If I have the equations of motion for a double pendulum (from https://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/DoublePendulum.pdf), can I include a time dependent damping and spring force term to model the system in the picture?

If so, where would I introduce them and can I still use the same methodologies describe in the link (Newtonian mechanics and Lagrangian mechanics)?

(The function describing how the damping force varies with theta and d(theta)/dt is known)Diagram of the system

  • $\begingroup$ Hi! - Unfortunately, you cannot add damping to the lagrangian formulation. But everything else a doable. I suggest you edit the post and narrow the focus a bit on the question. Start working through this problem on your own and ask a specific question where you have difficulty. As it stands the question is a tad too broad since the linked paper explains all the steps, and you just need to re-create these steps but with the additional forces from the force elements (springs & dampers). $\endgroup$ Commented Jan 26, 2023 at 13:30

2 Answers 2


$\def \b {\mathbf}$

with the components of the damper and spring forces ( John Alexiou) \begin{align*} &\b F_d(\phi_1)=\begin{bmatrix} Fd_x \\ Fd_y\\ \end{bmatrix}\quad, \b F_s(\phi_1,\phi_2)=\begin{bmatrix} Fs_x \\ Fs_y\\ \end{bmatrix} \end{align*} add to the differential equations $~(69,70)~$ \begin{align*} &\b \tau_q=\left[ \begin {array}{c} l_{{1}}\cos \left( \phi_{{1}} \right) {\it Fd}_{{x}}+l_{{1}}\sin \left( \phi_{{1}} \right) {\it Fd}_{{y}}+l_{{1}} \cos \left( \phi_{{1}} \right) {\it Fs}_{{x}}+l_{{1}}\sin \left( \phi_ {{1}} \right) {\it Fs}_{{y}}+\tau_d\\ l_{{2}}\cos \left( \phi_{{2}} \right) {\it Fs}_{{x}}+l_{{2}}\sin \left( \phi_{{2}} \right) {\it Fs}_{{y}}\end {array} \right] \end{align*}

where $~\tau_d~$ is the "damper torque" about $~(0,0)~$ \begin{align*} &\tau_d=\left[\b R_d\times \b F_d\right]\cdot\b e_z=l_d\,\left[\cos(\phi_1)\,Fd_x+\sin(\phi_1)\,Fd_y \right] \end{align*}


The generalized torques are \begin{align*} &\b\tau_q=\sum\left[\frac{\partial \b R_i}{\partial \b q_i}\right]^T\,\b F_i \end{align*}

  • $~\b R_i~$ position vectors to the CM
  • $~\b F_i~$ external forces at the CM
  • $\b q_i~$ generalized coordinates vector

$~\b\tau_q~$ appeared at the right hand side of the NEWTON and EL equations

Example double pendulum

\begin{align*} &\b R_1=\left[ \begin {array}{c} \sin \left( \phi_{{1}} \right) l_{{1}} \\ -\cos \left( \phi_{{1}} \right) l_{{1}} \end {array} \right] \quad, \b R_2= \left[ \begin {array}{c} \sin \left( \phi_{{1}} \right) l_{{1}}+\sin \left( \phi_{{2}} \right) l_{{2}}\\ -\cos \left( \phi_{{1}} \right) l_{{1}}-\cos \left( \phi_{{2}} \right) l_{{2}} \end {array} \right]\\ &\b F_1= \left[ \begin {array}{c} {\it Fd}_{{x}}\\ {\it Fd}_ {{y}}\end {array} \right] \quad, \b F_2=\left[ \begin {array}{c} {\it Fs}_{{x}}\\ {\it Fs}_ {{y}}\end {array} \right]\quad\Rightarrow\\ &\b\tau_q=\left[\frac{\partial \b R_1}{\partial \phi_1}\right]^{T}\,\b F_1+ \left[\frac{\partial \b R_2}{\partial \phi_1}+ \frac{\partial \b R_2}{\partial \phi_2}\right]^T\,\b F_2 \end{align*}

to obtain the "damper torque" you can put a dummy mass at the connection point between the damper and the pendulum thus \begin{align*} &\b R_d= \left[ \begin {array}{c} l_{{d}}\sin \left( \phi_{{1}} \right) \\ -l_{{d}}\cos \left( \phi_{{1}} \right) \end {array} \right]\, \b F_d= \left[ \begin {array}{c} {\it Fd}_{{x}}\\ {\it Fd}_ {{y}}\end {array} \right]\quad\Rightarrow\\ &\b\tau_d=\left[\frac{\partial \b R_d}{\partial \phi_1}\right]^{T}\,\b F_d= \left[ \begin {array}{c} l_{{d}}\cos \left( \phi_{{1}} \right) {\it Fd}_{{x}}+l_{{d}}\sin \left( \phi_{{1}} \right) {\it Fd}_{{y}} \\ 0\end {array} \right] \end{align*}

  • $\begingroup$ The torque balance equations should be summed up on each mass location with $(R_{\rm point} - R_{\rm COM}) \times F$. Also check for missing $l_1$ in third equation above. $\endgroup$
    – JAlex
    Commented Jan 27, 2023 at 16:52
  • $\begingroup$ @JAlex you mean that this "theory" is wrong ? what is $~R_{\rm point}~$ which equation is $~l_1~$ missing $\endgroup$
    – Eli
    Commented Jan 27, 2023 at 17:20
  • $\begingroup$ I think the 3rd equation should be $$\begin{align*} &\tau_d=\left[ R_d\times F_d\right]\cdot e_z=\,\left[l_1 \cos(\phi_1)\,Fd_x+l_1 \sin(\phi_1)\,Fd_y \right] \end{align*}$$ since the component units should be that of [Nm] and not just [N]. Actually the moment arm length isn't $l_1$, but $l_1 - b_3$ the distance between the attachment point of the damper and the first mass. $\endgroup$
    – JAlex
    Commented Jan 27, 2023 at 18:35
  • $\begingroup$ Which brings me to my first point. The rotational equation of motion for each mass should be $\tau_{\rm net} = I_i \dot{\omega} = 0$ where $\tau_{\rm net}$ is the torque of the damping force and the pivot force about the center of mass. $\endgroup$
    – JAlex
    Commented Jan 27, 2023 at 18:37
  • $\begingroup$ $~R_d~$ is the position vector from the joint to the damper force, I missed the the length $~l_d$ $\endgroup$
    – Eli
    Commented Jan 27, 2023 at 19:59

What is missing from your diagram is that in order to describe the forces generated by the spring and the damper, you need to track their endpoints.


Generally, figuring out what locations are important, and tracking them are the first step in such analysis. In the attached paper this is done in equations (10)-(20).

In your case, choose a designation for these points, and track their position and velocity.

Once you have that you can describe the force elements as

  • Damper $$\boldsymbol{F}_d = -b \left( \boldsymbol{e}_{34} \cdot ( \boldsymbol{\dot{r}_3}-\boldsymbol{\dot{r}_4})\right) \boldsymbol{e}_{34} $$ where $\boldsymbol{e}_{34}$ is the direction vector for the damper.

  • Spring $$\boldsymbol{F}_s = -k \left( \boldsymbol{e}_{25} \cdot ( \boldsymbol{{r}_2}- \boldsymbol{{r}_5}) - \ell_{25}\right) \boldsymbol{e}_{25} $$ where $\boldsymbol{e}_{25}$ is the direction vector for the spring, and $\ell_{25}$ its free length.

Then add these forces to equations (30)-(31) as needed.

But because the 0-1 rod has the damper attached in the middle of it, an additional equation is needed at this point, making sure the sum of forces on the red rod is zero (since it is assumed massless).


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