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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?

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  • $\begingroup$ It depends if the speed is constant or not. Decompose the forces as tangent and normal to the path to see which goes towards changing the speed and which towards following the path. $\endgroup$ – ja72 Apr 9 '15 at 18:43
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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.

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  • $\begingroup$ Thank you, sir. You have cleared a mistake in my reasoning. So, I will rephrase my question: how comes that the body still stays on the track and doesn't fall? Is it because the normal force is always greater than zero? So, the fact that the resultant points to the center isn't a requierement for us to have a nice motion in a vertical loop..so this is the only condition? N>0? $\endgroup$ – Bardo Apr 9 '15 at 16:41
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    $\begingroup$ "However, an object moving freely in a vertical loop will slow down on the way up and speed up on the way down." And of course individual points on a balanced system can be in uniform vertical circular motion without even needing power. All that said this is the answer I was coming to write. $\endgroup$ – dmckee Apr 9 '15 at 16:59
  • $\begingroup$ Thank you for your answer, but, it doesn't make it any clearer for me.. $\endgroup$ – Bardo Apr 9 '15 at 17:03
  • $\begingroup$ $N>0$ is a requirement if the body is constrained only by forces "underneath" it. Some objects, like some roller-coaster cars, are constrained to stay on the circular path by devices that provide forces in both directions: toward the center and away. (The car is "clamped" against motion perpendicular to the track either "up" or "down"; the car can move only "forward" and "backwards" along the track. So $N>0$ is not a necessary condition in all cases. $\endgroup$ – garyp Apr 9 '15 at 17:17
  • $\begingroup$ But if the only forces that acts on the body are the gravity and the normal force, the requierment of N>0 is the only one, right? I'm a little big dumb, but I think it goes like this: no normal force, means no contact, means rollercoaster "falls". There must be a normal force, no matter how small, but there must be, such that that our simple roller coaster stays on tracks, correct? $\endgroup$ – Bardo Apr 9 '15 at 17:30
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For any circular motion , whether uniform or non uniform, centripetal force is a must. Now the normal force is sum of centripetal force (resulting from the change in direction of velocity of the particle) and tangential force (resulting from the component of weight force and frictional force). The normal force points towards the centre of the orbit in uniform circular motion as tangential force is zero. In non uniform circular motion, the tangential force is non zero (except at the topmost point of the orbit in case of vertical circular motion). So, Normal force doesnot points towards the centre. But, its component centripetal force maintains the circular orbit.

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    $\begingroup$ I appreciate your answer. But, I find it a little bit odd. When you say that "the normal force is sum of centripetal force" it isn't quite right. The centripetal force is not some sort of "special force". It's just the resultant of all the other forces that act on the body, in the radial direction. $\endgroup$ – Bardo Apr 9 '15 at 18:19
  • $\begingroup$ If the path is circular, a normal force would by definition point radially (either towards the center of the circle, or away from it, although the latter is generally not physical). $\endgroup$ – Sean Apr 9 '15 at 18:31
  • $\begingroup$ for the particular problem, Normal forc is indeed the sum of centripetal force (resulting from the change in direction of velocity of the particle) and tangential force (resulting from the component of weight force and frictional force). $\endgroup$ – Tejas Rathod Apr 12 '15 at 12:43
  • $\begingroup$ Its just the matter how you see the problem. You are right, the centripetal force is not some sort of "special force". It's just the resultant of all the other forces that act on the body, in the radial direction. But it also means that centripetal force has a component of some other force and vice versa. for the present case you can see it to be normal force. $\endgroup$ – Tejas Rathod Apr 12 '15 at 12:50

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