Rolling a ball into a cone; what should the forces overall be? This is my solution to finding angular displacement/velocity/acceleration on cone so far.
Consider a cone, with an apex of half-angle $\psi$ pointing down, and a height of $h$. If I roll a ball into the tangent of the top of the cone (the top of the cone is parrallel with the ground, as is the initial path of the ball) with velocity $u$. We will call this 'initial velocity'.
I am using the relationship $$\alpha = {a_t \over r} = {F\over mr}$$ to start off. Where alpha is angular acceleration.
Radius is given by trigonometry as: $$r = (h-Z)\tan (\psi)$$ where $Z$ is the displacement from the top of the cone.

the expressions:
Given that the only tangential force should be caused by friction (and negative, because it is acting against the motion), and integrating with respect to time we get the three equations: $$\alpha = - {\mu N \over m(h-Z)\tan (\psi)}$$ $$\omega = - {\mu Nt \over m(h-Z)\tan(\psi)}+{u \over h\tan(\psi)}$$ $$\theta = - {\mu Nt^2 \over 2m(h-Z)\tan(\psi)} + {ut \over h\tan(\psi)}$$

Z:
$Z$ is displacement from the top of the cone at time $t$. I first imagine the overall acceleration down the cone, $a$, as the hypotenuse of a triangle, and the vertical acceleration $a_v$ as the adjacent when using $\psi$. Therefore $$a_v = a\cos(\psi)$$ and through integration, the vertical displacement $Z$ is ... $$Z = {1\over 2} at^2 \cos(\psi)$$
we can then place these in for $Z$ in the above expressions!

summary of what I don't understand:
The parts I don't understand are the parts I have left! an expression for $N$, the normal force, and for $a$ the acceleration towards the apex of the cone. At first I thought these would be simple, like when you get a question on a mass on an inclined plane, or a banked turn. As it turns out it is a lot harder because there should be a force going up the slope (on top of friction), induced by $u$! (right?)
When not considering this part, the equations were almost right, except the path of the ball (when viewed on the top) did a sort of half turn around the apex before hitting it. Whereas in a simulation I did in unity (and as you would expect in real life) went round the apex more and more as it got closer, until terminating. (I'll get some pictures to show you).
So if I add this part to the overall force/acceleration then it should keep rolling for longer. I am also not sure, does the tangential velocity $v_t$ ever change from $u$? this may be useful to know if the force going up is proportional to $v_t$.
Thank you in advance!
The answer that @ja72 gave me (which was fantastic, but not sure where exactly to take it, may end up using some of it to solve this :D) was for an older version of this question. (I don't know if you can view my edits for this, if you can you should be able to see the old version).
 A: I set a coordinate system on the cone apex, with $+\hat{y}$ pointing up and $+\hat{x}$ pointing radially out at an azimouth (location) angle $\theta=0$. The position of the ball is defined by the distance from the apex up the slope to the ball $r$ and the longitude around the cone $\theta$
$$\vec{p} = {\rm Rot}(\hat{y}, \theta) \begin{pmatrix} r \sin \psi \\ 0 \\ r \cos \psi \end{pmatrix} = \begin{pmatrix} r \sin \psi \cos \theta \\ r \cos \psi \\ -r \sin \psi \sin \theta \end{pmatrix} $$
In the above the angle $\psi$ is fixed and $r$, $\theta$ are variable. Hence
$$ \vec{v} = \dot{\vec{p}} = \frac{\partial \vec{p}}{\partial r} \dot{r} + \frac{\partial \vec{p}}{\partial \theta} \dot{\theta} \\
\vec{v} ={\rm Rot}(\hat{y}, \theta) \begin{pmatrix} \dot{r} \sin \psi \\ \dot{r}\cos\psi \\ -r \dot{\theta} \sin \psi \end{pmatrix}  = \begin{pmatrix} 
\sin\psi (\dot{r} \cos\theta - r \dot{\theta} \sin \theta)
\\ \dot{r} \cos \psi
\\ -\sin\psi (\dot{r} \sin \theta + r \dot{\theta} \cos\theta )  \end{pmatrix}$$
Similarly
$$ \vec{a} = {\rm Rot}(\hat{y}, \theta) \begin{pmatrix} 
  \sin\psi (\ddot{r}-r \dot{\theta}^2) \\
  \ddot{r} \cos\psi  \\
  -\sin\psi(r \ddot{\theta}+2 \dot{r} \dot{\theta})  \end{pmatrix} $$
If you ignore the rotational inertia and spin motion and focus on the path of the ball then $\sum \vec{F} = m \vec{a}$ with the net forces acting from gravity $\vec{W} = \begin{pmatrix} 0 \\ -m g \\ 0 \end{pmatrix}$ and contact normal force $\vec{N} = {\rm Rot}(\hat{y},\theta) \begin{pmatrix} -N \cos\psi \\ N \sin\psi \\ 0 \end{pmatrix} =\begin{pmatrix} -N \cos\psi \cos\theta \\ N \sin\psi \\ N \cos\psi \sin\theta \end{pmatrix} $
$$ \vec{W} + \vec{N} = m \vec{a} $$
$$ \boxed{ \begin{aligned} 
 N & = m g \sin\psi + m r \dot{\theta}^2 \sin\psi\cos\psi \\
 \ddot{r} & = r \dot{\theta}^2 \sin^2\psi - g \cos\psi \\
 \ddot{\theta} & = - \frac{2 \dot{r} \dot{\theta}}{r} 
\end{aligned} } $$
You initial conditions are at $t=0$, $\theta=0$, $r = \frac{h}{\cos\psi}$, $\dot{\theta} = \frac{u}{h \tan \psi}$ and $\dot{r}=0$. I do not know of an analytical solution, so I would use a numerical simulation to get results.

Notes
The $\dot{\boxed{\,}}$ notation is the time derivative, and ${\rm Rot}(\hat{y},\theta)$ is the standard 3×3 rotation matrix about the y axis.
