Frame of reference question: Car traveling at the equator

[John Mcrain]

Earth rotate from west to east with 460m/s at equator.
Car travel at equator from east to west at speed 460m/s.

1- What is car speed from inertial ref.frame?
zero?How is zero if car travel in curved line?

2-What is car speed from rotational frame ?

3-How much is car weight,compare to car travel to east at 460m/s and when not moving,speed=0?
Which reference frame I must choose to find out what is car weight?

 

[PeterDonis]

Moderator's note: Moved thread to homework forum.

 

[PeterDonis]

@John Mcrain have you attempted to answer these questions yourself?

 

[haruspex]

1- What is car speed from inertial ref.frame?
How is zero if car travel in curved line?

The question is meaningless. Inertial reference frames can have any constant velocity relative to each other. Unless a specific reference frame is given it would only make sense to ask what the acceleration is.
To get an answer zero, the frame would be "a non-rotating frame centred on the centre of the Earth". In such a frame, is the car moving on a curved line?

3-How much is car weight,compare to car travel to east at 460m/s and when not moving,speed=0?
Which reference frame I must choose to find out what is car weight?

You don't need a frame to assess the weight. It equals the normal force from the road,

 

[John Mcrain]

To get an answer zero, the frame would be "a non-rotating frame centred on the centre of the Earth". In such a frame, is the car moving on a curved line?
,

Yes looking for space car is not moving,and looking from Earth surface plane is moving in curved path.
So that mean centripetal force is zero looking form space.
But why both frames don't get same results?

Is Earth rotating frame for plane?

 

[Kyouran]

Yes looking for space car is not moving,and looking from Earth surface plane is moving in curved path.
So that mean centripetal force is zero looking form space.
But why both frames don't get same results?

Is Earth rotating frame for plane?

No acceleration doesn't mean the forces are zero. It simply means that the sum of the forces (i.e. the force resultant) is zero. That said, there is indeed no notion of a "centripetal force" in this case as you need to be moving on a curved path for that. You will simply have force balance between the gravitational force and the normal force from the surface.

In a rotating frame, you will have (among others, like e.g. the coriolis force, which is not important here) an additional force called the "centrifugal force", which is actually not a real force but acts to account for the reduction in apparent gravity due to the frame rotating, i.e. the apparent gravity is the vectorial sum of the real gravitational force and the centrifugal force. This is why g = 9.83 m/s^2 at the poles and only 9.78 m/s^2 at the equator. By adding such "inertial forces" we can transform a rotating frame into an inertial frame, as Newton's laws are only valid for such a frame.

 

[haruspex]

the coriolis force, which is not important here)

On the contrary, it is crucial here.

Yes looking for space car is not moving,and looking from Earth surface plane is moving in curved path.
So that mean centripetal force is zero looking form space.
But why both frames don't get same results?

Is Earth rotating frame for plane?

Let the gravitational attraction of the Earth on the car be W.
In the frame of the rotating Earth you have to consider the centrifugal and Coriolis forces: https://en.m.wikipedia.org/wiki/Coriolis_force.
The centrifugal force, $mR\Omega^2$, points up. Here, R is the radius of the Earth and $\Omega$ is its rotation rate.
The Coriolis force is given by $-2m\vec\Omega\times\vec v$, where v is the velocity of the car in the rotating frame.
If the car were going West to East, this would also point up, giving a net upward force in the rotating frame of $3mR\Omega^2$. To an observer in that frame, the car has an acceleration $R\Omega^2$ towards the Earth's centre. For all this to add up correctly, the normal force from the ground would have to be $W-4mR\Omega^2$. Back in the inertial frame, the car's speed is $2R\Omega$, so the centripetal force is $4mR\Omega^2$, giving the same result.

But here the car is going East to West, so the Coriolis force points down, and its magnitude is $2mR\Omega^2$. Gravitational force plus centrifugal plus Coriolis gives a downward total of $W-mR\Omega^2+2mR\Omega^2=W+mR\Omega^2$. To the observer on the ground, the car's acceleration is the same as in the W to E case, so the normal force must be just $W$. In the inertial frame the car is stationary, giving the same result.

 

[Kyouran]

On the contrary, it is crucial here.

My bad, you are right it also has to be taken into account.

 

[John Mcrain]

On the contrary, it is crucial here.

Let the gravitational attraction of the Earth on the car be W.
In the frame of the rotating Earth you have to consider the centrifugal and Coriolis forces: https://en.m.wikipedia.org/wiki/Coriolis_force.
The centrifugal force, $mR\Omega^2$, points up. Here, R is the radius of the Earth and $\Omega$ is its rotation rate.
The Coriolis force is given by $-2m\vec\Omega\times\vec v$, where v is the velocity of the car in the rotating frame.
If the car were going West to East, this would also point up, giving a net upward force in the rotating frame of $3mR\Omega^2$. To an observer in that frame, the car has an acceleration $R\Omega^2$ towards the Earth's centre. For all this to add up correctly, the normal force from the ground would have to be $W-4mR\Omega^2$. Back in the inertial frame, the car's speed is $2R\Omega$, so the centripetal force is $4mR\Omega^2$, giving the same result.

But here the car is going East to West, so the Coriolis force points down, and its magnitude is $2mR\Omega^2$. Gravitational force plus centrifugal plus Coriolis gives a downward total of $W-mR\Omega^2+2mR\Omega^2=W+mR\Omega^2$. To the observer on the ground, the car's acceleration is the same as in the W to E case, so the normal force must be just $W$. In the inertial frame the car is stationary, giving the same result.

Can you explain how determine direction of coriolis force, why coriolis force points down or up when go east or west?
Is Eotovos effect just vertical component of coriolis effect?

 

[haruspex]

Is Eötvös effect just vertical component of coriolis effect?

Yes.
For more details follow the link I posted.

 

[John Mcrain]

Yes.
For more details follow the link I posted.

Does non-equatorial latitudes when travel at east or west has horizontal coriolis component?
Also I have problem using right hand rule..

https://www.physicsforums.com/threads/direction-of-coriolis-force.994368/

 

[jbriggs444]

Does non-equatorial latitudes when travel at east or west has horizontal coriolis component?
Also I have problem using right hand rule..

https://www.physicsforums.com/threads/direction-of-coriolis-force.994368/

Let us get some basics straight before getting into the details.

The Coriolis force will be at right angles to the rotation vector. That vector points up out of the north pole. So the Coriolis force will be at right angles to that. Of course, this still leaves a whole plane full of angles where the Coriolis force vector can point -- a plane going through your current line of latitude.

The Coriolis force will be at right angles to the velocity vector. We've specified that your velocity vector is east-west. This restricts the Coriolis force to be acting either north-south or vertically toward the Earth's center or anywhere on the plane defined by those two.

The Coriolis force in your case must be in the intersection of those two planes. That means that it will be directly toward (or away from) the Earth's rotational axis along the plane defined by your latitude.

That [somewhat] vertical force will have a north-south component. If you are near the equator, the force will be almost purely vertical and the north-south component will be small. If you are near the pole, the force will be almost purely horizontal and the north-south component will be nearly the entire force.

The purely vertical component of Coriolis can usually be ignored. It amounts to a fraction of a percentage point variation in the apparent force of gravity.

As for the right hand rule, just remember that things always deflect anti-spinward under Coriolis. In the Northern hemisphere, the Earth rotates counter-clockwise as seen from above. So things moving on the Earth deflect clockwise. In the Southern hemisphere, the Earth rotates clockwise as seen from above. So things moving on the Earth deflect counter-clockwise.

There is an extra layer of reversal for [idealized] toilet bowls and low pressure zones. Fluid coming in from the edges is deflected clockwise in the northern hemisphere. But the result is a counter-clockwise rotation of the low pressure zone or idealized toilet bowl.

 

[John Mcrain]

Let us get some basics straight before getting into the details.

The Coriolis force will be at right angles to the rotation vector. That vector points up out of the north pole. So the Coriolis force will be at right angles to that. Of course, this still leaves a whole plane full of angles where the Coriolis force vector can point -- a plane going through your current line of latitude.

The Coriolis force will be at right angles to the velocity vector. We've specified that your velocity vector is east-west. This restricts the Coriolis force to be acting either north-south or vertically toward the Earth's center or anywhere on the plane defined by those two.

The Coriolis force in your case must be in the intersection of those two planes. That means that it will be directly toward (or away from) the Earth's rotational axis along the plane defined by your latitude.

That [somewhat] vertical force will have a north-south component. If you are near the equator, the force will be almost purely vertical and the north-south component will be small. If you are near the pole, the force will be almost purely horizontal and the north-south component will be nearly the entire force.

The purely vertical component of Coriolis can usually be ignored. It amounts to a fraction of a percentage point variation in the apparent force of gravity.

As for the right hand rule, just remember that things always deflect anti-spinward under Coriolis. In the Northern hemisphere, the Earth rotates counter-clockwise as seen from above. So things moving on the Earth deflect clockwise. In the Southern hemisphere, the Earth rotates clockwise as seen from above. So things moving on the Earth deflect counter-clockwise.

There is an extra layer of reversal for [idealized] toilet bowls and low pressure zones. Fluid coming in from the edges is deflected clockwise in the northern hemisphere. But the result is a counter-clockwise rotation of the low pressure zone or idealized toilet bowl.

1) Which finger represent what in right hand rule?
2) Why car travel due East or due West at any non equator latitude experience any North or South coriolis force,is any intuintive explanation why this happend?

 

[jbriggs444]

1) Which finger represent what in right hand rule?

Personally, I never use the right hand rule. I work out the deflection for something moving North and, having reasoned it all out in the past, remember that if something moving North deflects clockwise then something moving south, east or west will do so as well.

2) Why car travel due East or due West at any non equator latitude experience any North or South coriolis force,is any intuintive explanation why this happend?

Asked and answered.

For an east-west mover not at the equator, Coriolis points straight to the rotational axis, not straight down toward the center of the Earth. That means that it has a non-zero north-south component.

You might ask why a bug crawling along the grooves of a spinning phonograph experiences a Coriolis force. That is because his seemingly constant velocity along the phonograph is actually changing direction. That takes a real radial force. The bug feels the real force but sees no change in his direction. So he invents a Coriolis force that compensates for that real force.

Or just consider the math. If you have a free falling object with a non-zero velocity and you use a rotating coordinate system then that velocity vector will be constantly rotating. That takes a [fictitious] acceleration which is constantly rotating to arrange.

 

[John Mcrain]

Personally, I never use the right hand rule. I work out the deflection for something moving North and, having reasoned it all out in the past, remember that if something moving North deflects clockwise then something moving south, east or west will do so as well.

Asked and answered.

For an east-west mover not at the equator, Coriolis points straight to the rotational axis, not straight down toward the center of the Earth. That means that it has a non-zero north-south component.

You might ask why a bug crawling along the grooves of a spinning phonograph experiences a Coriolis force. That is because his seemingly constant velocity along the phonograph is actually changing direction. That takes a real force. The bug feels the real force but sees his path as being a straight line. So he invents a Coriolis force that compensates.

Do you agree with below explantion?

" Why wind in the East direction at non equator latitude is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation:

F=mv2/r

If we re-arrange the above equation, we can find r in terms of v, and we arrive at:

r=mv2/F

This tells us that as velocity increases, the radius required to maintain the orbit also increases.

Now let's apply this concept to winds on the Earth. If we feel no wind on the Earth, then the air in the atmosphere is traveling at the same velocity as the Earth. The Earth is naturally spinning towards the East.

In the case of an additional Eastward wind felt on the Earth, this wind has effectively increased its velocity, and therefore the above equation tells us that the radius of orbit must increase as well. Radius in this case is the distance, measured perpendicularly of the Earth's axis, between the axis and the wind.

In order for the radius to increase, the wind moves southwards, where the radius is larger.

Similarly, wind moving in the West direction, is moving in the direction opposite of that to the Earth, and therefore its velocity is decreased. Consequently this wind moves towards the North, where the radius is less.

The above image shows what happens. The wind moving East begins to expand its radius, thus moving outwards. Gravity pulls it back, and the wind moves South, in order to maintain the larger radius required for its increased velocity."

 

[jbriggs444]

I'd considered attempting an explanation that invoked centripetal acceleration and mv²/r. Yes, I agree that such an argument works.

 

[John Mcrain]

The Coriolis force will be at right angles to the rotation vector. That vector points up out of the north pole. So the Coriolis force will be at right angles to that. Of course, this still leaves a whole plane full of angles where the Coriolis force vector can point -- a plane going through your current line of latitude.

The Coriolis force will be at right angles to the velocity vector. We've specified that your velocity vector is east-west. This restricts the Coriolis force to be acting either north-south or vertically toward the Earth's center or anywhere on the plane defined by those two.

The Coriolis force in your case must be in the intersection of those two planes. That means that it will be directly toward (or away from) the Earth's rotational axis along the plane defined by your latitude.

In the Southern hemisphere, the Earth rotates clockwise as seen from above.

Do you have some diagram or better animation of coriolis force as prepedicular vector to omega and v?

 

[jbriggs444]

Do you have some diagram or better animation of coriolis force as prepedicular vector to omega and v?

No. I simply know that the cross product of two things is always at right angles to both things.

 

[Kyouran]

You can always define a coordinate system and work out the cross products. Put the origin of the frame at the center of earth, with the z-axis pointing along the rotation axis towards Polaris, the x-axis pointing towards the intersection of your local meridian and the equatorial plane, and the y-axis pointing east to complete the right handed coordinate system. Then something like $- 2m\omega \vec{k} \times v\vec{j}$ (eastward motion) would give you a coriolis force in the $\vec{i}$ direction (i.e. this gives a force radially outward). Motion towards the axis of rotation (as in when moving from the equator to one of the poles) would then give you $- 2m \omega \vec{k} \times v (\vec{-i})$, so that would be a force in the $\vec{j}$ direction (eastward).

 

[John Mcrain]

No. I simply know that the cross product of two things is always at right angles to both things.

I still don't undestand how I can us my fingers to detremine coriolis direction,can you explain me with example below?I am doing something wrong...

On the Northern Hemisphere, the direction of the Coriolis force can be derived with the "three-finger-rule", where the thumb points towards the rotational axis, the index finger in the direction of the velocity and the middle finger in the direction of the force. The blue vector displays the centrifugal force.

 

[jbriggs444]

In the example shown there, v does not appear to be an east-west velocity. I thought we were supposed to be talking about an east-west velocity.

That drawing appears to depict a southward velocity in the Northern hemisphere. The green arrow for velocity points down and out. It is, of course, tangent to the Earth's surface.

The rotation vector is up out of the north pole.

The cross product of the two is at right angles to both -- it points due east. That is at 90 degrees to horizontally due south. It is also at 90 degrees to the north-south polar axis.

The velocity is south. The acceleration is east. This amounts to a clockwise deflection in the northern hemisphere as I had advised you to expect.

For the purposes of Coriolis force, the gray "r" vector and the blue centrifugal force vector can both be ignored.

 

[John Mcrain]

In the example shown there, v does not appear to be an east-west velocity. I thought we were supposed to be talking about an east-west velocity.

That drawing appears to depict a southward velocity in the Northern hemisphere. The green arrow for velocity points down and out. It is, of course, tangent to the Earth's surface.

The rotation vector is up out of the north pole.

The cross product of the two is at right angles to both -- it points due east. That is at 90 degrees to due south. It is also at 90 degrees to the north-south polar axis.

The velocity is south. The acceleration is east. This amounts to a clockwise deflection in the northern hemisphere as I had advised you to expect.

I strugle with concept "prependicular to axis of rotation".So that mean coriolis force is allways lies at latitude plane? Because latitude plane is allways prependicular to axis of rotation?

 

[jbriggs444]

I strugle with concept "prependicular to axis of rotation".So that mean coriolis force is allways lies at latitude plane? Because latitude plane is allways prependicular to axis of rotation?

Yes. That's right. The Coriolis force will always be in the latitude plane of the object.

 

[haruspex]

I strugle with concept "prependicular to axis of rotation".So that mean coriolis force is allways lies at latitude plane? Because latitude plane is allways prependicular to axis of rotation?

Yes.

 

[John Mcrain]

Yes.

How then right hand rule can works?
If velocity is form to east west, index finger is to the west,thumb points towards the rotational axis and middel finger (coriolis force)is prependicular to the thumb which represent latitude plane.
That can be true ,because coriolis force must be parallel to the latitude plane=prependicular to axis of rotation.

 

[haruspex]

thumb points towards the rotational axis

No, parallel to the axis, and pointing North (but note also the minus sign).
https://phys420.phas.ubc.ca/p420_12/tony/Coriolis_Force/Home.html

 

[John Mcrain]

No, parallel to the axis, and pointing North (but note also the minus sign).
https://phys420.phas.ubc.ca/p420_12/tony/Coriolis_Force/Home.html

Then middel finger point in direction of coriolis force or opposite of corioilis force?

 

[haruspex]

Then middel finger point in direction of coriolis force or opposite of corioilis force?

Given the minus sign, it’s opposite.
I don't know why it is given as $-2m\vec{\Omega}\times\vec v$ instead of $2m\vec v\times\vec{\Omega}$. Maybe there's some standard about writing the rotation vector first in cross products.

 

[John Mcrain]

No, parallel to the axis, and pointing North (but note also the minus sign).
https://phys420.phas.ubc.ca/p420_12/tony/Coriolis_Force/Home.html

But then if object velocity(index finger) is travel to the north,how can I set thumb parallel to axis of rotation?
It seems to me that right hand rule can apply only when velocity(index finger) is east or west...

 

[haruspex]

But then if object velocity(index finger) is travel to the north,how can I set thumb parallel to axis of rotation?
It seems to me that right hand rule can apply only when velocity(index finger) is east or west...

We seem to be having the same conversation on two concurrent threads. See my reply on the other.

 

[John Mcrain]

We seem to be having the same conversation on two concurrent threads. See my reply on the other.

Yes I see,now is clear.
Can object feels coriolis force ,same as you can feel centifugal force when driving car in curve?
When birds flight to the north,they bend to the right(east),they path is curved looking from satelite which rotate with Earth speed.
Do birds body feel centrifugal force because they travel at curve,if they fly with accelometer,will he notice Fc?
If we assume that birds flight is fast enough ..

Or this is just optical ilusion which can be seen only from rotating frame,so birds fly in straight line and their body don't feel centifugal force caused by curve path?

 

[jbriggs444]

Can object feels coriolis force ,same as you can feel centifugal force when driving car in curve?

No. You cannot feel centrifugal force either. Nor can you feel gravity. Coriolis is the same.

What you can feel are the strains in your body as the car pushes you into the curve, or as the surface of the Earth pushes you upward against your otherwise downward trajectory.

hen birds flight to the north,they bend to the right(east),they path is curved looking from satelite which rotate with Earth speed.

The effect of Coriolis on birds is entirely negligible. Just as it is on you when you walk down the aisle at the grocery store.

If Coriolis deflects you a half millimeter to the right, you correct course to avoid bumping into the tomato sauce and get to the end of the aisle without incident.

If Coriolis deflects a goose a couple of inches to the right, the goose corrects course to keep on its path. It takes a leftward bank angle too small to notice in order to maintain a straight-line flight path. Birds can deal with flying in a cross-wind; Flying with Coriolis is vastly easier.

Do birds body feel centrifugal force because they travel at curve,if they fly with accelometer,will he notice Fc?
If we assume that birds flight is fast enough ..

Are we talking centrifugal or Coriolis here? Either way, the effect is too small to be noticed.

We can walk on a tilting floor with no problem. We can walk away from the table after eating a big meal without any problem. We can walk away from the toilet without any problem. Those are all things that are equivalent to centrifugal or Coriolis forces -- minor differences in the magnitude or direction of the inertial force of gravity. Things that you deal with without bothering to think about it.

 

[John Mcrain]

The effect of Coriolis on birds is entirely negligible. Just as it is on you when you walk down the aisle at the grocery store.

Instead birds ,use missile.
Accelometer in missle will not show acceleration caused by curved path to the right?
If I am in missile which travel to the north ,will I feel same force as when car going in right turn?

 

[jbriggs444]

Instead birds ,use missile.
Accelometer in missle will not show accelartion caused by curved path to the right?
If I am in missile which travel to the north ,will I feel same force as when car going in right turn?

Again, you cannot feel any centrifugal force in a right turn. None. Nada. Zilch. Zip. Never. Ever.

What you can feel is the car forcing you to deviate from your natural free-fall trajectory. The car supports you against gravity and pushes you around the turn. The stresses from those forces are what you can feel.

An accelerometer in a cruise missile can sense the acceleration resulting when the missile's guidance system commands a slight left bank to compensate for the rightward deflection from Coriolis.

An accelerometer in a ballistic missile would feel nothing. The operator would have commanded a launch angle enough leftward so that Coriolis would result in a hit without any mid-course maneuvering.

 

[John Mcrain]

Again, you cannot feel any centrifugal force in a right turn. None. Nada. Zilch. Zip. Never. Ever.

What you can feel is the car forcing you to deviate from your natural free-fall trajectory. The car supports you against gravity and pushes you around the turn. The stresses from those forces are what you can feel.

Ok will I feel missile force my body to the right ,because missile path is curved to the right looking from rotation frame?

 

[jbriggs444]

Ok will I feel missile force my body to the right ,because missile path is curved to the right looking from rotation frame?

You feel the missile force your body to the left to maintain a "straight" course track over the rotating Earth. The natural un-forced trajectory would be a rightward curve.

This assumes a cruise missile. A ballistic projectile would be in free fall -- nothing to feel.

 

[haruspex]

You cannot feel centrifugal force

It depends what you mean.
Necessarily you work in your own reference frame; in that frame you have no velocity nor acceleration. If I feel an unbalanced horizontal normal force from the side of the car I'm in, my brain invents a countervailing force to explain the lack of acceleration. This I feel as centrifugal force.

For Coriolis, we can use the problem in this thread to create a theoretical example. Suppose I were on the equator of a small planet spinning so fast that it appreciably lowers the normal force from the ground. If I were to run at a comparable tangential speed in the same direction I would notice the normal becoming even less, i.e. Imwould feel as though I weighed less, but if Inwere to run the other way I would feel myself get heavier. You could say that I feel the Coriolis force.

 

[jbriggs444]

You mean with mid-course maneuvering ?

Yes. My assumption (not being a cruise missile expert) is that they follow a straight-line track relative to the rotating Earth and continually adjust their heading so as to keep the target on a fixed bearing. This will involve corrections for cross-wind and Coriolis, at least.

But if missile is shoot slighlty leftward to target then during flight _I don't feel nothing inside?

Yes.

A typical ballistic missile burns all of its fuel promptly and then free-falls toward the target. Hence the term "ballistic". The result is very much like a naval gun. In the northern hemisphere you have to aim left to compensate for Coriolis. A passenger on the [short] flight would be in free fall all the way after engine burnout.

From the rotating frame, the picture is of a missile curving right to hit the target. From the non-rotating frame, the picture is of a missile on a straight-line path where the target rotates left to arrive under the missile at the time of impact.

There is also a vertical component to Coriolis which needs to be managed. Naval gun tables have corrections for both the right-left deflection and the up-down deflection.

 

[John Mcrain]

From the rotating frame, the picture is of a missile curving right to hit the target. From the non-rotating frame, the picture is of a missile on a straight-line path where the target rotates left to arrive under the missile at the time of impact. [There is also a vertical component to Coriolis which needs to be managed. Naval gun tables have corrections for both the right-left deflection and the up-down deflection]

Ok that mean curved path is just "ilusion" which appear in rotating frame..

But what about wind,is low pressure counterclok wise rotation can be seen from inertial(non-rotating) frame too or this is just appeare in rotating frame or in other words ,is wind really rotate counterclok wise?

How is cyclone at north hemisphere look like in inertial frame?

 

[jbriggs444]

Ok that mean curved path is just "ilusion" which appear in rotating frame..

Yes, that is correct.

But what about wind,is low pressure counterclok wise rotation can be seen from inertial(non-rotating) frame too or this is just appeare in rotating frame or in other words ,is wind really rotate counterclok wise?

How is cyclone at north hemisphere look like in inertial frame?

The Earth is really rotating. The air on the Earth, by and large, rotates with the Earth.

This rotation can be amplified. Like the whirlpool in your bathtub or a skater twirling on the ice. If water is drawn to the center or if the skater pulls in her arms, any pre-existing rotation increases in rate in order to conserve angular momentum. That's a real rotation which can be seen from both inertial and rotating frame.

This is the inertial frame's explanation for why low pressure zones rotate in the same direction as the Earth (clockwise in the northern hemisphere). The rotating frame notes that Coriolis deflects the inward-bound air. The air in the south is deflected east. The air in the east is deflected north. The air in the north is deflected west. The air in the west is deflected south. All of those are clockwise deflections. The result is a counter-clockwise circulation. [Edit, after correction by @John Mcrain]

 

[John Mcrain]

This is the inertial frame's explanation for why low pressure zones rotate in the same direction as the Earth (clockwise in the northern hemisphere).

Low pressure-cyclone rotate in anti clock wise direction at north hemisphere.
This picture below is from rotating frame,camera is rotate with earth.
Perhaps this piture will looks like something different if camera is fixed at one point in space(inertial frame)

Agree?

 

[jbriggs444]

Low pressure-cyclone rotate in anti clock wise direction at north hemisphere.
This picture below is from rotating frame,camera is rotate with earth.
Perhaps this piture will looks like something different if camera is fixed at one point in space(inertial frame)

Agree?

View attachment 270361

Right you are. Earth rotates counter-clockwise and so do low pressure zones. Brain fade on my part.

 

[John Mcrain]

Right you are. Earth rotates clockwise and so do low pressure zones. Brain fade on my part.

How woould cyclon picture looks like in inertial frame(if camera is fixed at space)?

 

[jbriggs444]

How woould cyclon picture looks like in inertial frame(if camera is fixed at space)?

You have the picture of a hurricane. A snapshot is a snapshot. Reference frame is irrelevant.

[At least until we get to special relativity and the relativity of simultaneity]

 

[John Mcrain]

You have the picture of a hurricane. A snapshot is a snapshot. Reference frame is irrelevant.

[At least until we get to special relativity and the relativity of simultaneity]

Ok does video of hurricane from fixed camera at space(inertail frame ) will looks the same as when camera is rotating with earth?

 

[jbriggs444]

Ok does video of hurricane from fixed camera at space(inertail frame ) will looks the same as when camera is rotating with earth?

I am not sure that I understand the question. You understand that rotating the camera makes the video rotate. And that that is all that it does?

You understand that the rotation rate of the hurricane exceeds that of the earth?

 

[John Mcrain]

I am not sure that I understand the question. You understand that rotating the camera makes the video rotate. And that that is all that it does?

You understand that the rotation rate of the hurricane exceeds that of the earth?

You can record video of ball throwing between two person at merry go around with camera moving with him(rotating frame) or you can set camera fixed at Earth (inertial frame)and then record ball at merry go around.you will get different video of ball paths

Same with Earth and hurricane?

 

[jbriggs444]

You can record video of ball throwing between two person at merry go around with camera moving with him(rotating frame) or you can set camera fixed at Earth (inertial frame)and then record ball at merry go around.you will get different video of ball paths

Same with Earth and hurricane?

What, exactly, would you be tracking on the hurricane that is the analogue of a ball? No, it is not the same.

 

[John Mcrain]

What, exactly, would you be tracking on the hurricane that is the analogue of a ball? No, it is not the same.

My question is would hurrican rotate like this if we watch at Earth from inertial frame?

 

[jbriggs444]

My question is would hurrican rotate like this if we watch at Earth from inertial frame?

Would it rotate like what? Yes, it rotates. Yes, it rotates faster than the Earth. Yes, the inside rotates faster than the outside.

What is it that you are trying to ask?