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Lagrangian Problem

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    1. Two identical blocks A and B with mass m are joined together by a taut string B of length `. Block A moves on a frictionless horizontal table and block B hangs from the string which passes through a small hole in the table as shown in the figure.
    (a) Using polar coordinates for the block on the table, obtain the Lagrangian for the complete system. Take g to be the acceleration due to gravity. (5 Marks)
    (b) Obtain the equations of motion for the radial coordinate of A and its angular coordinate φ. (5 Marks)
    (c) Show that there is an equilibrium, constant-r solution with r = r0 = (p^2/m^2g)^1/3 , where pφ is the angular momentum of A. (4 Marks)
    (d) Suppose A is nudged slightly inwards whilst executing circular motion on the table with r = r0, show that the trajectory of A executes small oscillations with frequency ω = sqrt(3/2) p/mr^2. (6 Marks)

    2. Relevant equations

    The Lagrangian is L=m(dr/dt)^2+(1/2)mr^2(dφ/dt)^2-mgr
    Equations of Motion d^2r/dt^2=(1/2)(r(dφ/dt)^2-g)
    d^2φ/dt^2=-((dr/dt)(dφ/dt))/(r)

    3. The attempt at a solution

    The only part of this I have a problem with is (d), taking the radial E.O.M and equating it to the centripetal force gives mw^2r=-((dr/dt)(dφ/dt))/(r), however as it is executing circular motion, is the radial velocity dr/dt not=0? In which case leaving mw^2r=0, which can't be solved to give the above?
    The only way I can seem to get the factor of 3/2 from anywhere is by rewriting the Lagrangian, as dr/dt=0 and dφ/dt=p/mr^2 to give 3/2mgr, then I have an unwanted factor of g, and have no idea what this could be equated to in order to get an equation that makes sense to find w.

    Any pointers would be much appreciated :)
     
    Last edited by a moderator: Apr 16, 2015
  2. jcsd
  3. Apr 16, 2015 #2

    Orodruin

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    You have two possible routes:

    Yes, mw^2r = 0 for the circular solution. However, for solutions which are close to r0, you can expand the solution for small deviations from r0.

    Or:

    If you know how to compute effective potentials, compute the effective potential and deduce the second derivative at the minimum, which is at r0.
     
  4. Apr 16, 2015 #3
    What would be the function I would be expanding though? I don't see what makes sense to expand in order to find the angular frequency
     
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4

    Orodruin

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    The function r. Around the point of circular orbit.

    The assumption is that the deviation from the circular orbit is small. If it is large you will get nonlinearities and talking about an angular frequency becomes moot.
     
  6. Apr 16, 2015 #5
    and computing the effective potential gives me veff=(p^2/2mr^2)-mgr, still giving me an unwanted 'g'
     
  7. Apr 16, 2015 #6

    Orodruin

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    Did you try finding the second derivative at the minimum?
     
  8. Apr 16, 2015 #7
    That will still leave me with a g there wont it? The term mgr will just become mg(dr^2/dt^2)
     
  9. Apr 16, 2015 #8

    Orodruin

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    You need the derivative with respect to r, not t!
     
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