Q: Terminal velocity of ball rolling down stair? I am having a problem with a particular concept. 
  Here is where I have gotten, since the ball never loses contact with the stair, it will rotate around through the edges, the edges being the pivot, (used parallel axis theorem and all that), and since all collisions are inelastic,i figured the ball will lose the normal component of its velocity.But I still couldn't figure out how to find the terminal velocity(average). How and why?
I have posted an image for clarity.


A: Wow, I was able to find a solution, but it is super complicated. 
I can offer though that the change in velocity when it hits the next step is
$$ \Delta v = \frac{m r^2 (\cos\varphi-1)}{I+m r^2} v_1$$
where $v_1$ is the impact (drop final) speed, $I$ is the mass moment of inertia, $r$ is the sphere radius and $\varphi$ is the sweep angle going from step to step. The sweep angle if $\varphi = 2 \sin^{-1} \left( \frac{h}{\sqrt{2} r} \right)$.
This comes from the contact condition at point B on the next step.
Those act to change the linear and angular velocity of the center of mass C in such a way as the linear velocities at point B is zero. Two equations and two unknowns.
$$(\vec{v}_C + \Delta \vec{v}) + (\vec{\omega} + \Delta \vec{\omega} ) \times ( \vec{r}_B-\vec{r}_C) = (0,0,0)$$
$$(\vec{v}_C + \frac{\vec{J}_A}{m} ) + (\vec{\omega} + \frac{(\vec{r}_A-\vec{r}_C) \times \vec{J}_A}{I} ) \times ( \vec{r}_B-\vec{r}_C) = (0,0,0)$$
solving for the impulse $\vec{J}_A =(A_x,A_y,0)$ and used $\Delta v = | \vec{v}_C + \frac{\vec{J}_A}{m} | - |\vec{v}_C|$
The kinematics are driven by the locations
$$
\vec{r}_A = (0,0,0) \\
\vec{r}_C = (r \sin\left(\frac{\pi}{4}+\theta\right), r \cos \left(\frac{\pi}{4}+\theta\right),0) \\
\vec{r}_B = (r \sin \left(\frac{\pi}{4}+\theta\right) - r \cos \left(\frac{\pi}{4}+\varphi-\theta\right), r \cos \left(\frac{\pi}{4}+\theta\right)-r \sin \left(\frac{\pi}{4}+\varphi-\theta\right),0)$$
where the motion is described by the orientation angle $\theta = -\frac{\varphi}{2} \ldots \frac{\varphi}{2}$ which is the deviation from 45° of the line connecting the contact point A_ and the center C. Gravity acts to make $\theta$ positive. The angular velocity of the ball is $\vec{\omega} = (0,0,-\dot\theta)$.
If the initial speed is $v_0$ at orientation $\theta = - \frac{\varphi}{2}$ and final speed is $v_1$ at $\theta = \frac{\varphi}{2}$ then the fall is described by
$$ \frac{1}{2} \left( v_1^2 - v_0^2 \right) = \sqrt{2} r g \sin \left(\frac{\varphi}{2}\right) $$
when $v_0 = v_1 + \Delta v$ we have reached the terminal speed. To find the average, you need the time $t$ it takes to sweep the arc and the distance the center travels $s = r \varphi$ for an average speed of $v_{ave} = \frac{r \varphi}{t}$.
Good luck.
