Here's what I know: If a body moves in a circular trajectory, then the resultant of all the forces must point to the center of the circle it describes in its movement. If a body moves in a vertical loop, there will be two forces acting on it: The normal forces exerted by the track and the force of gravity. I have solved the "loop the loop problem".
But, what guarantees that, at any point, the resultant of the normal force and the gravity force will always point to the center of the circle? It is clear that the normal force will be greater than zero, so the body doesn't loose contact with the tracks (if it loops the loop). But is this enough to keep it in a circular motion?
So, what assures us that if we compose the vectors graphically, the resultant will always point to the center?
Best Answer
They don't point toward the center.
You are thinking of the special case for circular motion with constant speed. In that case, the velocity and acceleration vectors are perpendicular, implying the acceleration points toward the center.
However, an object moving freely in a vertical loop will slow down on the way up and speed up on the way down. This is not constant speed, so one shouldn't expect the acceleration to be perpendicular to the velocity. In fact, it isn't.
You can always break up (decompose) the acceleration vector into a component parallel to the velocity ($a_\perp$, which changes speed) and perpendicular to the velocity ($a_{||}$, which changes direction). When you're slowing down, the component parallel to the velocity will be opposite to the velocity; when you're speeding up, the parallel component will be in the same direction.
This can be applied to the vertical hoop situation. Slowing on the way up, speeding on the way down. Only when the speed is not changing (e.g., at the very top of the loop, or at the very bottom) will there be no parallel component, and hence the acceleration vector at those special points would only have a perpendicular component.