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## Homework Statement

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

## Homework Equations

Lagragian

L = T - V

Lagrange Equation

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''

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