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Centripetal Force vs Centrifugal Force 
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#1
Oct1812, 10:22 AM

P: 3

Hi, thank you for reading this topic.
This is something that really bothers me...First of all let me say that I know what the terms centripetal and centrifugal stand for. What's the problem then? The problem is that I've always read that the centripetal force is an actual force and that the centrifugal force is a fictitious one. Why does this bother me? Let's say that a satellite orbits around a planet. Why doesn't this satellite fall towards the planet? You would say that it doesn't fall because of the centripetal/centrifugal force. ( I don't know which of the two terms I should use...That's why I am here...) Now, if I say that the satellite doesn't fall because of the centripetal force I think I am saying something that doesn't make sense. Since the centripetal force is directed trowrds the center, the satellite should fall with greater speed towards the planet, since both centripetal and gravitational acceleration are directed towards the center. If I say that the satellite doesn't fall because of the centrifugal force everything would look fine to me. Except for one thing. This force would be real and not fictitious. To make things even more complex I think that the centripetal force is the force that actually doesn't exist and not the centrifugal one. Why? Because when there's something spinning it always moves away from the center. I can think about a lot of examples where there is the so called centrifugal force, but I can't think of anything when we talk about the centripetal force. To me this centripetal force moves always away from the center...which is a paradox of course. This thing is really counterintuitive to me, much more counterintuitive that the theory of relativity, which actually seems pretty logical once you think about it. I am sure I am wrong about all this. That's why I need your help. Thanks! 


#2
Oct1812, 10:33 AM

P: 834

Objects want to go in straight lines. It takes a force to make an object deviate from a straight path. For a satellite in orbit, it is the centripetal force inward that makes the satellite deviate from a straight path and bend into an orbit.
Centrifugal "forces" appear in, for example, a polar coordinate system. An object on a straight path has a changing distance to the origin over time, and the centrifugal "force" is supposed to be responsible for that. But the object is still in reality on a straight path and not experiencing any true force. 


#3
Oct1812, 10:47 AM

P: 3

Dear Muphrid, thank you for your answer.
In the way I think about it, the object wants to go in a straight and the force that makes it deviate from the straight path is the gravitational one. Why do I think it this way? Because if I accelerate the object, and so increase the centripetal force, the gravitational force won't be strong enough to keep the object near the planet. And if I slow down the object, and so decrease the centripetal force, the gravitational force will be stronger than the centripetal force and the object will get closer to the planet. But if both forces are directed towards the center why am I subtracting them from each other? Let's say that I put a person inside of a circular spinning room. That person will get pushed against the wall, right? Is this due to centripetal or centrifugal force? I am sorry but I really get frustrated when I don't understand something. Thank you very much. 


#4
Oct1812, 10:58 AM

P: 1,212

Centripetal Force vs Centrifugal Force
Neither are forces in the sense that neither have the tendency to cause masses to accelerate.
The centripetal "force" is actually a critical condition for circular motion. If the physical force is equal to the centripetal force, then the object will follow a circular trajectory. The centrifugal force is a manifestation of a noninertial coordinate system. If you have an inertial coordinate system, there is no centrifugal force. If you demand that your coordinate system is rotating, for example, then you're going to start wondering why, in that system, masses appear to accelerate away from the axis of rotation. Consider the EarthSun system, both bodies orbiting a common barycentre in (let's say) circular paths. If you pick an inertial coordinate system, then you'll find the gravitational forces acting on each body are equal to the centripetal forces. This condition being met, they orbit in circular paths. Now, if you chose to rotate your coordinate system so that the two bodies are permanently at rest on the xaxis, you're going to start wondering why these two "apparently stationary" bodies don't accelerate towards one another due to gravity. But if you work through the equations for acceleration in a rotating coordinate system (x' = x*cos(t)  y*sin(t), y' = x*sin(t) + y*cos(t), differentiate twice to find a') then you get all these weird new acceleration terms popping up. One is the centrifugal force, and as it happens, it will exactly balance the gravitational force and keep the two bodies at a constant orbital radius. It's a direct consequence of the mathematics. The ONLY real force in this example is the gravitational force. Everything else is just mathematics. 


#5
Oct1812, 11:13 AM

P: 907

It it is in a _rotating_ coordinate system where centrifugal forces appear. This would be, for instance, a coordinate system painted on a merrygoround or tied to lines of longitide and latitude on the earth.
A _polar_ coordinate system has to do with the way the coordinates are encoded and is something entirely different. Polar versus Cartesian. Rotating versus inertial. Since the coordinates are moving in a rotating coordinate system, an object that stays at a fixed coordinate position is really moving in a circle. This means that the object is really accelerating. It takes a real force to cause a real acceleration. That force is measurable. So for instance if you stand on the equator and adopt a rotating coordinate system in which both you and the communications satellite 22,000 miles above your head are stationary then there is a real (*) centripetal force on the satellite due to gravity and a ficticious (*) centrifugal force on the satellite due to your choice of coordinate system. This accounts for the satellite staying [apparently] motionless overhead. If, instead, you adopt a coordinate system in which the stars are motionless overhead and in which the earth is moving eastward at 1000 mph beneath your feet then the satellite is also moving eastward over your head. The real centripetal force accounts for the acceleration of the satellite around its circular path. (*) See SophieCentaur's .sig for some apt and pithy commentary on the distinction between real and ficticious. 


#6
Oct1812, 11:46 AM

P: 5,462

Yes the centripetal force is real and measurable with real forcemeters. Yes the centrifugal force is a convenient fiction to make the maths easy. How come? Velocity is a vector quantity and change of velocity may be effected by change of magnitude or change of direction. Either constitute an acceleration. Now for centripetal force, this is always provided by the normal suppliers or agent of force. So for instance consider a weight whirling round on a string. The centripetal force is provided by the tension in the string  they are one and the same. However the weight is also in constant acceleration. This acceleration is directed towards the centre of of rotation (inwards). This is in accordance with Newtons second law force = mass times acceleration. So the weight is not in equilibrium. This is so fundamental and important I will repeat it So the weight is not in equilibrium. Shazam ! * # ! Enter the centrifugal force!!! This ficticious force is exactly equal in magnitude to but opposite in direction to the centripetal force. Let us pretend that such a force is acting on the weight. Now the weight is in equilibrium Yay! The weight is said to be in equilibrium under the action of the tension in the string and the opposing centrifugal force. This trick or transformation was first proposed by D'Alembert. It enables us to reduce a problem in dynamics ( using Newtons Laws of motion) to a simpler one in statics (using the laws of equilibrium). Note that you never see both the centripetal force and the centrifugal force appear in the same analysis you see either one or the other. When the centrifugal force is used the acceleration is zero, since the system is in equilibrium. When the centripetal force is invoked the system is not in equilibrium but accelerates. This acceleration is provided by the centripetal force. So as regards gravity and acceleration, the acceleration due to gravity = the centripetal acceleration or the force of gravity = the centripetal force. Here '= ' means the same as. Gravity and the centripetal force (acceleration) are not additional, they are one and the same. Does this help? Oh and welcome to Physics Forums 


#7
Oct1812, 02:12 PM

P: 3

Thank you very much for your answers, you've been just fantastic. Studiot, thanks for the welcome.
I think I am kind of understanding now how this thing works, but I want to make sure I understand well. If a person is in a merrygoround does and he gets pushed away from the center. Would you say this is due to the centripetal of centrifugal force? If I have a string with a rock attatched to its end, and I make this sping rotate would you say that this is due to the centripetal or centrifugal force? If I have a satellite rotating around a planet in a circular path, would you say that this is due to the centripetal or centrifugal force? I think I understood that you use the term centripetal or centrifugal according to the frame of reference, but I am not sure whether the centrifugal force is completely equal to the centripetal one, except for the fact that its direction is the opposite of the centripetal force. Because if they were two sides of the same coin then you could use them kind of interchangebly. What I mean by interchangebly is that I could say that if you go in a merrygoround you get pushed towards the outside because of the centrifugal or because of the centripetal force. Either way the only thing that would change would be the frame of reference and so I would always be right. So, when there's a centripetal force there's always a centrifugal one? And when there's a centrifugal force, there's always a centripetal one? I mean, of course the indicate two different things, but the only thing that would change would be the point of view, the frame of reference. Is that so, or am I saying something stupid? Basically, are both of these sentences correct? If I go in a merrygoround I get pushed towards the outside because of the centripetal force. If I go in a merrygoround I get pushed towards the outside because of the centrifugal force. Because of course the centripetal force would go towards the center of the merrygoround, but it is just thanks to the centripetal force if there's the centrifugal one. Or am I saying something that is just wrong? Thank you very much and sorry for bothering you. 


#8
Oct1812, 02:25 PM

P: 3,181

Most texts apply those terms inconsistently  again another Newspeak problem. That bothers you if you are a purist like I am, until you don't give a **** anymore but simply apply them consistently, with when needed a little warning to the public. And of course, this occurs because the wall is accelerating that person towards the centre (centripetal). Check out Newton's third law, in case nobody else thought to mention it yet. 


#9
Oct1812, 04:09 PM

P: 5,462

In the attachment I have shown the analysis of a mass m hanging by a string on the side of a cone of angle θ, and whirling round it.
I have done this for both the dynamic and static analysis methods which obviously must yield the same result, but in this case can be seen to have the same equations as well. Note the difference between the dynamic (newton) analysis and the static analysis after introducing the ficticious centrifugal force mrω^{2} to create equilibrium. The circular motion is in a horizontal plane parallel to the base of the cone and has radius r. The centripetal force providing a rightwards acceleration towards the vertical axis of the cone is the resultant of the resultant of the horizontal components of the real forces being the normal reation at the cone surface and the string tension. This is a real resultant providing a real acceleration. In the first analysis we equate this acceleration to the mass times the force. In the second we counter it with an equilibrating centrifugal force. 


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