# 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?

## 3 Answers

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.

• Well actually now when I think, its actually the case, problem is I am not confusing formulas I am confusing the linear velocity and angular velocity and thinking for wrong equation for the situation. – E.Berk Feb 19 at 7:39

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.
• Thats a nice example I did some math with formulas and this time considered the relationship between angular and linear velocity now I see the connection. I was thinking if I had to keep one constant I have to keep other one constant too but I forgot there is a v=ωR equation and because of this equation when I change the distance of the center I have to also change angular or linear velocity depending on which is constant so now I see the connection between two centripetal force formulas. – E.Berk Feb 19 at 8:06

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

• Unless I'm misunderstanding you, this isn't true. – Aaron Stevens Feb 19 at 1:29