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For small angles, a pendulum behaves like a linear system. When the angles are small in the Double Pendulum, the system behaves like the linear Double Spring. In the graph, you can see similar Lissajous curves being generated. This is because the motion is determined by simple sine and cosine functions.
For large angles, the pendulum is non-linear and the phase graph becomes much more complex. You can see this by dragging one of the masses to a larger angle and letting go.
We regard the pendulum rods as being massless and rigid. We regard the pendulum masses as being point masses. The derivation of the equations of motion is shown below, using the direct Newtonian method.
Kinematics means the relations of the parts of the device, without regard to forces. In kinematics we are only trying to find expressions for the position, velocity, & acceleration in terms of whatever variables we have chosen.
The variables we choose here are:
We place the origin at the pivot point of the upper pendulum. We regard
x
The velocity is the first derivative of the position.
x
The acceleration is the first derivative of the position.
x
We treat the two pendulum masses as point particles. Begin by drawing the free body diagram for the upper mass and writing an expression for the net force acting it.
The variables are as follows
The forces on the upper pendulum mass are the tension in the upper rod
m
For the lower pendulum, the forces are the tension in the lower rod
m
In relating these equations to the diagrams, keep in mind that in the example diagram
Now we do some algebraic manipulations with the goal of finding expressions for
m
Multiply equation 9 by
T
This leads to the equation
sin
Next, multiply equation 7 by
T
which leads to
sin
Next we need to use a program such as Mathematica to solve equations 13 and 16 for
-g (2 m_{1} + m_{2}) Sin θ_{1} - m_{2} g Sin(θ_{1} - 2 θ_{2})- 2 Sin(θ_{1} - θ_{2}) m_{2} (θ_{2}'^{2} L_{2} - θ_{1}'^{2} L_{1} Cos(θ_{1} - θ_{2})) θ_{1}'' = ----------------------------------------------------------------------------------------- L_{1} (2 m_{1} + m_{2} - m_{2} Cos(2(θ_{1} - θ_{2}))) 2 Sin(θ_{1} - θ_{2}) (θ_{1}'^{2} L_{1} (m_{1} + m_{2}) + g(m_{1} + m_{2})Cos(θ_{1}) + θ_{2}'^{2} L_{2} m_{2} Cos(θ_{1} - θ_{2})) θ_{2}''= --------------------------------------------------------------------------- L_{2} (2 m_{1} + m_{2} - m_{2} Cos(2(θ_{1} - θ_{2})))
The above equations are now close to the form needed for the Runge-Kutta method. The final step is convert these two 2nd order equations into four 1st order equations. Define the first derivatives as separate variables:
Then we can write the four 1st order equations:
θ_{1}' = ω_{1}
θ_{2}' = ω_{2}
-g (2 m_{1} + m_{2}) Sin θ_{1} - m_{2} g Sin(θ_{1} - 2 θ_{2})- 2 Sin(θ_{1} - θ_{2}) m_{2} (ω_{2}^{2} L_{2} - ω_{1}^{2} L_{1} Cos(θ_{1} - θ_{2})) ω_{1}' = ----------------------------------------------------------------------------------------- L_{1} (2 m_{1} + m_{2} - m_{2} Cos(2(θ_{1} - θ_{2}))) 2 Sin(θ_{1}-θ_{2}) (ω_{1}^{2} L_{1} (m_{1} + m_{2}) + g(m_{1} + m_{2})Cos(θ_{1}) + ω_{2}^{2} L_{2} m_{2} Cos(θ_{1} - θ_{2})) ω_{2}'= ------------------------------------------------------------------------ L_{2} (2 m_{1} + m_{2} - m_{2} Cos(2(θ_{1} - θ_{2})))
This is now exactly the form needed to plug in to the Runge-Kutta method for numerical solution of the system.