How is the Centrifugal Force Derived from Scalar Potentials?

[Dustinsfl]

The centrifugal force is
$$
\Omega\times r\times \Omega
$$

I paper I am reading then writes it as $\frac{1}{2}(r\Omega)^2 - \frac{1}{2}\Omega^2r^2$

How was this obtained?

Using the fact that $a\times b\times c = (ac)b - (ab)c$, I don't get what they are getting so there is something else that I am missing.

 

[SammyS]

The centrifugal force is
$$
\Omega\times r\times \Omega
$$

I paper I am reading then writes it as $\frac{1}{2}(r\Omega)^2 - \frac{1}{2}\Omega^2r^2$

How was this obtained?

Using the fact that $a\times b\times c = (ac)b - (ab)c$, I don't get what they are getting so there is something else that I am missing.

Is there something assumed about a relationship between Ω and r ?

Is Ω precessing about r, or vice-versa ?

$\Omega\times r\times \Omega\$ is a vector.

$\frac{1}{2}(r\Omega)^2 - \frac{1}{2}\Omega^2r^2$ is a scalar .

 

[D H]

The centrifugal force is
$$
\Omega\times r\times \Omega
$$

I paper I am reading then writes it as $\frac{1}{2}(r\Omega)^2 - \frac{1}{2}\Omega^2r^2$

How was this obtained?

Using the fact that $a\times b\times c = (ac)b - (ab)c$, I don't get what they are getting so there is something else that I am missing.

As written, that's a scalar, it has the wrong units for force, and it's zero.

However, $\frac 1 2 (\vec r\cdot\vec{\Omega})^2 - \frac 1 2 \Omega^2 r^2$ is something very different. It's still a scalar, it still has the wrong units for a force, but it is a rather special scalar. It's the centrifugal potential. Take the gradient with respect to $\vec r$ and negate and you'll get the centrifugal force.