Question 5.15: A Double Pendulum Consider the double pendulum in Figure 5.8...
A Double Pendulum
Consider the double pendulum in Figure 5.84, in which two point masses of equal mass m are attached to two rigid links of equal length L. The links are assumed to be massless. The motion of the system is constrained in the vertical plane. Neglecting friction, derive the equations of motion using Lagrange’s equations. Assume small angles for \theta_{1} and \theta_{2}.
The system is only subjected to gravitational forces, and it is a conservative system. The dynamics of the system can be described using two independent angular displacement coordinates, \theta_{1} and \theta_{2}. The kinetic energy of the system is
T=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} .
From kinematics,
v_{1}=L \dot{\theta}_{1}
and, as shown in Figure 5.85a,
v_{2}^{2}=\left(L \dot{\theta}_{1}\right)^{2}+\left(L \dot{\theta}_{2}\right)^{2}+2 L^{2} \dot{\theta}_{1} \dot{\theta}_{2} \cos \left(\theta_{2}-\theta_{1}\right),
which is obtained by applying the law of cosines. Thus,
T=\frac{1}{2} m\left(L \dot{\theta}_{1}\right)^{2}+\frac{1}{2} m\left[\left(L \dot{\theta}_{1}\right)^{2}+\left(L \dot{\theta}_{2}\right)^{2}+2 L^{2} \dot{\theta}_{1} \dot{\theta}_{2} \cos \left(\theta_{2}-\theta_{1}\right)\right] .
Note that no spring elements are involved in the system. This implies that V_{\mathrm{e}}=0 and V=V_{\mathrm{g}} Using the datum defined in Figure 5.85b, we can obtain the gravitational potential energy
V_{g}=-m g h_{1}-m g h_{2},
where
\begin{aligned} & h_{1}=L \cos \theta_{1}, \\ & h_{2}=L \cos \theta_{1}+L \cos \theta_{2} . \end{aligned}
Thus,
V=-2 m g L \cos \theta_{1}-m g L \cos \theta_{2}.
We then apply Lagrange’s equations
\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{q}_{i}}\right)-\frac{\partial T}{\partial q_{i}}+\frac{\partial V}{\partial q_{i}}=0, \quad i=1,2.
For i=1, q_{1}=\theta_{1}, \dot{q}_{1}=\dot{\theta}_{1}
\begin{aligned} \frac{\partial T}{\partial \dot{\theta}_{1}} & =2 m L^{2} \dot{\theta}_{1}+m L^{2} \dot{\theta}_{2} \cos \left(\theta_{2}-\theta_{1}\right), \\ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{\theta}_{1}}\right) & =2 m L^{2} \ddot{\theta}_{1}+m L^{2} \ddot{\theta}_{2} \cos \left(\theta_{2}-\theta_{1}\right)-m L^{2} \dot{\theta}_{2} \sin \left(\theta_{2}-\theta_{1}\right)\left(\dot{\theta}_{2}-\dot{\theta}_{1}\right), \\ \frac{\partial T}{\partial \theta_{1}} & =-m L^{2} \dot{\theta}_{1} \dot{\theta}_{2} \sin \left(\theta_{2}-\theta_{1}\right)(-1), \\ \frac{\partial V}{\partial \theta_{1}} & =2 m g L \sin \theta_{1} . \end{aligned}
Substituting into Lagrange’s equation results in
\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\partial T}{\partial \dot{\theta}_{1}}\right)-\frac{\partial T}{\partial \theta_{1}}+\frac{\partial V}{\partial \theta_{1}}=2 m L^{2} \ddot{\theta}_{1}+m L^{2} \ddot{\theta}_{2} \cos \left(\theta_{2}-\theta_{1}\right)-m L^{2} \dot{\theta}_{2}^{2} \sin \left(\theta_{2}-\theta_{1}\right)+2 m g L \sin \theta_{1}=0
Similarly, for i=2, q_{2}=\theta_{2}, \dot{q}_{2}=\dot{\theta}_{2}
\begin{aligned} \frac{\partial T}{\partial \dot{\theta}_{2}} & =m L^{2} \dot{\theta}_{2}+m L^{2} \dot{\theta}_{1} \cos \left(\theta_{2}-\theta_{1}\right), \\ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{\theta}_{2}}\right) & =m L^{2} \ddot{\theta}_{2}+m L^{2} \ddot{\theta}_{1} \cos \left(\theta_{2}-\theta_{1}\right)-m L^{2} \dot{\theta}_{1} \sin \left(\theta_{2}-\theta_{1}\right)\left(\dot{\theta}_{2}-\dot{\theta}_{1}\right), \\ \frac{\partial T}{\partial \theta_{2}} & =-m L^{2} \dot{\theta}_{1} \dot{\theta}_{2} \sin \left(\theta_{2}-\theta_{1}\right), \\ \frac{\partial V}{\partial \theta_{2}} & =m g L \sin \theta_{2} . \end{aligned}
Substituting into Lagrange’s equation gives
\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{\theta}_{2}}\right)-\frac{\partial T}{\partial \theta_{2}}+\frac{\partial V}{\partial \theta_{2}}=m L^{2} \ddot{\theta}_{2}+m L^{2} \ddot{\theta}_{1} \cos \left(\theta_{2}-\theta_{1}\right)+m L^{2} \dot{\theta}_{1}^{2} \sin \left(\theta_{2}-\theta_{1}\right)+m g L \sin \theta_{2}=0
For small motions \left(\theta_{1} \approx 0\right. and \left.\theta_{2} \approx 0\right), the two differential equations of motion are linearized as
\begin{aligned} & 2 m L^{2} \ddot{\theta}_{1}+m L^{2} \ddot{\theta}_{2}+2 m g L \theta_{1}=0, \\ & m L^{2} \ddot{\theta}_{1}+m L^{2} \ddot{\theta}_{2}+m g L \theta_{2}=0, \end{aligned}
or in second-order matrix form
\left[\begin{array}{cc} 2 m L^{2} & m L^{2} \\ m L^{2} & m L^{2} \end{array}\right]\left\{\begin{array}{l} \ddot{\theta}_{1} \\ \ddot{\theta}_{2} \end{array}\right\}+\left[\begin{array}{cc} 2 m g L & 0 \\ 0 & m g L \end{array}\right]\left\{\begin{array}{l} \theta_{1} \\ \theta_{2} \end{array}\right\}=\left\{\begin{array}{l} 0 \\ 0 \end{array}\right\} .
