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Lagrangian in rotating space without potential

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    I want to derive the centrifugal and Coriolis forces with the Lagrangian for rotating space. The speed of an object for an outside observer is dr/dt + w x r, where r are the moving coordinates. So m/2(dr/dt + w x r)^2 is the Lagrangian.


    3. The attempt at a solution

    Everything above makes intuitive sense to me. The problem is that I don't know how to use the chain rule on the w x r.

    When I derive it by dr/dt and then by d/dt I get m(d^2r/dt^2 + w x dr/dt), which is what I'm supposed to get.

    But when I do it by d/dr I just cannot get what I'm supposed to. I looked all over the internet and I can't find the chain rule for cross products and I've played with it for a long time and it just doesn't work.

    Please help.
     
  2. jcsd
  3. May 10, 2012 #2
    [tex]

    \frac{\partial}{\partial t} (A \times B) = \frac{\partial A}{\partial t} \times B + A \times\frac{\partial B}{\partial t}

    [/tex]

    Then just apply the chain rule as you normally would for each derivative term, i.e.
    [tex]
    \frac{\partial A}{\partial t} = \frac{\partial A}{\partial r}\frac{\partial r}{\partial t}
    [/tex]

    edit - I realize this might not answer your question, but I'm having a hard time following what you're asking. What do you mean by 'derive it by dr/dt'
     
  4. May 10, 2012 #3
    I mean derive it by r dot, the time derivative of of r.

    When I derive it by r I get m(dr/dt + w x r)(d/dr(w x r)) and I don't know what to do with that. It seems reasonable to think that it is equal to m(dr/dt + w x r)(w x 1), but what do I do now? When you use the chain rule like this is it multiplied by a dot product? A cross product?
     
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