What is the Lagrangian for a Bead Sliding on a Cycloid-Shaped Circle?

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SUMMARY

The discussion focuses on deriving the Lagrangian for a bead sliding on a cycloid-shaped circle defined by the parametric equations x=a(B-sinB) and y=a(1+cosB) for 0<=B<=2pi. Participants suggest using polar coordinates to simplify calculations, emphasizing that the Lagrangian L can be expressed as L=T-V, where T is the kinetic energy (T=(1/2)mv^2) and V is the potential energy. The consensus is that while Cartesian coordinates can be used, polar coordinates may provide a more straightforward approach to describe the bead's motion effectively.

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Homework Statement


A bead slides without friction on a circle in the shape of cycloid with equations x=a(B-sinB) and y=a(1+cosB) where 0<=B<=2pi.
Find the lagrangian.

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The Attempt at a Solution


I've tried to solve that problem by converting the positions x and y from cartesian coordinate to a new system of coordiantes defined by those equations. The problem is that am having many long calcucaltions. My new idea is to work it out in the polar coordinates. Am i going in the roght way?
 
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I'm a bit confused by your wording of the problem ("circle in the shape of a cycloid"), but I'm not sure you need to change coordinate systems; the lagrangian L=T-V should be straightforward to find in cartesian coordinates. T=(1/2)mv^2, etc...
 
Yes, I would use polar coordinates...the idea of the lagrangian is to compute an equation which can describe the motion of the particle (bead) anywhere that it could possibly be..since in this case it will always be somewhere on the outside of the circle the easiest way to describe its position is with r and theta.
 

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