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Mechanics particle problem

  1. May 7, 2005 #1
    I am trying to solve the following problem:

    A particle of mass m is constrained to move under gravity with no friction on the surface xy=z. What is the trajectory of the particle if it starts from rest at (x,y,z) = (1,-1,-1) with z-axis vertical?

    The lagrangian is

    [tex]L=T-V=\frac{1}{2}m\left(\dot{x}+\dot{y}+\dot{z}\right)-mgz[/tex]

    the constrain is

    [tex]g(x,y,z) = xy-z = 0[/tex]

    for the x-component,

    [tex]
    \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }

    \frac{d}{dt}\pd{L}{\dot{x}}{} - \pd{L}{x}{} + \lambda \pd{g}{x}{} = 0
    [/tex]

    where [tex]\lambda[/tex] is the lagrange multiplier.

    [tex]m \ddot{x}+\lambda y = 0
    [/tex]

    Similarly,
    [tex]m \ddot{y}+\lambda x = 0
    [/tex]

    for z-component,
    [tex]m\ddot{z}+mg-\lambda = 0[/tex]

    however, i find the system is very difficult to solve. I define
    [tex]\phi = x + y[/tex]
    then
    [tex]\dot{\phi} = \dot{x} + \dot{y}[/tex]
    [tex]\ddot{\phi} = \ddot{x} + \ddot{y}[/tex]

    then adding the first 2 equations,
    [tex]m\ddot{\phi} + \lambda \phi = 0[/tex]

    which is the same as simple harmonic equation. but this is wrong because the answer of the question is [tex]x = - y = \sqrt{-z}[/tex]

    so, how can I solve the system of equations? and, is [tex]\lambda[/tex] a constant, of a function of [tex]t[/tex]?

    thanks for your help
     
  2. jcsd
  3. May 7, 2005 #2

    OlderDan

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    I found a surface plotting applet online at

    http://www.fedu.uec.ac.jp/~yanto/java/surface/

    The attached figure illustrates the constraining surface, showing a saddle point at the origin. The surface runs uphill into the first and third quadrants and downhill into the second and fourth quadrants. The left panel shows contours of constant z (constant potential energy). Clearly there are lines of symmetry along y = x and y = -x. The zeros of z are the x and y axes.


    One thing is wrong in your solution

    [tex]L=T-V=\frac{1}{2}m\left(\dot{x}+\dot{y}+\dot{z}\right)-mgz[/tex]

    should be

    [tex]L=T-V=\frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)-mgz[/tex]

    but the derivatives that follow are OK

    I'm not convinced the rest of your solution is "wrong". If you move along the line y = x, the distance d from the origin is

    [tex]d = \sqrt{2}x = \sqrt{2}\frac{x+y}{2} = \frac{x+y}{\sqrt{2}} = \frac{\phi}{\sqrt{2}}[/tex]

    [tex]d^2 = \frac{\phi^2}{2} = \frac{2x*2y}{2} = 2xy = 2z[/tex]

    [tex]z = \frac{1}{2}d^2[/tex]

    On this line you are running the ridge on a parabolic path. If you could stay on the ridge without being pushed to the side you would have a quadratic potential energy, so you should expect harmonic motion into and out of the valley, passing through the origin. But if you deviate even slightly from this path, gravity will push you down the gully on either side of the ridge.

    In this problem, the boundary conditions are that you start on the line y = -x, with zero velocity. The force component in the xy plane is going to be directed along the line y = -x, and that line is going to be the projection of the path of motion in the x-y plane. The particle is simply going to run down the hill along the line of symmetry, and the surface constraint leads to the solution given for x, y, and z.

    For a different set of boundary conditions, starting near the line y = -x, your [itex]\phi = x + y[/itex] is zero on the line of symmetry. There will clearly be a restoring force down the hill pointing somewhat toward the line of symmetry for any deviation from that line, so your harmonic solution for [itex]\phi = x + y[/itex] makes sense in the more general case where the particle is released from some point off the line.

    I'm inclined to think [itex]\lambda[/itex] should not be time dependent if the constraining surface equation has no explicit time dependence. For a while I was bothered by what appeared to me to be a different [itex]\lambda[/itex] in different quadrants, but I think that has been resolved. For any starting point there is going to be a tendency to restore the particle to the line y = -x, which is equivalent to your variable [itex]\phi = x + y = 0[/itex]
     

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    Last edited: May 8, 2005
  4. May 8, 2005 #3

    OlderDan

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    I reworked my reply in #2 to make sense, I hope. I thought I should pop this thread into the recent arena in case anyone else has thoughts on this problem.
     
  5. May 8, 2005 #4
    I agree. I will make a couple suggestions though. First of all, lambda is a time dependent variable. What the lagrange multiplier technique does for this problem is give you a system of four equations in x, y, z, and lambda. So you need to include the original constraint equation as the fourth equation of the system. Still, I couldn't figure out how to solve that system. It probably requires an algebra trick or something.

    Normally the strategy is to eliminate lambda first thing. However, lambda does have a physical significance that makes the lagrange multiplier technique worthwhile. [itex](\lambda y, \lambda x, -\lambda)[/itex] are the components of the force of constraint (i may have the sign wrong). So if you decide to solve for lambda too, you will be solving for the time dependent force of constraint. Just using the regular lagrangian technique does not allow you to do this--it eliminates the forces of constraint by reducing the degrees of freedom.

    My suggestion would be to try to solve it in an easier way first, and then see if you can work toward that solution with the lagrange multiplier technique. What I would do is use xy=z to eliminate one variable right away. Then express the lagrangian in the two remaining variables. Now you can write down lagranges equations of motion for two variables. Not only that, but you can write down the constant energy of the system (constraints do no work). Hopefully there will be a way to solve that system.

    edit: two things....I should have said "this constraint does no work, since it is not time dependent." Time dependent constraints can do work. Also, when including the constraint equation in your system of four equations, you are free to differentiate it with respect to time as many times as you find is convenient.
     
    Last edited: May 8, 2005
  6. May 8, 2005 #5

    OlderDan

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    Now that you point it out, it does make sense that [itex]\lambda[/itex] is time dependent. From the original equations, there was the z-component equation

    Can you solve this for [itex]\lambda[/itex]?

    [tex]\lambda = m\ddot{z}+mg[/tex]

    At some point, I assume the particle could leave the surface when [itex]\lambda = 0[/itex] and the z component of acceleration becomes -g with the x and y components zero. For the initial conditions given, with the particle starting from rest on the line of symmetry of the constraining surface, the problem seems somewhat similar to the one discussed here

    http://www.lancs.ac.uk/depts/physics/teaching/py332/lecture_5_05.pps
     
  7. May 8, 2005 #6
    Well, I tried doing my "simplified" approach by eliminating one coordinate from the get go, but the resulting system of equations looks just as hard to solve :redface: . So I think OlderDan has right idea...solve it by "observation" based on the initial conditions they gave you. That doesn't mean you have to junk the Lagrange multiplier approach. You just look for a solution to those four equations such that y = -x for all t. That is, stick in -x everywhere you see a y in those equations and then try to solve it. That such a solution exists is perfectly reasonable, since you can show that the total force on the particle (normal force plus gravity) points "downhill" in the direction of the line y = -x as long as the particle is on that line. The particle is sliding down a parabolic hill. OlderDan has already pointed this out.

    I'm pretty sure the book/professor gave this particular initial condition because it would be darn near impossible to solve the problem otherwise. Even for this case, the resulting integral for x (or y) as a function of t looks pretty ugly.

    OlderDan: The example in that link you posted is very good. I had only seen the more "elementary" approach to that problem, which makes use of the constant energy. I think for the above problem we can assume that it stays on the surface just because they say it does, even if the normal force does go to zero somewhere (i haven't figured out if it does or not).
     
  8. May 9, 2005 #7
    thanks a lot, OlderDan and PBRMEASAP! :rofl: thanks!

    the applet is very useful coz i don't have mathematica in my computer. I will try each method very carefully.

    actually i am not sure whether it is a book problem or a homework given by a professor. i am self-learning lagrangian mechanics, which will be taught in next semester. so i just download the most relevant materials and try those questions. but it's really challenging :approve:
     
  9. May 9, 2005 #8
    That is great! And you're right, it is challenging to teach yourself Lagrangian mechanics. If you get stuck, don't hesitate to ask a question here. :smile:
     
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