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