I'll post the derivation using Lagrangian mechanics, then point out how one can interpret the results in terms of the coriolis and centrifugal pseudo-forces.
I'll use x' and y' as the non-rotating coordinates, and
x and y as the rotating ones.
In non-rotating coordinates, we have
L = T - V = .5*m*((dx'/dt)^2 + (dy'/dt)^2) - V(x',y')
here V(x',y') is the gravitational potential, which is equal to
-G m M1 /sqrt((x'-x1)^2 + (y'-y1)^2) - G m M2 / sqrt((x'-x2)^2+(y'-y2)^2)
x1 is the X coordiante of M1, y1 is the y coordiante of M1
x2 is the x coordiante of M2, y2 is the y coordinate of M2
To switch to rotating coordinates, we need to make the substitutiion
x(t) = x'(t)*cos(wt)-y'(t)*sin(wt)
y(t) = y'(t)*cos(wt)+x'(t)*sin(wt)
w being the angular frequency of rotation of the system.
The original expression for T was
T = .5*m*((dx'/dt)^2 + (dy'/dt)^2)
After much application of the chain rule, one gets the value of T in the new coordiante system
eq #1
T = .5*m*((dx/dt)^2+(dy/dt)^2 + 2*x*w*(dy/dt) - 2*y*w*(dx/dt)+w^2(x^2+y^2)
A change in notation becomes convenient at this point, we write
<br />
L = .5 m (\dot{x}^2 + \dot{y}^2 + 2 x \omega \dot{y} - 2 y \omega \dot{x} + w^2(x^2+y^2)) - V(x,y)<br />
V(x,y) remains essentially unchanged from its expression in the non-rotating system V(x',y'), except that x1,x2,y1,and y2 are now constants as desired, rather than varying with time as they did before we switched to the rotating coordinates.
Note that V(x,y) is a function of x and y only, it does not depend on dx/dt or dy/dt.
Now we apply Lagrange's equation to get the equations of motion
http://scienceworld.wolfram.com/physics/LagrangesEquations.html
<br />
\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}) = \frac{\partial L}{\partial x}<br />
<br />
\frac{d}{dt} (\frac{\partial L}{\partial \dot{y}}) = \frac{\partial L}{\partial y}<br />
This gives us the resulting two equations of motions
<br />
\ddot{x} - \omega \dot{y} = \omega^2 x + \omega \dot{y} - \frac{1}{m} \frac{\partial V}{\partial x}<br />
<br />
\ddot{y} + \omega \dot{x} = \omega^2 y - \omega \dot{x} - \frac{1}{m}\frac{\partial V}{\partial y}<br />
Let's look at the first equation. Moving the duplicate term so that it appears only on the right hand side of the equation, we can say that
m d^2x/dt^2 aka m \ddot{x} as the mass-acceleration term (in the rotating coordinate system x,y).
2*m*w*dy/dt as the coriolis force
m*w^2 * x as the centrifugal force
-dV/dx as the gravitational force
So we say that the acceleration is equal to the sum of the centrifugal force, the coriolis force, and the gravitational force in the rotating coordinate system.
Similarly, the second equation says
m*d^2 y/ dt^2 aka m \ddot{y} is the mass-acceleration term
-2*m*w*dx/dt is the coriolis force
m*w^2*y is the centrifugal force
-dV/dy as the gravitational force
and again, acceleration is the sum of tthe centrifugal force, the coriolis force, and the gravitational force.
The dervitaion on the webpage at
http://scienceworld.wolfram.com/physics/LagrangePoints.html
may or may not help you follow the one I did - it should at least serve as a double check for any typos which I may have missed in transcribe from my old archived worksheets on the problem to this post.
Without Lagrangian mechanics, it's probably just as easy to look at the final result for the equations of motion and not bother with the formalism of the derivation. Where Lagrangian mechanics shines is in its ability to keep the signs straight, something that's very easy to mess up when one does the problem without it, adding the separate terms together "by hand".
To further simplify the problem, when solving for the Lagrangian points, the coriolis terms drop out, because one assumes that x and y are constant, so dx/dt and dy/dt should be zero. The Coriolis force terms become important when one attempts to analyze the stability of the Lagrange points, something that's also done on the webpage I mentioned.
