centrifugal force

cmos

The reason I created the geosynchronous satellite example was because it is a simpler example to work with but still marvelously exemplifies the details we are trying to point out.

Newton's second law: F=ma. Since none of the bodies accelerate, we must conclude that all the forces are balanced to zero. The gravitational attraction between the observer and the satellite is balanced by the normal force between the two. But what balances the Earth-satellite gravitational attraction? This being a non-inertial reference frame we must invoke the use of the fictious centrifugal force to balance the Earth-satellite attraction. This is the second equation in my post #34.

That is why the force is termed fictious; it is because it is only manifested as a result of using a non-inertial reference frame. It has no physical manifestation in the sense that is gravity or electromagnetisim.

This is the view from an inertial reference frame. It requires that you view the system from a point in space where you see that, in reality, the satellite orbits the Earth with a period equal to the Earth's rotational period. Here, the Earth-satellite attraction is not balanced. This is the first equation in my post #34.

Bingo! This is precisely what we have been trying to say.

Wrong; the Coriolis force is another fictious force. It is manifested due movement with respect to a rotating (non-inertial) frame of reference. If you were to view the Earth from a point in space, you would see that the Earth rotates and that it orbits the sun. This would account for what would otherwise be the Coriolis force when viewed form the surface of the Earth. So in the inertial frame, there is no Coriolis force either.

rcgldr

It seems like the situation is being made far more complicated than it needs to be, Newton's 3rd law is simple enough, you can't apply a force to an object without that object responding with an equal and opposite force.

Going back to the simple example of a person twirling an object around while holding a string, it's clear that the person will experience a real outwards force due to tension in the string. The person applies countering inwards force on the string. Note that the person is in a non-rotating frame of reference.

At the string to object contact point, the string applies a centripetal force to the object, and the object applies a reactive centrifugal force to the string.

Doc Al

Anyone using "centrifugal force" to mean a real, reactive force--instead of the standard definition as a "fictitious" force--should add a disclaimer in bold red letters: This is not standard physics usage--read at your own risk!

Seriously, many of the posts in this thread confuse this very issue and that makes the discussion very hard to follow. If you want to use the term in that that fashion, make it clear that you are using a different definition than most of the other posters. (Otherwise we are bogged down in semantics.)

D H

A science fiction setting:

Two fiercely competitive species find an airless, spinning planet chock full of metals. Each builds an armed mining outpost, only to discover the other species has also found the planet. One species fires first, consistently missing their enemy. The commander asks the science officer whether the firing system models the Coriolis force. The science officer says it's not modeled and won't budge when told to incorporate it because "It's not a real force". The commanding officer finds another solution: Make the guns aim at where the enemy outpost will be when the missiles hit rather than where it is when the missiles are launched. The science officer agrees to this, but says that this will take some time.

Meanwhile, the other species retaliate. They come from a rapidly spinning world and have the Coriolis force factored in. It doesn't matter to them that the force is not real; the effect is real. Their guns are deadly accurate.

========================================================

All of this debate on whether the centrifugal and Coriolis effects are "real" misses the mark. Hurricanes form because of the Coriolis effect, and gravity varies with latitude in part because of the centrifugal force. These very real effects are explainable without the aid of fictitious forces from the point of view of an inertial observer. Sometimes an inertial perspective happens to be extremely inconvenient. Atmospheric modeling is one such example.

rcgldr

Well maybe you should update or delete the wiki reference to "reactive centrifugal force".

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

I for one prefer to use non rotational, non accelerating frame of references, and centrifugal force should have a meaning in a more common frame of reference.

Getting back to that person that is twirling an object, what do you call the outwards and rotating force that person applies at the contact patch between his feet and the ground (either a platorm on a frictionless plane, or the earth)?

Doc Al

In general, I would not use Wiki as a serious reference. Pick up any intermediate classical mechanics textbook instead.

Use a different word then. But don't use a word that already has a standard meaning in a nonstandard way. (And try analyzing large scale atmospheric effects from an inertial frame--good luck!)

Why does it need a special name? You twirl a rock on a string, the string pulls on you, you pull back on the string. (I'd call those forces string tension.) The ground exerts a friction force on you to prevent slipping (and you, of course, exert an equal and opposite friction force on the ground).

cabraham

So it appears to be that we agree that centrifugal is a fictitious or virtual force created to balance centripetal in a specific ref frame. I still take issue with those who say Coriolis is not a real force. My understanding is as follows.

If we regard earth as a ref frame and wish to fire a projectile from the equator northward, we must account for rotation of the earth. When the projectile is fired, it has the velocity eastward equal to that at the equator. As it traverses its path northward, it maintains its eastward linear velocity while the earth underneath maintains its eastward angular velocity. When the missile lands north of the equator, its eastward velocity is greater than the eastard *linear* velocity of that spot on earth north of the equator. Hence the missile lands *east* of its intended target spot had Coriolis not been accounted for.

I've always regarded Coriolis as a "correction term" that must be computed to account for the above phenomenon. While the missile is airborn, of course there is no literal "Coriolis force" acting on it to accelerate it eastwardly. From the earth ref frame, if we treated the earth as stationary then the missile won't land where we thought it should have. What "knocked the missile off course?" We add the Coriolis term as a correcting factor. The Coriolis component is not an actual force acting on the airborn missile deviating its path. Rather it is a correction term accounting for the fact that a rotating ref frame cannot be equated to a stationary one.

There is no literal Coriolis force actively influencing the missile trajectory, but rather it is a correction term accounting for the deviation AS IF THERE WAS a "real" Coriolis force. The force is not literal, but the missile's path deviation is absolutely real. Coriolis force is a virtual force mathematically defined to account for A VERY REAL NON-FICTITIUOS path deviation.

That's my understanding of the "Coriolis component" using my undergrad physics prof Dr. M terminology. Dr. M knew his stuff. In 3 decades of EE R & D, his teachings never once failed me. He was brilliant. Dr. M, wherever you are, "you da man!"

cmos

cabraham,

That's exactly why it is a fictious force. It is manifested only to correct for the fact that your frame of reference is non-inertial.

In firing a projectile (such as a missile) over long distances over the Earth surface, the Coriolis force will play a significant role. This is the non-inertial view.

But if your launch center was from a point in space, you wouldn't have to worry about the Coriolis force; you'd just have to account for the fact that you are firing at a moving target (the Earth). This is the inertial view.

D H

Bingo! The Coriolis force is a fictitious force. When viewed by an inertial observer, a balistic missile flies in a plane. There is no curvature of the path.

There is an easy test of whether a force is real or fictitious: Can you build a black box to measure the force? Fictitious forces arise from the observer's perspective rather than from some real force that truly does act on the body. Can we measure the normal force? Sure. You step on a scale every morning. Can we measure tension? Sure. Can we measure a centrifugal force or Coriolis force: Nope. These are observer-dependent. They aren't real.

One such measuring device is an accelerometer. The Newtonian view of a perfect accelerometer is a device that measures all real forces except gravitation acting on some body. Why that "except gravitation" clause? The GR view is "A perfect accelerometer is a device that measures all real forces acting on some body, period. There is no reason to exclude gravity because gravity is not a real force."

tiny-tim

Who … me??

I'm just a little goldfish who tries to make three-dimensional sense out of the two-dimensional images I see projected onto the boundary of the bowliverse.

oooh … I've just thought of a question …

forces do work …
can centrifugal force do work?

rcgldr

So when are they going to change the name of a centrifuge to a centripuge?

George Jones

I can't recall running into this, but it seems to me that work in non-inertial frames can be useful. For example, I think the work-energy theorem can be used to find a particle's change in speed in a non-inertial frame.

Still don't have time to respond like I want; on my out the door again.

I am going to need a vacation to recover from my current vacation at my in-laws!

cmos

Good question!


It seems counter-intuitive to me, but I want to say yes. Consider a particle placed on a rotating disc. The centrifugal force on the particle is
[tex]\vec{F}_{cent}=m\omega^2\vec{r}[/tex]
where r is directed radially. So in a particle being pushed from a point near the axis towards the edge of the disc, work must be performed:
[tex]W=\int \vec{F}_{cent} \bullet d\vec{r} [/tex]


My problem with this is where does the work go in the inertial frame? Also, clearly the Coriolis force cannot do work:
[tex]\vec{F}_{Cor}=-2m \vec{\omega} \times \vec{\dot r}[/tex]
where dr/dt is the velocity of the particle in the non-inertial frame. So I find it an oddity that one fictious force may do work while another may not.

George Jones

When considering work, usually only one frame is used. Even when restricted to inertial reference frames, a force can do zero work in one frame and non-zero work in another.

Consider the following example.

In an inertial frame, a particle is subject to a (net) force. Suppose that when the force starts acting, the velocity of the the particle is [itex]c \hat{e_1}[/itex], and that the velocity of the particle when the force stops acting is [itex]c \hat{e_2}[/itex], where [itex]c[/itex] is a constant. Because the initial and final speeds are the same, there is no change in kinetic energy, and, by the work-energy theorem, no work done on the particle by the force.

Now consider the same situation from the point of view of an inertial reference frame that moves with velocity [itex]c \hat{e_1}[/itex] with respect to the first inertial reference frame. In this frame, the initial velocity of the the particle is [itex]\vec{0}[/itex], and the final velocity of the particle is [itex]c \left( \hat{e_2} - \hat{e_1} \right)[/itex]. In this frame, the change in kinetic energy and work done is [itex]m c^2[/itex].

A "force" of this form can change the direction of a particle's motion, but cannot change a particle's speed. Again, this is true in both inertial and non-inertial frames. For example, a moving charged particle in a "real" magnetic field is subject to a force (proportional to) [itex]\vec{v} \times \vec{B}[/itex], which does no work.

cmos

Excellent! Thank you for the very clear explanation. Bravo, Prof. Jones.

rcgldr

Take the example of the person twirling an object. Change this to a person holding a very low friction pipe with a string going through it. The are objects attached to both ends of the string. The person twirls one of the objects, and the rotating object reacts to the centripetal force from the string by applying a "reactive centrifugal force" to the string, creating a tension, which in turn can lift the object dangling below the pipe at the other end of the string.

Work done is peformed on the hanging object, it's weight times the height the hanging object is raised.

Work is also done on the rotating object by the person, equal to it's change in kinetic energy.

Rockets in space and thier spent fuel are a good exception. Assume an environment with no external forces (free of gravity). Momentum is preserved. All the work done is internal, and increases the kinetic energy of fuel and/or rocket, depending on the reference frame, but it will turn out that, within reason, the frame of reference won't change the amount of work done, since it's source the the chemical energy of the fuel. Even if the frame of reference is the accelerating rocket itself, all the work is done to the fuel in this frame of reference, but it's the same amount of work done as observed in constant velocity frame of reference.

George Jones

As`Doc Al already said in post #54, reactive centrifugal force" is very non-standard terminology in physics.