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Imagine a point of mass `m` that moves in a circular trajectory. Since every mass has its inertia, (namely `m*a` ) it will tend to move in a straight line and keep its moving state as long as it is unperturbed by an external force. Therefore the act of curving the trajectory needs to correspond to a force applied to the mass, namely the centripetal force. The higher the curvature (smaller the radius of the curve), the higher the centripetal force needs to be.
`F_(cp) = A/R` , where A is a constant. (1)
Consider the point has a certain speed. To modify the direction of its constant speed (to make it move in a circle), one will need to create an acceleration, hence to apply a force (namely the centripetal force). Higher the speed, higher the acceleration that need to be created, thus higher the centripetal force. In practice
`F_(cp) = B*v^2` where B is a constant. (2)
Combining (1) and (2) one gets
`F_(cp) =C*v^2/R` , where C is a constant (3)
Now, to find the dependence on the frequency. Higher the speed, higher the number of complete circles he makes in the same time, thus a higher frequency. In practice
`v = 2*pi*F*R`
which combined with (3) gives
`F_(cp) =C*4*pi^2*(F^2*R^2)/R = C*4*pi^2*F^2*R `
`F_(cp) =D*F^2*R` , where D is a constant
Finally since frequency is just the inverse of period `F=1/T` one has
`F_(cp) =D*R/T^2` , where D is a constant
Answer: the relationship between the centripetal force, frequency, period and radius is `F_(cp) = D*F^2*R =D*R/T^2` where D is a constant.
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