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Centrifugal force as dot products

  1. Apr 29, 2013 #1
    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.
     
  2. jcsd
  3. Apr 29, 2013 #2

    SammyS

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    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 .
     
  4. Apr 30, 2013 #3

    D H

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    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.
     
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