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Pendulum with not homogeneous attraction

  1. Dec 28, 2013 #1
    I would like to study a pendulum in 2D, not on Earth but only with one fixed mass attract a disk. Considered this fixed mass like a point. No friction on this theoretical study. The disk can move only around a part of circle. There are only 2 forces, F the attraction and N the force from wall (circle). The disk is composed with homogenous material, same density everywhere. The center of mass is fixed and it is in the center of circle of the disk. The center of gravity change always:

    - it is more at center when the disk is far away the fixed mass
    - it is more at outer circle of the disk when the disk is closed to the fixed mass

    When I draw the trajectory of the center of gravity, it's not a circle. This would say the force from wall works.

    But N seems to reduce the attraction F due to the non circle trajectory. So, I thought the energy goes to the angular velocity but N anf F don't give a torque because they pass trought the center of mass.

    I tested an example with Algodoo software (2D software), and it confirm no rotation of the disk, it confirm too the lost of energy, sure it's not a professionnal software, maybe it's an error, like me the software see a non circle trajectory. I put an example, Algodoo is a free software.

    images:

    f1: the system
    f2: 3 positions of disk (examples)
    f3: N and F forces, orange point is the center of gravity

    So,

    1/ The disk have a torque or not ?
    2/ The trajectory of the center of gravity is perpendicular to the force from wall or not ?
     

    Attached Files:

    Last edited: Dec 28, 2013
  2. jcsd
  3. Dec 28, 2013 #2

    mfb

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    For objects with spherical symmetry, the center of gravity is the point of symmetry, so the center of gravity is identical to the center of mass.

    What does that mean?

    If the wall and the mass are fixed relative to each other and their orientation in space is fixed as well, there will be forces between ball and wall.

    That does not make sense. Forces between objects do not influence gravitational forces. And you just have two types of energies, potential energy and kinetic energy of the ball (if you assume wall and other mass to be fixed to something massive enough to not move significantly).

    Sure, without friction there is no reason why it should rotate.
    Loss where? The total energy is conserved.
     
  4. Dec 28, 2013 #3

    Dale

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    There are two equivalent ways of solving this. In both cases you will construct the Lagrangian and use the kinetic energy as ##T=1/2 \; m \; R\dot{\theta}##. In the first case you would construct the potential as a function of ##\theta## only by calculating ##r(\theta)##, the distance from the gravitating point mass. In the second case you would leave the potential as a function of r, but add a constraint term with a Lagrange multiplier.

    I think the first approach is more straightforward for this problem.

    EDIT: I forgot to square the velocity. The correct formula is ##T=1/2 \; m \; (R\dot{\theta})^2##
     
    Last edited: Dec 28, 2013
  5. Dec 28, 2013 #4

    mfb

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    This should be ##T=1/2 \; m \; \left(R\dot{\theta}\right)^2##.
    Apart from that, I agree, if you are interested in solving the system that is the right way.
     
  6. Dec 28, 2013 #5
    Thanks for your help, yes I would like to resolve this system.

    @mfb: are you sure the center of gravity is always the center of mass, expecially when the attraction is not homogeneous ? http://en.wikipedia.org/wiki/Centers_of_gravity_in_non-uniform_fields

    I read on Wikipedia the Lagrangian, with this method I can have the movement, but I know it, disk follow part of circle (I don't need a big angle). How can I use Lagragian for have the center of gravity and the angle between trajectory of center of gravity and force from wall ?
     
    Last edited: Dec 28, 2013
  7. Dec 28, 2013 #6

    mfb

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    See my previous post, for objects with spherical symmetry (in 1/r-potentials, gravity has this) they are the same. For other objects, they can be different.

    You can find the same statement in the article you linked:
     
  8. Dec 28, 2013 #7
    This would say in 2d (ball = disk) the center of gravity is not in the center of mass but in 3d (ball = sphere) it is ?
     
  9. Dec 28, 2013 #8

    mfb

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    Right.

    Anyway, I don't see the problem.
     
  10. Dec 28, 2013 #9
    Because Algodoo it's a 2D software ! and when I draw with my paper it's in 2D too. It's not possible to have 2D object in reality ? (I changed my first message because I would like to resolve in 2D not in 3D for show where the software bug).
     
  11. Dec 28, 2013 #10

    mfb

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    Even in 2D, there is no torque, as you still have the relevant symmetry left (circular in this case). The center of mass, center of gravity and the attracting mass are aligned. All forces are aligned with the center of mass of the [STRIKE]ball[/STRIKE] disk.
     
  12. Dec 28, 2013 #11
    No torque, I'm agree with that and the software too, I said that because the software lost energy and must go somewhere, it was an hypothesis. For me, the force from wall is not perpendicular to the trajectory of the center of gravity. Have you seen my images, the last showing center of gravity (orange point) ?
     
    Last edited: Dec 28, 2013
  13. Dec 28, 2013 #12

    Dale

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    D'oh! Yes, of course. I will go back and edit.
     
  14. Dec 28, 2013 #13

    Dale

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    No, you don't know the movement. You know that the movement is constrained to follow part of a circle, but the movement itself is the exact position on the circle as a function of time. In other words, in order to know the movement you need to know not only that it follows a circle but when it is at which point of the circle.

    If you need to determine the force from the wall then you should use the constraint method I mentioned above with the Lagrange multiplier. This can give you the generalized force from the wall:

    http://farside.ph.utexas.edu/teaching/336k/newton/node90.html [Broken]

    With that, you can calculate all of the angles you are interested in.
     
    Last edited by a moderator: May 6, 2017
  15. Dec 28, 2013 #14
    Thanks for links but I can't resolve that. It's not possible to calculate only 2 closed points of the center of gravity for show the trajectory and look of the angle ? Even, I don't know when the disk is on a precise part of circle I know all points. I don't know how to calculate the position of the center of gravity in the disk, can you explain this please ?

    NB: like the function of attraction can be other than quadratic, for example linear (in Algodoo it's possible to choose linear too), I can't imagine the trajectory of the center of gravity always perpendicular to the force from wall with all functions : linear, quadratic, cubic, etc. Even in a small part of circle the force from wall must be always perpendicular to the trajectory of the center of gravity. For a small part, all positions of disk are known, but trajectories of center of gravity can't be the same with all functions I think. For one position of the disk the force from wall is always the same (for all functions of attraction) but each function of attraction change the position of the center of gravity with different ways ("force"), each trajectory is different for each function.
     

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    Last edited: Dec 28, 2013
  16. Dec 28, 2013 #15

    mfb

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    If the software loses energy, it is a flaw in the software, or in your input.

    Energy is conserved, and you don't need the concept of a center of gravity for that. For each point (for each θ for example), you can calculate the potential energy, the kinetic energy is then just the difference between that and the (arbitrary, but constant) total energy. The type of potential does not matter at all for that.
     
  17. Dec 28, 2013 #16

    Dale

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    Then you should probably study a little more physics before attempting this problem. The first thing to do would be to solve this problem for a uniform gravitational field using this method. Once you have learned how to do a standard problem for a simpler case, then you can use a more complicated case.

    It is a messy formula. I don't know why you would want to calculate it, but it is the position, R, where ##\int_V (r-R)\times f(r) =0##.

    http://en.wikipedia.org/wiki/Center_of_mass#Center_of_gravity

    Sure, just put a different form for the potential energy in your Lagrangian.

    No, it doesn't. I don't know why you would think that. The force from the wall will be perpendicular to the trajectory of the center of mass, but not necessarily the center of gravity, particularly for non-uniform gravity potentials.
     
  18. Dec 28, 2013 #17
    for me it was the sum of each attraction force multiply (vector) by force from wall (N). For one force of attraction, N works in positive and for another force N works in negative, the sum is zero, so N don't works. For me it was logical to take center of gravity not center of mass. I hope you can understand what I think ?


    For study some points, it's possible to take a disk composed with 2 points (red and black in image). Fa and Fb are very differents in value.

    geometric example:

    angles are:

    black point: -32.4° for Fa, -46.15° for movement
    red point: -45.68° for Fb, -47.19° for movement

    the force from wall is 41.68 °

    Angle between force and wall is :

    black point: 87.83 °
    red point: 88.87 °

    With 2 points the disk turn, but with a lot of point around circle, the disk don't turn and for each point on the circle the angle is lower than 90°. It's possible to take smaller circle and at final if this work with all circles it works for a disk. In each case the force from wall is not perpendicalar to the movement.

    In this example, can you explain where is my error ?
     

    Attached Files:

    Last edited: Dec 28, 2013
  19. Dec 28, 2013 #18

    Dale

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    No, I don't understand why you would think that is logical. The center of gravity is relevant only for the gravitational force. It has no relevance whatsoever to the normal force.

    Regarding your drawing. It is entirely too cryptic for me to discern. You will have to troubleshoot it yourself. I have already told you how I would approach this problem, if you choose to ignore my advice and approach the problem in a more difficult way then you will undoubtedly run into many unnecessary problems.
     
    Last edited: Dec 28, 2013
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