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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?

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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.

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    $\begingroup$ 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. $\endgroup$ – E.Berk Feb 19 at 7:39
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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.
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  • $\begingroup$ 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. $\endgroup$ – E.Berk Feb 19 at 8:06
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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

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

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