Mass on a string.lagrange mechanics

[tmoan]

hi there i am studying lagrange and hamiltonian mechanics.
i came about this question from a previous test session please if anyone can help.

the question is about a mass rotating on a rectilinear string attached without friction from a point A on the z axis.
the length of the string is h = OA (O being the origin of the 3D system)

the mass is free to move on the string as a pearl on a necklace for example.

the rotational vel. "w" is given constant which makes the angle ф on the xy plane = wt
this means that ф is not a degree of freedom

i do not undersand however why the angle α between the string and the z axis is not.
in the solution r the position vector from A to the mass M is the only degree of freedom.

i think the answer would be simple and that if i scratched my head a bit i would know but i am still trying to figure it out in the mean time i tried asking here so..
if anybody can help me out it would be great thanks.

 

[Count Iblis]

So, the string is rotating at angular velocity w and you have a mass M on the string that can freely move along the string?

Then you can just write down the Lagrangian which only consists of a kinetic energy term:

L = 1/2 M [r-dot^2 + r^2 w^2]

Then write down the Euler-Lagrange equatons:

d/dt dL/dr-dot - dL/dr = 0

But it is easier to construct the Hamiltonian and then use that it is conserved. The canonical momentum corresponding to r is:

pr = dL/dr-dot = M r-dot (no surprise here)

The Hamiltonian is by definition:

H = pr r-dot - L = 1/2 M r-dot^2 - 1/2 M r^2 w^2

The total derivative of H w.r.t. t is zero as easily follows in general from Hamilton's equation (provided L does not depend explicitely on time). So, you can equate H to a consant and integrate the first order diff. equation.

 

[tmoan]

yes thanks for the reply.
i understand what you are sayying.
but my question is why isn't the angle alpha a degree of freedom which will require its own lagrange equation right.
secondly, 1/2M r-dot^2 is for the kinetic energy of a rotating body where r is the position from the origin while here r is from pt A to M (M is where the COM of the mass is) shouldn't i do some kind of mathematical transition to A.i didn go past that yet,
i know that a hamiltonian eq would be more in handy but i am answering a question here that requires a lagrange equation first,

i think there must be a relation between angle phi and angle alpha but i don't seem to find it out yet,
i know that i only need r using my common sense but i am required to state why alpha and phi are not degrees of freedom.
i hope i made myself clear and sorry if this area is not for course work i didn't realize that it is not when i posted.
thanks for your help

 

[Count Iblis]

I completely ignored the part about the angle alpha.
If the string were to make some angle alpha with the z-axis while it rotates, then the kinetic energy is:

1/2 M [r-dot^2 + r^2 sin^2(alpha) w^2 + r^2 alpha-dot^2]

I think this is quite obvious.

So, you then write down the Euler-Lagrange equations for this system. If alpha is not assumed to be constant, then you have to deal with the equation:

d/dt [dL/d alpha-dot] - dL/d alpha = 0

 

[tmoan]

point taken and true.
but the fact that w is constant makes wt =phi implies phi is no longer a degree of freedom.
while phi is not constant!
in the same context why isn't alpha a degree of freedom it was never given that it is constant
but alpha must change w.r.t. phi in some way which itself changes w.r.t. w i.e. a constant and that I think is the best answer. or the PhD holder who wrote this doesn't deserve his degree for not emphasizing enough i have solved 3 other questions that are much more tiresome than this and still didn't exactly figure out why it is not clear enough.

thanks again for the help

 

[Count Iblis]

Sometimes questions are not formulated in a clear way. You should simply focus on trying to master the theory and skip questions if they are unclear. Or just replace the unclearly formulated problem with a clearly defined problem and solve that.