Why did they add a Centripetal Force in the Free-Body Diagram? problem
http://img196.imageshack.us/img196/6610/problmus.jpg!
solution
http://img838.imageshack.us/img838/391/solnd.jpg!
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?
 A: 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.
A: 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.
A: 
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.
