Lagrangean and non inertial frame

  • #1
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...
 

Answers and Replies

  • #2
BvU
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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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  • #3
LCSphysicist
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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.
 
  • #4
Orodruin
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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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