 Solution

The coordinate $\psi$ is cyclic, so $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )$ is conserved. Initially the spin is zero, hence $\boxed{\dot{\psi }+\dot{\varphi }\cos \theta =0}$ throughout. The coordinate $\varphi$ is also cyclic. After imposing this zero spin, its conserved momentum becomes $(A+m{h}^2)\dot{\varphi }{\sin }^2\theta$. Initially $\theta =\pi /2$ and $\dot{\varphi }=\omega$, so $\boxed{\dot{\varphi }{\sin }^2\theta =\omega }$.

The equation for $z$ isThe quantity in parentheses is the vertical velocity of the centre of mass. Initially that velocity and its height are both zero: the initial rotation is horizontal and the vertex starts at rest at the reference height. Twice integrating givesIndependently, the frictionless vertical wire exerts only horizontal force. Gravity is therefore the only vertical external force on the body, and the centre of mass undergoes vertical free fall. This gives the same equation without the angular equations.