As per the given question,
x1=l1sinϕ1
y1=−l1cosϕ1
x2=l1sinϕ1+l2sinϕ2
y2=−(l1cosϕ1+l2cosϕ2),
Now, taking the differentiation ,
dtdx1=l1cosϕ1dtdϕ1
dtdy1=l1sinϕ1dtdϕ1
dty2=l1sinϕ1dtdϕ1+l2sinϕ2dtdϕ2
dtx2=(l1cosϕ1dtdϕ1+l2cosϕ2dtdϕ2)
L=T−V=2m1v12+2m2v22−V
T=2m1((dtx1)2+(dty1)2)+2m1((dtx2)2+(dty2)2)
Hence langrangian equation will be
L=2(m1+m2)l12(dtdϕ1)2+2(m2)l22(dtdϕ2)2+m2l1l2dtdϕ1dtdϕ2cos(ϕ1−ϕ2)+(m1+m2)gl1cosϕ1+m2gl2cosϕ1
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