Lagrange Equations of Motion for a particle in a vessel

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The discussion focuses on deriving the Lagrange equations of motion for a particle in a vessel using polar coordinates. The Lagrangian is formulated as (1/2)m(ṙ² + r²Θ̇² + b²r²ṙ²) - (1/2)mgbr², incorporating the time derivatives of the variables. A discrepancy arises when calculating the Lagrange equation of motion for 'r', leading to an incorrect result that lacks a negative term of b²ṙ²r. Participants are encouraged to clarify the Euler-Lagrange equation used in the derivation. The conversation highlights the complexities of applying Lagrangian mechanics in this context.
Wombat11
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I start out by substituting rcos(Θ) and rsin(Θ) for x and y respectively. This gives me z=(b/2)r^2. The Lagrangian of this system is (1/2)m(rdot^2+r^2⋅Θdot^2+zdot^2)-mgz. (rdot and such is the time derivative of said variable). I then find the time derivative of z, giving me zdot=br⋅rdot and plug it into the Lagrangian giving me (1/2)m( rdot^2 + r^2⋅Θdot^2 + b^2r^2⋅rdot^2) - (1/2)mgbr^2. (plugged in regular z too). Finding the Lagrange equation of motion for 'r' gives me (ddot will be the second time derivative of said variable) 0= rddot(1+b^2r^2)-r⋅Θdot^2 + b^2⋅rdot^2⋅r + gbr . This is not correct though, the right answer has a negative term of b^2⋅rdot^2⋅r. Any help would be greatly appreciated.
Sorry about the notation, I have no idea how to put the equations into the computer.
 

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Wombat11 said:
Lagrange equation of motion
The ##\mathcal L## you work out (*)
$$ (1/2)m( \dot r^2 + r^2\dot \theta^2 + b^2r^2\dot r^2) - (1/2)mgbr^2 $$
looks good to me. What is the Euler-Lagrange equation you then use ?(*) I 'typed':
$$ (1/2)m( \dot r^2 + r^2\dot \theta^2 + b^2r^2\dot r^2) - (1/2)mgbr^2 $$
 
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