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Hamiltonian for a particle moving on a plane tangent to the surface of the easth

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A point of mass m is placed on a frictionless plane that is tangent to the Earth’s surface. Determine Hamilton’s
    equations taking:
    (a) the distance x
    (b) the angle q
    as the generalized coordinate.


    2. Relevant equations



    3. The attempt at a solution

    Take the distance x as the distance from the point of contact of the plane with the surface
    angle q is the angle made with the normal to the plane through the center of the earth

    my soln. is L= 1/2*m*(x(dot)^2) - m*g*(sqrt(x^2+a^2)-a)

    (here a is the radius of the earth )
    Rest is elementary

    Is this correct ?
    Since its a standard problem in many textbooks does anyone know the exact solution ?
     
  2. jcsd
  3. Aug 3, 2011 #2
    You potential energy [itex]U(x) = m g (\sqrt{x^2 + a^2}-a)[/itex] looks fishy.

    One expects that [itex]\lim_{x\to\infty}U(x) = 0[/itex]. This is not satisfied. Also one expects that [itex]U(0) = - m g a[/itex], you have the wrong sign. Also gravitational potential energy should decays like [itex]1/x[/itex] asymptotically. I am guessing the correct potential should be:
    [itex]U(x) = -\frac{m g a^2}{\sqrt{x^2 + a^2}}[/itex].
     
  4. Aug 3, 2011 #3
    Hey thanks. I was calculating potential wrt ground level and for movements over small distances but i think your idea is probably more applicable here.

    i'm more doubtful about the kinetic energy term. Since the particle is moving on a plane can we ignore the y(dot)^2 term like i have done ? In that case we won't get the answer only in terms of x (is that required according to the question!!) ?
     
  5. Aug 3, 2011 #4
    What are the constraints on the particle? How many degrees of freedom does it possess?

    If you understand that, then you'll understand why your kinetic energy term is not sufficient.

    (Hint: why does the question mention an angle, yet you haven't included it in your Lagrangian)
     
  6. Aug 3, 2011 #5
    Actually a and b are two separate parts. Infact the angle q depends on x as tan q = x/a so it cant be used as another degree of freedom.
     
  7. Aug 3, 2011 #6
    JesseC what do you think about the potential energy term ?
     
  8. Aug 4, 2011 #7
    First: I'm sure you're misunderstanding the question. The particle is constrained to move in a plane, thus two degrees of freedom are required to describe its motion on the plane. The question states that the two generalised coordinates you should use are the radial distance from the point of contact of the plane with the earth, x, and some angle q, presumably about the point of contact. They're not supposed to be read as separate parts of the question. You need both coordinates in the kinetic energy term to fully account for the particles motion.

    Second: Since the earth is a a very good approximation to a sphere, the potential energy term will have no dependence on q. Can you write for me the potential energy of a small mass m, a height 'a' above the centre of a planet of mass M? Because the potential in your first post is wrong.

    As with all these questions it is best to sketch the problem. If you sketch, you may realise why your original idea for what the angle q represents is not correct. I'm not just going to give you the answer either!
     
    Last edited: Aug 4, 2011
  9. Aug 4, 2011 #8
    I agree that motion on a plane has two degrees of freedom. But here the question uses the term coordinate and not coordinates. Secondly in another version of the same question in a textbook it mentions clearly that write the hamiltonian using either x or q(angle) as the generalised coordinate !!!
     
  10. Aug 4, 2011 #9
    Second: Since the earth is a a very good approximation to a sphere, the potential energy term will have no dependence on q. Can you write for me the potential energy of a small mass m, a height 'a' above the centre of a planet of mass M? Because the potential in your first post is wrong.

    The potential in my first post is applicable only when 'a' is extremely small.
    I think for the general case we can use
    v = - G*M*m/(a+R) R is radius of earth.
    This is the same as v = -m*g*R^2/(a+R)
    now when a/R << 1 we'll get a potential similar to the one in my first post plus some constant

    (Here i have changed the meaning of 'a' according to your question )

    As with all these questions it is best to sketch the problem. If you sketch, you may realise why your original idea for what the angle q represents is not correct. I'm not just going to give you the answer either

    I can assure you that the figure verifies what i have written about the angle. If you want you can see the problem and the figure on Miguel Mostafa PH 621 course page (just do a google search). The problem is under homework # 8 and the clarifying figure is mentioned below the link for the homework under problem no. 6

    PS: i'm not a student of prof. Miguel Mostafa so i can't ask him
     
    Last edited: Aug 4, 2011
  11. Aug 4, 2011 #10
    Thanks for telling me where to find that figure, I can see where the confusion has come from. Just shows the wonders of what a figure can do! So the particle is confined to move along a wire tangent to the earth, thus it has one degree of freedom.

    All these different notations for things are getting confusing, so lets stick to the what Miguel uses here: http://hep.colostate.edu/~mostafa/PH621/fig8-1.pdf . So let [itex] a = l [/itex].

    You're pretty much right, the potential experienced by the particle (so long as [itex] l > R [/itex]) is just [itex] V = - G M m / l [/itex] and the kinetic energy is just [itex] T = \frac{1}{2} m \dot{x} ^2 [/itex].

    The trick is to write [itex] l [/itex] in terms of [itex] x [/itex] for the first part. Although I have to say, mathfeel has done that for you ;).
    The trick for the second part is to write [itex] \dot{x} [/itex] and [itex] l [/itex] in terms of [itex] \theta [/itex]. Have a go...
     
  12. Aug 4, 2011 #11
    Hey thanks ! I think that clears everything up
     
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