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raggle
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Homework Statement
A long light inflexible rod is free to rotate in a vertical plane about a fixed point O. A particle of mass m
is fixed to the rod at a point P a distance ℓ from O. A second particle of mass m is free to move along the rod, and is attracted to the point O by an elastic force of strength k.
Find any equilibrium points. (You may assume the rod is sufficiently long that the moving particle does not reach either end of the rod).
Homework Equations
The potential due to the elastic force is ##V = kx^2##
The Attempt at a Solution
Let ##X## be the distance of the second particle from the origin O. Then the Lagrangian I got is
##L = \frac{1}{2}m[\dot{X}^2 + (X^2 +l^2)\dot{\theta}^2] - mg[X(1-cos\theta) +l(1-cos\theta)] - kX^2##
This gives the equations of motion for ##X(t)## and ##\theta (t)## as
##\ddot{X} = X(\dot{\theta}^2 - 2k) -g(1-cos\theta)##
##\ddot{\theta}(X^2+l^2) + 2\dot{X}\dot{\theta} = -gsin\theta(X + l)##.
Equilibrium points occur when the second derivatives are both zero, so I have to solve
##0 = X(\dot{\theta^2} - 2k) -g(1-cos\theta)##
##2\dot{X}\dot{\theta} = -gsin\theta(X + l)##
And now I'm lost. If I try to solve these I will end up with the fixed points ##X## and ##\theta## as functions of ##t##, which doesn't seem right to me.
Any help would be appreciated, maybe the Lagrangian I got is incorrect.