MyPhysicsLab – Double Pendulum

Instructions: click near circles and drag with your mouse

vertical:    horizontal: graph type:


To purchase a professionally
engineered chaotic pendulum,
click here.

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 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:

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 θ12.

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

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) m22'2 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) m222 L2 - ω12 L1 Cos(θ1 - θ2))
ω1' = -----------------------------------------------------------------------------------------
                            L1 (2 m1 + m2 - m2 Cos(2(θ1 - θ2)))

      2 Sin(θ12) (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.