 Solution

By reflection symmetry, $G$ is the midpoint of the cylinder axis. With position $r$ measured from $G$, define $J_{ij}^G=\int (|r{|}^2{\delta }_{ij}-r_i{r}_j)dm$. Symmetry makes the off-diagonal entries zero. Uniform density gives $⟨x^2⟩=⟨y^2⟩=a^2/4$ and $⟨z^2⟩=b^2/3$, by integrating over a disk and the interval $[-b,b]$. HenceFor parallel axes based at $P$, the tensor form of the parallel-axis theorem is $J^P=J^G+M(|d{|}^{2}I-d{d}^T)$, where $d=\overrightarrow{PG}$. This follows by substituting $r+d$ in the defining integral and using $\int rdm=0$. Here $d$ is an axial vector of length $b$, soThe Euler angles $\psi ,\varphi$ are cyclic coordinates. Their conserved conjugate momenta are $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )=Cn$ and $p_{\varphi }=A\dot{\varphi }{\sin }^2\theta +Cn\cos \theta =h$. The Lagrangian has no explicit time dependence, so the conserved energy is kinetic plus gravitational potential, namely the expression given with constant $E$.

Eliminate the angular velocities using $\dot{\varphi }=(h-Cn\cos \theta )/(A{\sin }^2\theta )$ and $\dot{\psi }+\dot{\varphi }\cos \theta =n$. The energy equation becomesThusThe elimination is valid away from the coordinate poles; the multiplied equation extends by continuity to any pole reached by a regular motion.