Qualitative analysis of the equation and symmetry (point on sphere) A point moves on the surface of sphere ($R>0$ - radius) along the curve defined by the differential equation in spherical coordinate system:
$R^2(|\dot \theta|^2 + w^2 \sin^2 \theta)=(at)^2$,  azimuthal angle $\varphi=\omega t$, $a>0$,  $\omega>0$, $\theta(0)=0$, $\theta''(0)>0$.
I am trying to understand features of the solution and to have a qualitative analysis of the equation.
Hopefully it was possible to find solution for small $\theta$: $(at)^2 = R^2(|\dot \theta|^2 + \omega^2 \theta^2)$ ( it was done by Robert Bryant (What is symmetry group of non-linear equation?)
However, I am not sure what to do for $t \to \infty$ to get a general understanding of the solution.
According to computer simulation there is a 'loop' on the curve. My understanding is that the qualitive analysis should find out the reasons of the 'loop'.


Any hints and helps are welcomed.
 A: Well, here are a few comments that might be helpful:
First, if one sets $q = a/R>0$, then the equation the OP wants to study can be written in the form
$$
\dot\theta^2 +\omega^2\,\sin^2\theta - q^2\,t^2 = 0
$$
Using the same methods that I described in the MO question the OP referenced in this question, it is not hard to show that there is a unique smooth (in fact, real-analytic) solution satisfying $\theta(0) = \theta'(0) = 0$ and $\theta''(0) >0$.  Its Taylor series at $t=0$ starts out
$$
\theta(t) = \frac{q}2\,t^2 - \frac{q\,\omega^2}{32}\,t^4 + \frac{q\,\omega^4}{768}\,t^6-\frac{q\,\omega^2\,(\omega^4-64\,q^2)}{49152}\,t^8 
- \frac{q^3\,\omega^4}{5120}\,t^{10}-\cdots.
$$
Moreover, $\theta$ is defined for all time $t$ with $\theta(-t)=\theta(t)$ while $\theta'(t)>0$ for $t>0$ and, for $t>\!>0$, one has $\theta(t)/t^2\approx q/2$. In particular, $\theta(t)$ increases monotonically without bound as $t\to\infty$.
There is a monotonically increasing sequence $0<t_1<t_2<t_3<\cdots$ so that $\theta(t_k) = k\pi$.  Given a fixed $q>0$ one can choose $\omega>0$ sufficiently small, so that for small $k$, one will have $t_k\approx \sqrt{2\pi k/q}$.
If I understand the OP correctly, the red curve being sketched is the curve
$$
\gamma(t) = \left(\,R\cos(\omega t)\sin\theta(t),\,R\sin(\omega t)\sin\theta(t),\,R\cos\theta(t)\,\right).
$$
The time $t_1>0$ will be the first time that $\gamma(t_1) = (0,0,-R)$ (i.e., the 'south pole').  At that point, the velocity of $\gamma$ will be $\gamma'(t_1) = \left(R\cos(\omega t_1)\theta'(t_1),R\sin(\omega t_1)\theta'(t_1),0\right)\not=(0,0,0)$. (The velocity of $\gamma$ vanishes only at $t=0$.)
The question of whether $\gamma\bigl([t_1,t_2]\bigr)$ crosses  $\gamma\bigl([0,t_1]\bigr)$ other than at the poles, thus creating a 'loop' as depicted in the above images, is the question of whether there exist $s_1$ and $s_2$ satisfying $0<s_1<t_1<s_2<t_2$ so that $\cos(\theta(s_1))=\cos(\theta(s_2))$ while $s_2-s_1 = (2m{+}1)\pi/\omega$ for some integer $m\ge0$.  However, given a sufficiently large $q$ and sufficiently small $\omega$, this would yield a contradiction, since one has $\pi/\omega \le s_2-s_1 < t_2\approx\sqrt{4\pi/q}$.
Thus, while there can be a loop as depicted for some values of the positive parameters $(\,a/R,\,\omega\,)$, this does not always happen.
