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Why do they have the centripetal force in there? I understand that the normal force is the centripetal force here, but why would they say "or $\frac{mv^2}{r}$"? I thought it was wrong to include this.

Also, on centripetal force, how come there is a net force inwards to the center, but there is no REAL force counterbalancing? Why do we have a fictitious force?

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You have a fictitious force because you're working in a non-inertial frame. You're doing this because it makes the calculations much easier. Doing things the hard way should give exactly the same answer. I'm sure somebody has told you that you shouldn't use centripetal force because it doesn't really exist, but I think that as long as you understand that, there isn't any reason not to use it to do your calculations. –  Peter Shor Dec 23 '11 at 3:10
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@Peter: While you are absolutely correct philosophically speaking, they didn't do this in the problem, they just set the normal force to mv^2/r because it's true. –  Ron Maimon Dec 23 '11 at 4:45
    
@Ron: you're absolutely right; I answered too quickly. –  Peter Shor Dec 23 '11 at 14:47
    
Are you familiar with D'Alembert's theorem? en.wikipedia.org/wiki/D'Alembert's_principle. It states how to add acceleration forces into a free body diagram. –  ja72 Apr 25 '12 at 14:20

3 Answers 3

up vote 4 down vote accepted

The vector you see is not the centrifugal force, which would appear in a rotating frame. The book is working consistently in an inertial frame. This is just the normal force, and they are just noting that it is equal to $mv^2/r$.

What the answer did not do is include a left-pointing centrifugal force in the diagram to cancel out the right-pointing normal. Including a centrifugal force in a free-body diagram is considered a mistake in elementary mechanics, although as Peter Shor points out, it is not a mistake at all if you are working in a non-inertial frame (which is consistent, just considered slightly more advanced).

The reason that the normal is not cancelled by anything is that you need an acceleration to keep something moving in a circle. This is counterintuitive, because we tend to shift reference frame to move along with the object, so that if you have a force pointing to the left, intuition suggests that it must be balanced by a force to the right. This is true in the rotating frame, but not in the inertial frame description.

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It's not a mistake. When working in an inertial frame of reference there shouldn't be any left-pointing force. The net force must point to the right if the body is rotating. –  Arnoques Dec 23 '11 at 4:54
    
@Arnoques: I didn't say it was a mistake. I said exactly what you said. –  Ron Maimon Dec 23 '11 at 4:58
    
@Arnoques: I see where my wording was confusing. Thanks for pointing it out. I'll fix it. –  Ron Maimon Dec 23 '11 at 5:24
    
So what they really don't tell you in intro physics class is that all the free-body diagrams aer done in an inertia frame of reference? –  sidht Dec 23 '11 at 5:41
    
@Jak: I think they do tell you this explicitly--- Newton's laws are for inertial frames. They just don't tell you how to work in noninertial frames, which is also possible. –  Ron Maimon Dec 23 '11 at 6:28

Why do they have the centripetal force in there? I understand that the normal force is the centripetal force here, but why would they say "or $\frac{mv^2}{r}$"? I thought it was wrong to include this.

Well, I guess it's not technically wrong, since the normal force will have a value equal to $\frac{mv^2}{r}$ in this particular situation. But it is misleading. Students who are just learning about circular motion have a tendency to think that $\frac{mv^2}{r}$ is a force in its own right, separate from any other forces that may exist in the problem. If you let them write $\frac{mv^2}{r}$ on a free body diagram, it just reinforces that erroneous thought.

In reality, of course, the fact that any force (or sum of forces) equals $\frac{mv^2}{r}$ can only be concluded after applying Newton's second law. In particular, the quantity $\frac{mv^2}{r}$ comes from the $ma$ side of the equation. It's not supposed to appear in the sum of forces. And since, for convenience, you usually want to be able to copy the forces directly from a free-body diagram into that sum, it helps not to put $\frac{mv^2}{r}$ on the diagram.

Also, on centripetal force, how come there is a net force inwards to the center, but there is no REAL force counterbalancing? Why do we have a fictitious force?

There is no fictitious force in this diagram. The normal force is very real. But the reason there does not need to be any force counterbalancing it is that the object is accelerating.

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-1: If you don't "let" students right every correct statement on their diagram, you are not engaged in teaching, but in stupid power games. The type of "education" drives students nuts, because, in addition to learning the physics, they have to learn your psychology. How are they supposed to know that you consider trivial step #3 less trivial than trivial step #5? And if you don't think the students find it equally trivial as you, you don't know students. About 60% of the difficulties they have is wading through the morass of psychology and stupid conventions, not the physics. –  Ron Maimon Dec 23 '11 at 18:37
    
@Ron: so what is your basis for downvoting this? As a matter of fact, what in your statement has any bearing on my answer? –  David Z Dec 24 '11 at 4:11
    
You @David make an aught statement--- you say one aught not let students label normal forces with mv^2/r, even if the normal force is equal to mv^2/r. This means you would have downmarked me on all your exams, because I always did this, so I decided to grading you as arbitrarily as you grade your students. –  Ron Maimon Dec 24 '11 at 6:50
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I said no such thing. This post doesn't contain any statements about what one should or shouldn't do when grading exams, and it certainly does not contain anything that would allow you to judge how I grade students' papers. –  David Z Dec 24 '11 at 7:02
    
Here is the offending quote: "If you let them write mv^2/r on a free body diagram, it just reinforces that erroneous thought." This, and the surrounding context, makes it clear that you consider it bad practice to write the magnitude of the force on a diagram. This is not reasonable. –  Ron Maimon Dec 24 '11 at 8:25

The normal force is the force that the wall of the cylinder does on the person. It's there because it's a real force. If you are in the ride you can certainly feel the wall pressing on your back.

Also, since you know the person is performing a uniform circular motion, the laws of motion tell you that the net force on the person must point to the center of the circle, and be equal to $mv^2/r$. So the diagram should read "$N=mv^2/r$".

Notice that the centri_f_ugal force is the fictitious one (or the one that appears on a rotating frame of reference). The centri_p_etal force is simply a real foce (or sum of forces) that happens to point to the center of a circle.

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