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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?

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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