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Point particle Lagrangian paradox!

  1. May 15, 2007 #1
    Suppose, there's an inclined surface, and a sphere, with radius [tex]R[/tex], is rolling without slipping. The Lagrangian is

    [tex]L = I \frac{\dot \theta ^2}{2} + m \frac{\dot s ^2}{2} - V[/tex]

    where [tex]\theta[/tex] is the angle of rotation of sphere and [tex]s[/tex] is the curve length from top, and V is a potential which depends on coordinates. But coordinates [tex]\theta[/tex] and [tex]s[/tex] are interralated with [tex]s = R\theta[/tex] so Lagrangian can be equivalenty written as

    [tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{I}{mR^2} + 1 \right ) - V/m[/tex]

    I'd expect the solution will be reduced to point particle solution in the limit [tex]R \to 0[/tex] but...
    The point particle Lagrangian is

    [tex]L = \left ( \frac{ds}{dt} \right )^2 - V/m[/tex]

    so I assume this's what the solution will be reduced, but when I put [tex]I=(2/5)mR^2[/tex] in the previous Lagrangian, there are no longer any [tex]R[/tex]s, i.e. the Lagrangian is independent of radius and

    [tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{7}{5} \right ) - V/m[/tex]

    becomes. What's more painful is, the same weirdness occurs in any circular rolling object; if my object was a rotating disk in the first place, my Lagrangian would become

    [tex]L = \left ( \frac{ds}{dt} \right )^2 \left ( \frac{3}{2} \right ) - V/m[/tex]

    and in the point particle limit, where I imagine the point particle as a disk with infinitesimal radius and thickness...

    1. Can someone shed light on this "paradox"; what's going on here?!? :yuck:
    2. And how do i get point particle limit? Or is there a way to get point particle limit, which has a delta function mass density, starting from a continuous, smooth mass distribution??
     
    Last edited: May 15, 2007
  2. jcsd
  3. May 15, 2007 #2
    What is V? The potential? Wouldn't that somehow depend on R as well?. That is to say, should there not be two potentials. One for the linear kinetic energy and one for the angular kinetic energy?
     
  4. May 15, 2007 #3
    Fixed the post. There is only one potential: V, and it is a potential which depends only the position of the object.
     
  5. May 15, 2007 #4
    I certainly haven't checked your work closely, but I wouldn't consider "a point particle that still seems to have rotational inertia" too unphysical, because we live in a universe containing point particles that do have angular momentum.

    This also reminds me of a cute mechanics problem: after rolling down the same ramp, which flies further: a solid ball, or a hollow one? (The answer does not depend on masses nor radii.)
     
    Last edited: May 15, 2007
  6. May 16, 2007 #5

    Doc Al

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    It's certainly interesting that a "particle" with an internal degree of freedom (a ball rolling down an incline) is not the same as a "particle" with only translational freedom (a box sliding down an incline). Of course, these are only models--in reality, long before reaching the limit of zero size, both models break down as you approach the molecular level. The tiny "ball" won't be rolling down the incline; the tiny "box" won't be sliding.
     
  7. May 16, 2007 #6

    jambaugh

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    I think the answer to your post is that the rolling ball shouldn't
    reduce in the [tex]R\to 0[/tex] limit to a point particle of the same mass. The proper pre-limit case is that of a sliding sphere (or any other shape).

    The key is to think of mass as the constant representing how much energy is stored kinetically in the object due to its motion. (Or better still the ratio of the momentum to the velocity) Since the rolling object is storing energy both in its linear and its rotary motion it should have a larger effective mass. You see this in the fact that the formal momentum is:

    [tex] P_s= \frac{\partial L}{\partial \dot{s}} = \left(\frac{I}{R^2} + m\right) \dot{s}[/tex]
    (Note that I didn't divide out the mass to get an equivalent Lagrangian as I want to keep the dimensional units of the action.)
    Hence:
    [tex] M_{eff} = m + \frac{I}{R^2} = \frac{7}{5}m[/tex]
    for a sphere.

    Heck simply look at the solution to the kinetic equation. A ball sliding down a frictionless surface will get to the bottom faster than a ball will roll down the same inclined plane. Further the time to the bottom will be independent of the mass and size of the ball but will be less than that of a sliding object by a factor of:
    [tex] \frac{t_{roll}}{t_{slide}}= \sqrt{\frac{5}{7}}[/tex]

    You can intuit it in two ways:
    1.) More kinetic energy stored in the rotation equates to an increase in effective inertial mass over the gravitational mass.

    2.) The back reaction of the torque required to rotationally accelerate the ball must include a contact force opposing the ball's motion down the incline.
    This latter is the way a Masse shot works in pool.

    There is no paradox as your limiting case is not appropriate. There are similar "paradoxes" such as the case of the man falling on a heavy ladder. He will hit the ground faster than if he simply jumps off and falls.

    Trust the Lagrangian formulation. It works and you can get a lot of insight when you consider the form of the generalized momentum.

    Regards,
    James Baugh

    P.S. In case you didn't spot the obvious: To get the effective point particle case you simply use the effective inertial mass I mentioned while working with the the actual mass when calculating the gravitational potential. Recall that when we solve the two-body Kepler problem you get the equivalent of a one body in a central k/r potential but again with an effective mass.
     
    Last edited: May 16, 2007
  8. May 16, 2007 #7

    jambaugh

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    P.P.S. I'm trying to think up a good bar bet where one would race different objects down an incline. As I recall some cue balls for coin tables have metal cores so a magnet can divert it to its own slot when you scratch. Thus its moment of inertia should be a smaller multiple of its mass and it should beat the 8-ball to the bottom of a slope. And of course you could race say full vs empty beer bottles, or frozen vs liquid water bottles.

    Hmmm... I think one could cook up some nice class room demonstrations along these lines. Lead ball vs plastic ball ties. Solid ball beats hollow ball. Which will win? Hollow ball or solid cylinder? Can you test which egg is hard boiled vs which is raw by racing them down a slope? Et cetera.
    R.
    J.B.
     
  9. May 16, 2007 #8
    Thanks for the replies!

    I think i've developed a clear answer to the "paradox":

    The point particle configuration assumes no friction, the object freely slides under the gravity. However, in the rolling ball configuration, the is a friction, and it's responsible for the torque which causes the rotation, and which enables use to write [tex]R \dot \theta = \dot s[/tex].

    In other words, the two systems are not physically same, one with friction, other without. Therefore we cannot hope to reduce the solution of one to another.
     
  10. May 16, 2007 #9

    Doc Al

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    An excellent point. :wink:
     
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