For circular motion in a vertical plane, why does Net Force = Centripetal Force? I'm struggling with some of the concepts pertaining to the forces and acceleration associated with circular motion in a vertical plane (only concerned with what happens at the 'top' and 'bottom' of the loop for now though).
Let's say that we have a roller-coaster going around a circular (vertical) track. I understand that at the bottom of the loop, the $F_{Net}=F_{Normal}-mg$, and this makes sense. But what is confusing me is why $F_{Net}=ma={mv^2\over r}$. I realise that ${mv^2\over r}$ is the centripetal force acting inwards (towards the centre of the loop), and $a$ is the centripetal acceleration, but why does this equal the net force?! Shouldn't we take into account the acceleration due to gravity?
eg. If we are trying to find the Net Force, why don't we find the Net acceleration, which would be something like "Centripetal acceleration - acceleration due to gravity" (at the bottom of the loop)?
So if I was to find the Net Force on 60kg person at the bottom of a roller coaster of radius 9m, travelling at 18.8 $m/s^2$, I could just use $F_{Net}={mv^2\over r}$? This seems to me like we just ignored gravity :/
Any help much appreciated, I feel as though I know enough to harm myself but not enough to properly understand it
Smeato
 A: The question you bring up is a very common one, and the source of some difficulty to novices.  I know it's been addressed here, but I can't find a good presentation.  What follows is not a good one, but it might be enough to answer the question.
Centripetal force describes a force (often the net of several forces). It does not name a force.  Gravity is a force.   Friction is a force.  Centripetal is not a force.  "Centripetal" describes a net force that points toward the center of a circle.  In the event that the speed of the object is not changing, or if observing only over a very short period of time, then kinematics demands that $a_c=v^2/r$ and Newton's second law says that there must be a net force that produces that acceleration.   We know that there is a centripetal force by observing the motion of the object, not by studying the interactions between objects as we would for gravity or friction.  In a sense, "centripetal force" is more a statement of kinematics than dynamics. 
The motion tells us that there must be a centripetal force.  What the nature of that force is presents a different question, one that is answered by analyzing the actual real forces, those due to interactions between objects. 
Kinematics tells us that the following must be true: $$F_\mathrm{net} = \frac{mv^2}{r}$$
Analysis of forces at the bottom of the loop tells us that $$F_\mathrm{net}=N-mg$$
A: The thing to understand about circular motion is that the centripetal acceleration is a purely kinematic fact. 


*

*Something moving along a curved path is accelerating because it's velocity vector is not constant. This is just kinematics

*If that curved motion has a known, constant speed $v$ and known radius of curvature $r$ than the magnitude of the acceleration is $a_c = v^2/r$. This is still just kinematics
It's a lot like looking at something that is not moving or is moving at constant velocity and saying "Hey, that thing is in equilibrium!". When we look at an object following a curved path we know it is accelerating just as we know that uniform motion implies equilibrium.
And just like the case of equilibrium we then use Newton's third law to connect net force to that acceleration.
\begin{align*}
F_\text{net,equilibrium} 
&= m a_\text{equilibrium}\\ 
&= 0\\
\\
F_\text{net,circular motion} 
&= m a_c \\
&= m\frac{v^2}{r} \;.
\end{align*}
In other words, for objects in uniform circular motion it is always true that the net force acting on that is the "the centripetal force".

Something to watch out for: if the speed is not constant there is a component of acceleration tangential to the path as well and the net force is no longer the centripetal force.

I encourage students to think of "the centripetal acceleration" as a primary thing, and the forces that cause that acceleration as a kind of intermediate step in the work that we don't dignify with a title in order to reinforce the fact that finding which forces (or components thereof) that combine to cause the centripetal acceleration as a to-be-worked-out feature of the problem.
A: 
eg. If we are trying to find the Net Force, why don't we find the Net acceleration, which would be something like "Centripetal acceleration - acceleration due to gravity" (at the bottom of the loop)?

Forget about the idea of having contributions to acceleration. An object has only one acceleration, not a sum of several. 


*

*There can be many forces, and they all combine into a net force. The object is being pushed a bit from this side, a bit from that side, a bit from behind, a bit from above and all in all, all these are summed up to a net force $F_{net}$.

*But you are not accelerating a bit sideways and a bit to the other side and a bit backwards, which should then combine into one acceleration. Acceleration is not made of contributions. There can be several forces, but they together result in one acceleration - they don't cause an acceleration each which is then summed up to a "net" acceleration.


Newton's 2nd law does indicate this: The formula is not $\sum F=m\sum a$ but only $\sum F=ma$; the $F$ is a sum of many $F$'s, but the $a$ is not a sum of many $a$'s.
Therefore there is no such thing as "net" acceleration - it is just acceleration. And in the case of uniform circular motion, where this acceleration must be pointing inwards, this acceleration has been named: centripetal acceleration. It is not another "type" of acceleration - just a name we call it, when it causes a circular motion.
And for such inwards-pointing acceleration to be present, all forces acting on the object combined must cause it. At the bottom of the vertical loop, there is weight pulling down and normal force upwards, and those must together cause upwards acceleration. (You don't subtract the "acceleration contributions" from each other, you just combine the forces and see what acceleration that sum causes.)
If it wasn't upwards, it wouldn't be a uniform circular motion.


why $F_{Net}=ma=\frac{mv^2}r$

People have fond out that if a motion is uniformly circular, the acceleration is always $a=\frac{v^2}{r}$ and pointing towards the centre. This can be proven seperately and has gotten nothing to do with forces.
Newton's law always holds, $F_{net}=ma$, and so in the case of uniform circular motion, the $a$ in this law can be replaced with $a=\frac{v^2}{r}$. 


So if I was to find the Net Force on 60kg person at the bottom of a roller coaster of radius 9m, travelling at 18.8 $m/s^2$, I could just use $F_{Net}=\frac{mv^2}r$? This seems to me like we just ignored gravity :/

This isn't ignoring gravity - you just aren't done. Gravity will surely enter as a part of the net force $F_{Net}$. As you even correctly wrote yourself, the net force does consist of gravity downwards and normal force upwards, $F_{Net}=F_n-mg$. So plug this into the expression and the gravity influence (the weight $mg$) is indeed included:
$$F_{Net}=\frac{mv^2}r\quad\Leftrightarrow\quad F_n-mg=\frac{mv^2}r$$
