Question: I was trying to solve this problem, I got the right answer with the wrong interpretation:

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> At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers will not fall out? Assume a radius of curvature of 7.4 m.
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So I drew up my free body diagram and I assumed that the forces acting on this scenario were:

$netforce = weight$

So I assumed the only force acting on the passengers was weight because they did not mention any friction. However, when my teacher explained this, he drew up his diagram including the normal force as well. I understood, of course there must be a normal force, due to the contact of the passenger with the seat, but I did not understand when he said that normal force was going the in direction of weight. How could normal force go in the direction of weight when the passenger is touching the seat, allowing them to avoid falling down, and pushing them upwards. So he said that the summation of forces was:

$netforce = weight + F\_{normal}$

However, now there are two unknowns, how would you solved for velocity if we have the unknown of normal as well?

Answer: 
The question in the problem is about the **minimum** speed at which the passengers will not fall out. Therefore, you may assume that in the upper point the normal force is zero (otherwise you can decrease the speed without the passengers falling out). Or, more precisely, it is possible to show that if the passengers do not fall in the upper point of the circle, they do not fall in other points of the circle, so, to find the minimum speed you can require that the normal force is zero at the upper point of the circle.

As for the normal force going in the direction of the weight... I cannot confidently say that this statement is wrong, because there are actually two very different definitions of weight (<https://en.wikipedia.org/wiki/Weight>): 1) "the weight of an object is usually taken to be the force on the object due to gravity" 2) "There is also a rival tradition within Newtonian physics and engineering which sees weight as that which is measured when one uses scales. There the weight is a measure of the magnitude of the reaction force exerted on a body. Typically, in measuring an object's weight, the object is placed on scales at rest with respect to the earth, but the definition can be extended to other states of motion. Thus, in a state of free fall, the weight would be zero. In this second sense of weight, terrestrial objects can be weightless. Ignoring air resistance, the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, is weightless."

Under the second definition, the normal force is indeed going in the direction of the weight, but the formula in your question for the summation of forces is correct under the first definition:-) So you may wish to find out (or tell us:-) ) which definition your teacher uses.
For any object to move in a circle, the forces toward the center minus the forces away from the center--the $net force$--must at every instant be equal to $m\frac{v^2}{r}$.   This is true in all situations, including a race car going around a curve, a ball being swung in a circle at the end of a string, an airplane doing a vertical loop, or a roller coaster doing a vertical loop.   $m\frac{v^2}{r}$ is called the centripetal force.

For the roller coaster passenger, his weight, $mg,$ and the normal force both point toward the center of the circle only when the passenger is at the very top of the circle.   Apparently the instructor didn't make that clear to you.   And yes, at that instant $net force= weight+F\_{normal}$

We can see that $m\frac{v^2}{r}$ gets smaller as the speed $v$ gets smaller, but the weight is always the same, so it must be $F\_{normal}$ that gets smaller if circular motion is maintained as $v$ gets smaller.   So the minimum speed where the passenger can make it through the top of the loop without falling out is when $F\_{normal}$ becomes zero.   At any speed lower than that, $mg$ will be bigger than $m\frac{v^2}{r}$, and circular motion can't be maintained.

But at the speed where $F\_{normal}$ becomes zero, $$mg=m\frac{v^2}{r}$$
which we can easily solve for $v$ if we know $r$