Circular Motion Explanation Of Normal Force 
How can we prove that a particle (ball) moving along a circular path with the only forces acting being its weight and the normal force (forming the centripetal force) will result in uniform (constant speed) circular motion. Furthermore why is the normal force present in point D and why is it bigger in magnitude in point B compared to the y-axis component of the weight?
 A: You can't prove uniform circular motion happens for those two forces. Uniform circular motion is usually a constraint which the author of a problem imposes on a system. In fact, for a vertical circle with only weight and normal force (from some track or rail), uniform circular motion isn't possible. 
The normal force acts perpendicular to the circular motion of the particle, so it does zero mechanical work on the ball. Only the weight does work, and we can account for that by considering a constant mechanical energy problem including the gravitational potential energy of the ball/Earth system:
$$ E_{\mathrm{top}}=E_{\mathrm{bottom}}$$
$$ \frac{1}{2}mv_{\mathrm{top}}^2+mg(2R)=\frac{1}{2}v_{\mathrm{bottom}}^2$$
where $R$ is the radius of the circle, $m$ is the ball mass, $g$ is the gravitational field strength, and the reference point for the gravitational potential energy is chosen to be a the bottom of the circle.
From this it's clear that the speed at the top is different from the speed at the bottom. We could generalize this to angular positions (you can do the maths if you're curious), but we clearly see this is not, nor can it be, uniform circular motion.
A: Consider circle of radius $R$ with the origin at its center, where $\vec{r} = R(cos\theta \hat{i} + sin\theta \hat{j})$ is the radius vector of a particle moving on this circle.
If you differentiate this vector twice wrt time, you will find that the centripetal acceleration has to, at every instant be equal to $v^2/R$
Now if you can show that the net force on a particle is always directed towards the center and has a magnitude of $v^2/r$(which won't change because the net force is perpendicular to the disp) , it means that the particle will travel in a circle of radius $R$ with constant velocity $v$
