Lagrangian for rolling disk inside a ring

Posted 6 months ago
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 In the discussion for equation for motion for "rolling disk inside a ring" in webpage http://demonstrations.wolfram.com/DiskRollingInsideARotatingRing/ by Mr. Erik Mahieu Lagrangian for the system is not getting displayed. I tried getting it. First, thing I felt that the first equation of motion should r^-2 instead of r^-1. By reverse engineering and my own knowledge I could get the Lagrangian as (K - V) where K = (1/2 I_1 φ'^2) +(1/2 I_2 ψ'^2) + (1/2 m_2 (R-r)^2 θ'^2 ) + (m_2 * (R-r) )(Cos θ) θ' (R φ' ) + (1/2 (m_1 + m_2 ) R^2 φ'^2) V = m_1 g R + m_2 g (R-r)(1-Cos θ) But the problem is first equation of motion should have the term (m_2 (R-r) )(Sin θ) θ' (R φ' ) I am not able to get the clear picture and complete physical interpretation of the last two terms of K. I will be grateful if I can get the Lagrange of the system be displayed with a bit of explanation. Regards, Ravi Shankar Gautam Attachments:
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Posted 6 months ago
 There is Lagrangian. To see this you need to download the demo and open the file using Mathematica. In section DETAILS there is an expression \[ScriptCapitalL]=1/2(2 g Subscript[m, 2] (-r+R) cos \[Theta](t)+r^-2((Subscript[I, 2]+Subscript[m, 2] r^2) (r-R)^2 \[Theta]^\[Prime](t)^2+2 (r-R) R (Subscript[I, 2]+Subscript[m, 2] r^2 cos \[Theta](t)) \[Theta]^\[Prime](t) \[Phi]^\[Prime](t)+(Subscript[I, 1] r^2+(Subscript[I, 1]+(Subscript[m, 1]+Subscript[m, 2]) r^2) R^2) \[Phi]^\[Prime](t)^2)) 
Posted 6 months ago
 Thank you very much Alexander Trounev for your reply. It is of great help for me. I could not arrange for Mathematica. I don't know if it is version specific. I will try to get access from my friend.In the mean while I could deduce that kinetic energy part seems to match except that in the last bracket there seems to be a typo. The second term should be I2 instead of I1.If the above expression posted by you is the Lagrangian then in that case my previous observation that one term in the first equation of motion is missing seems to hold true. This expression arises due to partial derivative of lagrangian with respect to theta.Please let me know if I am missing out something.Regards,Ravi Shankar Gautam
 You possibly did not realize that there is no slipping between the disks, therefore $r\psi =R \phi -(R-r) \theta$. Then $(\psi ')^2=(R \phi '-(R-r) \theta ')^2/r^2$.