Normal Force in Circular Motion 
In the above diagram, sec 1 (on the left side), an object of mass $m$, after releasing from rest from a slant track, continues into a vertical circular track. At a random position on the circular track, I have shown the forces acting on the object. Likewise in the sec 2 (on the right side), an object of mass $m$ climbs the inclined track with a initial velocity $v$. At a random position on the inclined plane, I have shown the forces acting on the object.All of the surfaces have no friction.
Hoping the forces which have shown above are correct, I have below questions:


*

*In sec 1, $N1 - mg\cos(a)$ acts as centripetal force which is required for changing of direction of velocity along the circular track. But what does $mg\sin(a)$ do? Does it acts as the decelerating force which changes the tangential velocity and hence leading this object into a non-uniform circular motion? If so, Is vertical uniform circular motion even possible?

*In sec 2, $N2 = mg\cos(a)$ since there is no acceleration in that direction. Also the object decelerates along the inclined plane, with a magnitude of $g\sin(a)$ until $t = v/g\sin(a)$, after that it accelerates back down the inclined plane. Is this analysis correct?

*Why are the normal reaction forces are different in the two scenarios? On what does the normal reaction force depend?
 A: 
  
*
  
*Is vertical uniform circular motion even possible?
  

No, it isn't. Because magnitude of velocity isn't constant and we know that in a uniform circular motion the object moves with constant speed. $\large{\frac {\mathrm d}{\mathrm dt}}v=g\sin\alpha\neq 0$ ($v$ is the speed (magnitude of the velocity vector $\vec v$) of the object)


  
*Is this analysis correct?
  

Yes, it is.


  
*Why are the normal reaction forces are different in the two scenarios?
  

Because the object experiences different motions in the two scenarios. Equation of motion for a particle with constant mass is $\Sigma\vec F=m\vec a$. If the right side of motion's equation is different for two scenarios; then, the left side of that will certainly be different. So, in the instant that angle $\alpha$ is same for two scenarios, the normal reaction forces will be different. Because in the first case, we have $N=mg\cos\alpha+m\large{\frac{v^2}R}$ and in the second case we have $N=mg\cos\alpha$


  
*On what does the normal reaction force depend?
  

Normal reaction force depends on the pressure that two surfaces exert on each other and area of contact surface $\mathrm dN=P\mathrm dA$
A: 1.mgsin(a) stands for the projection of the gravity on the x axis, if you take the axis like, y-axis is N1 and mgcos(a) and x-axis  magsin(a). It is the force that stops the mass, so it can´t be uniform because if it has no acceleration and it is being dragged down it eventually stops.
2.I would write using Newton 
$$m \cdot a = g \cdot sin (a)$$
then integrate on both sides using that a is the derivative of v
$$v-v_0=g \cdot sin (a) \cdot t \rightarrow t=\frac{v-v_0}{sin(a)}$$.
If it onlye has the downward force acting on it, it goes up with the starting velocity and then down.
3.So the projections $mgsin(a)$ and $mgcos(a)$ depend on a parameter called $a$ that is the angle between the vector $mg$ and $mgcos(a)$. in sec 1 that angle is changing along the curve. If your see the picture if the mass goes a but up that angle is inferior than before because the normal force is always perpendicular tu the surface. On the other hand in sec 2 that angle is always the same so $mgcos(a)$ is a constant.
