Solving for Angular momentum without mass

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SUMMARY

The discussion centers on solving a Lagrangian problem involving a bead rolling on a vertical cone, specifically addressing how to compute angular momentum without mass. The key equations derived include d^2r/dt^2 = (L^2*sin^2α)/(m^2*r^3) - gsinα*cosα and d/dt(mr^2*dθ/dt) = 0, indicating that angular momentum is conserved. The participant questions the validity of substituting L = mr^2*dθ/dt into the differential equation for r'', expressing concern about the implications of mass on the equations of motion.

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thestrangequark
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Homework Statement


The problem is a Lagrangian problem that solves for a differential equation. I need to write a program to solve the Lagrangian numerically. My professor said you do not need mass for the program, but I'm not sure how. The problem is a vertical cone with a bead rolling around the cone. I drew it:
http://puu.sh/d4sN4/37c910ca4f.png
In the diagram, θ is the angle displacement of the bead. Here is a vertical view to show θ.
http://puu.sh/d4t13/e7de35cb2e.png
the variables are r and θ. x_0, dx/dt, dy/dt are all given. α=45 degrees. Assume θ_0 =0, implying y_0=0

Homework Equations


Solving for the Lagrangian gives:
d^2r/dt^2 =(L^2*sin^2α)/(m^2*r^3)-gsinα*cosα
d/dt(mr^2*dθ/dt)=0 showing angular momentum is conserved.

The Attempt at a Solution


Is it as simple as plugging L=mr^2*dθ/dt into the differential equation for r''? This would cancel out the masses in that equation, but I feel like that is too simple and I am missing something.
 
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It looks like you're using L for angular momentum, which is a bad idea when you also have a Lagrangian L.

L = T - U, right? Is T proportional to mass? Is U? If the answer to both is yes, what effect will there be on the equations of motion?
 

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