If I understand the problem correctly there is a bowl (like a large salad bowl) and a ball rolling inside of it under the influence of gravity. Here is how to derive the equations of motion. I parametrize the position of the ball center using the quasi-coordinates $q_1$ and $q_2$ as:
$$ \vec{r}(q_1,q_2) = [0,0,R]-{\rm Rot}_{X}(q_{1})\cdot{\rm Rot}_{Y}(q_{2})\cdot[0,0,R]$$
where $R$ is the radius of the bowl minus the radius of the ball, equals the pitch radius of motion. My coordinate system has the bowl laying on the $x$-$y$ plane and up being the +$z$ direction. ${\rm Rot}_X()$ and ${\rm Rot}_Y()$ are the $3\times3$ rotation matrices respectively.
Differentiating I get the velocity vector of the center of gravity
$$ \vec{v}=\begin{bmatrix}0 & \cos q_{2}\\
\cos q_{2}\cos q_{1} & -\sin q_{2}\sin q_{1}\\
\cos q_{2}\sin q_{1} & \sin q_{2}\cos q_{1}\end{bmatrix}\begin{bmatrix}R\,\dot{q}_{1}\\
R\,\dot{q}_{2}\end{bmatrix} $$
and differentiating again, I get the acceleration of the c.o.g.
$$ \vec{a}=\begin{bmatrix}0 & \cos q_{2}\\
\cos q_{2}\cos q_{1} & -\sin q_{2}\sin q_{1}\\
\cos q_{2}\sin q_{1} & \sin q_{2}\cos q_{1}\end{bmatrix}\begin{bmatrix}R\,\ddot{q}_{1}\\
R\,\ddot{q}_{2}\end{bmatrix}+\begin{bmatrix}R\sin q_{2}\,\dot{q}_{2}^{2}\\
\mbox{-}R(\dot{q}_{1}^{2}+\dot{q}_{2}^{2})\cos q_{2}\sin q_{1}-2R\dot{q}_{1}\dot{q}_{2}\sin q_{2}\cos q_{1}\\
R(\dot{q}_{1}^{2}+\dot{q}_{2}^{2})\cos q_{2}\cos q_{1}-2R\dot{q}_{1}\dot{q}_{2}\sin q_{2}\sin q_{1}\end{bmatrix} $$
Now for the fun part. The forces acting on the ball are gravity $\vec{W}=[0,0,-m\,g]$ and the contact force
$$ \vec{N}=F\,\begin{bmatrix}\sin q_{2}\\
\cos q_{2}\sin q_{1}\\
\cos q_{2}\cos q_{1}\end{bmatrix} $$ where $F$ is the magnitude of the force (unknown), $m$ is the mass of the ball and $g$ is acceleration of gravity.
To verify that $\vec{N}$ is a reaction force check with $\vec{N}\cdot\vec{v}=0$, thus providing zero power to the system.
The equations of motion (ignoring the rotational components and friction) are
$$ \vec{W}+\vec{N} = m\;\vec{a} $$ The solution of which yields $F$, $\ddot{q}_2$ and $\ddot{q}_2$ as
$$\ddot{q}_{1}=2\dot{q}_{1}\dot{q}_{2}\tan q_{2}-\frac{g}{R}\,\frac{\sin q_{1}}{\cos q_{2}}$$
$$\ddot{q}_{2}=\mbox{-}\dot{q}_{1}^{2}\sin q_{2}\cos q_{1}-\frac{g}{R}\,\sin q_{2}\cos q_{1}$$
$$F=m\, g\,\cos q_{2}\cos q_{1}+R\, m\,\left(\dot{q}_{1}^{2}\cos^{2}q_{2}+\dot{q}_{2}^{2}\right)$$
Whats left to be done, is put it through a numerical integrator scheme like Runge-Kutta
[1] [2] and watch the variables $q_1$ and $q_2$ evolve over time.
To add friction, add a component of force in the direction opposite of $\vec{v}$ with magnitude $\mu\,F$ where $\mu$ is the coefficient of friction (like 0.02
-0.10
for rolling).