Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 7D / b / Solution

Eliminate the cyclic velocities using $\dot{\varphi }=(L_z-L_3\cos \theta )/(I_1{\sin }^2\theta )$. The conserved energy becomesDifferentiating the effective potential giveswhich is also the $\theta$ Euler-Lagrange equation. Hence $I_1\ddot{\theta }=-V_{\mathrm{eff}}^{\prime }(\theta )$. Deriving it directly from the Euler-Lagrange equation ensures it also holds at turning points, where division by $\dot{\theta }$ in an energy argument would be invalid.