= Solution
For a body-fixed <mass> element at position $\mathbf r$ relative to the fixed tip, $\mathbf v=\boldsymbol\omega\times\mathbf r$. Therefore
$$
\mathbf L=\int\mathbf r\times(\boldsymbol\omega\times\mathbf r)\,dm=\mathsf I\boldsymbol\omega,\qquad I_{ij}=\int(r^2\delta_{ij}-r_ir_j)\,dm.
$$
The <inertia tensor> is real symmetric. Its orthogonal <eigenvectors> define <principal axes> and its <eigenvalues> are <principal moments of inertia>. <angular momentum> is parallel to <angular velocity> when rotation is along a <principal axis>, but generally not for arbitrary rotation.
Write $\mathbf n=\hat{\mathbf e}_3$ and $s=I_3\omega_3$. Axial symmetry gives $\mathbf L=I_1\boldsymbol\omega+(I_3-I_1)\omega_3\mathbf n$. Since $\dot{\mathbf n}=\boldsymbol\omega\times\mathbf n$, the vector triple-product identity gives $\mathbf n\times\dot{\mathbf n}=\boldsymbol\omega-\omega_3\mathbf n$. Hence
$$
\boxed{\mathbf L=I_1\mathbf n\times\dot{\mathbf n}+s\mathbf n.}
$$
Differentiate in the <inertial frame>, using constant $s$ and $\dot{\mathbf n}\times\dot{\mathbf n}=0$. The gravitational <torque> about the fixed tip is $(\ell\mathbf n)\times(-mg\hat{\mathbf z})$, yielding the <symmetry-axis equation of a heavy symmetric top>:
$$
\boxed{s\dot{\mathbf n}+I_1\mathbf n\times\ddot{\mathbf n}=-mg\ell\mathbf n\times\hat{\mathbf z}.}
$$
The tip reaction exerts no moment about that tip.
For steady precession, $\mathbf n=(\sin\theta\cos\Omega_Pt,\sin\theta\sin\Omega_Pt,\cos\theta)$. At azimuth zero, put $s_\theta=\sin\theta$ and $c_\theta=\cos\theta$. Then $\dot{\mathbf n}=(0,\Omega_Ps_\theta,0)$, $\ddot{\mathbf n}=(-\Omega_P^2s_\theta,0,0)$, and the $y$ component gives $s\Omega_Ps_\theta-I_1\Omega_P^2s_\theta c_\theta=mg\ell s_\theta$. For a nonvertical axis this becomes
$$
I_1\cos\theta\,\Omega_P^2-s\Omega_P+mg\ell=0,
$$
$$
\boxed{\Omega_P=\frac{I_3\omega_3\pm\sqrt{I_3^2\omega_3^2-4mg\ell I_1\cos\theta}}{2I_1\cos\theta}.}
$$
Real steady branches require a nonnegative discriminant. In particular, for a top pointing above the horizontal there is a spin threshold for these fixed-angle solutions. At $\cos\theta=0$ use the quadratic's linear limit $\Omega_P=mg\ell/s$, rather than dividing by zero. For a vertical axis, azimuth does not describe a physical conical motion and the division by $\sin\theta$ is inappropriate.
For rapid positive spin $s$ with $s^2\gg4mg\ell I_1\cos\theta$, the <slow and fast steady precession> roots are
$$
\boxed{\Omega_{\rm slow}\simeq\frac{mg\ell}{s},\qquad \Omega_{\rm fast}\simeq\frac{s}{I_1\cos\theta}-\frac{mg\ell}{s}.}
$$
The approximation assumes a nonsingular fixed inclination. For $g=0$, the quadratic instead gives \b[$\Omega_P=0$ or $\Omega_P=s/(I_1\cos\theta)$]. The zero branch has a stationary symmetry axis and pure axial spin. On the other branch <angular momentum> is constant and parallel or antiparallel to the chosen $z$ axis; $L_z=I_1\Omega_P$ and $s=L_z\cos\theta$. This is torque-free precession in space, not the different rate at which transverse components rotate in body coordinates.
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