# Frame of reference question: Car traveling at the equator

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?

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PeterDonis
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Moderator's note: Moved thread to homework forum.

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

haruspex
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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,

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 dont get same results?

Is earth rotating frame for plane?

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 dont 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
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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 dont 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.

On the contrary, it is crucial here.
My bad, you are right it also has to be taken into account.

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
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Is Eötvös effect just vertical component of coriolis effect?
Yes.

jbriggs444
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Does non-equatorial latitudes when travel at east or west has horizontal coriolis component?
Also I have problem using right hand rule..

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.

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
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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?

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.

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• PeroK
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.

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

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
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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.

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).

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

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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
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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
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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.

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.