# include the stretch of the spring own weight in potential energy for spring pendulum?

we are given a problem with spring with its own mass $m$. I am confused how to set up the PE term in the Lagrangian.

Assume the spring has length of $L_{0}$ when it is laying on a table horizontally. Now when we hold the spring upright and hang it vertically from a pivot in the ceiling to make a pendulum, it will stretch a little due to its own weight by $\frac{(m/3)g}{k}$ where $k$ is the spring constant. I use $m/3$ there since that is the common factor used to generate an effective mass used for a spring with its own weight.

http://en.wikipedia.org/wiki/Effective_mass_%28spring-mass_system%29

Hence now the length of the spring is $L_{0}+\frac{(m/3)g}{k}$ before we even put the bob mass at the end of it. Should one use $L_{0}+\frac{(m/3)g}{k}$ as the relaxed length of the spring or use $L_{0}?$

Will the PE for the spring be $\frac{1}{2}k\left( y-L_{0}\right) ^{2}$ or $\frac{1}{2}k\left( y-(L_{0}+\frac{(m/3)g}{k})\right) ^{2}$?

Assuming one is measuring $y$ here from the ceiling where the spring is hanging from

                 +---------
\     |
/     |
\     |
/     | y(t)
\     |
/     |
M     V


Which term should be used for PE?

I know there will be additional PE term for the spring due to its own mass, which is $-mg\frac{y}{2}\cos\theta$

Hence total PE from the spring is

$PE=\frac{1}{2}k\left( y-L_{0}\right) ^{2}-mg\frac{y}{2}\cos\theta -Mgy\cos\theta$

or

$PE=\frac{1}{2}k\left( y-(L_{0}+\frac{(m/3)g}{k})\right) ^{2}-mg\frac{y} {2}\cos\theta-Mgy\cos\theta$

where $\theta$ is angle of rotation of the spring pendulum.

not sure which to use.

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