Do centripetal and centrifugal forces cancel in a frame moving with a body undergoing uniform circular motion? There is something I don't quite undestand about the role the centrifugal term plays when describing motion in a non-inertial reference frame. In most Classical Mechanics books, you can find similar discussions on non-inertial reference frames. The "effective acceleration" felt by a body in one of those reference frames would be:
$$\vec{a}_{ef} = \vec{a}_{real} - \ddot{\vec{R}} - \dot{\vec\omega}\times\vec{a}-\vec\omega \times (\vec\omega \times \vec r) - 2\vec\omega\times v_{r}$$
Where $\vec{a}_{real}$ are the real forces acting upon the body, like grativational forces, and the rest of the terms correspond to fictitious forces. My doubt is the following:
For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we have to account for. It is a real force, after all; making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this?
I think I am confused about something and would be thankful if someone could clarify this for me.
 A: 
For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we (I suppose) have to write. It is a real force, after all, making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this?

If you are in a frame rotating with the child, then yes, you would want the centripetal force to cancel out the centrifugal force. In your rotating frame you see a child at rest, and this child has a net force of $0$ acting on them. Therefore, you have $F=ma\to0=0$, and everything is consistent.
A: The confusion is not realizing that real forces are real, and fictitious forces are not real. Therefore, they can not "cancel each other out" in real life. 
Using a rotating frame of reference and introducing fictitious forces is just adding and equal quantities (a force, and a mass $\times$ acceleration) to both sides of the math equations. It has nothing to do with the real forces acting on the system. It is no different from solving $x + 3 = 7$ by saying $x + 3 - 3 = 7 - 3$, and therefore $x + 0 = 7 - 3$ and $x = 4$.
The physics of the situation (i.e. the interaction between the various parts of the system which we call "force", "stress", etc) does not depend on what coordinate system you choose to model it. You get the same values of the physical quantities in any coordinate system. 
The only reason for choosing one coordinate system or a different one is because the math is easier in one than in the other, not because it changes the physics.
A: You put considerable effort on merry go rounds (Even when seated) in order to subconsciously show there is no such thing/s. Or you are just wired up to face the rotating system you are in and your brain is taking charge. and $a=0$. You can't fool the Coriolis part is a lot more sneaky. 
Just look at that term with the angular velcoity vector crossed with linear velocity. Move your hand back and forth, facing the center, perpendicular to the axis of rotation. Then we'll see some $ma$. :)
When you are moving in a planar circle then by construction there is an acceleration vector perpendicular to the linear velocity at each instant, and it keeps rotating the velocity vector while keeping it fixed in size. Otherwise, you would not be moving in a circle.
If you are part of the rotating system then mathematically: 


*

*You modify with an acceleration vector in the other direction. That's what is happening.


Now if you take your make believe coordinates and formally differentiate twice, nothing happens. Clearly not what you should expect.
On paper it is often much cleaner and easier to restrict yourself to one particular non-inertial system and get a correct answer. In reality you would trip and fall most probably if you walk around as if it is not a restricted non-inertial system you are in.
But, if you recall Newton's postulates are meaningful for your prior coordinates, then you do the correct thing and substitute back to inertial coordinates into the expression, and only then do you satisfy Newtonian postulates. What you have written is a nicely packed vector expression of correct Newtonian physics in an inertial frame. The accelerations are all real. You apply force in order to nullify them.
Again try to toy around really with the Coriolis expression. It's real. Like bold letters! And like proper books on Newtonian Mechanics.
