MyPhysicsLab – Double Pendulum

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For small angles, a pendulum behaves like a linear system (see Simple Pendulum). 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 of the Double Pendulum

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:

• x = horizontal position of pendulum mass
• y = vertical position of pendulum mass
• θ = angle of pendulum (0=vertical downwards, counter-clockwise is positive)
• L = length of rod (constant)

We place the origin at the pivot point of the upper pendulum. We regard y as increasing upwards. We indicate the upper pendulum by subscript 1, and the lower by subscript 2. Begin by using simple trigonometry to write expressions for the positions x₁, y₁, x₂, y₂ in terms of the angles θ₁,θ₂.

x₁ = L₁ sin θ₁
y₁ = -L₁ cos θ₁
x₂ = x₁ + L₂ sin θ₂
y₂ = y₁ - L₂ cos θ₂

The velocity is the first derivative of the position.

x₁' = θ₁' L₁ cos θ₁
y₁' = θ₁' L₁ sin θ₁
x₂' = x₁' + θ₂' L₂ cos θ₂
y₂' = y₁' + θ₂' L₂ sin θ₂

The acceleration is the first derivative of the position.

x₁'' = -θ₁'² sin θ₁ + θ₁'' L₁ cos θ₁    (eqn 1)
y₁'' = θ₁'² L₁ cos θ₁ + θ₁'' L₁ sin θ₁    (eqn 2)
x₂'' = x₁'' - θ₂'² L₂ sin θ₂ + θ₂'' L₂ cos θ₂    (eqn 3)
y₂'' = y₁'' + θ₂'² L₂ cos θ₂ + θ₂'' L₂ sin θ₂    (eqn 4)

Forces in the Double Pendulum

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

• T = tension in the rod
• m = mass of pendulum
• g = gravitational constant

The forces on the upper pendulum mass are the tension in the upper rod T₁, the tension in the lower rod T₂, and gravity -m₁ g. We write separate equations for the horizontal and vertical forces, since they can be treated independently. The net force on the cart is the sum of these. Here we show the net force and use Newton's law F = m a.

m₁ x₁'' = -T₁ sin θ₁ + T₂ sin θ₂    (eqn 5)
m₁ y₁'' = T₁ cos θ₁ - T₂ cos θ₂ - m₁ g    (eqn 6)

For the lower pendulum, the forces are the tension in the lower rod T₂, and gravity -m₂ g.

m₂ x₂'' = -T₂ sin θ₂    (eqn 7)
m₂ y₂'' = T₂ cos θ₂ - m₂ g    (eqn 8)

In relating these equations to the diagrams, keep in mind that in the example diagram θ₁ is positive and θ₂ is negative, because of the convention that a counter-clockwise angle is positive.

Direct Method for Finding Equations of Motion

Now we do some algebraic manipulations with the goal of finding expressions for θ₁'', θ₂'' in terms of θ₁, θ₁', θ₂, θ₂'. Begin by solving equations 7, 8 for T₂ sin θ₂ and T₂ cos θ₂ and then substituting into equations 5 and 6.

m₁ x₁'' = -T₁ sin θ₁ - m₂ x₂''    (eqn 9)
m₁ y₁'' = T₁ cos θ₁ - m₂ y₂'' - m₂ g - m₁ g    (eqn 10)

Multiply equation 9 by cos θ₁ and equation 10 by sin θ₁ and rearrange to get

T₁ sin θ₁ cos θ₁ = - cos θ₁ (m₁ x₁'' + m₂ x₂'')    (eqn 11)
T₁ sin θ₁ cos θ₁ = sin θ₁ (m₁ y₁'' + m₂ y₂'' + m₂ g + m₁ g)    (eqn 12)

This leads to the equation

sin θ₁ (m₁ y₁'' + m₂ y₂'' + m₂ g + m₁ g) = - cos θ₁ (m₁ x₁'' + m₂ x₂'')    (eqn 13)

Next, multiply equation 7 by cos θ₂ and equation 8 by sin θ₂ and rearrange to get

T₂ sin θ₂ cos θ₂ = - cos θ₂ (m₂ x₂'')   (eqn 14)
T₂ sin θ₂ cos θ₂ = sin θ₂ (m₂ y₂'' + m₂ g)    (eqn 15)

which leads to

sin θ₂ (m₂ y₂'' + m₂ g) = - cos θ₂ (m₂ x₂'')   (eqn 16)

Next we need to use a program such as Mathematica to solve equations 13 and 16 for θ₁'', θ₂'' in terms of θ₁, θ₁', θ₂, θ₂'. Note that we also include the definitions given by equations 1, 2, 3, 4, so that we have 2 equations (13, 16) and 2 unknowns (θ₁'', θ₂''). The result is somewhat complicated, but is easy enough to program into the computer. Here are the resulting equations of motion.

       -g (2 m1 + m2) Sin θ1 - m2 g Sin(θ1 - 2 θ2)- 2 Sin(θ1 - θ2) m2 (θ2'2 L2 - θ1'2 L1 Cos(θ1 - θ2))
θ1'' = —————————————————————————————————————————————————————————————————————————————————————————
                            L1 (2 m1 + m2 - m2 Cos(2(θ1 - θ2)))

      2 Sin(θ1 - θ2) (θ1'2 L1 (m1 + m2) + g(m1 + m2)Cos(θ1) + θ2'2 L2 m2 Cos(θ1 - θ2))
θ2''= ———————————————————————————————————————————————————————————————————————————
                           L2 (2 m1 + m2 - m2 Cos(2(θ1 - θ2)))

Numerical Solution

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:

ω₁ = θ₁'
ω₂ = θ₂'

Then we can write the four 1st order equations:

θ₁' = ω₁
θ₂' = ω₂

       -g (2 m1 + m2) Sin θ1 - m2 g Sin(θ1 - 2 θ2)- 2 Sin(θ1 - θ2) m2 (ω22 L2 - ω12 L1 Cos(θ1 - θ2))
ω1' = —————————————————————————————————————————————————————————————————————————————————————————
                            L1 (2 m1 + m2 - m2 Cos(2(θ1 - θ2)))

      2 Sin(θ1-θ2) (ω12 L1 (m1 + m2) + g(m1 + m2)Cos(θ1) + ω22 L2 m2 Cos(θ1 - θ2))
ω2'= ————————————————————————————————————————————————————————————————————————
                           L2 (2 m1 + m2 - m2 Cos(2(θ1 - θ2)))

This is now exactly the form needed to plug in to the Runge-Kutta method for numerical solution of the system.