# Finding Centripetal Force

*Most recent answer: 10/04/2012*

Q:

How do you derive the equation F=mv^2/r?

- Beth (age 18)

Norfolk, England

- Beth (age 18)

Norfolk, England

A:

Formulas express how particular things act. Usually, we can't read back from a formula to figure out what the situation was. Here, however, you're most likely referring to a mass

So here's an argument. Draw a picture of a circle of radius

Now use exactly the same picture but let the points on it be velocity vectors, so the radius is just |

Now you can just use algebra to get |

For those who like arguments simple, the point is that the relation between

Since

Mike W.

*m*moving at fixed speed*v*in a circle of radius*r*. In order to keep going in a circle, the object has to change the direction of its velocity. A changing velocity is the same thing as an acceleration.So here's an argument. Draw a picture of a circle of radius

*r*. Take an arrow from the circle to a point to represent the position at some time. You know that over one rotation that point just goes around the circle in time*T*. You can write a relation between the speed and*r*and*T*: |*v*|=2π*r*/*T*since the distance around the circle is 2π*r*.Now use exactly the same picture but let the points on it be velocity vectors, so the radius is just |

*v*|. You get by exactly the same reasoning that the magnitude of the acceleration, |*a*|, is 2π|*v*|/*T*.Now you can just use algebra to get |

*a*|=*v*^{2}/*r*.For those who like arguments simple, the point is that the relation between

*a*and*v*is the same as the relation between*v*and*r: a*/*v=v*/*r*, so*a*=*v*^{2}/r.Since

*F*=*ma*, you get*F*=*mv*/^{2}*r*.Mike W.

*(published on 10/04/2012)*