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What exactly is centrifugal force 
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#19
Jan2913, 05:48 AM

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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 centrefleeing 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
Jan2913, 06:20 AM

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#21
Jan2913, 06:53 AM

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#22
Jan2913, 07:05 AM

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#23
Jan2913, 07:07 AM

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To me, "fictitious forces" are very ugly, because they spoil Newton's laws of motion. In a rotating coordinate system, there are "centrifugal forces" acting on absolutely every object in the universe. Where's the equal and opposite forces? There are none.
So some people would take this to mean "Newton's laws only apply in an inertial frame". I don't like that conclusion. If you view them as vector equations, then they apply in all circumstances, not just inertial frames. If [itex]\stackrel{\rightarrow}{U}[/itex] is the velocity vector, you write [itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex] this equation is always valid, regardless of what coordinate system you are using, and the force on an object always has an equal and opposite force. What changes when you go to a noninertial or curvilinear coordinate system is not Newton's laws, but the idea that you can compute the time derivative of a vector by only taking the derivative of its components. That is, it is not the case that [itex](\frac{\stackrel{\rightarrow}{dU}}{dt})^j = \frac{ dU^j}{dt}[/itex] 


#24
Jan2913, 07:19 AM

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#25
Jan2913, 07:34 AM

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[itex]\frac{\stackrel{\rightarrow}{dU}}{dt} = 0[/itex] It's always the case that in presence of a (real, nonfictitious) force [itex]\stackrel{\rightarrow}{F}[/itex], the velocity satisfies [itex]m \frac{\stackrel{\rightarrow}{dU}}{dt} = \stackrel{\rightarrow}{F}[/itex] It's always the case that if one object exerts a (real, nonfictitious) force [itex]\stackrel{\rightarrow}{F}[/itex] on another, then the second exerts a force [itex]\stackrel{\rightarrow}{F}[/itex] on the first. Newton's laws, as vector equations are not changed at all by using noninertial or curvilinear coordinates. Fictitious forces are not needed to extend Newton's laws to noninertial frames. What's needed is a more sophisticated notion of what [itex]\frac{\stackrel{\rightarrow}{dU}}{dt}[/itex] means when the basis vectors themselves are nonconstant. 


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Jan2913, 07:36 AM

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#27
Jan2913, 07:55 AM

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With curvilinear coordinates, the basis vectors become functions of position. With noninertial coordinates, the basis vectors become functions of time. Curvilinear coordinates can be handled with the usual Newton's laws of motion, viewed as vector equations. Noninertial coordinates can also be handled with the usual Newton's laws of motion, but only if you adopt a 4D spacetime view. 


#28
Jan2913, 08:01 AM

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#29
Jan2913, 08:04 AM

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hms.tech
My advice is to stick with one the two methods I outlined for you, either will always get you the right answer, D'Alembert's often leads to rather less computational effort. I'm sure you can see that a by product of introducing reference frames is to cause experts to squabble amongst themselves as a result of a great increase in complexity. 


#30
Jan2913, 08:28 AM

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AM 


#31
Jan2913, 08:34 AM

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Words like "fictitious force" and "particle" should have a warning sign on them when they're used for public consumption. If you can feel the effect then it's not really fictitious so a different word should really be used to describe it. Likewise, calling a photon a particle is just asking for the sort of misinterpretations we read every day. A term like quasiforce or quasiparticle would at least ring warning bells for people who might want to rush off and think in terms of conventional meanings.



#32
Jan2913, 08:41 AM

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#33
Jan2913, 08:46 AM

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Centrifugal force (from Latin centrum, meaning "center", and fugere, meaning "to flee") is the apparent outward force that draws a rotating body away from the center of rotation. There is nothing about the "centrifugal reaction force" that causes anything to flee from the centre. Nothing. The only thing that causes a rotating object and rotating rope to "flee" the centre is inertia, not a force. AM 


#34
Jan2913, 08:48 AM

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#35
Jan2913, 09:27 AM

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