Motion Equations by Newton's Formalism for a Double Pendulum

[Daniel Boy]

By Lagrange's formalism, the motion equations for double pendulum are:

Using Newton's formalism I can't obtain the second equation. Anyone can help?

 

[BvU]

Hello @Daniel Boy , $\qquad$ $\qquad$ !

Usually, at this point we ask 'what have you got so far ? ' (for the 2nd eqn, I mean)
You do have a Lagrangian already ? Or do you try a Newton approach from scratch ?

Perhaps a comparable thread (with $l_2 = l_3$) is :

View attachment 272888

 

[etotheipi]

If you want to use the Newton formalism, there are some different approaches you can take. Maybe the easiest would be to write one $\boldsymbol{\tau} = \dot{\boldsymbol{L}}$ equation for the whole system in coordinate system with origin at the topmost hinge, and then another $\boldsymbol{\tau}' = \dot{\boldsymbol{L}}'$ equation for the lower rod only in a coordinate system with origin at the position of an $m_1$ [N.B. this latter coordinate system will be accelerating, so you need to include a further 'fictitious' force $-m_2 \boldsymbol{a}_1$ acting on $m_2$ in your analysis].

I think the algebra will be a little nasty, whichever way you go about it. Good luck!

 

[Daniel Boy]

Hello @Daniel Boy , $\qquad$ $\qquad$ !

Usually, at this point we ask 'what have you got so far ? ' (for the 2nd eqn, I mean)
You do have a Lagrangian already ? Or do you try a Newton approach from scratch ?

Perhaps a comparable thread (with $l_2 = l_3$) is :

I already have the Lagragian. I want to obtain the same equations using Newton's formalism, so I did the free body diagrams:

For m2, I found F2 (analysis in x) and I found the following equation (analysis in y):

So, using F2 in the free body diagram for m1, the resultant equation does not coincide with: