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Pendulum on Pendulum, velocity in non inertial frame

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data

    A double pendulum consists of two particles of equal mass m suspended
    by massless rods of equal length l. Assuming that all
    motion is in a vertical plane:
    1. Find the Lagrangian of this system.
    2. then find the equations of motion and,
    3. linearize these equations, assuming small motion.


    a = angle between rod1 and vertical
    b = angle between rod2 and vertical
    l = length of each rod
    m = mass of the point masses
    g = acceleration due to gravity
    v = velocity of the point mass


    2. Relevant equations

    Lagragian
    L = T - V

    Lagrange Equation
    d/dt(∂L/∂n1) - ∂L/∂n1 = 0 for n=1 through number of variables



    3. The attempt at a solution

    I started by looking at an easier example, a single pendulum with a mass less rod and point mass on the end. I used the same approach to find the lagragian and eqn of motion for the first rod as follows:


    T = kinetic energy = 1/2 m v^2
    v can be expressed in terms of a.
    v = angular velocity * radius
    v = l * a' <- ' denoting derived wrt to t

    so T = 1/2 m (l*a')^2

    V (potential) is described as mg*height above the lowest possible height so:

    mg * (l - cos a)

    which gives the Lagrangian
    L = 1/2 m l^2 a'^2 - mgl (1 - cos(a))

    Using the Lagrangian equation:
    d/dt(∂L/∂a') - ∂L/∂a = 0

    d/dt(∂L/∂a') = l^2 m a''
    and
    ∂L/∂a = mgl sin(a)

    which solves to 0 = a'' - g/l sin(a)

    assuming small motion, sin(a) is approx equal to a so

    0 = a'' - g/l a

    Which seems to be fine for a single pendulum. The tricky bit comes with adding the second onto the end of the first. The way I think of it either I can:

    1) Treat the second as if it were the first and ignore that it takes place inside a non inertial frame

    This gives me a second equation that looks like: 0 = b'' - g/l b


    2) When calculating the Lagrangian add the T and V terms of the first rod to the second to try and take the non inertial frame into account:

    This evaluates to the same as above: 0 = b'' - g/l b , as the extra terms are removed when passed through the Lagrangian equation.

    Both of these evaluate to equations that do not include a so i don't think they are right.

    3) When calculating the kinetic energy, instead of simply using v = a' l, somehow describe the velocity in terms of the entire frame. I feel this would be the correct way to do it but is where I get stuck as I don't know how to go about it.

    so T = 1/2 m (velocity relative to inertial frame)
    V = mg (2l - lcos(a) - lcos(b))

    So in summary how can I represent the actual velocity of the second point mass on the end of the second rod? I think that once I have that I should be able to find the eqn of motion and linearize it. Thanks
     
  2. jcsd
  3. Oct 26, 2011 #2
    I've found the places where I have gone wrong in this problem and what I needed to do. I'll post the solution for reference soon.
     
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