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

\[ \begin{align}\begin{aligned}\pder{\potential}{\pos_{\ia}} & = - m g l \sum_{\ib = 0}^{N-1} \left( N - \ib \right) \pder{}{\pos_{\ia}} \sin{\pos_{\ib}}\\& = - m g l \sum_{\ib = 0}^{N-1} \left( N - \ib \right) \pder{\pos_{\ib}}{\pos_{\ia}} \cos{\pos_{\ib}}\\& = - m g l \sum_{\ib = 0}^{N-1} \left( N - \ib \right) \delta_{\ib \ia} \cos{\pos_{\ib}}\\& = - m g l \left( N - \ia \right) \cos \pos_{\ia}.\end{aligned}\end{align} \]