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 *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.

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.

*(published on 10/04/2012)*