Can the double pendulum equations not be derived solely using Newtonian Mechanics (Newton's Laws of Motion)?

The double pendulum can be solved directly using Newton's laws of motion. However, I would imagine that it is easier to solve it using the Lagrangian formulation because the constraints (the lower pendulum is attached to the upper one) are more easily introduced. If you just wanted to use $F=ma$, you would have to figure out what force is being exerted at every moment in time by the joint between the two pendulums to keep them from coming apart.

The EOM's with NEWTON method are:

$$J^T,M,J,vec{ddot{q}}=J^T,left(vec{f}_a-M,vec{f}_zright)tag 1$$

where:

• $vec q$ generalized coordinates
• $J=frac{partialvec R}{partialvec q}$ the Jacobi matrix
• $vec R$ Position vector
• M Mass Matrix
• $vec{f}_a$ Applied force vector
• $vec{f}_z=frac{partial(J,vec{dot q})}{partialvec{q}},vec{dot q}$ fictitious force vector

Double Pendulum equations

generalized coordinates

$$vec q=left[ begin {array}{c} varphi _{{1}} varphi _ {{2}}end {array} right] $$

position vectors

$$vec R_1=left[ begin {array}{c} L_{{1}}sin left( varphi _{{1}} right) L_{{1}}cos left( varphi _{{1}} right) end {array} right] ~,vec R_2=left[ begin {array}{c} L_{{2}}sin left( varphi _{{2}} right) +L _{{1}}sin left( varphi _{{1}} right) L_{{2}} cos left( varphi _{{2}} right) +L_{{1}}cos left( varphi _{{1}} right) end {array} right] $$

Mass Matrix:

$$M_1= begin{bmatrix} m_1 & 0 0 & m_1 end{bmatrix}~,M_2= begin{bmatrix} m_2 & 0 0 & m_2 end{bmatrix}$$

Applied Forces:

$$vec f_{a1}=begin{bmatrix} 0 -m_1,g end{bmatrix}~,vec f_{a2}begin{bmatrix} 0 -m_2,g end{bmatrix}$$

combine $$vec R=begin{bmatrix} vec{R}_1 vec{R}_2 end{bmatrix}~,M=begin{bmatrix} M_1 & 0 0 & M_2 end{bmatrix}~,vec f_a=begin{bmatrix} vec{f}_{a1} vec{f}_{a2} end{bmatrix} $$

with equation (1) you obtain the equations of motion

edit

$$J=left[ begin {array}{cc} L_{{1}}cos left( varphi _{{1}} right) &0 -L_{{1}}sin left( varphi _{{1}} right) &0 L_{{1}}cos left( varphi _{{1}} right) &L_{{2} }cos left( varphi _{{2}} right) -L_{{1}}sin left( varphi _{{1}} right) &-L_{{2}}sin left( varphi _{{2}} right) end {array} right] $$

$$vec{f}_z= left[ begin {array}{c} -L_{{1}}sin left( varphi _{{1}} right) { dotvarphi _{{1}}}^{2} -L_{{1}}cos left( varphi _ {{1}} right) {dotvarphi _{{1}}}^{2} -L_{{1}}sin left( varphi _{{1}} right) {dotvarphi _{{1}}}^{2}-L_{{2}}sin left( varphi _{{2}} right) {dotvarphi _{{2}}}^{2} -L_{{1}}cos left( varphi _{{1}} right) {dot varphi _{{1}}}^{2}-L_{{2}}cos left( varphi _{{2}} right) {dot varphi _{{2}}}^{2}end {array} right] $$

$$J^T,M,J=left[ begin {array}{cc} {L_{{1}}}^{2}m_{{1}}+{L_{{1}}}^{2}m_{{2}}&L _{{1}}m_{{2}}L_{{2}}cos left( varphi _{{1}}-varphi _{{2}} right) L_{{1}}m_{{2}}L_{{2}}cos left( varphi _{{1}}- varphi _{{2}} right) &{L_{{2}}}^{2}m_{{2}}end {array} right] $$

$$J^T,vec f_a=left[ begin {array}{c} L_{{1}}sin left( varphi _{{1}} right) m_ {{1}}g+L_{{1}}sin left( varphi _{{1}} right) m_{{2}}g L_{{2}}sin left( varphi _{{2}} right) m_{{2}} gend {array} right] $$

$$J^T,M,vec f_z=left[ begin {array}{c} L_{{1}}m_{{2}}L_{{2}}{dotvarphi _{{2}}}^{2} sin left( varphi _{{1}}-varphi _{{2}} right) -L_{{2}}m_{{2}}L_{{1}}{dotvarphi _{{1}}}^{2}sin left( varphi _{{1}}-varphi _{{2}} right) end {array} right] $$