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Lagrange qustion, a partilcle confined to a spherical cone

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle is confined to move on the surface of a circular cone with its axis
    on the vertical z axis, vertex at origin (pointing down), and half-angle α(alpha)

    a) write down the lagrangian in terms of spherical coordinates r and ø (phi)

    2. Relevant equations

    x=rsinθcosø y=rsinθsinø z=rcosθ
    the constraint for a circular cone is z=( x^2 + y^2)^1/2

    3. The attempt at a solution

    So using this constraint and some definitions of cartesian--> spherical coordinates one can show
    that θ is constant, i.e θ=α (alpha)

    My problem here is setting up the Kinetic Energy, as the Lagrangian (L) is
    L= T (kinetic) - U(potential) energies.
    In cartesian T= 1/2m(d/dt(x)^2+d/dt(y)^2+d/dt(z)^2)
    My problem is now converting this to spherical polar coordinates, keeping in mind all time derivatives of θ=zero because theta is constant (θ=α)
    I've found a solution online and it gives the kinetic Energy as
    T=1/2m(d/dt(r)^2+(rsinαø^(dot))^2) ...so the 1/2m( rdot^2 + (rsinαø(dot)^2)
    where ø(dot) is time derivate w.r.t phi...If anyone could help me get to this conclusion it would be appreciated. I've tried substituting directly for d/dt (x^2+y^2+z^2) but i do not get this answer,
    i think it is just perhaps my math (algrebra) screwing me up.

    Thanks in advance.
     
  2. jcsd
  3. Apr 8, 2014 #2

    BvU

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    d/dt (x^2+y^2+z^2) is not the same as (dx/dt)^2 + etc.
     
  4. Apr 8, 2014 #3

    ehild

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    You need to know the conversion between Cartesian and spherical polar coordinates.

    See http://en.wikipedia.org/wiki/Spherical_coordinate_system (scroll down page).

    [tex]x=r\sin(\theta)\cos(\phi)[/tex]
    [tex]y=r\sin(\theta)\sin(\phi)[/tex]
    [tex]z=r\cos(\theta)[/tex]

    Find the derivatives with respect time when θ=constant=α. Substitute for ##\dot x##, ##\dot y##, ##\dot z## in the formula for the KE. It simplifies to
    [tex]KE = \frac{m}{2} \left(\dot r^2+(r \sin(\theta)\dot {\phi} )^2\right)[/tex]
     
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