# A Connetion between centrifugal force and weight force

1. Apr 3, 2016

### Gabriele Pinna

Can someone explain me what is the connetion between centrifugal force and weight force thanks ? In which situation should I use the centrifugal force (that is a fake force) ?

2. Apr 3, 2016

### jamie.j1989

Centrifugal force isn't a fake force, tie a mass to the end of a string and spin around, the string gets taught and stops the mass from flying off, the string applies a force in the radial direction towards you, this is centrifugal force. Weight is a gravitational force.

3. Apr 3, 2016

### mathman

The force in the radial direction is call centripetal. Centrifugal "force" is the inertia which is trying to keep the object going in a straight line.

4. Apr 3, 2016

5. Apr 3, 2016

### hackhard

weight force = net gravitional force on body
normal force(apparent weight) = mass * gforce = projection of contact force on body in dir normal to contact surface
centrifugal force = (-1) * (centripetal force)
=(-1) * (mass) * (projection of resultant acceleration in dir normal to dir of resultant velocity of body )
= (-1) * ( (radial proj of normal force) + (radial proj of force 2) + (radial proj of force 3) ..... )
op must be referring to normal force ---
there is no special connection between centrifugal force and normal force
op must be refering to case where normal force or gravitational force is acting only in radial dir and no other force acts
gravitional force - supppose satallite orbits earth . net force acting is gravity towards centre . so centrifugal force on sat = mag of gravity away from centre
normal force - suppose lolipop spins about axis . net force on ball of lolipop is normal force (or tension) due to stick inwards. so centrifugal force on ball of lolipop = mag of tension outwards
add centrifugal force when in rotating frame
you (at rest) see that lolipop spinning (net force = centripetal force = normal force !=0)
. ant on lolipop (at rest ) says lolipop at rest .both correct
so ant adds centrifugal force (due to change of reference frame)(net force = centrifugal force + centripetal = normal force + (- normal force) = 0

6. Apr 4, 2016

### vanhees71

I'd rather call the centrifugal force an inertial force rather than a fictitious force, because then it's immediately clear that it occurs exclusively in non-inertial (rotating) reference frames.

7. Apr 7, 2016

### Andrew Mason

I think this kind of terminology just makes it more confusing. There is no centrifugal "force" in the Newtonian sense. There is a tension that exists if there is a mechanical connection but that does not mean that there is a centrifugal "force". Nothing can be accelerated outward with such an "effect". If the mechanical connection fails, there is not even a centrifugal motion - there is uniform motion of the fleeing mass in a direction that is at right angles to the radial direction when the connection is lost.

The introduction of the concept of a "centrifugal reaction force" makes this very confusing. It makes it very difficult for students to distinguish between a "centrifugal reaction force" and a centrifugal force (the non-Newtonian force or inertial effect). Just look at the Wikipedia discussion on Reactive Centrifugal Force.

The problem is with the term "centrifugal". Every real force has a reaction force to which it is paired. Centripetal forces are not excepted. That is just Newton's third law. But it is confusing to refer to this reaction force as "centrifugal". It is not necessarily, if ever, centrifugal. If the centripetal force on a rotating body is supplied by a force acting at a distance - e.g. gravity, electro-magnetic force - all forces are centripetal: the reaction to the centripetal force force on one body is a centripetal force on the other bod(y)(ies). In the case where the force between masses is supplied by a mechanical connection, there are tensions in two (eg. tethered rotating mass) or many directions (eg. a rotating rigid body) and all masses experience centripetal accelerations. In a rotating system, the integral of all point masses multiplied by their accelerations sum to 0. So all net forces are centripetal. All tensions are bi- or multi-directional.

In any event, nothing can ever accelerate outward in a radial direction by virtue of a rotation. So I am not sure why anyone wants to refer to a centrifugal real force as a reaction to a centripetal force. In my view, it is clearer and much less confusing to say that in any rotating system, all net forces are centripetal and all forces (which includes "reaction" forces) are either purely centripetal forces and, in cases where there are mechanical connections, a combination of centripetal net forces and bi or multi-directional tensions.

AM

Last edited: Apr 7, 2016
8. Apr 8, 2016

### A.T.

Now, this sounds very unclear and confusing to me. In my view, it is better not to obsess so much about the "centrifugal"/"centripetal" terminology, or always provide a precise mathematical definition of those terms (which you didn't).

9. Apr 8, 2016

### Andrew Mason

No one is "obsessing" about anything. It is a matter of being clear on the physics. If one must talk about centrifugal "force", it is the apparent force that appears in a non-inertial (rotating) frame of reference. To suggest that the reaction force to a centripetal force should be called "the centrifugal reaction force" is just wrong in many cases and confusing where one is referring to tensions that necessarily always operate in at least two directions.

As far as mathematical treatment goes, the mathematical treatment of centripetal force, of course is trivial: $\vec{F_c} = -m\omega^2r \hat r$. For a rotating mass element dm subject to tensions, $d\vec{F} = \sum T_i = -dm\omega^2r \hat r$. How to apply to a particular case depends on the geometry of the system being considered.

AM

Last edited: Apr 8, 2016
10. Apr 8, 2016

### Andrew Mason

Just to follow up on my last couple of posts, this is a perfect example of why it is so confusing to speak of the "reactive centrifugal force". The string applies a force to both the rotating mass and to "you". The real forces acting on the rotating mass and "you" are centripetal, not centrifugal. In the non-inertial frame of "you", there seems to be some force pulling the mass away from you so it seems centrifugal. That is not a real (Newtonian) force. "You" (the person who is holding the other end) are actually rotating (kind of wobbling) about an inertial point somewhere in between you and the rotating mass. The rope is supplying a centripetal force to both (the rotating) you and the (rotating) mass. So the string or rope is supplying centripetal forces at both ends.

Even if the mass is rotating about a post fixed to the earth, the centre of the post is not really an inertial point. It, and the earth to which it is fixed, has to be actually rotating or wobbling a tiny bit about a point (very close to the centre of the post) in between.

AM

11. Apr 8, 2016

### A.T.

What about the real forces on the string?

12. Apr 8, 2016

### Andrew Mason

The real force on any string element dm is the sum of the tensions acting on it: $\vec{dF} = \sum T_i = -dm\omega^2 r \hat r$. It is always toward an inertial point, which is the centre of rotation. r is the distance from the mass element to that centre of rotation. For the string as a whole, the tensions vary along the length of the string itself but the total net force is determined by integrating $\vec{F} =\int \vec{dF}= -\omega^2 \hat r \int r dm$. That is equivalent the centripetal force on a point mass located at the centre of mass of the rope.

AM

Last edited: Apr 8, 2016
13. Apr 8, 2016

### A.T.

That is the net force on a string element. I'm asking about the force exerted by the mass on the string.

14. Apr 8, 2016

### Staff: Mentor

To make this more obvious, suppose you and a friend (with roughly equal mass) are holding opposite ends of the rope, and "orbiting" around each other while ice-skating or riding frictionless air-carts.

15. Apr 8, 2016

### jbriggs444

To make this less obvious, consider a hub, as on a carnival ride with chains leading to the seats swinging around in a circle. The force exerted on the hub by the chains is in the centrifugal direction.

16. Apr 8, 2016

### Andrew Mason

It is equal and opposite to the force of the string on the mass.

Whether that force is toward or away from the centre of rotation depends on where the centre of rotation is and what you mean by the string.

If you are asking: "what is the direction of the force that the man is applying to the string as a whole" you have to consider where the centre of mass of the string is in relation to the centre of rotation.

If you are asking: "what is the direction of the force that the man is applying to the mass element of the string of length dL at the end of the string" you have to consider where the centre of mass of the string element is in relation to the centre of rotation. If the centre of rotation is not within the string itself, the force is toward the centre of rotation. For example, where a 100 kg man is swinging a 1 kg rope and mass with his outstretched arms holding the rope, the centre of rotation may well be somewhere between the man's centre of mass and his outstretched hands. In that case, the force on the rope by the man is toward the centre of rotation no matter how you analyse it. And the rope provides a centripetal force to the man as a whole (i.e his centre of mass), even though the tension feels like it is trying to rip his arms away from his body.

If the man ties the rope around his waist and lets the rope swing around him, the force direction depends on which part of the rope you are speaking about. But the only reason he is able to exert any force at all on the rope is because he is accelerating the rope as a whole, and the mass attached to it, centripetally about an inertial point.

AM

17. Apr 8, 2016

### Andrew Mason

Only if you think the hub centre is the inertial centre of rotation. To determine the inertial centre of rotation you have to take into account the mass of the earth. The hub/earth actually rotates about that point. So the chain supplies tension to the hub, which distributes that tension throughout the apparatus and the earth, causing the hub/earth to rotate or wobble (albeit imperceptibly) around that inertial point and accelerate centripetally toward that point.

This is why it is not a good idea to talk about the direction of the "reaction" force to a centripetal force supplied by tension as always having a particular direction (eg. "centrifugal"). It depends on the geometry and mass distribution in the rotating system and what parts you are considering. It also makes it very difficult to distinguish the real reaction force from the pseudo (non-Newtonian) force (apparent only in the non-inertial frame of reference) if you call both "centrifugal".

AM

18. Apr 8, 2016

### jbriggs444

As long as the center of rotation is anywhere in the interior of the hub, the force of the chains on the hub will be at least primarily in the centrifugal (aka outward) direction.

Yes, the unbalanced component of seats and riders on the carnival ride and the ride and earth are mutually orbitting a common barycenter and are all undergoing centripetal acceleration around that hypothetical point. But that's hardly a useful or even a particularly accurate description of the situation.

19. Apr 8, 2016

### A.T.

So it's opposite to the "centripetal" force, and has the same point of application (just acts on a different object).

It's the same center that the centripetal force point towards. Since the two forces have the same point of application, but opposite directions, they can hardly both point toward the center of rotation.

20. Apr 8, 2016

### Andrew Mason

So how useful is it to always refer to the reaction force to a centripetal force as centrifugal if it is sometimes not centrifugal (even when the force is tensile) and, in the case of force acting at a distance, never centrifugal? And, regardless of how the tension may be directed, that tension can never accelerate a mass away from the centre.

AM