Analytical mechanics problem

  • Thread starter Azael
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  • #1
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As in the attached picture.
A cylindrical shell with mass M can roll without gliding on a horizontal plane

In the cylindrical shell a particle ,p, with mass m can glide without friction.

At the begining there is no motion and the angle to the particles position is [tex] \phi=\frac{\pi}{2} [/tex]


I am suposed to find the movement of the center of the circle as a function of the angle [tex] \phi [/tex]

Im not sure how I should start.

The potential energy of the system is, if I place the plane of reference on the level of the horizontal plane.

[tex] V=mgR(1-cos\phi) [/tex]

Now this problem obviously only has one degree of freedom and that is the angle [tex] \phi [/tex]

But if I want to construct a lagrangian I must use two degres of freedom. The rotation angle of the cylinder [tex] \alpha [/tex] and the angle to the particle p [tex] \phi [/tex]. Because I dont know how [tex] \alpha [/tex] and [tex] \phi [/tex] are connected. Is this the correct thinking?

In that manner I get this equation

[tex] L=T-V=\frac{3}{4}MR^2\dot{\alpha}^2+\frac{1}{2}m[\dot{\alpha}^2 R^2 + R^2\dot{\phi}^2 + \dot{\alpha}^2 R^2 sin\phi ] - mgR(1-cos\phi ) [/tex]

Should I use this and solve the two lagrande equations

[tex] \frac {d}{dt} \frac{dL}{d\dot{\alpha}}-\frac{dL}{d\alpha}=0 [/tex]

[tex] \frac {d}{dt} \frac{dL}{d\dot{\phi}}-\frac{dL}{d\phi}=0 [/tex]

Im not sure if this will give me any answere though? It feels like I should express [tex] \alpha [/tex] in [tex] \phi [/tex] before I even do the lagrangian. Am I on the right track?
 

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Answers and Replies

  • #2
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I assume that I in the same manner can get the hamiltonian since its a conservative force situation and holonomic constrains.

[tex] H=T+V=\frac{3}{4}MR^2\dot{\alpha}^2+\frac{1}{2}m[\dot{\alpha}^2 R^2 + R^2\dot{\phi}^2 + \dot{\alpha}^2 R^2 sin\phi ] + mgR(1-cos\phi ) [/tex]

But Im not quite sure what I would do with that one either. I know that

[tex] \dot{\alpha}=\frac{dH}{dP_{\alpha}}[/tex]
and
[tex] -\dot{P_{\alpha}}=\frac{dH}{d\alpha} [/tex]

But that doesnt seem to be much help either??
 
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  • #3
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I don't know...I wish I could help...

It seems like you really do need two generalized coordinates. I mean, the particle is free to move; it doesn't seem like it is constrained in any way such that [tex]\phi[/tex] and [tex]\alpha[/tex] would be related.
 
  • #4
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I think that there is an error in your Lagrangian: Since you are interested in the movement of the cylinder and watch it in the "labatory system", so to say, you can't say that the velocity of the particle in y-direction is [itex] R \dot \phi \cos(\phi)[/itex]. [itex] \phi [/itex] is fixed in the "cylinder system", so you have you add the y-velocity of the cylinder. If you square this, you should get a "mixed term" so that the both velocities are related.
 
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  • #5
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I managed to solve it. my expression for the Y position of the particle was

[tex] Rsin\phi + R\alpha [/tex]

So velocity was

[tex] R\dot{\phi}cos\phi + R\dot{\alpha}[/tex]

I used the fact that the conjugated momentum is constant for the alpha coordinate since there is no alpha, just alpha dot, dependance :) tricky problem.
 

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