Double pendulum

Analysis and interpretation

Several variants of the double pendulum may be considered; the two limbs may be of equal or unequal lengths and masses, they may be simple pendulums or compound pendulums (also called complex pendulums) and the motion may be in three dimensions or restricted to one vertical plane. In the following analysis, the limbs are taken to be identical compound pendulums of length ℓ and mass m, and the motion is restricted to two dimensions.

In a compound pendulum, the mass is distributed along its length. If the double pendulum mass is evenly distributed, then the center of mass of each limb is at its midpoint, and the limb has a moment of inertia of I = ⁠1/12⁠mℓ² about that point.

Although it is possible to derive the equations of a double pendulum with Newtonian mechanics, it is considered to be cumbersome to work with as it would require resolving vectors with respect to constraint forces. So it is more convenient to use the angles between each limb and the vertical as the generalized coordinates defining the configuration of the system. These angles are denoted θ₁ and θ₂. The position of the center of mass of each rod may be written in terms of these two coordinates. If the origin of the Cartesian coordinate system is taken to be at the point of suspension of the first pendulum, then the center of mass of this pendulum is at:$${\displaystyle \begin{array}{cc}{x}_1& =\frac{1}{2}\ell \sin {\theta }_1\\ y_1& =-\frac{1}{2}\ell \cos {\theta }_1\end{array}}$$

and the center of mass of the second pendulum is at $${\displaystyle \begin{array}{cc}{x}_2& =\ell \left(\sin {\theta }_1+\frac{1}{2}\sin {\theta }_2\right)\\ y_2& =-\ell \left(\cos {\theta }_1+\frac{1}{2}\cos {\theta }_2\right)\end{array}}$$ This is enough information to write out the Lagrangian.

Lagrangian

The Lagrangian is given by $${\displaystyle \begin{array}{cc}L& =\text{kinetic energy}-\text{potential energy}\\ & =\frac{1}{2}m\left(v_1^2+v_2^2\right)+\frac{1}{2}I\left({\dot{\theta }}_1^2+{\dot{\theta }}_2^2\right)-mg\left(y_1+y_2\right)\\ & =\frac{1}{2}m\left({\dot{x}}_1^2+{\dot{y}}_1^2+{\dot{x}}_2^2+{\dot{y}}_2^2\right)+\frac{1}{2}I\left({\dot{\theta }}_1^2+{\dot{\theta }}_2^2\right)-mg\left(y_1+y_2\right)\end{array}}$$ The first term is the linear kinetic energy of the center of mass of the bodies and the second term is the rotational kinetic energy around the center of mass of each rod. The last term is the potential energy of the bodies in a uniform gravitational field. The dot-notation indicates the time derivative of the variable in question.

Using the values of ${\displaystyle x_1}$ and ${\displaystyle y_1}$ defined above, we have $${\displaystyle \begin{array}{cc}{\dot{x}}_1& ={\dot{\theta }}_1\left(\frac{1}{2}\ell \cos {\theta }_1\right)\\ {\dot{y}}_1& ={\dot{\theta }}_1\left(\frac{1}{2}\ell \sin {\theta }_1\right)\end{array}}$$ which leads to $${\displaystyle v_1^2={\dot{x}}_1^2+{\dot{y}}_1^2=\frac{1}{4}{\dot{\theta }}_1^2{\ell }^2\left({\cos }^2{\theta }_1+{\sin }^2{\theta }_1\right)=\frac{1}{4}{\ell }^2{\dot{\theta }}_1^2.}$$

Similarly, for ${\displaystyle x_2}$ and ${\displaystyle y_2}$ we have $${\displaystyle \begin{array}{cc}{\dot{x}}_2& =\ell \left({\dot{\theta }}_1\cos {\theta }_1+\frac{1}{2}{\dot{\theta }}_2\cos {\theta }_2\right)\\ {\dot{y}}_2& =\ell \left({\dot{\theta }}_1\sin {\theta }_1+\frac{1}{2}{\dot{\theta }}_2\sin {\theta }_2\right)\end{array}}$$

and therefore

$${\displaystyle \begin{array}{cc}{v}_2^2& ={\dot{x}}_2^2+{\dot{y}}_2^2\\ & ={\ell }^2\left({\dot{\theta }}_1^2{\cos }^2{\theta }_1+{\dot{\theta }}_1^2{\sin }^2{\theta }_1+\frac{1}{4}{\dot{\theta }}_2^2{\cos }^2{\theta }_2+\frac{1}{4}{\dot{\theta }}_2^2{\sin }^2{\theta }_2+{\dot{\theta }}_1{\dot{\theta }}_2\cos {\theta }_1\cos {\theta }_2+{\dot{\theta }}_1{\dot{\theta }}_2\sin {\theta }_1\sin {\theta }_2\right)\\ & ={\ell }^2\left({\dot{\theta }}_1^2+\frac{1}{4}{\dot{\theta }}_2^2+{\dot{\theta }}_1{\dot{\theta }}_2\cos \left({\theta }_1-{\theta }_2\right)\right).\end{array}}$$

Substituting the coordinates above into the definition of the Lagrangian, and rearranging the equation, gives $${\displaystyle \begin{array}{cc}L& =\frac{1}{2}m{\ell }^2\left({\dot{\theta }}_1^2+\frac{1}{4}{\dot{\theta }}_1^2+\frac{1}{4}{\dot{\theta }}_2^2+{\dot{\theta }}_1{\dot{\theta }}_2\cos \left({\theta }_1-{\theta }_2\right)\right)+\frac{1}{24}m{\ell }^2\left({\dot{\theta }}_1^2+{\dot{\theta }}_2^2\right)-mg\left(y_1+y_2\right)\\ & =\frac{1}{6}m{\ell }^2\left({\dot{\theta }}_2^2+4{\dot{\theta }}_1^2+3{\dot{\theta }}_1{\dot{\theta }}_2\cos ({\theta }_1-{\theta }_2)\right)+\frac{1}{2}mg\ell \left(3\cos {\theta }_1+\cos {\theta }_2\right).\end{array}}$$

The equations of motion can now be derived using the Euler–Lagrange equations, which are given by $${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\theta }}_i}-\frac{\mathrm{\partial }L}{\mathrm{\partial }{\theta }_i}=0,i=1,2.}$$ We begin with the equation of motion for ${\displaystyle {\theta }_1}$. The derivatives of the Lagrangian are given by $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{\theta }_1}=-\frac{1}{2}m{\ell }^2{\dot{\theta }}_1{\dot{\theta }}_2\sin ({\theta }_1-{\theta }_2)-\frac{3}{2}mg\ell \sin {\theta }_1}$$ and $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\theta }}_1}=\frac{4}{3}m{\ell }^2{\dot{\theta }}_1+\frac{1}{2}m{\ell }^2{\dot{\theta }}_2\cos ({\theta }_1-{\theta }_2).}$$ Thus $${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\theta }}_1}=\frac{4}{3}m{\ell }^2{\ddot{\theta }}_1+\frac{1}{2}m{\ell }^2{\ddot{\theta }}_2\cos ({\theta }_1-{\theta }_2)-\frac{1}{2}m{\ell }^2{\dot{\theta }}_2({\dot{\theta }}_1-{\dot{\theta }}_2)\sin ({\theta }_1-{\theta }_2).}$$ Combining these results and simplifying yields the first equation of motion, $${\displaystyle \frac{4}{3}\ell {\ddot{\theta }}_1+\frac{1}{2}\ell {\ddot{\theta }}_2\cos ({\theta }_1-{\theta }_2)+\frac{1}{2}\ell {\dot{\theta }}_2^2\sin ({\theta }_1-{\theta }_2)+\frac{3}{2}g\sin {\theta }_1=0.}$$

Similarly, the derivatives of the Lagrangian with respect to ${\displaystyle {\theta }_2}$ and ${\displaystyle {\dot{\theta }}_2}$ are given by $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{\theta }_2}=\frac{1}{2}m{\ell }^2{\dot{\theta }}_1{\dot{\theta }}_2\sin ({\theta }_1-{\theta }_2)-\frac{1}{2}mg\ell \sin {\theta }_2}$$ and $${\displaystyle \frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\theta }}_2}=\frac{1}{3}m{\ell }^2{\dot{\theta }}_2+\frac{1}{2}m{\ell }^2{\dot{\theta }}_1\cos ({\theta }_1-{\theta }_2).}$$ Thus $${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }L}{\mathrm{\partial }{\dot{\theta }}_2}=\frac{1}{3}m{\ell }^2{\ddot{\theta }}_2+\frac{1}{2}m{\ell }^2{\ddot{\theta }}_1\cos ({\theta }_1-{\theta }_2)-\frac{1}{2}m{\ell }^2{\dot{\theta }}_1({\dot{\theta }}_1-{\dot{\theta }}_2)\sin ({\theta }_1-{\theta }_2).}$$

Plugging these results into the Euler-Lagrange equation and simplifying yields the second equation of motion, $${\displaystyle \frac{1}{3}\ell {\ddot{\theta }}_2+\frac{1}{2}\ell {\ddot{\theta }}_1\cos ({\theta }_1-{\theta }_2)-\frac{1}{2}\ell {\dot{\theta }}_1^2\sin ({\theta }_1-{\theta }_2)+\frac{1}{2}g\sin {\theta }_2=0.}$$

No closed form solutions for ${\displaystyle {\theta }_1}$ and ${\displaystyle {\theta }_2}$ as functions of time are known, therefore the system can only be solved numerically, using the Runge Kutta method or similar techniques.