Finding equation of motion from lagrangian

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SUMMARY

The discussion focuses on deriving the equation of motion from the Lagrangian for a spring-mass system. The Lagrangian is defined as L = T - U, where T is the kinetic energy and U is the potential energy. The Euler-Lagrange equation is applied, leading to the equation of motion: (\omega^2 - \frac{k}{m})r = \frac{d^2r}{dt^2}. This confirms that \omega^2 equals the natural frequency of the spring system, \omega^2_{spring}.

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Hi, I am trying to solve this problem here:

http://img201.imageshack.us/img201/7006/springqo9.jpg

We're supposed to find the equation of motion from the lagrangian and not Newton's equations.

Attempted solution:

[tex]L = T - U = \frac{I\omega^2}{2} + \frac{mv^2}{2} - \frac{kr^2}{2}[/tex]
[tex]I = m(r^2 + l^2)[/tex]
[tex]v = \frac{dr}{dt}[/tex]

From the euler-lagrange equation [tex]\frac{\partial L}{\partial r} = \frac{d}{dt}\frac{\partial L}{\partial v}[/tex] I get:

[tex]m\omega^2r - kr = m \frac{d^2r}{dt^2}[/tex]
[tex](\omega^2-\frac{k}{m})r = \frac{d^2r}{dt^2}[/tex]

If anyone can see any mistakes i'd appreciate it if they could let me know. Thanks
 
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Your equation

[tex](\omega^2-\frac{k}{m})r = \frac{d^2r}{dt^2}[/tex]

appears correct. Recall that

[tex]\frac{k}{m} = \omega^2_{spring}[/tex]

So, this shows the special value for

[tex]\omega^2[/tex]

Particularly,

[tex]\omega^2=\omega^2_{spring}[/tex]
 

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