 Solution

With $m_1=m_2=m_3=m$, the explicit energy and ordinary rotation generators just discussed areThe energy is constant because the drive factor no longer depends on time. For each rotation charge, differentiating and using ${\ddot{x}}_i=-{\omega }^2{x}_i$ gives ${\dot{J}}_{ij}=m(x_i{\ddot{x}}_j-x_j{\ddot{x}}_i)=0$.

These are angular momenta in the auxiliary three-dimensional space whose coordinates are the particle labels $(x_1,x_2,x_3)$. They are not the physical spatial angular momentum ${\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$; generic $U_{12},U_{13},V$ do not have physical rotational invariance. They are precisely the equal-mass representatives of the weighted charges already present in solution. The new energy integral supplies the autonomous relative-$y$ dynamics not present in the driven case.