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Reactive centifugal force.

  1. Apr 26, 2009 #1
    In Goldstein's 'Classical Mechanics' second edition at the top of page 30, the author directs the reader to page 102 of Osgood's 'Mechanics'. In the edition of Osgood that I looked at (I forgot to note which edition) on that page is the continuation of a discussion that begins on page 101 and concerns the misuse of the term 'centrifugal force' to describe circular motion. Osgood makes an impassioned argument (I think it was the passion that impressed Goldstein) but one that I think is incorrect. Please read the following carefully as there are two incorrect arguments here. There is the incorrect argument of the crank. Then there is the incorrect argument that Osgood makes against the crank.

    Crank: I agree that the particle on a string revolving around the central peg is subject to centripetal force, not centrifugal. However, I speak of the reactive force against the central peg. That force is centrifugal is it not?

    Osgood: There is a reactive force against the central peg, but it is not exerted by the particle.

    I don't understand what Osgood is driving at. The source of the reactive force is not the crank's issue, the mere existence of the force is. Here is a wiki article that, in my opinion, makes the same argument as the crank.

    http://en.wikipedia.org/wiki/Reactive_centrifugal_force" [Broken]

    I don't think the crank is correct, I don't think Osgood is correct, and I don't think the wiki page is correct. I think that the reactive force is centripetal. It points from the center of the peg to the center of the revolution of the revolving particle. Consider, for instance, the motion of a binary star system. One star is revolving about the other under the influence of a (somewhat) centripetal force. The reactive force on the other star is also centripetal and both stars revolve around a (somewhat) central point. Am I wrong?

    Edit: I googled osgood centrifugal and the first hit brought me to page 105 of Osgood's book and allowed me to page back to pages 101 and 102.
     
    Last edited by a moderator: May 4, 2017
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  3. Apr 26, 2009 #2

    rcgldr

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    "Reactive centrifugal force" would be just like it reads, the outward reactive force to a centripetal force (Newtons 3rd law force pairs). In the case of the string and whirling ball, the string exerts a centripetal force on the ball, and the ball exerts a "reactive centrifugal force on the string". Tension in the string results in a Newton 3rd pair of outwards and inwards forces at the peg, but I'm not sure if I would call these centripetal or centrifugal.

    Most physicists only consider "centrifugal force" to be the apparent force as observed from a rotating frame of reference. I don't have a problem with the usage of "reactive centrifugal force", since centripetal force is just a special name given to a force perpendicular to the direction of travel, similar to "lift" in aerodynamics, and "reactive centrifugal force" would just the other half of the Newton 3rd law pair.
     
  4. Apr 26, 2009 #3
    I don't agree that there is a 'Newton pair' at the center. I think there is just a single centripetal force causing the center of the peg to revolve around the center of rotation of the particle. Just as there is no pair of forces at the particle, just the centripetal force. I ask you once again to consider the behavior of a binary star system.
     
  5. Apr 26, 2009 #4

    D H

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    The problem with that view is that it is wrong. Suppose you are flying a spacecraft and following a circular path. Your spacecraft is not exerting a force on the center of the path, nor is the center of the path exerting a force on your spacecraft. There is nothing at the center of the path with which your craft interacts.

    The same applies to Osgood's example of a mass attached to a string that in turn is attached to a peg. The mass does not exert a force on the peg anymore than the peg exerts a force on the mass. The peg exerts a force on the end of the string attached to the peg, and the string exerts an equal-but-opposite force on the peg. The particles in the string exert equal-but-opposite forces on each other, eventually transmitting the force to the end of the string attached to the mass. Finally, this end of the string and the mass exert equal-but-opposite forces on each other.

    As Osgood mentioned, the explanation results from a confusion of ideas, kinematics versus dynamics. The acceleration required to follow a curved path is a kinematic concept. Centripetal force is simply this kinematic acceleration times mass. This is a derived force. The mistake is assuming this force comes from the center. This is not the case, as the spacecraft and mass on a string examples show. Centripetal force and central force are distinct concepts.
     
  6. Apr 26, 2009 #5

    rcgldr

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    If the spacecraft is orbiting, the there are equal and opposing forces due to gravity between spacecraft and the object the space craft orbits. I'm not sure I'd call these centripetal or centrifugal. If the environment is free of gravity and the space craft is using rocket power to follow a circular path, then the thrust generates an inwards force on the spacecraft, which might be called centripetal force, while I'm not sure what to call the outward force on the exhaust.

    For a more classic example, if a puck is sliding around a frictionless cylinder, then the walls of the cylinder would be exerting a "centripetal force" on the puck, and the puck would be exerting a "reactive centrifugal force" on the walls of the cylinder. If you're the oustide passenger in a twirling ride you feel a compression due to the centripetal force from the side of the compartment and the "reactive centrifugal force" from the passenger on inside of the compartment, although in this case you're in a rotating frame of reference, an outside observer could see that you were getting squished between the wall's of the V8 powered (yes some used automotive engines) "scrambler" and the person on the inside.
     
    Last edited: Apr 26, 2009
  7. Apr 26, 2009 #6
    our teacher said that centrifugal force is not a force but a direction is that true?
     
  8. Apr 26, 2009 #7

    D H

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    Consider a pair of stars of nearly equal mass orbiting their common center of mass. This alone makes a complete mess of the concept of "reactive centrifugal force." It is a bad idea.

    That is exactly the situation of which I was thinking, and I should have been clearer in that regard. This really makes a mess of the concept of "reactive centrifugal force." There is absolutely nothing in the center of the path with which the vehicle interacts. Reactive centrifugal force is a very bad idea.
     
  9. Apr 27, 2009 #8

    rcgldr

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    Center of the path interaction isn't required. There's nothing in the center of the path of a puck sliding on the wall of a frictionless cylinder. Even in the spaceship case, Newtons 3rd law about forces only existing in equal an opposing pairs still applies, an inwards force on the spaceship, an outwards force on the spent fuel. I understand many don't use the term centrifugal to refer to the outwards force, but that force still exists regardless of the terminology used to describe it. In the case of the space ship, it would seem that the "real" force is the outwards force from the rocket engine to the exhaust, and the reaction force is the inwards force from the the exhaust to the rocket engine.

    I don't think anyone is disputing Newton's 3rd law, just the usage of the term "centrifugal force" to describe a reaction force. Since linear reactions are often described as accelerations or decelerations, to imply a direction of the reactions, I don't have a problem with angular reactions being described as centripetal (inwards) or centrifugal (outwards).
     
  10. Apr 27, 2009 #9

    D H

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    The problem is that the forces are not always outward. Consider a pair of binary stars orbiting each other. There are two forces, both inward (toward the center of mass).

    I'm not even all that fond of the term "centripetal force", let alone "reactive centrifugal force". What makes a curved path so special that the a new name is invented for the net force needed to make an object follow said curved path? I like kinematic force a lot better, because it
    • Indicates that the force is derived from kinematics rather than from dynamics.
    • Eliminates the confusion between centripetal force and central force.
    • Generalizes to linear acceleration.

    Newton's third law covers individual dynamic forces. Suppose multiple dynamic forces act on a body. The net force acting on a body does not have a third law counterpart. It has a bunch of third law counterparts, which may act on different objects.

    Example: A spacecraft can orbit an asteroid at some distance r from the asteroid just by gravity alone. Suppose the spacecraft instead uses its thrusters to keep this distance r from the asteroid such that the orbital period is reduced to 1/4 that of the gravitationally induced period at this distance. Dynamically, the thrusters and gravity are equally contributing to the net force. While there are third-law counterparts to the thruster force and to the gravitational force, there is not a single third law counterpart to the net force. As the kinematically-derived centripetal force must be equal to this net force, the centripetal force has no third law counterpart.
     
  11. Apr 27, 2009 #10

    rcgldr

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    True, but those forces are still equal and opposing (opposite directions). From each star's frame of reference, the "other" stars gravity is exerting an "inwards" on the "reference", and the "reference" stars gravity is exerting an "outwards" force on the "other" star. Neither of these are reaction forces though. In this case, all inertia is doing is preventing the orbital paths from collapsing over time by limiiting the rate of acceleration in response to force. However both the force and the inertia are relative to the amount of mass involved, making gravity a special case. In this example of binary star system, no "reactive" forces are involved, so there can't be "reactive centrifugal force" either. "Reactive centrifugal force" would be better applied to the puck sliding on the interior wall of a frictionless cylinder, or a weight being twirled on a string.
     
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