Consider a heavy spinning symmetric top with moments of inertia $I_1=I_2=I$ and $I_3$ in an inertial frame which experiences two forces:
🧠 Remember: we will use the following formulas derived
$$ \begin{aligned} \vec\omega&=\dot \phi[\cos(\theta)\hat e_r-\sin(\theta)\hat e_\theta]+\dot \theta\hat e_\phi \\ \dot{ \hat e}r&=\dot\theta\hat e\theta+\dot\phi \sin(\theta)\hat e_\phi \\ \dot{\hat e}\theta &= -\dot\theta \hat e_r+\dot\phi\cos(\theta)\hat e\phi \\ \dot {\hat e}\phi&=-\dot\phi[\sin(\theta)\hat e_r+\cos(\theta)\hat e\theta] \end{aligned} $$
We can start by deriving the torque imparted by the gravitational force
$$ \vec M=\vec r\times m\vec g=-mg R(\hat e_r\times \hat e_z)=mgR\sin(\theta)\hat e_\phi $$
where $\theta$ and $\phi$ are used to describe the orientation of the principal axis, the 3$^{\text{rd}}$-axis points in the direction $\vec e_r$
🗒️ Note: torque only acts in $\vec e_\phi$ direction, thus there are 2 conserved components of the angular momentum
We will call the rotational degree of freedom of the top $\psi$ and the angles $\theta,\phi,\psi$ are known as Euler angles
We can derive the angular velocity and angular momentum
$$ \begin{aligned} \vec \omega &=\left [\dot\phi\cos(\theta)+\dot \psi\right ]\hat e_r-\dot \phi\sin(\theta)\hat e_\theta+\dot\theta\hat e_\phi \\ \vec L&=I_3 \left [ \dot \psi \cos ( \theta ) +\dot \psi \right ]\hat e_r+I \left [ \dot \theta \hat e_\phi -\dot \phi \sin( \theta ) \hat e_\theta \right ] \\ \frac 12 \frac {\text d}{\text dt} \left | \vec L \right |^2&=\vec M \cdot \vec L \end{aligned} $$
🗒️ Note: in the case where $\theta$ is constant: $\vec M\cdot\vec L=0$ and hence $|\vec L|$ is conserved. In this case the angular momentum vector will evolve so as to maintain a constant modulus ( it will rotate ).
We can find the rate of change of the angular momentum vector
$$ \begin{aligned} \dot{\vec L}&=I_3\left [ \ddot \phi \cos(\theta)-\dot \phi \dot \theta\sin(\theta)+\ddot \psi \right ]\hat e_r \\ &+\left \{ I_3 \left [ \dot\phi\cos(\theta)+\dot\psi \right ]\dot \phi \sin(\theta)+I \left [ \ddot \theta-\dot \phi^2 \sin(\theta)\cos(\theta) \right ] \right \} \hat e_\phi \\ &+\left \{ I_3 \left [ \dot \phi\cos(\theta)+\dot \psi \right ]\dot\theta-I\left [ \ddot \phi\sin(\theta)+2\dot\phi\dot\theta\cos(\theta) \right ] \right \}\hat e_\theta \end{aligned} $$
Using N2 we can show that
$$ \begin{aligned} \ddot \phi \cos(\theta)-\dot \phi \dot \theta\sin(\theta)+\ddot \psi &=0 \\ I_3 \left [ \dot \phi\cos(\theta)+\dot \psi \right ]\dot\theta-I\left [ \ddot \phi\sin(\theta)+2\dot\phi\dot\theta\cos(\theta) \right ] &=0 \\ I_3 \left [ \dot\phi\cos(\theta)+\dot\psi \right ]\dot \phi \sin(\theta)+I \left [ \ddot \theta-\dot \phi^2 \sin(\theta)\cos(\theta) \right ]&=mgR\sin(\theta) \end{aligned} $$
We find motion $\dot\phi$ using the above equation we can write
$$ \frac{\text d}{\text dt}\left [ \dot\phi\cos(\theta)+\dot\psi \right ]=0 $$
which implies that the component of angular velocity along the spin axis of the top $\omega_3\equiv \dot\phi\cos(\theta)+\dot\psi$ is constant.
$$ \begin{aligned} I_3\omega_3\dot\theta-\frac{I}{\sin(\theta)} \frac{\text d}{\text dt}\left [ \dot \phi \sin^2(\theta) \right ]&=0 \\ \Rightarrow \quad I_3\omega_3\cos(\theta)+I\dot\phi\sin^2(\theta)&\equiv bI
\end{aligned} $$
where $b$ is a constant, this can be re-aranged to find
$$ \dot \phi=\frac{bI-I_3\omega_3\cos(\theta)}{I\sin^2(\theta)}=\frac{b-a\cos(\theta)}{\sin^2(\theta)} $$
where $a=I_3\omega_3/I$. The parameters $a$ and $b$ can be used to define the dynamics of the system
We can define angular momentum of the system about the spin axis ( $3^\text{rd}$-axis) as $L=I_3\omega_3=Ia$ and the conservation of angular momentum in the direction $\hat e_\theta$ as $K=Ib$. Thus we can write the final equation of motion as
$$ \ddot\theta=\sin(\theta)\left [ \frac 12 \beta-a\dot\phi+\phi^2\cos(\theta) \right ] $$
where $\beta=2mgR/I$
Consider the case where $\theta=\theta_0$ is a constant, where there can be a spinning and precessing solution with constant $\dot\psi=\Omega_S$ and $\dot \phi=\Omega_P$ with $\Omega_S=\frac{I}{I_3}a-\Omega_P\cos(\theta_0)$ and $\Omega_P$ is the solution to the following quadratic equation:
$$ \cos(\theta_0)\Omega^2_P-a\Omega_P+\frac 12 \beta=0 $$
where the solution is
$$ \Omega_P=\frac{a}{2\cos(\theta_0)}\left [ 1 \pm \left ( 1-\frac{2\beta\cos(\theta_0)}{a^2} \right )^\frac 12 \right ] $$
Cases:
When $\cos(\theta_0)<a^2/(2\beta)=I^2_3\omega^2_3/(4mgRI)$ if $a^2>2\beta$ then there appears to always be a solution
$a^2<2\beta$ then there exists a critical angle $\theta_\text{crit}=\cos^{-1}\left [ a^2/(2\beta)\right ]$ where $\theta_0>\theta_\text{crit}$, there is also no solution with $\theta_0=0$. we can also see that
$$ \omega_3>\frac{\sqrt{4mgRI\cos(\theta_0)}}{I_3} $$
is necessary to prevent the body from falling down.
To better understand the behavior we can look when $2\beta\cos(\theta_0)/a^2 \ll 1$ which is the limit when gravity force is unimportant
$$ \begin{aligned} \text{when}\; a^2<2\beta \quad \Omega^\text{fast}_P&\approx \frac{a}{\cos\theta_0}=\frac{I_3\omega_3}{I\cos\theta_0}=\frac{L}{I\cos\theta_0} \\ \text{when}\;a^2>2\beta \quad \Omega^\text{slow}_P&\approx \frac{\beta}{2a}=\frac{mgR}{I_3\omega_3}=\frac{mgR}{L} \end{aligned} $$
These solutions are known as fast and slow solution respectively because $\Omega_P^{\text{fast}}>\Omega_P^{\text{slow}}$ when $\cos(\theta_0)>0$.
The rotational kinetic energy $T$ satisfies $\dot T_R=\vec \omega\cdot\vec M$ for a general rotation system in our case
$$ \vec \omega\cdot\vec M=mgR\dot\theta \sin(\theta)=-mgR\frac{\text d}{\text dt}\cos(\theta) $$
We can then deduce that $E=T_R+V=T_R+mgR\cos(\theta)$ is constant and can be interpreted as the energy of the system
$$ E=\frac 12 I_3\left [ \dot\phi\cos(\theta)+\dot \psi \right ]^2+\frac 12 I \left [ \dot \theta^2+\dot\phi^2\sin^2(\theta)\right ]+mgR\cos(\theta) $$
Finally we can write
$$ \begin{aligned} \bar E&=\frac 12 \dot\theta^2+U_\text{eff}(\theta) \\ \text{where:} \quad U_\text{eff}(\theta)&=\frac 12 \left [ \left ( \frac{b-a\cos(\theta)}{\sin(\theta)}\right )^2+\beta \cos(\theta)\right ] \end{aligned} $$
We see that the general motion of the body can be described by an effective 1D motion in the $\theta$ direction described by the effective potential $U_\text{eff}(\theta)$.