Lagrangean and non inertial frame

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Homework Help Overview

The discussion revolves around the application of the Lagrangian approach in the context of a non-inertial frame, specifically when using polar coordinates to describe a system. The original poster is exploring the implications of their calculations and questioning the assumptions made regarding the reference frame used in their analysis.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how their use of polar coordinates in the Lagrangian formulation may imply a non-inertial frame. They question the assumptions made during their calculations and the role of potential energy in their results.

Discussion Status

Participants are engaging in a dialogue about the validity of the Lagrangian in different reference frames. Some suggest that the Lagrangian can be applied universally across various coordinate systems, while others are clarifying the specific conditions under which the calculations were made. There is an exploration of the implications of radial and tangential components of motion.

Contextual Notes

The original poster notes a potential confusion regarding the presence of potential energy in inertial versus non-inertial frames and the implications of using polar coordinates in their analysis.

LCSphysicist
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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
.
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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