- #1

adjklx

- 13

- 0

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

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

Last edited by a moderator: