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A question about classical dynamic

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Assume a ball has mass "Me" and wrapped by water like ocean and earth ,the water has mass "Mw" and density [itex]\rho[/itex] ,when the ball spin with angular velocity [itex]\omega[/itex] ,and the shade of system goes to ellipsoid, please expres the water depth in spherical coordinate system

    2. Relevant equations



    3. The attempt at a solution

    i was thinking to use calculus of variation to solve this problem , i assume radius of ball is "R" and shell to the water's surface is "a" , when ball spin , the new distant between shell and surface is "a+[itex]\eta[/itex] "

    from lagrange equation L= T-V

    L=[itex]\frac{1}{2}[/itex][itex]\rho[/itex][itex]\int[/itex][([itex]\frac{\partial\eta}{\partial t}[/itex])[itex]^{2}[/itex]+(a+[itex]\eta[/itex] )[itex]^{2}[/itex][itex]\omega[/itex] [itex]^{2}[/itex]]r[itex]^{2}[/itex]sin[itex]\theta[/itex]drd[itex]\theta[/itex]d[itex]\varphi[/itex]+GMe[itex]\rho[/itex] [itex]\int[/itex][itex]\frac{1}{r}[/itex]r[itex]^{2}[/itex]sin[itex]\theta[/itex]drd[itex]\theta[/itex]d[itex]\varphi[/itex]

    am i right?
     
    Last edited: Jul 28, 2011
  2. jcsd
  3. Jul 28, 2011 #2

    BruceW

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    [tex] {(a + \eta )}^2 \ \omega^2 [/tex]
    I don't think this is the correct velocity, because it should use distance from the axis of rotation, not distance from the origin. I think instead it should be:
    [tex] {((a + \eta)sin(\theta))}^2 \ \omega^2 [/tex]

    Edit: Also, I would define [itex] \eta [/itex] as constant with time, if you want to get the steady shape of the water.
     
  4. Jul 28, 2011 #3
    if [itex]\eta[/itex] is defined constant with time , so that [itex]\frac{\partial \eta}{\partial t}[/itex] = 0 ???

    and whether to set [itex]\eta[/itex]=[itex]\eta[/itex] (r,[itex]\theta[/itex],[itex]\varphi[/itex],[itex]\alpha[/itex])=[itex]\eta[/itex] (r,[itex]\theta[/itex],[itex]\varphi[/itex],0)+[itex]\alpha[/itex][itex]\zeta[/itex](r,[itex]\theta[/itex],[itex]\varphi[/itex]) ???
     
  5. Jul 28, 2011 #4

    BruceW

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    Actually, it should have been:
    [tex] {((R + a + \eta)sin(\theta))}^2 \ \omega^2 [/tex]
    since R+a+eta is distance from origin to where the water level is.

    Ok, so R+a+eta is the total radius at which the water level is (which is some non-spherical surface). Just to make it clear, there is no variational parameter yet (which is fine). So were you thinking of using alpha as a variational parameter? You could instead just use the Euler-Lagrange equations, without having to go through the derivation.
     
  6. Jul 28, 2011 #5
    yes, i'm using alpha as a variational parameter,

    i don't understand???
     
  7. Jul 29, 2011 #6

    BruceW

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    Well, you're trying to find [itex]\eta[/itex] as some function of [itex]\theta[/itex]. (I assume that the radius of the water surface doesn't depend on [itex]\phi[/itex] by symmetry considerations). (Also, [itex]\omega[/itex] Is just a constant, equal to [itex] \frac{2 \pi }{1 day} [/itex] ).
    You've written down the Lagrangian, so I'm guessing you're thinking of minimising the action, i.e. finding the stationary point of:
    [tex] A = {\int_{t_1}}^{t_2} L dt [/tex]
    But our independent variable is [itex]\theta[/itex] (not time), so I'm not sure how to find the action of this system.
    Another possible methods: Guess that the correct function will result in the system with least energy, so maybe you could try to find the stationary point of the energy, while applying the constraints of the system?
     
  8. Aug 2, 2011 #7
    do you mean Hamilton's equation???
     
  9. Aug 2, 2011 #8

    BruceW

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    Yes, or simply T + V = energy, then try to minimise this while keeping constraints.

    Or alternatively, going back to the lagrangian, I guess you could say that t is the independent variable, and [itex]\phi[/itex] is the dependent variable, and [itex]\omega = \frac{d \phi }{dt} [/itex] and then use the Euler-Lagrange equation:
    [tex] \frac{\partial L}{\partial \phi} = \frac{d}{dt} \frac{\partial L}{\partial \omega} [/tex]
    And maybe use this to find out what [itex]\eta[/itex] would be?
    Sorry if i've not been a lot of help, I'm just trying to think of how I would try to do the problem, but I'm not certain myself.
     
  10. Aug 4, 2011 #9
    i don't understand this sentence???could you teach me more detail?? thank you^^
     
  11. Aug 4, 2011 #10

    BruceW

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    Speaking completely generally, T is the total kinetic energy of all particles in a system (which includes the rotational KE of a rotating body made up of smaller objects). V is the total potential energy of the system. This will include the potential energy between two objects in the system. And also, if there is an external field it includes the potential energy of each object due to this field.
    In your case, I guess there is one rotating, rigid body (the earth and the sea, since I assume they are rigidly attached at their boundary).
    The kinetic energy in your system is the rotational KE due to the individual particles which make up the earth and sea. And the potential energy is the gravitational potential energy between each particle and each other particle.
    So now, we should probably start making assumptions to make the question easier. First, assume the earth is spherical (or oblate spheroid). Maybe we could also assume the gravitational force between sea particles is negligible, so the only gravitational force is between the earth and the sea particles. We might also assume the sea particles are rotating around the earth's axis at the same angular speed as the earth. and that the sea particles aren't moving radially or in the other angular direction.
     
  12. Aug 4, 2011 #11
    i got it

    thank you very much ~~~~ orz
     
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