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

Question

A Double Pendulum

Consider the double pendulum in Figure 5.84, in which two point masses of equal mass [latex]m[/latex] are attached to two rigid links of equal length [latex]L[/latex]. 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 [latex] heta_{1}[/latex] and [latex] heta_{2}[/latex].

Accepted Answer

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, [latex] heta_{1}[/latex] and [latex] heta_{2}[/latex]. The kinetic energy of the system is

[latex]T=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} .[/latex]

From kinematics,

[latex]v_{1}=L \dot{ heta}_{1}[/latex]

and, as shown in Figure 5.85a,

[latex]v_{2}^{2}=\left(L \dot{ heta}_{1} ight)^{2}+\left(L \dot{ heta}_{2} ight)^{2}+2 L^{2} \dot{ heta}_{1} \dot{ heta}_{2} \cos \left( heta_{2}- heta_{1} ight),[/latex]

which is obtained by applying the law of cosines. Thus,

[latex]T=\frac{1}{2} m\left(L \dot{ heta}_{1} ight)^{2}+\frac{1}{2} m\left[\left(L \dot{ heta}_{1} ight)^{2}+\left(L \dot{ heta}_{2} ight)^{2}+2 L^{2} \dot{ heta}_{1} \dot{ heta}_{2} \cos \left( heta_{2}- heta_{1} ight) ight] .[/latex]

Note that no spring elements are involved in the system. This implies that [latex]V_{\mathrm{e}}=0[/latex] and [latex]V=V_{\mathrm{g}}[/latex] Using the datum defined in Figure 5.85b, we can obtain the gravitational potential energy

[latex]V_{g}=-m g h_{1}-m g h_{2},[/latex]

where

[latex] \begin{aligned} & h_{1}=L \cos heta_{1}, \ & h_{2}=L \cos heta_{1}+L \cos heta_{2} . \end{aligned} [/latex]

Thus,

[latex]V=-2 m g L \cos heta_{1}-m g L \cos heta_{2}.[/latex]

We then apply Lagrange's equations

[latex]\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{q}_{i}} ight)-\frac{\partial T}{\partial q_{i}}+\frac{\partial V}{\partial q_{i}}=0, \quad i=1,2.[/latex]

For [latex]i=1, q_{1}= heta_{1}, \dot{q}_{1}=\dot{ heta}_{1}[/latex]

[latex] \begin{aligned} \frac{\partial T}{\partial \dot{ heta}_{1}} & =2 m L^{2} \dot{ heta}_{1}+m L^{2} \dot{ heta}_{2} \cos \left( heta_{2}- heta_{1} ight), \ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{ heta}_{1}} ight) & =2 m L^{2} \ddot{ heta}_{1}+m L^{2} \ddot{ heta}_{2} \cos \left( heta_{2}- heta_{1} ight)-m L^{2} \dot{ heta}_{2} \sin \left( heta_{2}- heta_{1} ight)\left(\dot{ heta}_{2}-\dot{ heta}_{1} ight), \ \frac{\partial T}{\partial heta_{1}} & =-m L^{2} \dot{ heta}_{1} \dot{ heta}_{2} \sin \left( heta_{2}- heta_{1} ight)(-1), \ \frac{\partial V}{\partial heta_{1}} & =2 m g L \sin heta_{1} . \end{aligned} [/latex]

Substituting into Lagrange's equation results in

[latex] \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\partial T}{\partial \dot{ heta}_{1}} ight)-\frac{\partial T}{\partial heta_{1}}+\frac{\partial V}{\partial heta_{1}}=2 m L^{2} \ddot{ heta}_{1}+m L^{2} \ddot{ heta}_{2} \cos \left( heta_{2}- heta_{1} ight)-m L^{2} \dot{ heta}_{2}^{2} \sin \left( heta_{2}- heta_{1} ight)+2 m g L \sin heta_{1}=0 [/latex]

Similarly, for [latex]i=2, q_{2}= heta_{2}, \dot{q}_{2}=\dot{ heta}_{2}[/latex]

[latex] \begin{aligned} \frac{\partial T}{\partial \dot{ heta}_{2}} & =m L^{2} \dot{ heta}_{2}+m L^{2} \dot{ heta}_{1} \cos \left( heta_{2}- heta_{1} ight), \ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{ heta}_{2}} ight) & =m L^{2} \ddot{ heta}_{2}+m L^{2} \ddot{ heta}_{1} \cos \left( heta_{2}- heta_{1} ight)-m L^{2} \dot{ heta}_{1} \sin \left( heta_{2}- heta_{1} ight)\left(\dot{ heta}_{2}-\dot{ heta}_{1} ight), \ \frac{\partial T}{\partial heta_{2}} & =-m L^{2} \dot{ heta}_{1} \dot{ heta}_{2} \sin \left( heta_{2}- heta_{1} ight), \ \frac{\partial V}{\partial heta_{2}} & =m g L \sin heta_{2} . \end{aligned} [/latex]

Substituting into Lagrange's equation gives

[latex] \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{\partial T}{\partial \dot{ heta}_{2}} ight)-\frac{\partial T}{\partial heta_{2}}+\frac{\partial V}{\partial heta_{2}}=m L^{2} \ddot{ heta}_{2}+m L^{2} \ddot{ heta}_{1} \cos \left( heta_{2}- heta_{1} ight)+m L^{2} \dot{ heta}_{1}^{2} \sin \left( heta_{2}- heta_{1} ight)+m g L \sin heta_{2}=0 [/latex]

For small motions [latex]\left( heta_{1} \approx 0 ight.[/latex] and [latex]\left. heta_{2} \approx 0 ight)[/latex], the two differential equations of motion are linearized as

[latex] \begin{aligned} & 2 m L^{2} \ddot{ heta}_{1}+m L^{2} \ddot{ heta}_{2}+2 m g L heta_{1}=0, \ & m L^{2} \ddot{ heta}_{1}+m L^{2} \ddot{ heta}_{2}+m g L heta_{2}=0, \end{aligned} [/latex]

or in second-order matrix form

[latex] \left[\begin{array}{cc} 2 m L^{2} & m L^{2} \ m L^{2} & m L^{2} \end{array} ight]\left\{\begin{array}{l} \ddot{ heta}_{1} \ \ddot{ heta}_{2} \end{array} ight\}+\left[\begin{array}{cc} 2 m g L & 0 \ 0 & m g L \end{array} ight]\left\{\begin{array}{l} heta_{1} \ heta_{2} \end{array} ight\}=\left\{\begin{array}{l} 0 \ 0 \end{array} ight\} . [/latex]

Question 5.15: A Double Pendulum Consider the double pendulum in Figure 5.8...

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