Solve Lagrangian Problem: Pendulum Spring System

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lagrangeian problem...

Homework Statement


in a spring pendulum system the spring is initially in equilibrium at upright position with theta = 0 and elongation (mg)/k, (k is spring constant). total length of spring in equilibrium condition is a with respect to which we define additional deviations in terms of a variable r. write dwn the lagrangian for this system

Homework Equations



L = kinetic energy - potential energy
kinetic energy = 0.5 mv^2
potential = mgh

The Attempt at a Solution


i know that the l = kinetic energy - the potential energy...but not exactly sure how to form my equations..please help...

i think potential energy would be 0.5(k*x^2) ?
 
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clueless_rosh said:

Homework Statement


in a spring pendulum system the spring is initially in equilibrium at upright position with theta = 0 and elongation (mg)/k, (k is spring constant). total length of spring in equilibrium condition is a with respect to which we define additional deviations in terms of a variable r. write dwn the lagrangian for this system


Homework Equations




The Attempt at a Solution


i know that the l = kinetic energy - the potential energy...but not exactly sure how to form my equations..please help...

i think potential energy would be 0.5(k*x^2) ?

The first thing here is picturing the whole problem schematically! So graph the pendulum with given data and show it to us!

AB
 


ok from what i can guess

potential energy = mgrcos(theta) - 0.5k*(r^2)
kinetic energy = 0.5(m(dr/dt)^2) + 0.5(mr^2)(dtheta/dt)^2
 


clueless_rosh said:
ok from what i can guess

potential energy = mgrcos(theta) - 0.5k*(r^2)
kinetic energy = 0.5(m(dr/dt)^2) + 0.5(mr^2)(dtheta/dt)^2

You got it right!

AB
 


ok thnx.. :)
 

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