What exactly is centrifugal force

  1. I think that there is no such thing as centrifugal force .

    Am I right ? is this force fictitious ?
     
  2. jcsd
  3. For some systems that are not in equilibrium ie moving, D'Alembert introduced a method of applying a fictitious force to bring the system to equilibrium (rest). The equations of equilibrium could then be employed to solve the system. These equations then included the fictitious force.

    Centrifugal force is just such a force applied to objects executing circular motion.
     
    Last edited: Jan 28, 2013
  4. What is the source of this force ?

    For example, in gravitational circular motion, where does this force originate from ?
     
  5. It doesn't exist.
    That is why it is fictitious.

    However we can do the maths as though it did exist.
    It is a mathematical trick.

    The modern alternative is to use Newton's Laws of Motion, which I think came after D'Alembert, but I will check that.
     
  6. If something as a source of centrifugal force exists, then I think it is the inertia.
     
  7. rcgldr

    rcgldr 7,451
    Homework Helper

    Depends on how you're using the term centrifucal force. There's the fictitious centrifugal force in a rotating frame. A less common (for physicists) usage is "reactive centrifugal force", which is part of a pair of Newton third law forces: say a string exerts a centripetal force on an object, then the object exerts a reactive outwards force on the string. Note that this "reactive centrifugal force" is not exerted on the object itself, but on the string. In a rotating frame, the fictitious forces, centripetal and coriolis (and euler if there is angular acceleration of the frame) all appear to act on the object. Wiki articles:

    http://en.wikipedia.org/wiki/Reactive_centrifugal_force

    http://en.wikipedia.org/wiki/Centrifugal_force_(fictitious)

    http://en.wikipedia.org/wiki/Rotating_frame
     
  8. A.T.

    A.T. 5,597
    Gold Member

    All forces are just mathematical models invented by humans.
    "Fictitious force" is just an unfortunate name choice. It gives people the wrong idea that some forces are more "real" than others. But that is just philosophy, irrelevant to physics. That is why I prefer the terms "inertial forces" and "interaction forces".
     
  9. Yes, I read this word "inertial" a lot on the Wikipedia link provided by rcgldr.

    Can you explain this notion ?
     
  10. hms.tech

    Before you get confusing explanations as to the different ways of regards motion under a central force or circular motion perhaps you would tell us what your course says ie which approach they take. Your exams will be based on this.
     
  11. Centripetal and centrifugal forces

    I couldn't find the term centrifugal force in any of my physics books, hence I checked out PF. Can you provide some links (non Wikipedia) about centrifugal force .

    Here is what I understand about this force :
    I think this term was self created by humans for their own better understanding of the "motion" of objects in circular orbit.
    We, the modern physicists would call it nothing but a consequence of the fact that any object undergoing circular motion tends to move in a straight line (tangential to its path) . So it can be thought of as "inertia" , but I would stick to my claim that this force is non existent.
     

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    Last edited: Jan 28, 2013
  12. rcgldr

    rcgldr 7,451
    Homework Helper

    An "inertial" frame of reference means the frame of reference is not accelerating (no linear acceleration and no rotation). Newton's laws of physics apply in an inertial frame without any modification.

    If the frame was accelerating, such as a rocket in space, then to an observer inside the rocket, there would be an apparent force on any object within the spacecraft, similar to gravity on earth.

    The wiki article on this goes into more detail:

    http://en.wikipedia.org/wiki/Inertial_frame_of_reference

    http://www.answers.com/topic/centrifugal-force

    In your diagram of the car in a turn, there is a Newton third law pair of forces, an inwards centripetal force exerted by the road onto the car, and an equal in magnitude but opposing outwards reaction force that the car applies to the road. That outwards force is real, the issue for some people is using the term "reactive centrifugal force" to describe that outwards force that the car exerts to the road.
     
    Last edited: Jan 28, 2013
  13. A more simple example might serve to make this clear:

    consider a train that is accelerating with acceleration a. You are on this train and you drop a ball. The only force due to a physical interaction acting on the ball is that of gravity which is due to the ball's interaction with the Earth.

    However, you see the ball accelerating away from you with the same magnitude of acceleration as that of the acceleration of the train. Thus if we consider the acceleration of your reference frame (keep in mind that since you're measuring the motion of the ball with your position as the origin, you are at rest in your reference frame) as the source of a force on the ball, it applies a force of magnitude ma, where m is the mass of the ball. This force isn't due to any physical interaction that you can observe, that's why we call it fictitious. I agree that calling it fictitious is confusing and that inertial is less confusing and also an apt name in that the magnitude of the force is proportional to the mass of the object

    A similar thing happens when you're in a rotating reference frame (for instance say someone does this while on a merry go round) except that now this fictitious force is directed outwards from the center of the merry go round. This type of fictious force due to being on an accelerated (or noninertial reference frame) is called centrifugal.

    A centripetal force is the net force in the radial direction due to real forces and is generally talked about in an inertial or non-accelerated reference frame. Hope that clears it up a bit
     
  14. A.T.

    A.T. 5,597
    Gold Member

    http://en.wikipedia.org/wiki/Fictitious_force

    In short: Newton based his laws of motion for inertial reference frames on interaction forces. To generalize some of the laws so they apply to non-inertial frames inertial forces were introduced.
     
  15. Gosh I'm glad I didn't meet that wiki as my first introduction to centrifugal force.

    It's one saving grace is that at the very bottom it refers to the D'alembert's principle of inertia wiki.

    That is much more digestible.

    http://en.wikipedia.org/wiki/D'Alembert's_principle

    Look towards the end of the much shorter page.
     
  16. stevendaryl

    stevendaryl 2,929
    Science Advisor

    Here's a mathematical way to think about fictitious forces.

    Suppose you have a velocity vector [itex]\stackrel{\rightarrow}{U}[/itex] which is constant (no forces acting on it). That means

    [itex] \dfrac{d}{dt}\stackrel{\rightarrow}{U} = 0 [/itex]

    Now, this is a vector equation. It's common for people to denote vectors by ordered pairs:
    [itex]\stackrel{\rightarrow}{U} = (U^x, U^y)[/itex], where [itex]U^x = \dfrac{dx}{dt}[/itex] and [itex]U^y = \dfrac{dy}{dt}[/itex].

    But that's actually a BAD notation, because what it really means to write a vector as components is that we've chosen certain vectors called basis vectors, and we've expressed [itex]\stackrel{\rightarrow}{U}[/itex] as a combination of those basis vectors. So a better way to write it is

    [itex]\stackrel{\rightarrow}{U} = U^x\ \hat{e_x} + U^y\ \hat{e_y}[/itex]

    When you write it this way, you can see that
    [itex] \dfrac{d}{dt}\stackrel{\rightarrow}{U} = \dfrac{dU^x}{dt} \hat{e_x} + U^x \dfrac{d \hat{e_x}}{dt} + \dfrac{dU^y}{dt} \hat{e_y} + U^y \dfrac{d \hat{e_y}}{dt}[/itex]

    Of course, the most important fact about Cartesian coordinates is that the basis vectors are constant. So the terms [itex]\dfrac{d\hat{e_x}}{dt}[/itex] and [itex]\dfrac{d\hat{e_y}}{dt}[/itex] are both zero.

    But if you switch to polar coordinates [itex]r,\theta[/itex], it's no longer the case that the basis vectors are constant. So you get additional terms having to do with the rate of change of the basis vectors. Those additional terms are the "fictitious forces".

    I'm going to spare the details, but it turns out to be, in polar coordinates:

    [itex] \dfrac{d}{dt}\stackrel{\rightarrow}{U} = (\dfrac{dU^r}{dt} - r (U^\theta)^2) \hat{e_r} + (\dfrac{dU^\theta}{dt} + \dfrac{2}{r} U^r U^\theta) \hat{e_\theta}[/itex]

    where [itex]U^r = \dfrac{dr}{dt}[/itex] and [itex]U^\theta = \dfrac{d \theta}{dt}[/itex].

    So, if there are no forces acting on an object, then [itex]\stackrel{\rightarrow}{U}[/itex] will be constant, but that doesn't mean the components will be constant. Instead, the components will obey:

    [itex]\dfrac{dU^r}{dt} = r (U^\theta)^2[/itex]
    [itex]\dfrac{dU^\theta}{dt} = - \dfrac{2}{r} U^r U^\theta[/itex]

    So it seems that there are "fictitious" forces acting on the object, causing the components of the velocity to change.

    Nonconstant basis vectors give rise to nonconstant components of velocity, even in the absence of any real forces.
     
  17. If you think about it, the reactive force is just the force exerted by the car on to the road, but this does not relate (even remotely) to the diagram .
    The diagram shows that there "seems" to be a force acting on the objects which tend to move outward (in the opposite direction of centripetal force). So the force you talked about is the force exerted by the car on the road and not the objects.

    Think about this :

    You are sitting on the backseat of a car, and this car is turning around a corner at high speed. You would have noticed yourself moving outwards (right ?) until you reach the door of the car which then would provide the appropriate centripetal force to your body.
    Notice that this situation is based on the fact that friction could not provide the centripetal force. Notice also that at all times you would be "FEELING" some outward push , (again , which in my opinion is just inertia but not centrifugal force)

    Am I right in the above mentioned aspect ?
     
    Last edited: Jan 28, 2013
  18. A.T.

    A.T. 5,597
    Gold Member

    You switch frames back and forth and confuse the issue. Stick to an inertial frame, and there are no inertial forces. Stick to a non-inertial frame, and you always have inertial forces which act just like interaction forces (except Newtons 3rd).

    Both is correct and leads to the same quantitative predictions, which is all that matters in physics. Musings about "realness of forces" is philosophy.
     
  19. Andrew Mason

    Andrew Mason 6,846
    Science Advisor
    Homework Helper

    The centripetal force produces centripetal acceleration. The force that causes centripetal acceleration produces mechanical tensions (ie chains of electromagnetic forces at the atomic level). These mechanical tensions operate radially. The net sum of these tensions is the centripetal force that produces the centripetal acceleration.

    Somebody had the idea that we should call the reaction to a centripetal force that is the result of the application mechanical forces a "centrifugal reaction force". As far as I can tell, this is a concept that serves no purpose and simply causes confusion. It can produce no acceleration so it is hardly a centre-fleeing force as the term "centrifugal" would imply. It is a mechanical tension so it does not exist where the centripetal force is applied by a force operating at a distance, such as gravity or an electric field.

    Calling it a centripetal reaction force also confuses the physics. The true reaction to a centripetal force is another centripetal force. One does not appreciate this in the situation of a tethered ball rotating about a pole fixed to the earth but that is only because we do not take into account the movement of the earth. In the case of two balls tethered to each other by a rope and rotating in space about their centre of mass, the rope has a tension. Each ball pulls on the rope and, through the rope, pull on each other ie toward the centre. The ball pulls on the rope but produces no acceleration of the rope radially outward. Rather this pull is applied through the rope, via tensions, to accelerate the balls and rope toward the centre.

    AM
     
  20. A.T.

    A.T. 5,597
    Gold Member

    Because you are considering only the most trivial cases, like:
    In the other thread we gave you plenty examples where a reactive centrifugal force exists.

    The term "centrifugal" has nothing to do with producing acceleration. It simply means that the force points away from the center of rotation. Acceleration is a matter of net force, not just one centrifugal force.

    Sometimes it is, and sometimes it is not.
     
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