Aitor,It doesn’t matter. What you see now is centrifugal force, a real force.
Your Mr. Knowledge needs to study this site What is a Pseudo force? and understand what he is trying to get across in his introduction.
We are dealing with acceleration terms not force terms in the rotating field. The pendulum does not swing to the right because there is some mysterious "real force" acting on it. In the same context a stone does not fall when it is dropped because there is a real force pushing it down. That is why he says:Today we will ask a few questions such as
What is a pseudo force?
What is a fictitious force?
Are pseudo forces real?
Is gravity a pseudo force?
Alas, we will not answer these questions!
As far as I can tell, nobody knows, and nobody cares. Or at least nobody should care. If these questions have answers at all, each answer is a matter of opinion, and opinion is divided. There is no consensus. When I asked several hundred physics teachers via the phys-l mailing list, nobody seemed to have any firm opinions. I don’t have a firm opinion, either.
There is no experiment you can do that will tell you the answers to these questions.
For example, as far as I can tell, gravity can be classified as a “pseudo force” – or not. It’s six of one and half-a-dozen of the other. You can choose to do it either way, and the choice has no observable consequences.
The fact that there is no easy answer suggests that we are asking the wrong questions, and that we should ask more physically-meaningful questions instead.
How do we compute what experience has told us is the force we can expect for gravitational acceleration on earth? We use Newton's second law and come up with F=mg where g is the gravitational acceleration you get on the surface of the earth.For example, as far as I can tell, gravity can be classified as a “pseudo force” – or not. It’s six of one and half-a-dozen of the other.
How do we compute the equivalent centrifugal force (call it real, pseudo, fictitious or what ever you want. It is 6 of one a half dozen of the other just like gravity) we use the same second law and say F=m*(centrifugal acceleration) where the centrifugal acceleration term is equal to omega.^2 (the square of the angular velocity in radians/sec) times the distance (lets call it r) you are away from the axis of rotation of the rotating field. Thus f_centrifugal=m*omega.^2*r.
Work out the numbers and you will find that f_centrifugal is the same magnitude as f_centripetal which is m*v.^2/r when you are going around a circular path with a tangential velocity (v) having a radius (r0. By convention they point in different directions; one out, one in.
Gordy