 Solution

For a holonomic conservative system the Lagrangian is $L(q,\dot{q},t)=T-V$, the difference between kinetic energy and potential energy. The Euler-Lagrange equations areIf $\partial L/\partial q_j=0$, this gives $d{p}_j/dt=0$ for the generalized momentum $p_j=\partial L/\partial {\dot{q}}_j$. Such a coordinate is a cyclic coordinate.

In cylindrical coordinates the squared speed is ${\dot{r}}^2+r^2{\dot{\theta }}^2+{\dot{z}}^2$. The holonomic constraint $z=r^2/a$ gives $\dot{z}=2r\dot{r}/a$, while gravity has potential energy $V=mgz$. ThereforeThere is no explicit $\theta$ dependence, so $\theta$ is a cyclic coordinate andThis is angular momentum about the vertical symmetry axis. At $r=0$ the polar angle is singular, but the Cartesian angular momentum $m(x\dot{y}-y\dot{x})$ remains well defined and has the same conserved value.