# The force arrow for centripetal force points in the opposite direction of my calculation?

I need someone to clarify a conceptual problem I can't seem to surpass. Image there is a rollercoaster loop and a rollerbcoaster car enters the loop at high speed.

Once the car completes the full loop and is then at the last point at the bottom of the loop, I am trying to understanding the forces at play.

I can see that there is gravitational force: $F_g$ = $mg$ and centripetal force: $F_c = \frac{mv^2}{r}$, and then there is the normal force $F_N$.

Let's say $F_g = 10$ ,and $F_c = 25$, wouldn't $F_N = 35$?

And if so, I am very confused about the force diagram. We have $F_N$ pointing up and $F_g$ pointing down, but what about $F_c$? My calculation has it pointing down, but in reality I thought centripetal force always points to the center of a circle?

Thanks for the help!

• $F_c$ always points to the center of the loop
– Jim
Commented Jun 13, 2013 at 17:53
• in this experiment, $F_N$ is the centripetal force
– Jim
Commented Jun 13, 2013 at 17:56
• wouldn't $F_N$ equal $F_c$ + $mg$? Commented Jun 13, 2013 at 18:06

The centripetal force isn't a separate force. It is the vector sum of the forces that are present. In the case of a roller coaster, the normal force and the force of gravity will add vectorically to produce a net force, and that net force must apply the centripetal force to keep you moving in a circle. Because the force of gravity is constant, the normal force will change (both in magnitude and direction) to provide the remainder of the centripetal force.

When you are at the top of the loop, both the normal force and gravity point in the same direction. When you are at the bottom of the loop, the normal force points up, the opposite direction as gravity (down). So the normal force will have to be larger in magnitude when you are at the bottom of the loop than it is when you are at the top of the loop.

The word "centripetal" comes from the latin, and means "seeking the center". To seek or desire is one of the meanings of the latin verb "petere". The word "petition" has the same origin. Hence that force is always towards the center, upward in your case.

The Normal force is the (total) reaction of the rails on the car (in the absence of friction or lateral forces).

The simpler way to analyze it is to first ignore gravity. Your rollercoaster is lost in space. It goes on its rails in a straight line, at constant speed, just by inertia (no engine, no friction). At some point, there is this loop on the rails. You pass the loop on your inertia and reach the end of it. Analyse what happens. What are the forces (no gravity).

This analysis is simpler because you have uniform motion speed (not uniform velocity), no gravity and no falling cars.

Then you bring it back to earth. Horizontal linear motion works the same (forget the air and other friction). Almost the same analysis, except that you have all the time an extra force produced by gravity, which must be balanced too. That is all.

I am saying almost the same because your motion speed through the loop is no longer uniform. But you can forget it, because the speed that you loose going up, you get back going down, so that there is no change in the case you analyze. And you are not asked about that anyway. More importantly for you, you analyze at a point where the angular acceleration caused by gravity is null. Forget what I just wrote, if you do not get it. The effect of gravity is null because all it can do at the bottom is make the car go through the rails, which is necessarily compensated by a normal force (unless the rails are allowed to break).

( Fun question : assuming the loop is in the middle of a straight horizontal stretch, where are the rails most likely to break ? i.e., where do they get the greater strain ? )

The conceptual difficulty may be that the centripetal force here is the reaction of the rails balancing the inertial force of the car in circular motion (it may be only a part of that reaction of the rail, if other forces are to be balanced). That is the part which I do not know how it is taught. As long as the motion stays circular, this inertial force looks like an outward force (centrifugal) that balances the centripetal force. But go to your books for that part.

This inertial force results from the circular motion. If the car stops (or slows too much) at the top of the loop, then the inertial force disappear, and so does the centripetal force. The car falls because of gravity, not because of the lost centripetal force, now (close to) null. It is the inertial force that keep the car on the rails. And that is why I say that the centripetal force is a reaction in this case. (in the case of the moon it is earth gravity that causes it)

Now, in this problem, the normal force is the force that keeps the car from going through the rails, i.e. the global reaction of the rails which may balance several other forces.

Concepts and Misconceptions :

The difficulty is that forces can be combined in different way and take different names according to the role you are considering in the system. But they must balance.

A centripetal force is not any force going through the center. It is defined with respect to some motions around a center. Uniform circular motion is one of them (actually the only one if you apply strictly what I say at the end of this paragraph). A body may well be submitted to other forces going through the center (rocket push for example), but that does not necessarily make them centripetal forces. Differentiation allows to consider it for other non linear motions. It is exactly balanced by the part of the inertial force orthogonal to the motion.

The Normal force is a reaction from a surface preventing a body from entering that surface, and is orthogonal to that surface ("orthogonal"= "normal"="perpendicular" in this context). It balances the other forces that would tend to make the body enter the surface. Since There may be several such forces, the normal force can be decomposed into a sum of colinear (i.e., parallel) forces corresponding to the reaction to each of these other forces.

The reaction from a surface is not always orthogonal to that surface. It can then be decomposed into a force orthogonal to the surface (the normal force) and a force tangential to the surface ( the friction force in general ... but it is actually a bit different when you talk of rails).

A centripetal force is any force (possibly resulting from a sum of forces) that maintain the motion around the center as said above. It may be gravity, the pull of a string, the push of a rocket engine (not obvious), the reaction from an outer surface (normal force, or friction, or both), or even combinations of these. It may be only a part of the Normal force (or of the combination of other forces) as explained above.

I hope I am not too much at odds with the way these things are taught.

The centripetal force is always going to be supplied by some other force in the problem. You shouldn’t have a force labeled as the centripetal force but one of the forces in a problem will be serving as the centripetal force. What you should do is draw all the forces that are acting on the cart and determine which force is pointing towards the center of the circular path that the cart is taking.

In this case at the bottom of the loop you only have the gravitational force and the normal force acting on the car. The normal force is the centripetal force because it is pointing towards the center of the circle. For example at the top of the loop the gravitational force would serve as the centripetal force and there would be no normal force. Anyway at the bottom of the loop since the normal force is serving as your centripetal force you should have:

$F_N = m\frac{V^2}{r}$

Hopefully that helps to clear up your calculations.

• This is wrong in several respects. A force pointing to the center need not be the centripetal force. Vectorially, the centripetal force is the sum of normal force and radial component of the gravity force. At loop top, the centripetal force is the sum of the normal force and the gravity force since they have the same orientation. At the bottom it is their difference since both the normal force and the centripetal force are reversed. Hence the answer is not $F_N = m\frac{V^2}{r}$ - (after 39 views, one upvote and some comments). Commented Jun 14, 2013 at 13:16
• I understand the error in my original answer. However why would the centripetal force be only the sum of the normal force and the radial component of gravity. Since the motion woulden't be uniform (the cart would accelerate on the way down and decelerate on the way up) the centripetal force (which I think is the sum of $F_N$ and $F_g$ would not be pointing directly towards the center of the circular path of motion since this isn't uniform circular motion. However if you only considered the radial component of gravity then it would always point towards the center. Is this correct? Commented Jun 14, 2013 at 14:27
• I must repeat my answer disclaimer that I do not know how this is currently taught in the USA. Presentations may differ, if only for pedagogical reasons. As I understand it, the concept of centripetal force has meaning only with respect to uniform circular motion (though one could want to consider extension to elliptical motion). It is the motion orthogonal part of the total force on the concerned body, the part that preserve the circular motion. The very name means it has to point to the center. The rest modifies the speed on the trajectory. I did not mean to pick on you, but on the system. Commented Jun 14, 2013 at 15:55