3. The attempt at a solution

    Alright, so I'm essentially lost in this problem, and my last calculus class was approximately 2 years ago.
    I understand that the total energy should be the sum of the potential and kinetic energies of the pendulum, so it seems that E = 1/2mv^2 +mgh. But it seems that i am stuck here. I observe that the change between the kinetic energy portion of the equation is different in that v^2 is now l^2 (d(theta)/dt)^2, and the potential mgh now appears as 1/2mgl (theta)^2, but I cannot think of how to determine how to get to that point. And therefor I am unable to get to the main portion of the problem.

    {a} One more problem. use mathematica to solve (d(theta)/dt)^2 +g/lsin(theta)=0 . And, show a graph of period vs. (theta 0).

    Before this course I have not used mathematica, and now I am facing difficulties. I tried using the dsolve and manipulate functions many times over the past week in an attempt to graph the problem but I have been unsuccessful.

    Please note that I am NOT asking for any answers just for guidance.

    Attached Files:

  2. jcsd
  3. Sep 12, 2010 #2
    I managed to figure out, after drawing a simple diagram (of course!), that h in the potential equation has got to be equal to L-Lcos(theta). Also, the velocity is going to be equal to L(dtheta/dt), the length multiplied by the rate of change of the center angle. Knowing this,
    k= 1/2 m (L(dtheta/dt))^2, and
    U= mgh= mg (L-Lcos(theta))
    E= 1/2 m L^2 (dtheta/dt)^2 + mg(L-Lcos(theta)), which appears to be very similar to the original equation that is given, but I am again stuck. I cant seem to find why the first given equation is equal to the second.
  4. Sep 12, 2010 #3
    So... anybody understand how to graph in mathematica or have any words of wisdom on the above problems? Anything would be useful... :D
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