Lagrangean and non inertial frame

Click For Summary
SUMMARY

The discussion centers on the application of the Lagrangian approach to a problem involving polar coordinates in a non-inertial frame. The user successfully derived the correct answer using the Lagrangian, $$L = T - V = m((\dot r)^2 + (r \dot \theta)^2)/2 - 0$$, but questions the assumptions made regarding the reference frame. It is established that while the Lagrangian can be applied in any coordinate system, the interpretation of the results must consider the nature of the reference frame, particularly when dealing with non-inertial frames and the implications of potential energy.

PREREQUISITES
  • Understanding of Lagrangian mechanics and the Euler-Lagrange equations
  • Familiarity with polar coordinates and their application in physics
  • Knowledge of kinetic and potential energy concepts
  • Basic principles of inertial and non-inertial reference frames
NEXT STEPS
  • Study the derivation and application of the Euler-Lagrange equations in various coordinate systems
  • Explore the implications of using polar coordinates in non-inertial frames
  • Learn about the role of potential energy in Lagrangian mechanics
  • Investigate examples of problems involving rotating reference frames and their solutions
USEFUL FOR

Physics students, researchers in classical mechanics, and anyone interested in the application of Lagrangian mechanics in different reference frames.

LCSphysicist
Messages
644
Reaction score
163
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...
 
Physics news on Phys.org
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.

##\ ##
 
  • Haha
Likes   Reactions: LCSphysicist
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.
 
  • Like
Likes   Reactions: LCSphysicist

Similar threads

Replies
24
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
19
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
13
Views
3K