Potential energy¶
The potential energy of the \(\ib\)-th object \(U_{\ib}\) is given by
\[U_{\ib}
\equiv
-
m g
\sum_{\ic = 0}^{\ib}
l \sin \pos_{\ic}.\]
Thus the total potential energy \(U\) is
\[ \begin{align}\begin{aligned}U
&
\equiv
\sum_{\ib = 0}^{N-1}
U_{\ib}\\&
=
-
m g l
\sum_{\ib = 0}^{N-1}
\sum_{\ic = 0}^{\ib}
\sin{\pos_{\ic}}\\&
=
\pene.\end{aligned}\end{align} \]
Derivative¶
First,
\[\tder{}{t} \pder{U}{\vel_{\ia}}
=
0\]
since \(U\) is not a function of \(\vel_{\ia}\).
On the other hand, I have
\[ \begin{align}\begin{aligned}\pder{U}{\pos_{\ia}}
&
=
-
m g l
\sum_{\ib = 0}^{N-1}
\left( N - \ib \right)
\pder{}{\pos_{\ia}} \sin{\pos_{\ib}}\\&
=
-
m g l
\sum_{\ib = 0}^{N-1}
\left( N - \ib \right)
\pder{\pos_{\ib}}{\pos_{\ia}}
\cos{\pos_{\ib}}\\&
=
-
m g l
\sum_{\ib = 0}^{N-1}
\left( N - \ib \right)
\delta_{\ib \ia}
\cos{\pos_{\ib}}\\&
=
-
m g l
\left( N - \ia \right)
\cos \pos_{\ia}.\end{aligned}\end{align} \]