# Lagrangian in rotating space without potential

1. May 10, 2012

### ShamelessGit

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.

2. May 10, 2012

### Clever-Name

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

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

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'

3. May 10, 2012

### ShamelessGit

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?