Why $g$-force increase with radius and centrifugal force decreases? Trying to understand the 2 formulas and the relations between radius. Common sense will tell me that centrifugal force increases while we are on a curve with a short angle , as newtons law formula suggests, but checking g-force formula it is directly proportional to radius.
A bit confuse on which one should be used on centrifugal fluids. What happens to rotating fluids with increasing radius and same rpm or angular velocity, and what happens with fluids with decreasing radius and same rpm or angulsr velicity.
 A: In fact the two equations given are essentially identical. The first:$$F=\frac{mv^2}{r}$$
gives the centripetal force $F$ needed to keep an object of mass $m$ moving at linear velocity $v$ in a circle of radius $r$.  For this equation to work as written, the units for these quantities must be related in a specific way. For example, in the SI system of units, F is in Newtons, $m$ is in kilograms, $v$ is in metres/sec, and $r$ is in metres.
It is also apparent that by deviding both sides of this equation by $m$, we can arrive at $$a=\frac{v^2}{r}$$where $a$ is the centripetal acceleration.  If we continue to use SI units, the acceleration will be in units of $\text{metres/second}^2$
However, it is possible to define another quantity, angular velocity, $\omega$.  This is essentially how fast something is spinning. The relationship between angular velocity $\omega$ and linear velocity $v$ is given by $$v=\omega r$$ 
The SI unit for angular velocity is radians/second.  The radian is a unit of angle;  there are $2\pi$ radians in a complete revolution.
If we substitute $\omega r$ for $v$ in the first equation for acceleration we obtain: $$a=\frac{v^2}{r}=\omega ^2 r$$which is the second of the two original equations.
As for the arbitrary(?) factor of $0.00001118$, this comes of using RPM (revolutions per minute) instead of radians/second, and from using $r$ in either metres or centimetres, and $g$ in metres/second$^2$ or centimetres/second$^2$
A: It is important to make explicit your implicit assumptions (espeiclly if you don't recognize that you've made them.


*

*In the comments to the question you quote the formula $$F = m v^2 / r\ \;. $$ 
If you are looking at that and saying saying "if I increase radius the required force goes down" you are implicitly assuming that $v$ remains constant.

*In the question itself you talk about "rotating fluids". That's a lot less clear than the comment, but if you are imagining a spinning container after in initial turbulence has dissipated then you are assuming that everything turns through a circle in the same amount of time. That is a constant period or constant angular velocity assumption. The formula that treats that most obviously in 
$$F = m \omega^2 r$$ 
where $\omega$ id the angular velocity. Now the force increases linearly with radius.

*While you don't mention it I'll give a third example that is applicable to an introductory class.  
The planets of the solar system don't obey either constant velocity or constant angular velocity. More distance question have both lower linear velocity and lower angular velocity (Kepler's 3rd law, right?).  
Working the relationship starting from Kepler's law would be difficult, but in the case of circular orbits we can use the universal gravitation $$F = m (G M_\text{sun} / r^2) \;.$$
(Notice that here the product $GM_\text{sun}$ is a constant, so this is not more complicated than the others.)
All of these situation express the same centripetal physics, but they result in different behaviors as a function of radius because the other variables in the problem are behaving differently.
