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
L = T - V
d/dt(∂L/∂n1) - ∂L/∂n1 = 0 for n=1 through number of variables
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''
∂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