Lagrangean and non inertial frame

AI Thread Summary
The discussion revolves around the application of the Lagrangian approach in non-inertial frames, specifically when using polar coordinates. The original poster questions how their calculations, which yielded correct results in the context of a rotating frame, imply an assumption of being in a non-inertial frame. They express confusion over the absence of potential energy in their inertial frame analysis, despite arriving at a non-inertial frame solution. The response clarifies that while the Lagrangian can be valid in any coordinate system, the presence of tangential velocity components in polar coordinates indicates the influence of non-inertial effects. Ultimately, the conversation highlights the complexities of transitioning between inertial and non-inertial frames in Lagrangian mechanics.
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Homework Statement
Smooth rod OA of length l rotates around a point O in a horizontal plane with a
constant angular velocity $\delta$. The bead is fixed on the rod at a distance a from
the point O. The bead is released, and after a while, it is slipping off the rod. Find its
velocity at the moment when the bead is slipping?
Relevant Equations
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I have tried to solve this problem using the lagrangean approach: $$L = T - V = m((\dot r)^2 + (r \dot \theta)^2)/2 - 0 = m((\dot r)^2 + (r \delta)^2)/2 - 0 $$

The problem is that the answer i got is the right answer at the smooth rod referencial, that is, at the non inertial frame.

Now we can easily translate my answer of v to the inertial frame, but that is not the point of my question. The point is, when did i have assumed i was at the rod frame? That is, i have just written the kinect energy in polar coordinates, so instead of $$\sum m x_i x^{i} / 2$$ i have written it in polar language. All of sudden, am i now in a non inertial frame?

Do polar coordinates are by definition non inertial?

Should have a potential here? I can't see where does it come from. In fact, another point i can't understand is, in a inertial frame there is not a potential energy, and in fact i have used (V=0). But yet, my answer is given in a non inertial frame...
 
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Herculi said:
i have written it in polar language
I don't speak Inuit, but fortunately a Lagrangian is valid in any language. And in any coordinate system: inertial, non-inertial, constrained, whatever. As long as you use 'something' correctly to express kinetic energy and potential energy, Euler-Lagrange equations hold.

##\ ##
 
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BvU said:
I don't speak Inuit, but fortunately a Lagrangian is valid in any language. And in any coordinate system: inertial, non-inertial, constrained, whatever. As long as you use 'something' correctly to express kinetic energy and potential energy, Euler-Lagrange equations hold.

##\ ##
Oh yes, i know it works in any reference frame. In fact, as i said in my question and probably was not clear, i got the right answer using my approach (but my answer was right with respect to the rod frame, that is, the rotating frame). My question is: When did i have assumed that i was at the rod frame?. In another words: I have just written the lagrangian i thought was right, got the answer, but the answer i got indicates i have used the legrangian in a non inertial frame. Even so i don't know where did i have assumed that i was at the non inertial frame itself, since i have just write the lagrangian in polar coordinates.
 
There is no non-inertial frame at work here. What you have done is to compute ##\dot r##, the radial velocity of the bead, but the bead also has a component in the tangential direction that you cannot ignore.
 
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