= Solution
Write the position of a mass element relative to the <centre of mass> as $r$. Expand its squared <velocity> and integrate:
$$
T=\frac12\int|v_G+\omega\times r|^2\,dm
=\frac12M|v_G|^2+\frac12\int|\omega\times r|^2\,dm.
$$
The mixed term vanishes because $\int r\,dm=0$. Also
$$
L_G=\int r\times(\omega\times r)\,dm,\qquad
\omega\cdot L_G=\int|\omega\times r|^2\,dm,
$$
using the scalar triple-product identity. Therefore
$$
\boxed{T=\tfrac12M|v_G|^2+\tfrac12\omega\cdot L_G.}
$$
In <principal axes> the <inertia tensor> is diagonal, so the second term is $\tfrac12(A\omega_1^2+B\omega_2^2+C\omega_3^2)$. Substituting the three Euler-angle components given in the source gives its displayed kinetic-energy expression directly.
For $A=B$, the first two component squares have opposite mixed terms and sum to $\dot\theta^2+\dot\phi^2\sin^2\theta$. Put $n=\dot\psi+\dot\phi\cos\theta$. Using the orientation-dependent potential printed in the source therefore gives the formal model
$$
L=\tfrac12 A\dot\theta^2+\tfrac12 A\dot\phi^2\sin^2\theta+\tfrac12 Cn^2
+\tfrac12M(\dot x^2+\dot y^2+\dot z^2)-Mgz\cos\theta.
$$
The <cyclic coordinates> $x,y,\psi$ have conserved momenta $M\dot x,M\dot y,Cn$. Consequently $\dot x,\dot y,n$ are constant. The coordinate $\phi$ is also cyclic, with
$$
p_\phi=A\dot\phi\sin^2\theta+Cn\cos\theta.
$$
For imposed fixed tilt $\theta=\alpha\in(0,\pi/2)$, conservation of $n$ and $p_\phi$ forces $\dot\phi=\Omega$ constant. Impose the fixed-height condition $z=h$ in this formal constrained model. Its $\theta$ equation is
$$
A\ddot\theta-A\dot\phi^2\sin\theta\cos\theta
+Cn\dot\phi\sin\theta-Mgz\sin\theta=0.
$$
<Set> the tilt constant and divide by its nonzero sine. This gives
$$
\boxed{A\Omega^2\cos\alpha-Cn\Omega+Mgh=0.}
$$
A real <polynomial root> requires its <quadratic discriminant> to be nonnegative:
$$
\boxed{C^2n^2\ge4AMgh\cos\alpha.}
$$
When this condition holds the two algebraic rates are $(Cn\pm\sqrt{C^2n^2-4AMgh\cos\alpha})/(2A\cos\alpha)$.
There is a physical inconsistency in the printed data that must be kept distinct from this algebra. If $(x,y,z)$ are literally <Cartesian coordinates> of the <centre of mass> in uniform <gravity>, the <gravitational potential> is $Mgz$, not $Mgz\cos\theta$. The printed <Lagrangian> consequently does not follow from that Cartesian-height interpretation. Moreover its unconstrained $z$ equation is $\ddot z=-g\cos\theta$, so constant height at the indicated tilt requires a constraint reaction. The requested rate and <discriminant> follow from the printed orientation-dependent potential with fixed height imposed; a standard heavy-top interpretation instead uses a height proportional to an axis distance times $\cos\theta$, with the reference point and corresponding inertia specified consistently. The kinetic-energy derivation is independent of this source inconsistency.
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