What is the Concept of Centrifugal Force and Who Introduced it?

[anmolnanda]

when we travel around a circular path we exert force on the path(ground)can be considered centrifugal force and petal which acts inwards,as a reaction to it...it violates the rule that force of centrifugal is pseudo force...
then what exactly is centrifugal force who gave this concept??seeing what??
i have read all examples of man sitting in a car and force exerted towards right when driver turns towards left...but still not that clear...

 

[tiny-tim]

welcome to pf!

hi anmolnanda! welcome to pf!

when we travel around a circular path, the force we exert on the ground is friction …

since it is centripetal (which simply means that it points towards the centre of the circle), we can call it the centripetal force …

but there's only one force … call it "friction" or call it "centripetal", that's just a name!

centrifugal force only exists in a rotating frame …

if you're sitting in the car, using the car's frame, you can see that the ground is exerting friction (or centripetal) force on the tyres, and since you consider that the car is stationary (that's what using the car's frame means), the only way that can be consistent with a real centripetal force (which exists either in the car frame or in the ground frame) is if it is balanced by a "fictitious" centrifugal force (which of course exists only in the car frame)

 

[anmolnanda]

that means the force exerted by car on road is not centrifugal,so it is only the action force on the ground and nothing else?
but,
i encountered a image:- http://img818.imageshack.us/img818/4782/anmol.png
it states that reaction force on the car is mg-mv^2/r so here it is considering centrifgual force so how is the concept applicable here
it is from a book by hc.verma

 

[rcgldr]

The force the ground exerts on the tires at the contact patch is centripetal resulting in inwards acceleration. The force the tires exert on the ground is outwards, and results in a tiny outwards acceleration of the surface of the Earth (the Earth's momentum is huge compared to the car so it only accelerates by a very tiny amount). The outwards force is a reaction force to the inwards acceleration of the car, and some call this a reactive centrifugal force. From the Earth surface perpective (where it accelerates a tiny amount), the outwards force is the force applied by the tires, and the inwards force is the reactive force due to outwards acceleration of the Earth's surface.

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

 

[anmolnanda]

The force the ground exerts on the tires at the contact patch is centripetal resulting in inwards acceleration. The force the tires exert on the ground is outwards, and results in a tiny outwards acceleration of the surface of the Earth (the Earth's momentum is huge compared to the car so it only accelerates by a very tiny amount). The outwards force is a reaction force to the inwards acceleration of the car, and some call this a reactive centrifugal force. From the Earth surface perpective (where it accelerates a tiny amount), the outwards force is the force applied by the tires, and the inwards force is the reactive force due to outwards acceleration of the Earth's surface.

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

cannot get what u want to say tough reactive centripetal force was useful..how can he force exerted by contact path be inward?

 

[rcgldr]

how can the force exerted by contact path be inward?

The contact patch includes the tire and the pavement the tire is in contact with. The pavment genererates an inwards force onto the tire at the contact patch, and coexists with the tire generating an equal an opposing outwards force onto the pavement, the Newton third law pair of equal and opposing forces at the contact patch.

 

[anmolnanda]

but
force on the car on the pavement as per the diagram is inward force and path on car is outward what r u trying to say?

 

[vish_al210]

i have read all examples of man sitting in a car and force exerted towards right when driver turns towards left...but still not that clear...

When u r traveling in a car. moments after the car started ur body starts acting as one with the car and is in inertia inside the car. But the car carries a momentum. You do not experience it as u are in the car. Once the car starts to deviate from its path, ur body tries to adjust to the change in momentum and direction, but it takes a few moments. So ineffect ur body continues to go straight while ur car moves towards right. So sitting inside the car u experience or observe that ur body is moving to the left of the car (relatively).
Also if you turn at a comfortably slow speed u would not experience that as the momentum change that ur body needs to adjust to is very less.
So walking in a circular path is not considered exerting sufficient forces, and you cannot observe them on yourself. Now consider you are casually jogging along a circular//eliptical track carrying a handbag, pouch or goggles or just a bottle on a sling.. If you let go of the object on you (such that u do not exert force to throw it but only jus let go) you will continue to run in the circular path but the object would have gone off the circular path tangentially... but the momentum exerted or the centrifugal force depends on the momentum of the carrier (i.e., you. If you run faster the force experienced by the object you let go is much higher). The path of the object is affected by many otehr factors once you let go of it.. So i am not discussing the exact path...taken,.. ok.

 

[anmolnanda]

i guess u guys are not getting the question,
see the diagram
it represents a car on 2 mountainous track in both the cases gravity acts downwards
a force mv^2/r acts upwards according to books so from Newtons second law of motion we subtract both finding out the reaction force on the car by the path...my question is why don't we consider inward force which will add both forces to give reaction force why it has to act upwards...

 

[Doc Al]

i guess u guys are not getting the question,
see the diagram
it represents a car on 2 mountainous track in both the cases gravity acts downwards
a force mv^2/r acts upwards according to books so from Newtons second law of motion we subtract both finding out the reaction force on the car by the path...my question is why don't we consider inward force which will add both forces to give reaction force why it has to act upwards...

There's a net downward force on the car as it goes over the top of each hill. That net downward force, per Newton's 2nd law, equals mv^2/r.

There are two forces that act on the car: gravity acts downward and the normal force from the ground acts upward. Using Newton's 2nd law you can compute the value of the normal force.

 

[vish_al210]

(the Earth's momentum is huge compared to the car so it only accelerates by a very tiny amount).

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

Hope you are not referring to the Earth's momentum... or acceleration... ;)

 

[anmolnanda]

@Doc Al
the net downward force is mv^2/r is there any circular concept hidden behind it?

 

[Doc Al]

@Doc Al
the net downward force is mv^2/r is there any circular concept hidden behind it?

Yes, concepts of circular motion are involved but they are not hidden. As the car goes over each hill, it must execute circular motion. A purely kinematic analysis tells you that the car must have a downward (centripetal) acceleration equal to v^2/r. That, plus Newton's 2nd law, is all you need to find the reaction force. (There is no need to invoke the concept of centrifugal force here.)

 

[rcgldr]

In the diagrams showing 2 circular hills and a car traveling over the hills, two of the external forces acting on the car is the downwards force of gravity and the opposing upwards force from the road. Assume the car is just coasting so that there is zero friction force related to the hill.

At the peak of each hilll, gravity attempts to accelerate the car downwards at g (9.8 m / s²), but the road applies just enough upwards force for the car to maintain it's circular path over the hill. At the peak of each hill, the difference of the downwards force from gravity minus the upwards force from the road will results in a net force on the car equal to m v² / r, and the car acclerates downwards at v² / r.

In this case there is no centrifugal force, because no object is experiencing an outwards force relative to the center of mass of the car + Earth system. Between the tires and the surface of the road, you have equal and opposing forces with magnitude m (g - v²/r), the road pushing upwards on the car, and the car pushing downwards on the road. The gravitational forces have magnitude mg, pulling downwards on the car and upwards on the earth. The net force magnitude (downwards on car, upwards on earth) = mg - m(g - v² / r) = m v² / r.

It might be easier to understand the idea of no centrifugal force if you consider 2 objects in space orbiting each other due to gravity about a common center of mass. If the orbits are circular, each object experiences a centripetal force equal to the gravitational force, and there is no reactive centrifugal force.

For an example where reactive centrifugal force does exist, asssume the objects are small enough and distant enough to that gravity can be ignored and instead a string is used to keep the objects moving in a circle about a common center of mass. At each end of the string, the end of the string exerts a centripetal force on the object, and the object exerts and equal and opposing reactive centrifgual force on the end of the string.

 

[tiny-tim]

wow! i go away for fishmas, and there's a dozen posts when i get back!

… i encountered a image:- http://img818.imageshack.us/img818/4782/anmol.png
it states that reaction force on the car is mg-mv^2/r so here it is considering centrifgual force so how is the concept applicable here
it is from a book by hc.verma

yes, you can caluclate this either in the inertial frame or in the car's frame

in the car's frame, there are three forces: gravity and reaction force and centrifugal force … since in the car's frame the car has zero acceleration, we apply good ol' Newtons's second law (in the vertically upward direction) to get N - mg + mv²/r = 0

but in the inertial frame, there are only two forces: gravity and reaction force … since in the inertial frame the car has acceleration mv²/r downward, we apply good ol' Newtons's second law (in the vertically upward direction) to get N - mg = - mv²/r

that means the force exerted by car on road is not centrifugal,so it is only the action force on the ground and nothing else? …

"by car on road"? sorry, but considering the acceleration of the road is just bizarre!

if you mean "by road on car", the force N exerted by the road on the car is away from the centre (of curvature), so in that sense it could be described with the adjective "centrifugal" …

but that's not what we mean by "the centrifugal force" … the centrifugal force is a force (mv²/r) which only exists in a rotating frame such as the frame of the car … N is a real force, the same in either frame, and it isn't mv²/r

 

[NEILS BOHR]

to understand this properly u can read H C verma...

 

[anmolnanda]

Thanks everyone rcgldr,tiny-tim,Doc Al and vish_al210
for more kind of conceptual approach...really helpful of u guys...

 

[siranwar]

when we travel around a circular path we exert force on the path(ground)can be considered centrifugal force and petal which acts inwards,as a reaction to it...it violates the rule that force of centrifugal is pseudo force...
then what exactly is centrifugal force who gave this concept??seeing what??
i have read all examples of man sitting in a car and force exerted towards right when driver turns towards left...but still not that clear...

No when we move in a circular path there is force which rotates the object in circle actually that is known is the type u disscuss but there is another force which try to break the path and free to move tentionally is fugal force .there is no direct origin ot this forec centripetal is a origin of thais force if Fc is vanished Fg automatically vanished
Note :there is no concern with ficitious force directly