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What exactly is centrifugal force 
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#1
Jan2813, 12:51 PM

P: 247

I think that there is no such thing as centrifugal force .
Am I right ? is this force fictitious ? 


#2
Jan2813, 01:09 PM

P: 5,462

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. 


#3
Jan2813, 01:11 PM

P: 247

What is the source of this force ?
For example, in gravitational circular motion, where does this force originate from ? 


#4
Jan2813, 01:14 PM

P: 5,462

What exactly is centrifugal force
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. 


#5
Jan2813, 01:31 PM

P: 49

If something as a source of centrifugal force exists, then I think it is the inertia.



#6
Jan2813, 01:44 PM

HW Helper
P: 7,126

http://en.wikipedia.org/wiki/Reactive_centrifugal_force http://en.wikipedia.org/wiki/Centrif...e_(fictitious) http://en.wikipedia.org/wiki/Rotating_frame 


#7
Jan2813, 02:20 PM

P: 4,049




#8
Jan2813, 02:25 PM

P: 247

Can you explain this notion ? 


#9
Jan2813, 02:40 PM

P: 5,462

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. 


#10
Jan2813, 02:57 PM

P: 247

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. 


#11
Jan2813, 03:01 PM

HW Helper
P: 7,126

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/Inertia...e_of_reference 


#12
Jan2813, 03:10 PM

P: 5

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 nonaccelerated reference frame. Hope that clears it up a bit 


#13
Jan2813, 03:40 PM

P: 5,462

Look at post#9 of this thread.
http://www.physicsforums.com/showthr...trifugal+force It puts what I said in post#2 here into equations. 


#14
Jan2813, 04:50 PM

P: 4,049

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 noninertial frames inertial forces were introduced. 


#15
Jan2813, 06:23 PM

P: 5,462

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%27Alembert%27s_principle Look towards the end of the much shorter page. 


#16
Jan2813, 08:53 PM

Sci Advisor
P: 2,108

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
Jan2813, 11:57 PM

P: 247

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 ? 


#18
Jan2913, 04:06 AM

P: 4,049

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


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