Solving Cylinder+Bead System Problem | Hello.

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Homework Statement
A uniform cylinder of radius a and density ρ is mounted so as to rotate freely
about its axis, which is vertical. On the outside of the cylinder is a rigidly fixed uniform spiral
or helical track along which a point mass m can slide without friction. Suppose the point mass
starts at rest at the top of the cylinder and slides down under the influence of gravity.
FInd the lagrangian of the (cylinder+bead) system
Relevant Equations
.
Hello. I am having some difficulties regarding this problem.

I am not sure, actually, how to relate the angle rotated by the cylinder ##\phi## and the angle drawed by the bead itself ##\theta##.
I have been thinking in two alternatives:

a) The first one is that, since there are no friction forces, the angle are independent. So the Kinect energy of the whole system would be $$\frac{I \dot \phi ^2}{2} + \frac{m(\dot z ^2 + (r (\dot \theta + \dot \phi) ) ^2)}{2}$$

b) The second alternative would be to consider that, since gravity itself acts only at the z direction, and the rest of forces are internal. ##L_{z (cylinder)} = L_{z( bead)}##

I am not sure which one is correct, if neither.
 
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Orodruin said:
Is this the full problem formulation? There seems to be no actual question involved.

Your LaTeX is also missing a brace.
I just forgot to add the question hahah. I should need the lagrangian. Since the kinect term is the main problem here, i am not even talking about lagrangian itself.

My Latex seems normal here, please att the page to see if it changes.
 
a) You need to introduce an appropriate number of generalised coordinates for your system. How many coordinates would be required in order to specify the system’s configuration?

b) I am unsure what you are trying to say here. It is unclear if L refers to some Lagrangian or to angular momentum about the z axis.

Herculi said:
My Latex seems normal here, please att the page to see if it changes.
For some reason it was displayed as a yellow box saying brace missing. It seems ok upon reload..
 
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Orodruin said:
a) You need to introduce an appropriate number of generalised coordinates for your system. How many coordinates would be required in order to specify the system’s configuration?

b) I am unsure what you are trying to say here. It is unclear if L refers to some Lagrangian or to angular momentum about the z axis.For some reason it was displayed as a yellow box saying brace missing. It seems ok upon reload..
a) I think there are three independent coordinates, supposing that the cylinder and the bead rotates independently. So the coordinates would be:
Angle rotated by the cylinder ##\theta##
Angle rotated by the bead ##\phi##
z distance traced by the bead ##z##

b)If this condition is right, there would be just two coordiantes. L means angular momentum

The problem is that i am not sure which, if any, is right
 
Herculi said:
a) I think there are three independent coordinates, supposing that the cylinder and the bead rotates independently. So the coordinates would be:
Angle rotated by the cylinder θ
Angle rotated by the bead ϕ
z distance traced by the bead z
Think about the following: What if the cylinder was fixed, how many independent coordinates would you have? How many are added by allowing the cylinder to rotate?

Herculi said:
b)If this condition is right, there would be just two coordiantes. L means angular momentum
Angular momentum conservation comes in as a first integral of the equations of motion. That comes a couple of steps later unless you are looking for effective descriptions which you do not seem to be.
 
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Orodruin said:
Think about the following: What if the cylinder was fixed, how many independent coordinates would you have? How many are added by allowing the cylinder to rotate?Angular momentum conservation comes in as a first integral of the equations of motion. That comes a couple of steps later unless you are looking for effective descriptions which you do not seem to be.
Oops. Forgot a thing: The curve itself described probably has the shape ##r = r_{o}, f(z,\theta)=0##. So tecnically, we have just two coordinates if the cylinder is rotating? ##\theta, \phi##?
I am sorry, i am always confused when i come across problems involving two angles. I am not sure when the angles are independent or not. But since here there are no friction, i guess they are independent?
So is a) right?, if we replace ##\dot z## by ##\frac{dz}{d \theta} \dot \theta##?