# Centripetal force

by John78
Tags: centripetal, force
 P: 21 1. The problem statement, all variables and given/known data Find the centripetal force required to rotate 6 kg object in a circle at a radius of 6m at one revolution per second. 2. Relevant equations C.F. = mv2/r [b]3. The attempt at a solution[/b Circumference=2 pi r=2*3.14*6=37.68 C.F. = 6*(37.68)2/6=1419.78 I just want to check whether my working is correct or not.
 P: 614 that's about what I got too ^^
 P: 343 Looks like you wrote down CF = m v2/r but then you calculated the circumference, C = 2 π r and then you said, CF = r C2/6 = C2 Why would you expect the centripetal force to be equal to the square of the circumference? It really is not. Try it again, starting with CF = m r ω2 and see how that turns out for you. You failed to get the time information into the problem.
 P: 21 Centripetal force so CF = m r ω2 CF = 6*6*(37.68)2=51112???
P: 66
 Quote by OldEngr63 Looks like you wrote down CF = m v2/r but then you calculated the circumference, C = 2 π r and then you said, CF = r C2/6 = C2 Why would you expect the centripetal force to be equal to the square of the circumference? It really is not. Try it again, starting with CF = m r ω2 and see how that turns out for you. You failed to get the time information into the problem.
That's not really what he is doing, I think... He is doing CF = m v^2 / r, it just so happens that v is one circumference per second and that m and r have the same numerical value...
 P: 21 now i am confused :C
P: 66
 Quote by John78 now i am confused :C
You shouldn't be, you got it right the first time around.
However, to get it right with the second formula, you have to use the proper value for ω, which in this case is 2π/s.

It would probably clear things up, if you would carry the units throughout your calculation and not just the numbers, that way you can always check if your result makes sense.
 P: 12 I think your working is correct

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