Is Centripetal Force Affected by the Earth's Rotation?

[lappong999]

Supposed that a man is standing on the edge of a spinning merry-go-round which rotates at an constant angular speed with respect to a stationary fame of reference(the ground), there will be a centripetal force pulling him toward the centre of the marry-go-round and prevents him from sliding off it. Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary, and it is the Earth rotating in the opposite direction at the same angular speed, will there still be any force pulling you toward the centre of the 'stationary' merry-go-round? If we consider the Earth as stationary and it is the universe rotating around it, how can we define the centripetal force? As the Earth is not spinning, the gravitational pull of the poles will be identical to that of the equator if we consider the Earth as an uniform sphere, which contradicts with the reality that in the poles there will be a theoretically greater gravitatinal pull since there is no centripetal force acting. Could anyone explain it?

 

[A.T.]

will there still be any force pulling you toward the centre of the 'stationary' merry-go-round?

Yes, the same force that pulls you in the non-rotating frame.

If we consider the Earth as stationary and it is the universe rotating around it, how can we define the centripetal force?

Gravity, which provides the centripetal force in the non-rotating frame, still acts in the rotating frame of the Earth. But since the Earth doesn't rotate in that frame, the concept of centripetal force is not useful for the Earth. But the universe rotates in that frame, and here the centripetal force is providied by the inertial Coriolis force.

 

[Doc Al]

Supposed that a man is standing on the edge of a spinning merry-go-round which rotates at an constant angular speed with respect to a stationary fame of reference(the ground), there will be a centripetal force pulling him toward the centre of the marry-go-round and prevents him from sliding off it. Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary, and it is the Earth rotating in the opposite direction at the same angular speed, will there still be any force pulling you toward the centre of the 'stationary' merry-go-round?

Since the man is moving in a circle, and thus accelerating in the inertial frame of the earth, there must be a net centripetal force on him. That centripetal force is given by the merry-go-round exerting a force on the man.

In the non-inertial rotating frame of the merry-go-round, Newton's 2nd law must be modified to include non-inertial forces such as an outward centrifugal force acting on the man. Including that force, you can apply Newton's 2nd law. Since the man is not accelerating, the net force must be zero. The inward force from the merry-go-round (which is a 'real' force and thus appears in any frame) is canceled by the outward centrifugal force.

 

[lappong999]

Yes, the same force that pulls you in the non-rotating frame.


Gravity, which provides the centripetal force in the non-rotating frame, still acts in the rotating frame of the Earth. But since the Earth doesn't rotate in that frame, the concept of centripetal force is not useful for the Earth. But the universe rotates in that frame, and here the centripetal force is providied by the inertial Coriolis force.

Thanks for replying!

But why in the non-rotating frame, no corilis effect could be observed, as the Earth rotating around the table and the table rotating with respect to the Earth in an opposite direction are literally the same thing? Consider the swing carousel, if we we say it is the Earth rotating around the passengers, why people have nausea. There is also a similar question that bothers me, when we are traveling in a accelerating car we would feel a force exerting on us by the seat, as F=ma, but how when we consider it is the road accelerating backward? Will we have the same feeling?

 

[Nugatory]

Thanks for replying!

But why in the non-rotating frame, no corilis effect could be observed, as the Earth rotating around the table and the table rotating with respect to the Earth in an opposite direction are literally the same thing? Consider the swing carousel, if we we say it is the Earth rotating around the passengers, why people have nausea. There is also a similar question that bothers me, when we are traveling in a accelerating car we would feel a force exerting on us by the seat, as F=ma, but how when we consider it is the road accelerating backward? Will we have the same feeling?

It's not the exact same thing, because the acceleration produces measurable phyical effects.

Let's say I'm in a windowless room. I cannot tell whether I'm moving at or at rest (no windows, so I can't look out and see if my position relative to other objects in the universe is changing); all I know is that I am at rest relative to the room. However, I can tell whether that frame is inertial or not - if I feel a force pulling towards floor, ceiling, or one of the walls then I know that the frame is not inertial.

 

[Andy Resnick]

Supposed that a man is standing on the edge of a spinning merry-go-round which rotates at an constant angular speed with respect to a stationary fame of reference(the ground), there will be a centripetal force pulling him toward the centre of the marry-go-round and prevents him from sliding off it.

No, friction is what keeps the person from sliding off- if the merry-go-round were made of ice, the person would indeed slide off. In this case, the friction force balances the cerntripetal force.

Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary, and it is the Earth rotating in the opposite direction at the same angular speed, will there still be any force pulling you toward the centre of the 'stationary' merry-go-round?

There is still no net force, and no contradiction.

 

[A.T.]

In this case, the friction force balances the cerntripetal force.

No, the friction on the person acts inwards, towards the center, and thus provides the centripetal force necessary to keep the person in circular motion. In the co-rotating frame, where the person is at rest, there is an inertial centrifugal force on the person that balances the friction, and assures zero net force.

Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary,

There is still no net force

What do you mean by "still no net force"? In the non-rotating there was a net force.

 

[lappong999]

I'm still confused. What does it mean by acceleration? How can we tell whether an object is accelerating or not? Let's take linear motions as example, as I have learned from my alevel physics that if a frame is accelerating with respect to another frame, from the stationary frame's point of view there is a net force acting on the objects in the accelerating frame. But if we consider the accelerating frame as stationary, and the objects inside the frame have no relative motion to it, we imagine that there is an inertial force and it is balanced by an opposite force (We are 'pushed' by the seat when we sit in an acelerating car). If the "opposite force' is removed, we will accelerate in the same direction as the inertial force (If we throw a ball in an accelerating train it accelerates backward.) If we consider the accelerating frame stationary, how could it happens? We could say in an accelerating car we are not accelerating with respect to another car which accelerates at the same rate, so why do we experience the inertial force? In a stationary car which other cars accelerate with respect to it, why don't we experience the inertial drag?

 

[Andy Resnick]

No, the friction on the person acts inwards, towards the center, and thus provides the centripetal force necessary to keep the person in circular motion. In the co-rotating frame, where the person is at rest, there is an inertial centrifugal force on the person that balances the friction, and assures zero net force.

What do you mean by "still no net force"? In the non-rotating there was a net force.

The video "frames of reference" does a very good job of explaining why we 'invent' forces to reconcile observations between accelerating frames of reference.

 

[AlephZero]

There is still no net force, and no contradiction.

There are two equal and opposite friction forces, one acting on the man, one on the merry-go-round. So there is a net force on the man, which is obviously true. Otherwise he would be moving with constant velocity in an inertial frame, not in a circle.

There is also a reaction force on the axis of the merrry-go-round, balancing the outward friction force on it. Or, if you consider the combined system of the merry-go-round plus the man, its center of mass is not on the axis of rotation, and you will calculate the same force on the axis doing it that way.

 

[A.T.]

The video "frames of reference" does a very good job of explaining why we 'invent' forces to reconcile observations between accelerating frames of reference.

The friction acting inwards on the man is a real force, not an 'invented' one. It acts as the centripetal force, not to "balance the centripetal force" as you claimed.

 

[cabraham]

The friction force *IS* the centripetal force. It keeps the man accelerating towards the center. This is the view from the earth, a rest frame. Centripetal force is real, but it is always, AFAIK, a specific force like friction or gravity. Friction in the merry go round provides the centripetal force/acceleration, whereas gravity does so in an orbit such as moon/earth system. The term "centripetal" simply infers a center oriented force provided by one of several types of force like gravity or friction. Did I help at all?

Claude

 

[A.T.]

Centripetal force is real,

Not in general. In rotating frames the inertial Coriolis force can act as the centripetal force in circular motion. See post #2.

 

[cabraham]

Not in general. In rotating frames the inertial Coriolis force can act as the centripetal force in circular motion. See post #2.

I did not say "in general". Please reread my post. I said in the reference frame of the earth, a stationary frame. Now you invoke rotating frames, which I did not include. Do you get a rush correcting people? You're rebuking a point I never made.

Claude

 

[A.T.]

I did not say "in general".

You said:

Centripetal force is real, but it is always, AFAIK, a specific force like friction or gravity.

It wasn't clear what you meant by "specific force like friction or gravity", so I merely clarified that it be can a coordinate effect like Coriolis.

 

[cabraham]

Coriolis and centripetal are 2 differing entities. Please clarify so that I may understand any situation where centripetal is a mere coordinate effect. I realize that "centrifugal" force is coordinate related, a pseudo-force, a math construct, but I am not aware that centripetal can have that type of status. An example would be appreciated. Thanks.

Claude

 

[olivermsun]

Well suppose you are in a rotating coordinate system, e.g., on the earth.

Viewed from within your frame:

You fire an artillery shell far and fast enough that you observe the trajectory curving to one side.

The acceleration is "centripetal" because it involves a movement on a curved path.

The cause of the acceleration is the Coriolis force.

 

[A.T.]

I realize that "centrifugal" force is coordinate related, a pseudo-force, a math construct, but I am not aware that centripetal can have that type of status. An example would be appreciated.

See post #2. In the rest frame of the Earth distant starts move in circles around the Earth. The centripetal force associated with that circular motion is the Coriolis pseudo-force (reduced by the centrifugal pseudo-force).

 

[jbriggs444]

The example I like is from the point of view of a fellow on a carousel who uses the rotating frame in which the carousel is at rest. He observes a child standing on the ground outside the carousel.

In the rotating coordinate system, the child is revolving around the carousel with a centripetal acceleration of ω²r and is subject to a centrifugal force whose magnitude is also ω²r. The centripetal force required to produce that centripetal acceleration and negate the centrifugal force is provided by the Coriolis force with a magnitude of 2ωv = 2ω²r.

 

[cabraham]

The example I like is from the point of view of a fellow on a carousel who uses the rotating frame in which the carousel is at rest. He observes a child standing on the ground outside the carousel.

In the rotating coordinate system, the child is revolving around the carousel with a centripetal acceleration of ω²r and is subject to a centrifugal force whose magnitude is also ω²r. The centripetal force required to produce that centripetal acceleration and negate the centrifugal force is provided by the Coriolis force with a magnitude of 2ωv = 2ω²r.

Bold underlined statement - agreed.

Bold italicized statement - disagree. In the rotating frame R, the child is revolving with centripetal acceleration rω², as we agree on that. But this centripetal force is NOT counter-balanced by any centrifugal force. Let me elaborate.

In R, the rotating frame, the man incurs both a centripetal and centrifugal force, they effectively cancel. In this frame R, Newton's laws hold by creating the math construct known as centrifugal force. Since the man is not moving in R, the net force on him must be 0 in the R frame. Hence his centripetal component of force accelerating him towards the center must be canceled by an equal and opposite force directed away from the center. This force is centrifugal. In R, the man incurs both, which he must since he is at rest. At rest in R, his net forces incurred must sum to 0. Centripetal and centrifugal forces are equal in magnitude and opposite in direction, therefore summing to 0.

But the child (assume a girl) outside R, in the stationary frame S, when viewed from R, differs from the man in R as far as incurred forces go. In R, the girl is accelerating, not at rest. Since she is accelerating as viewed from R, her net forces incurred are non-zero, wrt R. The force accelerating her towards the center when viewed in R, is indeed centripetal as you stated. But unlike the man, the girl is not at rest wrt R, she is accelerating.

Therefore the centripetal force on the girl in R frame is not balanced by centrifugal. Unlike the man who is at rest in frame R, the girl incurs a centripetal force unbalanced, resulting in a centripetal acceleration. The man, OTOH, in the frame R, is at rest, not accelerating. Thus any force on him must be counter-balanced to keep him at rest. So the man in R incurs both forces, summing to 0, which agrees with his rest state in R.

Coriolis is not the same as centripetal nor centrifugal. The 2ωv is not equivalent to rω². Please recheck your references. Best regards.

Claude

 

[Doc Al]

Coriolis is not the same as centripetal nor centrifugal. The 2ωv is not equivalent to rω². Please recheck your references. Best regards.

You might want to reread jbriggs444's post.

 

[A.T.]

But the child (assume a girl) outside R,

You cannot be "outside" of a reference frame, as it extends to infinity. All objects exist in all reference frames.

Therefore the centripetal force on the girl in R frame is not balanced by centrifugal.

The Coriolis magnitude is twice the centrifugal magnitude, so it cannot be completely balanced. The remainder of the Coriolis force provides the centripetal force.

 

[jbriggs444]

In the rotating frame R, the child is revolving with centripetal acceleration rω², as we agree on that. But this centripetal force is NOT counter-balanced by any centrifugal force.

Half of it is. Half of it is not.

Let me elaborate.

I accept without comment your paragraph concerning the man. Let us concentrate instead on the child.

But the child (assume a girl) outside R, in the stationary frame S, when viewed from R, differs from the man in R as far as incurred forces go. In R, the girl is accelerating, not at rest. Since she is accelerating as viewed from R, her net forces incurred are non-zero, wrt R. The force accelerating her towards the center when viewed in R, is indeed centripetal as you stated. But unlike the man, the girl is not at rest wrt R, she is accelerating.

Yes. Her acceleration is given by ω²r. We agree on this.

Therefore the centripetal force on the girl in R frame is not balanced by centrifugal. Unlike the man who is at rest in frame R, the girl incurs a centripetal force unbalanced, resulting in a centripetal acceleration.

Yes. The centripetal force is given by 2ωv = 2ω^r which is greater than ω²r. The two forces are not in balance. We agree on this.

The net force on the girl is 2ω²r centripetal plus ω²r centrifugal. The net is ω²r centripetal.

Coriolis is not the same as centripetal nor centrifugal. The 2ωv is not equivalent to rω². Please recheck your references. Best regards.

2ωv = 2ω²r is what I said. For an object moving tangentially in the rotating frame of reference, the resulting Coriolis force will be radial. For an object moving tangentially and anti-spinward, the resulting Coriolis force will be centripetal. For an object moving tangeitially and spinward, the resulting Coriolis force will be centrifugal.

"Coriolis force" is the inertial pseudo-force that is used to explain the curved path of moving objects as viewed from a rotating frame of reference.

"Centripetal" means "toward the center". A "centripetal force" is any force toward the center of rotation.

"Centrifugal" means "away from the center". I have used the term by itself once in this very post.

"Centrifugal force" can refer to:

1. Any force that is directed away from the center of rotation.

2. The specific inertial pseudo-force that causes motionless objects to accelerate away from the center in a rotating frame of reference. This is the most common meaning of the term and the one that I have primarily used in this thread.

3. The third law partner force corresponding to a real centripetal force on an object in rotary motion. For clarity, this is often called "centrifugal reaction force".

The "centrifugal force" in the second sense above has no third law partner force.
The "Coriolis force" never has a third law partner force.

 

[Buckleymanor]

Supposed that a man is standing on the edge of a spinning merry-go-round which rotates at an constant angular speed with respect to a stationary fame of reference(the ground), there will be a centripetal force pulling him toward the centre of the marry-go-round and prevents him from sliding off it. Then comes the question, what happens if you take the rotating fame of reference of the marry-go-round as stationary, and it is the Earth rotating in the opposite direction at the same angular speed, will there still be any force pulling you toward the centre of the 'stationary' merry-go-round? If we consider the Earth as stationary and it is the universe rotating around it, how can we define the centripetal force? As the Earth is not spinning, the gravitational pull of the poles will be identical to that of the equator if we consider the Earth as an uniform sphere, which contradicts with the reality that in the poles there will be a theoretically greater gravitatinal pull since there is no centripetal force acting. Could anyone explain it?

So the man stands on the edge of the spinning merry go round.
How will there be a force pulling him towards the centre of the merry go round.
Surely there will be a force acting in the opposite direction which will wan't to throw him off.
If he is leaning towards the centre of rotation this might be so but this is not made clear in the OP.

 

[sophiecentaur]

So the man stands on the edge of the spinning merry go round.
How will there be a force pulling him towards the centre of the merry go round.
Surely there will be a force acting in the opposite direction which will wan't to throw him off.
If he is leaning towards the centre of rotation this might be so but this is not made clear in the OP.

If you don't provide a force to keep the man on the ride, he will just carry on moving in a straight line (leaving the merrygoround and landing on the ground) because Newton 1 applies. The fact that he is moving in a circle proves that there is a force acting on him.

Anyone who has a conceptual problem with all of this, needs to realize that the details of how the force is actually measured or felt affects their perception of it. To make a ball on a string move in a circle, you need to provide a centripetal tension (which you can 'measure' with your hand) at the centre of the circle. To keep someone on a fairground ride there must be a centripetal force on them which they will 'feel' (measure) as a force pushing them inwards. But, if they are holding an object, they will 'feel' a force, pushing the object outwards - very much like the weight of the same object when they are stationary, on the ground. They are actually providing the centripetal force on that object.

It's down to Frames of Reference, which need some thinking about.

 

[A.T.]

So the man stands on the edge of the spinning merry go round.
How will there be a force pulling him towards the centre of the merry go round.

The friction on his shoes acts towards the centre.
http://en.wikipedia.org/wiki/Friction#Static_friction

Surely there will be a force acting in the opposite direction which will wan't to throw him off.

Only in the rotating frame of reference:
http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame)

If he is leaning towards the centre of rotation this might be so but this is not made clear in the OP.

It has nothing to with leaning. Circular motion requires a net force towards the center:
http://en.wikipedia.org/wiki/Centripetal_force

 

[olivermsun]

The friction on his shoes acts towards the centre.
http://en.wikipedia.org/wiki/Friction#Static_friction
...
It has nothing to with leaning. Circular motion requires a net force towards the center:
http://en.wikipedia.org/wiki/Centripetal_force

Well, similarly to what was discussed in a recent, lengthy thread about a tug-of-war, one of the ways to produce the proper balance of "forces" would be to have the man lean inward and rely on the friction between his shoes and the deck of the merry-go-round. Another way would be to hang on to something.

 

[A.T.]

one of the ways to produce the proper balance of "forces" would be to have the man lean inward and rely on the friction between his shoes and the deck of the merry-go-round.

It doesn't matter if he leans or not, there is always the same net force acting on him inwards.

 

[cabraham]

Centripetal and Coriolis are being discussed, but I seem to remember Coriolis as 2*omega*v, where "v" is the velocity of the person moving wrt the rotating frame R. If the man on the merry go round was moving at a velocity of 'V', on the frame that's rotating, his Coriolis acceleration is 2*omegaXv (cross product). Since the man is stationary in frame R, Coriolis is 0. Likewise in frame R, the girl standing on the ground observing the merry go round has a 0 Coriolis acceleration as she is at rest wrt to her frame S. That seems to be the only point of disagreement. Comments welcome.

Claude

 

[A.T.]

Coriolis as 2*omega*v, where "v" is the velocity of the person moving wrt the rotating frame R
...
Likewise in frame R, the girl standing on the ground observing the merry go round has a 0 Coriolis acceleration as she is at rest wrt to her frame S.

Wrong. As you said above v is the velocity wrt the rotating frame R. And the girl is moving wrt the rotating frame R, so Coriolis is not zero.

 

[Dale]

Likewise in frame R, the girl standing on the ground observing the merry go round has a 0 Coriolis acceleration as she is at rest wrt to her frame S.

She is at rest with respect to S, she is moving with respect to R.

The Coriolis acceleration is $2 v \times \omega$ where v is measured in R, not S. In R, she is moving perpendicular to $\omega$, and her speed is $|r \omega|$ so the total Coriolis acceleration is $2 r \omega^2$ inwards.

 

[cabraham]

She is at rest with respect to S, she is moving with respect to R.

The Coriolis acceleration is $2 v \times \omega$ where v is measured in R, not S. In R, she is moving perpendicular to $\omega$, and her speed is $|r \omega|$ so the total Coriolis acceleration is $2 r \omega^2$ inwards.

Ok, but how is it that the total Coriolis acceleration is $2 r \omega^2$ inwards, when we know a priori that the net acceleration of the girl in frame R has to be $r \omega^2$ inwards? In frame R, the girl rotates with uniform angular speed $\omega$, is that correct? So her net acceleration must be directed towards the center, and have a magnitude of $r \omega^2$ inwards, is that correct? Hence if the Coriolis component is twice that value, $2 r \omega^2$ inwards, then there would have to be another acceleration equal to $r \omega^2$ directed outward, centrifugal? Please explain. Thanks.

Claude

 

[D H]

Hence if the Coriolis component is twice that value, $2 r \omega^2$ inwards, then there would have to be another acceleration equal to $r \omega^2$ directed outward, centrifugal? Please explain.

You've explained it yourself. The net apparent force in the rotating frame is $m r\omega^2$ directed inward. The centrifugal force is $m r\omega^2$, directed outward. That's the right magnitude, but exactly the wrong direction. To yield that apparent net force, there must be some other force (or forces) with twice this magnitude but directed inward. This is the Coriolis force.

 

[Khashishi]

You can find a frame of reference with no absolute rotation. If you want to have the simplest physical model for multiple rotating objects, you better use a non-rotating frame. You can only cancel out the Earth's rotation with a merry-go-round if your merry-go-round has the same axis as the Earth, so this is only possible at the poles. Away from the poles, you can find a parallel axis, but a parallel axis isn't good enough to cancel out the rotation.

 

[cabraham]

You've explained it yourself. The net apparent force in the rotating frame is $m r\omega^2$ directed inward. The centrifugal force is $m r\omega^2$, directed outward. That's the right magnitude, but exactly the wrong direction. To yield that apparent net force, there must be some other force (or forces) with twice this magnitude but directed inward. This is the Coriolis force.

I was merely pointing out to Dale, that a $2 r\omega^2$ acceleration on the girl in frame R cannot be all the acceleration. There has to be another component acting such that the NET acceleration on the girl in frame R is $r\omega^2$, oriented inward, i.e. centripetal acceleration. Since Coriolis in frame R, wrt the girl, computes to be ##2 r\omega^2## inward, then there has to be an acceleration of ##1 r\omega^2## outward. The net result would be ##1 r\omega^2## inward. This agrees with what we already know about Newton's 3rd law. Since the girl's net acceleration is centripetal, we expect ##1 r\omega^2## inward as the net final result. We also know that in the R frame, a centrifugal force is added to keep Newton's laws intact. This force has the same magnitude as the centripetal, i.e. ##1r\omega^2## inward, but oriented outward. The Corioils computes to ##2 r\omega^2## inward, so that subtracting $1r\omega^2$ outward, the centrifugal component, we get what I believe is the correct final result, ##1 r\omega^2## inward, which is expected as it equals the centripetal component.

Anyway, that's how I see it, I'll accept correction if I erred. I originally was working in frame S for the girl, so naturally in that frame her Coriolis component would be 0. I generally use frame R for the man on the merry-go-round, and frame S for the girl on the ground. But there is no reason why we cannot view the girl as being in frame R. We must add the centrifugal component, which we have. Best regards.

Claude

 

[jbriggs444]

I was merely pointing out to Dale, that a $2 r\omega^2$ acceleration on the girl in frame R cannot be all the acceleration.

Of course we all agree that there is another acceleration on the girl in frame R: Centrifugal, as you had suspected.

This agrees with what we already know about Newton's 3rd law

Newton's 3rd law applies only to real forces. It does not apply to inertial forces that arise from a choice of reference frame. Newton's 2nd law is the one that is important here.

[Adopting a positive = inward convention]

$F=ma$
$Coriolis + Centrifugal + Real = ma$
$2mr\omega^2 - mr\omega^2 + 0 = mr\omega^2$

Anyway, that's how I see it, I'll accept correction if I erred. I originally was working in frame S for the girl, so naturally in that frame her Coriolis component would be 0. I generally use frame R for the man on the merry-go-round, and frame S for the girl on the ground.

If one is going to even speak about Coriolis force the implication is that one has adopted a rotating frame that is not tied to the motion of the object of interest. Talking about the Coriolis force on an object using a frame of reference in which it is at rest is not often useful. Of course it will be zero. The object is not moving in that frame. Talking about the Coriolis force on an object using a non-rotating frame is not often useful. Of course it will be zero. There is no Coriolis force in a non-rotating frame.

 

[A.T.]

I generally use frame R for the man on the merry-go-round, and frame S for the girl on the ground.

When you analyse a scenario you should stick with one frame for all analysed objects, otherwise you will just confuse yourself. Of course can you do multiple analyses from different frames, but that means considering all objects in each analysis .

 

[Dale]

Ok, but how is it that the total Coriolis acceleration is $2 r \omega^2$ inwards, when we know a priori that the net acceleration of the girl in frame R has to be $r \omega^2$ inwards? In frame R, the girl rotates with uniform angular speed $\omega$, is that correct? So her net acceleration must be directed towards the center, and have a magnitude of $r \omega^2$ inwards, is that correct? Hence if the Coriolis component is twice that value, $2 r \omega^2$ inwards, then there would have to be another acceleration equal to $r \omega^2$ directed outward, centrifugal? Please explain. Thanks.

Claude

Yes, Coriolis acceleration is $2 r \omega^2$ in and centrifugal acceleration is $r \omega^2$ out for a net acceleration of
$r \omega^2$ in, as expected.