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ian2012

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Hi, Hope someone can help me clear up this question. I know the answer but I am unsure of the reasoning behind it, so here it is:

Question:A simple pendulum of mass

(a)Using the position

[tex]L=\frac{1}{2}(M+m)\dot{x}^{2}+ml\dot{x}\dot{\theta}cos\theta+\frac{1}{2}ml^{2}\dot{\theta}^{2}+mglcos\theta[/tex]

Answer:Use [tex]X=lsin\theta[/tex] and [tex]Y=lcos\theta[/tex] as the Cartesian coordinates of the pendulum relative to the trolley.

[tex]L=\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m[(\dot{X}+\dot{x})^{2}+\dot{Y}^{2}]+mgY[/tex]

And the answer falls out when you follow this through...

I don't understand why the first term is [tex]1/2M\dot{x}^{2}[/tex]

and not [tex]1/2M(\dot{X}+\dot{x})^{2}[/tex]. And I also don't understand why the last potential energy term is only [tex]mgY[/tex] and not [tex]mg(l-Y)[/tex]?

Would appreciate if someone could help me out on this, or if there is an easier way to go about the answer! Many thanks!

Question:A simple pendulum of mass

*m*and length*l*hangs from a trolley of mass*M*running on smooth horizontal rails. The pendulum swings in a plane parallel to the rails.(a)Using the position

*x*of the trolley and the angle of inclination*θ*of the pendulum as generalised coordinates, show that the Lagrangian may be written as[tex]L=\frac{1}{2}(M+m)\dot{x}^{2}+ml\dot{x}\dot{\theta}cos\theta+\frac{1}{2}ml^{2}\dot{\theta}^{2}+mglcos\theta[/tex]

Answer:Use [tex]X=lsin\theta[/tex] and [tex]Y=lcos\theta[/tex] as the Cartesian coordinates of the pendulum relative to the trolley.

[tex]L=\frac{1}{2}M\dot{x}^{2}+\frac{1}{2}m[(\dot{X}+\dot{x})^{2}+\dot{Y}^{2}]+mgY[/tex]

And the answer falls out when you follow this through...

I don't understand why the first term is [tex]1/2M\dot{x}^{2}[/tex]

and not [tex]1/2M(\dot{X}+\dot{x})^{2}[/tex]. And I also don't understand why the last potential energy term is only [tex]mgY[/tex] and not [tex]mg(l-Y)[/tex]?

Would appreciate if someone could help me out on this, or if there is an easier way to go about the answer! Many thanks!

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