Centripetal Force Formula Confusion After I finished studying and trying to test my knowledge to see what I have learned, I realized I am confused about the centripetal force formula:
$$F_c= \frac{mv²}{R}$$
which I know is also equal to 
$$F_c= m\omega^2R$$
because $v= \omega R$, so
$$F(c)= \frac{m\omega^2R^2}{R} = m\omega^2R$$
My problem is that if I had to guess how the length $R$ of, say a rope, effects the centripetal force, I'm confused if it decreases or increases as length increases because in one formula the distance$R$ is at a dividing position while in other it is multiplying. What am I missing here?
 A: As mentioned in the comments, it all depends on what is constant in your system. For example, if the linear velocity $v$ is constant as you move outward from the center of rotation, then $\omega=v/R$ must be decreasing. So, if you wanted to see how the centripetal force changed in this case you would want to look at $F_c=mv^2/R$, since the only varying value is $R$. A similar thought process can be done for constant $\omega$.
Therefore, you need to determine what is actually constant in the system you are considering. I will leave this for you to think through.
A: An example might be helpful: consider a car going in circles. 


*

*Assume the car has constant speed $v$. In sharp curves (small $R$), the centripetal force that is required to keep the car on the road is large and for large $R$, the force is small. That's the $1/R$ equation. 

*Now consider the car has to keep the "lap time" for each full circle constant. This is a movement with constant angular velocity $\omega$. Now the car must speed up for larger radii, and because the force depends on $v^2$, the radius hops up in the numerator. 

A: It's because angular velocity (w) = v / r ... so if you substitute v/r for w in your second equation, you get your first equation again. It's a 1/r relationship either way
