Hi there thanks for your response!
I've updated my eq. of motion based on your feedback, please let me know what you think
##\ddot{X} = X(\dot{\theta}^2 -k/m) -g(1-cos\theta) ##
I'm struggling to change this one though as I thought the 1/2 cancels with the squared term when you differentiate...
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...
Hi there, I am also struggling on this question? I get the same Lagrangian except the potential of the spring is 1/2*k*x^2. Can anyone explain why this L is wrong?
So would it be correct to use polar coordinates then?So then r would be the distance for the 2nd particle and it is related to the kinetic energy by x=rcosθ y=rsinθ and the first particle has fixed r=l?
Thanks for your swift response!
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...