Modeling a Small Ball Rolling Off a Bigger ball A friend gave a challenging math problem to solve for fun, but since I'm a high school calc student, it's too hard for me to figure out. 

If a small ball is at the top of a larger
  stationary sphere (radius = $1$) and it starts to roll down
  the side, at what point will the smaller ball lose contact with the
  larger sphere?

I got off to a good start, but then got stuck. Here's my work. If you don't care to read it all, you can skip to the last paragraph:
I first aim to find the velocity. I start with the acceleration of the ball: $a = g\dot{}\sin(\theta)$ where theta is the angle of inclination the ball is rolling at any given instant. The derivative of the circle equation gives the slope of this incline. The circle is modeled by $f(x) = \sqrt{1-x^2}$ and $f'(x)= -x\dot{}(1-x^2)^{-1/2} \ $Therefore: $\ \theta = \tan^{-1}(f'(x))$. 
Substituting this back into the original equation, this is where it gets ugly: 
$$a = g\dot{}\sin\left(\tan^{-1}(f'(x))\right)$$ 
I figured I needed to put all of this is terms of time, so I change $x$, horizontal displacement of the ball, to $s(t)$, and $a$ to $a(t)$. Using my current knowledge of kinematics, I can relate displacement to acceleration such that
$s(t) = \int_{}v(t)~dt \ $ and $\ v(t) = \int_{}a(t)~dt$
so if I'm not mistaken
$$s(t) = \int_{}\int_{}a(t)\ dt\ dt$$
So, substituting this all back into my original equation, I get this:
$$a(t) = g\dot{}\sin\left(\tan^{-1}\left(f'(\int_{}\int_{}a(t)\ dt\ dt)\right)\right)$$ 
Not sure if this is even correct syntax at this point. Anyways, I got everything in terms of $a$! But, I've never formally learned to solve DE's, so I'm not sure what to do next or if this DE is solvable, and if it is, it must be so complex that it's not a practical solution. There must be a simpler way that I'm missing, what do I do? After solving for the velocity, how would I use it to find when the ball loses contact with the larger sphere?
 A: If we ignore the rolling motion of the small ball with mass $m$ (it slides over the spherical surface), we can write these equations:
\begin{cases}mg\cos\theta-R=m{v^2\over r}\hspace{3cm}\text{Newton's law along the radius}\\mgr+0=mgr\cos\theta+{1\over2}mv^2\hspace{1,3cm}\text{Conservation of energy}\end{cases}

The small mass loses contact with the sphere when the reaction $R$ is zero, so from the first equation:
$$mg\cos\theta=m{v^2\over r}$$
$$mv^2=mgr\cos\theta$$
where $\theta$ is the angle between the radius $r$ of the sphere and the vertical line passing through the center.
Thus, replacing into the second equation: $$mgr=mgr\cos\theta+{1\over2}mgr\cos\theta\longrightarrow 1=\cos\theta+{1\over2}\cos\theta$$
$$\cos\theta={2\over3}$$
$$\theta=\arccos\left({2\over3}\right)$$
Rolling motion
If we take into account the rotation motion of the small ball (we'll consider simple rolling motion, i.e without slip along the surface), we can write equations similar to the previous ones:
\begin{cases}mg\cos\theta-R=m{v_{cm}^2\over r+r'}\hspace{7cm}\text{Newton's law along the radius}\\mg(r+r')+0=mg(r+r')\cos\theta+{1\over2}mv_{cm}^2+{1\over2}I\omega^2\hspace{1,3cm}\text{Conservation of energy}\end{cases}
where $v_{cm}=\omega r'$ is the speed of the center of mass, $r'$ is the radius of the small ball, $\omega$ is the angular velocity of the small ball and $I={2\over5}mr'^2$ is the moment of inertia of the small ball respect to its center.

Solving the system we get:
$$\cos\theta={2m\over 3m+{I\over r'^2}}$$
If $I=0$, as in the first case, we obtain the previous solution.
