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The Myth of Fictitious Force

  1. Jul 22, 2010 #1
    It is standard pedagogy in our educational system to teach centrifugal force is a fictitious force, a force that arises within a non-inertial, rotating reference frame. High school students, for example, are brainwashed in believing this. Brainwashed because it perpetuates a false dogma about the nature of inertia. It is a myth and should be eliminated from the curriculum of our school system. Why do I say this? Hello, has anyone heard of Albert Einstein? Back in the last century he formulated the theory of special relativity. He made this profound observation: the laws of nature are the same for all inertial reference frames. He believed in a universe, where no matter what inertial frame you were in, the laws would be invariant. They would be the same for everyone. Pretty deep stuff. But then, years later, he took it to the next level, and proposed a theory of general relativity. In this theory he postulated that the laws of nature would be the same for any type of reference frame, inertial or non-inertial-- an even deeper insight. Thus, no matter what frame you measured phenomenona in, what you see is what you get. Labeling a force as "fictitious" is silly and meaningless. What does "fictitious" mean? What does real mean? Fictitious is a label that is used for convenience in analyzing the dynamics of bodies in non-inertial frames, the problem is, people start "really" believing they are fictitious.
    If you stand on a spinning stool that is spinning fast enough, your arms could be pulled out of your sockets. Hmmm, was this due to a fictitious force or a real force? Did you have a fictitious injury? Did you pay a fictitious doctor bill to have your arms popped back into your sockets? Fictitious is an arbitrary convenience placed by classical theorists before Einstein's time because frankly, they really did not have a clue what caused inertia. Einstein had a good definition of force. It went something like this: Any force that you can feel is a real force. It doesn't matter what type of frame you feel or observe the action of this force, inertial or non-inertial, the force is real. Labeling it as fictitious because you are in a rotating system is violating the principle of general relativity. You are implying that only forces observed in an inertial reference frame are real forces. You are therefore, implying that the laws of physics are only real when observed in inertial frames, and that inertial frames have a preferred status over non-inertial frames-- a blatant violation of the theory of general relativity. Look at it this way. From the theory of general relativity, using the geodesic equation, one can predict the path of a body in "free fall" in inertial space. Any departure of the body from this natural path would result in a manifestation of inertial force-- not a fictitious inertial force-- a force that arises out of a deviation from a geodesic path.
    Centrifugal forces are real no matter which frame you observe the forces from. This can be demonstrated by empirical means. For example, through the action of an inertial centrifugal force acting on a rotating body which rotates around an axis not through its center of mass, the inertial centrifugal force can do work on a system to increase the speed and kinetic energy of the center of mass of a system with respect to an inertial reference frame. This experiment has been done and the observational facts have confirmed this. Only a real force can increase the speed of the center of mass of a system with respect to an inertial frame.
     
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  3. Jul 22, 2010 #2
    An additional thought. The geodesic equation is expressed in covariant, tensor form. That is, it is frame independent. The geodesic path or deviation from the path is not dependent from which frame you analyze the dynamics of a body in its deviation. That means the manifestation of inertia is independent from which frame you observe the inertial force. General relativity, clearly, sends the concept of "fictitious" to a black hole, where it should remain, removed forever from our universe.
     
  4. Jul 22, 2010 #3

    K^2

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    Riiigh... So first of all, you really need to learn a bit more about accelerated reference frames. And centrifugal/centripetal forces. And about relativity.

    Fictitious forces are forces necessary to be added to the F=ma for it to work out correctly in your reference frame.

    Take an object. Hold it in front of you. Release. Observe object accelerating down. Newton said, "Gravity!" Einstein said, "Pants to that! That's a fictitious force!"

    Basically, Einstein's General Relativity doesn't say anything new about forces we already understood to be fictitious, like centrifugal force, but instead he said that if it looks like a fictitious force, and it walks like a fictitious force, it is a fictitious force. The only force that got affected was Gravity, of course, but a lot of funky math had to happen in between.

    Anyways, back to centrifugal force. Say you are sitting in a spinning chair and holding two heavy objects in your hands. Your friends decide to play a joke on you, and give that chair a good spin. Why is there a force acting on your shoulder sockets? Well, pick a reference frame.

    1) You chose the inertial frame of reference. One relative to people who you used to think of as your friends and who are playing this mean joke on you. Relative to them, the weights in your arms are not sitting still. They are going around in circles. Furthermore, they are not traveling in a straight line. And you know what Newton said about that. Where there is motion along a curve, there is a force. And so something must apply a force to keep the weights going in circles. That force is called the centripetal force, and it is not a fictitious one. It is a very real force, which if found lacking, would result in weights traveling in a straight line, taking your arms with you.

    Notice how there isn't a force that's trying to break your arms off? There is instead a force that's trying to keep your arms in place. That's something you might need to sit down and think about. If there are no forces acting between parts of your body, they'll be able to float away on their own. The forces are there trying to keep it all together. And depending on what your body is doing, there might not be enough force to do so.

    2) You chose a rotating frame. Because your friends are not your friends anymore, and who cares about their inertial frame. We have our own frame. One that's fixed to the chair. In that coordinate frame, the weights are not moving. Neither do your arms. So there really shouldn't be a need for a force in your shoulder socket trying to keep the arm from flying off. But there is. Aha! It must be balancing something. If you are applying a force on your arms, and they are not accelerating, something else must be applying a force on them. So there must be a centrifugal force in play.

    It is a fictitious force, because it is always perfectly proportional to gravitational mass of your arms. For any such force, there exists a coordinate system in which that force is exactly zero. That means the force in question is fictitious. And hence the gravity and centrifugal force are fictitious forces.

    If you'd like some references for further reading, I can look around.
     
  5. Jul 22, 2010 #4

    HallsofIvy

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    Do you really think it helps to use provocative words?

    Einstein wrote his paper on General Relativity the same year he wrote his first paper on Special Relativity (as well as papers on Brownian Motion and the photo-electric effect that also changed physics), 1905, not "years later".

    Do you have any evidence that people, other than yourself, believe that? You do understand, don't you, that words used in physics have meanings specific to physics that may not be the same as their meaning in general conversation. If you stand for an hour holding a 100 weight at exactly 5 feet above the ground how much "work" have you done?

    Essentially, then,what you are saying is that you do not understand the phrase "ficticious force".
     
    Last edited by a moderator: Jul 22, 2010
  6. Jul 22, 2010 #5

    Doc Al

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    Slight correction here: Einstein published his general theory of relativity in 1915, 10 years after his annus mirabilis.
     
  7. Jul 22, 2010 #6
    Lets take your example one step further. I am spinning around in a chair, counter-clockwise, and in each hand I have equal masses. Lets say my chair has wheels and it is sitting on a railway track, such that it can only move in a straight line, and assume negligible friction. The initial velocity of the center of mass of my system-- my body, the two masses, and the chair-- is zero. My "friends" are observing my spinning motion from above somewhere, laughing, having a good time. Lets say at some point in time, relative to the friends' frame, they see that I release just one of the masses in my hand.
    I release it when my arms are parallel to the x-axis with respect to an x-y coordinate system as viewed from the friends' frame. In the friends' frame they will see the mass fly off at a tangent in the negative y-direction. Now my friends are sophisticated, educated people, who know a few fundamental laws of physics. For example, they know that linear momentum must always be conserved. So they expect and observe that my body-chair-one-mass system begins to move in the positive y-direction. They are not surprised that this happens, and they further assert that the speed of the center of mass of my system must always be some value such that the momentum of the center of mass of my system is always equal and opposite to the momentum of the mass that is moving in the negative y-direction. They all agree to this, but after some thought they begin to question something. What caused my system to move in the positive y-direction? Was it the earth? One friend conjectures the earth through the rails somehow pushes me in the positive y-direction, but the other friends eventually convince him that this would be impossible and would violate Newtonian physics. This is because a free-body diagram of the reaction forces, acting on the chair from the rail, would reveal that these forces would always be orthogonal to the motion of the chair in the positive y-direction, and therefore, by the definition of work, the work of the rail on the chair would be zero. Another friend ventures friction is causing
    the motion of the chair. After a some cold stares from his other friends, he backs off, realizing the impossibility of that explanation. Friction is present, but it is a reaction to the rubbing of the wheels against the rails, it is not the cause of the motion of the chair. Finally, one friend ventures that an inertial force from the imbalance of my system is causing the motion. He states that an inertial "centrifugal" force, acting on the remaining mass that I hold in my hand, is "pulling" my system in the positive y-direction. Immediately, this friend is overwhelmed with verbal rebuttals from his other friends. "Centrifugal force is not a real force," they say, " it is a fictititous force!!" "It can't do work on the system because the force is not real! From the frame of our sucker friend, centrifugal force is pulling on the mass, but from our frame, such a force does not exist!" The discussion goes on, until they all agree to do an experiment. First, they determine what the center of mass of my system is when I am holding only one mass. They place a sticker on my arm, where this center of mass is. Then they get a high-speed video camera to record my motion. They convince me to do the experiment again on my chair, exactly as I did it before, and I do it because they promise to give me my favorite chocolate ice cream cone if I do. Sure enough, I get spun around and around until I release the one mass so that it flies off in the negative y-direction.
    At that point they turn on the camera to record the speed of the center of mass of my system, and they record me all the way up to the point where my arms are parallel to the y-axis. So, another words, they record my action for 90 degrees with respect to my x-axis. Afterwards, they play back the video and to the surprise of most of my friends, they find that the speed of the center of mass of my system increased with respect to their inertial frame! They are baffled! How can this be? Now, they know by Euler's first law this can only happen if an external force acted on my system, but they cannot think of what this external force was. The one friend states this proves it was inertial force that caused the increase in the speed of the center of mass, but the other friends cannot accept this, because the inertial force manifested with respect to their frame is not a real force they insist, therefore, it could not do work on the center of mass of the system to increase its speed or kinetic energy. But the truth is, I have performed such an experiment, and a video of the motion of the system clearly shows the speed of the center of mass of the system increases! Inertia must be viewed as a real force with respect to an inertial frame, capable of doing work on a system! Anyone interested in seeing the video, let me know, and I will e-mail the video to you.
     
  8. Jul 22, 2010 #7
    You're right, using the word "brainwashed" is a little too heavy. Maybe I should say the concept of ficticious force can confuse people into misunderstanding the nature of inertia. Please see my response to k^2. Often, the idea of "fictitious force" is used to dismiss any potential work capability of inertia.
    We are in some ways talking semantics. Richard Feynmann sometimes talked about the superficiality of languare, how it often does not get to the heart of a matter.
    No, holding an object steady above the ground against gravity does not require work as far as the specific definition of work used in physics. The work would be zero, just as work would be zero if the force was alway perpendicular to the motion of a body.
     
  9. Jul 22, 2010 #8
    One important addendum to the rotating chair experiment. The video reveals the speed of the center of mass of the system increases, but the momentum of the center of mass of the system with respect to the y-axis is still equal and opposite to the momentum of the first mass that is moving in the negative y-direction. The total momentum of the center of mass, has increased, but momentum with respect to the y-axis is the same. How can this be? Simple, the direction of the center of mass momentum vector has changed, "twisted" counter-clockwise, such that its magnitude has increased (thus, the observed increase in speed of the center of mass), but the y-component of the center of mass stays rock-solid the same value with respect to the friend's inertial reference frame.
     
  10. Jul 22, 2010 #9

    Cleonis

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    Well, the expression 'centrifugal force' is used for several things.

    An example:
    Take a human centrifuge (the ones that jetfighter pilots ride to get used to pulling G's.)
    Set up a stack of bathroom weighing scales. Let's say you have stack of 10 pieces, that stack will be about half a meter high. Let's say each scale weighs one kilogram in 1 G gravity. Let's say the centrifuges is pulling 2 G.
    Then the bottommost scale will read 18 kg; the force that the 9 scales above it are exerting upon the bottom scale. The higher up the stack, the lower the compression.
    Important: what the stack of scales registers is compression, and the fact that there is a gradient in the amount of compression.

    Likewise, when a human is riding a centrifuge the physical sensation is a sensation of compression (and the gradient in that compression). The brain automatically translates that sensation of compression to being subject to an outward directed force. (In fact the human is subject to an inward directed force. Inertia gives rise to the compression.)


    It is clear of course that for physics taking place the phenomenon Inertia is just as important as the Four Fundamental Forces of Nature.

    We have that the concept of 'Force' is defined by Newton's third law: We call something a force if and only if it is a mutual interaction between two objects, with each object trackable through time.
    This definition implies that inertia cannot be categorized as a force. It follows that Inertia must be regarded as belonging to a category of its own.

    The expression 'Fictitious force' is an amazingly ill choice of name. 'Fictitious force' refers to Inertia; as we know, Inertia is not a force, and it's not fictitious.

    I have seen physics textbooks in which the concept of inertia is not mentioned. Inertia is not discussed, and it's not in the index. I find that absolutely weird. That's like a genetics textbook in which DNA is not discussed anywhere.


    e2m2a,
    my hunch is that your anger comes from this weirdness.

    We have that the General Theory of Relativity unifies the description of gravitation and the description of inertia. GR is presented most clearly when Inertia is always referred to by it's own name: Inertia.
     
  11. Jul 22, 2010 #10
    Cleonis,

    You understand where I am coming from completely. Thank you. My anger or maybe I should say my frustration comes from this "play on words", this ill-defined fictitious force that can send many people onto confusing paths. Yes, Einstein did present a strong argument for the equivalence between gravitation and inertia, and I suspect he would consider "fictitious" an unnecssary adjective to stick on to inertial force, being that he envisoned the laws of nature being expressed in covariant form, where measurement with respect to any kind of reference system was not necessary or maybe without meaning. The term fictitious arises out of reference to different frames. I believe covariance implies laws can be expressed in forms that are independent of any reference system. Take away the arbitrary conventions of man, measurement systems, scales, reference systems, etc., the laws of nature will go on doing their thing even if no one refers to them with respect to a reference system. Of course, since we live among other humans, we are social creatures, we need to communicate with universally-accepted symbols that represent these laws. And these laws must always must be expressed in terms of the undefinables time, space, and charge.
    I agree with you that inertia is in a category by itself. Its non-Newtonian because where is the equal and opposite effect happening when inertia is manifested? Well...there are some theorists who give a strong argument that acceleration with respect to the vacuum energy causes a net Poynting vector effect on an accelerated object, opposite to the acceleration vector. That is, inertia arises by a "radiation reaction" with the vacuum, zero-point photons that comprise the vacuum energy. Thus, the reaction of the photons-- the decrease in frequency and energy and momentum-- would account for the "other force" required by Newton's third law.
    Whatever the cause of inertia, an experiment I performed clearly implies a field exists in space that can effect changes in the speed of the center of mass of a system when acceleration is present. The experiment showed that the speed increased. And the only effect acting at the time of the acceleration of the center of mass was an inertial effect. Although inertia is not a force in the traditional sense of the word, the experiment suggests that it can do the same thing that a traditional force can do-- work, by increasing the kinetic energy of the center of mass of a system. And yes, before the experiment was performed we measured the system to insure it was level, so that gravity would not play a part in the increase in the speed of the center of mass.
     
  12. Jul 22, 2010 #11
    Actually, concerning my comment about theorists who conjecture a link between inertia and the vacuum energy. They actually hypothesize that mass is consequence of this link as found in Newton's second law.
     
  13. Jul 22, 2010 #12
    Centrifugal force is called fictitious because it is fictitious. Draw a free body diagram in the inertial frame & there is no centrifugal force. If a car I'm in makes a sudden left turn & I bang into the right door panel with my shoulder, is my pain "fictitious"? Of course not. But the force I feel is not directed outward, i.e. centrifugal. The auto is accelerating inward, centripetally, & my momentum tends to carry me along a tangent.

    High school students are not brainwashed. The centrifugal force in my accelerated frame feels real to me. I think I'm being thrust "outwards", when in reality, the car is thrusting inwards. It's all relative. As long as we analyze the problem accounting for all criteria, there should be no confusion. Centrifugal is a "virtual" force. As long as we account for all facts, we should get the right answer.

    In college physics, we solved motion problems w/o centrifugal force & got the right answers. Can't argue with verifiably correct answers, can we?

    Claude
     
  14. Jul 22, 2010 #13
    I advise caution here. Virtual has an entirely different meaning in mechanics qv virtual work, as done by virtual forces.

    e2m2a
    You rely too much on 'believing' as compared to logical deduction (mathematical or otherwise) from axioms or postulates for my taste. Perhaps if you put forward some of the latter ?
     
  15. Jul 22, 2010 #14
    Claude, if you had a chance to read all of the posts on this thread and if you read about the rotating chair experiment, I mentioned that I have performed an experiment that is similar
    to the rotating chair experiment. In my experiment it is clearly observable that the speed of the center of mass of the chair-body-mass system increases with respect to an inertial frame. This experiment is surprisingly simple to do, maybe a little expensive to construct, depending on the materials you use, but regardless, if you did the same experiment with the same masses, etc., I gurantee you would replicate the same results. (This replication is a necessary requirement of good experimental science. This experiment could be performed by a high school student.) Now, this increase in speed can only happen if an external force acts on the system. (This follows from Euler's first law and the conservation of linear momentum.) The interesting twist to this experiment is that the only "effect", acting simultaneously during the acceleration of the center of mass, was an inertial effect. From the point of view of an inertial frame, this inertial effect is doing work on the center of mass because the speed of the center of mass increases. By the very definition of kinetic energy, this means the kinetic energy of the center of mass increases. This is not a trivial result. The experiment, not I-- I am only the messenger, the conveyor of the facts, so don't stone me--demonstrates that there is an increase in the kinetic energy of the center of mass of the system soley by the action of inertia with respect to an inertial frame. There must be some model of inertia, either within the context of Machian relativity or non-Machian relativity or within the frame-dragging effect demonstrated in the Gravity Probe B experiment or Dennis Sciama's formalism or some model that relates inertia to the vacuum energy that would explain the results of this experiment.
     
  16. Jul 22, 2010 #15
    studiot, I am not sure what you mean, but here goes a logical syllogistic declaration of what I have observed in the experiment. Major statement: The conservation of linear momentum states only an external force can cause the momentum of the center of mass of a system to increase. Minor statement: The empirical results of the experiment showed the momentum of the center of mass of the system increased. Conclusion: An external force acted on the center of mass of the system observed in the experiment.
     
  17. Jul 22, 2010 #16
    Moment of momentum?
     
  18. Jul 22, 2010 #17

    Doc Al

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    You need to find some friends with some physics knowledge. :smile:

    For one thing, you seem to be using 'centrifugal force' in some non-standard way. In standard physics usage, centrifugal force is a fictitious force that only appears when analyzing things from a rotating frame of reference. It plays no part in your story.

    As you rotate in your chair while holding onto the two weights, those weights are being centripetally accelerated. You must exert a real force on them to keep them going in a circle. And, of course, those weights also exert a force on you per Newton's 3rd law. Some folks call that force that the weights exert on you a reactive centrifugal force. But there's nothing fictitious about it.

    Before you let one of the weights go, there is no net force on you. Each weight is pulling you the same amount. Of course, the second you let one weight go, the other weight--which is moving--begins to drag you along for a ride. (No different than if someone threw the weight and you caught it.) No mystery. No 'fictitious' forces.
     
  19. Jul 22, 2010 #18
    Again I would urge caution here.

    The conservation principle actually states that the totality of momentum before an event equals the totality of that momentum after. It does not provide any information directly about how any changes occur during the event. In particular it does not mention or require forces of any description at all.

    Most conservation laws work like this.
     
    Last edited: Jul 22, 2010
  20. Jul 22, 2010 #19
    Others have analysed your description, here is my version, which is equivalent to Doc Al's.

    I think you are making the same mistake that some do when considering how a rocket works in space and trying to take a momentum balance, without considering the exhaust.

    When you are sat spinning on your chair, there is one (combined) object.

    Once you let go of a ball there are two, but the centre of mass of the system still includes the lost ball, just as the rocket sytem includes both the forward travelling rocket and the exhaust. And the momentum of the system includes that of the lost ball.
    You cannot suddenly exclude the lost ball from the system as you have done.
     
  21. Jul 22, 2010 #20
    I'll have to make some computations, but in the meantime, you seem to be suggesting that letting go of one weight resulted in a change of energy/momentum that suggests a force acting on the system. The net force exerted by the person in the chair has indeed changed, but it is the person who changed it.

    When holding both weights, the person is exerting a force vertically to counter the gravity, and horizontally to keep the weight from literally flying off on a tangent. When the person drops one weight, 2 forces formerly exerted are now gone. Hence the person in the chair altered the amount of force on the system.

    An ice skater spinning comes to mind. With her arms extended she develops angular momentum/kinetic energy. When she draws her arms inward, her angular speed increases dramatically, as well as her KE. The increase in KE came from the skater doing work moving her arms inward, exerting force through a distance. No "centrifugal force" acting here, just the skater changing her net force.

    I just gave a crude explanation, & with time I'd like to compute the system energy/momentum for the chair problem.

    Claude
     
  22. Jul 22, 2010 #21
    "Of course, the second you let one weight go, the other weight--which is moving--begins to drag you along for a ride. (No different than if someone threw the weight and you caught it.) No mystery. No 'fictitious' forces."

    Doc Al, you are on the right track of understanding what I am trying to explain. Now let me try to give a more exact description of what the chair experiment or my experiment is demonstrating. Cabraham, Studiot, this description should answer your objections and give you an understanding of what I am asserting. I am going to simplify the experiment into an idealized thought experiment. Instead of a person sitting on a chair holding two objects, lets just have a rod with two equal masses on each end of the rod. The rod is rotating counter-clockwise around an axis which is through the middle of the rod. Assume the rod's mass is negligible. The axis is attached to a second body which can slide along a near frictionless linear track. The track is aligned along the y-axis of an x-y coordinate system. The body that can slide is denoted as the slider. The axis of rotation is initally located at the origin of the x-y coordinate system. Now two other definitions. When I say the system, I mean the two objects, the rod, and the slider. When I say the sub-system, I mean the one object still attached to the rod, the rod, and the slider. Now, with respect to an inertial frame the initial linear momentum of this system is zero, therefore, the final momentum of the system hereafter must be equal to zero. This is required by the conservation of momentum. Now, at some point in time when the rod is parallel or coincident with the x-axis, the object on the left flies off at a tangent. It is not given a push, lets say the bond that was holding it to the end of the rod breaks. With respect to our laboratory frame, the object, which I will refer to as the flying object for now on, flies off in the negative y-direction. The momentum of the center of mass of the sub-system from this point on with respect to the y-axis must always be equal and opposite to the momentum of the flying object in the negative y-direction. The conservation of linear momentum requires this. But now lets examine what the sub-system is doing at this time. The object attached to the rod continues to rotate counter-clockwise, this means, the center of mass of the sub-system is starting to turn in the counter-clockwise direction. We break the momentum vector of the center of mass of the sub-system into its x and y components. Clearly, as the rotator-rod continues its counter-clockwise direction the x-compoment of the center of mass momentum vector of the subsystem increases in the negative x-direction and the y-component of the center of mass momentum vector decreases in the positive y-direction. Wait a minute. If the y-component of the center of mass momentum vector is decreasing, than momentum of the "system" (remember, the system includes the sub-system and the flying object) is no longer being conserved with respect to the y-axis. The y-component of the center of mass momentum vector of the sub-system is begining to decrease in magnitude, and therefore is not equal in magnitude to the magnitude of the y-component of the momentum vector of the flying object. Adding the two momentum vectors together will result in a non-zero sum. The conservation of linear momentum forbids this. How is this paradox resolved? Nature always insures that momentum is conserved with respect to a given axis. How nature accomplishes this is through the action of inertia in this case. The only way that the y-component of the momentum vector of the center of mass of the sub-system can remain constant in magnitude with respect to the y-axis, since the vector is "twisting" to the left, the total magnitude of the momentum vector of the center of mass of the subsystem must be increasing. This will ensure that even though the center of mass momentum vector of the sub-system is turning counter-clockwise that its y-component stays the same magnitude, equal and opposite to the magnitude of the momentum vector of the flying object. This is why in the experiment the video shows that the speed of the center of mass of the sub-system increases. Inertia is doing work on the subsystem, such that the total momentum and kinetic energy of the center of mass of the subsystem increases. This increase in the total momentum of the center of mass of the subsystem must happen in order for momentum of the system to be conserved with respect to the y-axis. If you doubt this, think of what happens when the object-rod is now parallel with the y-axis at 90 degrees with respect to the x-axis. At that instant if the second remaining object broke away, with respect to our inertial, laboratory frame, the y-component of the momentum of the second object plus the y-component of the momentum of the slider must be equal and opposite to the momentum of the flying object which is moving in the negative y-direction. The conservation of linear momentum requires this. It is inescapable and not subject to debate.
     
  23. Jul 22, 2010 #22

    K^2

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    e2m2a, you really just need to learn some mechanics. That's all there is to it. Pick up a descent graduate level text and start reading.
     
  24. Jul 22, 2010 #23
    If you have digested what I have just posted, now consider this. At the point where the rod has rotated 90 degrees, where it is parallel to the y-axis, we know the total momentum of the sub-system with respect to the y-axis must be equal to the total momentum the subsystem had with respect to the y-axis right after the point the flying object broke away. If we only included motion with respect to the y-axis, the kinetic energy of the center of mass of the subsystem will not have changed. The kinetic energy of the center of mass of the subsystem at the 90 degree point will be the same as the kinetic energy of the center of mass of the subsystem when the rod was parallel with the x-axis. But in truth, we now have motion of the remaining second object still attached to the rod in the negative x-direction at the 90 degrees point. When we factor this component in, the total kinetic energy of the center of mass of the subsystem must be greater than the total kinetic energy of the center of mass of the subsystem had when the rod was parallel with the x-axis. Again, this is why the video reveals the speed of the center of mass of the subsystem increases!
     
  25. Jul 22, 2010 #24
    k^2, please give me your explanation why the video shows the speed of the center of mass of the sub-system increases with respect to a laboratory frame. I am interested in your explanation.
     
  26. Jul 23, 2010 #25

    vanesch

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    Religious wars :smile:

    You are entirely right that fictitious forces are the result of inertia. But why is there a very good reason to call them fictitious ? What *are* fictitious forces ?

    Fictitious forces are the forces you need to pretend there to be if you are working in a non-inertial frame of reference and you want to pretend it to be an inertial reference frame.

    That is, you are working in a non-inertial reference frame, and you want to use Newton's laws AS IF they were applied in an inertial reference frame. If you do that bluntly, you make a mistake. Newton's law: F = m a is NOT valid in a non-inertial frame.
    That is to say, if you take on the left side all INTERACTION forces, and on the right, the acceleration as calculated in your (non-inertial) reference frame, then the equation simply isn't valid.

    But there's a trick. It is sufficient to ADD some terms to the left hand side of this equation, and then the equation DOES work again.

    To keep track of things, let us consider first an inertial frame of reference, Oxyz, and let us look at a specific point particle.

    In the system Oxyz, being an inertial frame, Newton's law is valid. What's newton's law ?

    We calculate the acceleration of our point particle in the frame Oxyz (so we do [itex] \frac{d^2 x}{dt^2},\frac{d^2 y}{dt^2},\frac{d^2z}{dt^2} [/itex] ), which we call a, and we consider all INTERACTION forces acting from the outside on our particle: gravity, electromagnetic forces, friction, material pressure, ...

    Their sum is F_int.

    Newton tells us that F_int = m a

    In the special case of a free particle, F_int = 0, so m a = 0, and hence a = 0. The velocity vector, calculated in Oxyz, is a constant.

    Now we go to our non-inertial frame (say, a rotating frame)O'x'y'z'. We do the same exercise, and this time we have to calculate the acceleration in the frame O'x'y'z', so we do [itex] \frac{d^2 x'}{dt^2},\frac{d^2 y'}{dt^2},\frac{d^2z'}{dt^2} [/itex], call it a'.

    Note that a' is NOT the same vector as a. Not only its components are different, it is a different vector.

    We calculate again the interaction forces, which are the same. F_int is the same vector, although it might have different components.

    Well, F_int = m a' is NOT valid. Too bad.

    BUT there's a trick. It can be shown that there is a term you can add to the left hand side of this equation, which is independent of any interaction, which makes the equation work again. In fact, the dirty secret is that there is a term on the RIGHT HAND SIDE which gives us the transformation of a into a' (as a vector)

    a = a' + A where A only depends on the non-inertial frame O'x'y'z' and not on any specific interaction.

    So from Newton's valid equation in Oxyz, F_int = m a, we can write:

    F_int = m (a' + A) and hence F_int = m a' + m A.

    Bringing the right hand term to the left, we find:

    F_int - m A = m a'.

    Now let us CALL F_fic = - m A

    We then have:

    F_int + F_fic = m a'

    So this looks a lot like our non-valid equation of Newton in the frame O'x'y'z'
    F_int = m a' (which is false) There's just a term on the left missing.

    So we see the trick now: it is sufficient to ADD A TERM to the left, and Newton's equation becomes valid. We call this added term a fictitious force, and it only depends on the non-inertial frame.

    So we can apply Newton's equation in a non-inertial frame, and "pretend" it to be an inertial frame, on the condition that we add some forces to the interaction forces. These forces are called (whether you like it or not) fictitious forces.

    But why do we call them fictitious ? Simply because they do not result from any physical interaction.

    We come back to our case of a free particle. Remember that a free particle has no interaction forces working on it. F_int = 0. In an inertial frame of reference, Newton's equation reads then m a = 0, or a = 0. The velocity vector is constant, and (in an inertial frame of reference), its path is a straight line.

    Consider that free particle now in our non-inertial frame O'x'y'z'

    In that frame, a' of our particle is NOT 0. Its path is a complicated spiraling motion. We can calculate that path using Newton's (modified) equation:

    F_fic = m a.

    It is as if funny forces pull on the particle to make it describe such a strange spiraling path, while actually nothing is "pulling" on the particle. It is a free particle.

    Very special example: a particle AT REST in an inertial frame, and our non-inertial frame doing a pure rotation.

    In the non-inertial frame, the particle is describing a circle around the origin. But to have a particle describe a circle and pretend to apply Newton's law as if it were happening in an inertial frame, there has to be a force "pulling" towards the center of the frame. So if you want to pretend your rotating frame to be an inertial frame, you have to pretend also that there is a force pulling towards the center of the frame on the particle. You will then find that a free particle will describe a perfect circle if it had the right initial condition (which corresponds to a particle at rest in the true inertial frame).

    So you have to pretend there to be a force on a particle that's actually at rest, in order to find the correct motion in the non-inertial frame (pretending it to be inertial and hence applying Newton's law)

    That's why one calls this term, which acts as forces fictitious forces. They represent inertial motion in a non-inertial frame, and they behave as if they pull on a free particle (if we understand by "pulling": applying Newton's law as if it were in an inertial frame)

    EDIT: btw, you might be surprised in my example that the fictitious force is pulling INWARD on our particle (externally at rest). You might be used to having a centrifugal fictitious force, away from the center.

    The reason here is that there is not only the centrifugal force, but also the coriolis force, and in this particular example, the coriolis force is towards the center and twice as large as the centrifugal term, so the term flips.

    If you want to see this in detail: (see wiki for fictitious forces)

    centrifugal force: - m omega x (omega x X)
    coriolis force: - 2 m omega x V

    Now, if X is the radius R, and let us pretend that at t=0 the particle is on the positive X axis and omega is along the positive Z axis:
    omega x X points to the positive Y axis, omega x (omega x X) points towards the negative X axis, and there is a minus sign, so the centrifugal force points towards the positive X axis with magnitude (m omega^2 R) (away from the center).

    Now, V is in the direction of the negative Y axis with magnitude (R omega), because it is a particle at rest, as seen from a rotating frame.
    omega x V is hence along the positive X-axis, and there's a minus sign, so the coriolis force points towards the NEGATIVE X axis (so inward), with magnitude (2 m omega^2 R).

    Together, we have (centrifugal) m omega^2 R pointing to the positive X axis, and (coriolis) 2 m omega^2 R towards the negative X axis, so the result is m omega^2 R pointing towards the negative X axis, inward.
     
    Last edited: Jul 23, 2010
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