Frame of reference question: Car traveling at the equator

[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?

 

[John Mcrain]

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?

In inertial frame coriolis effect don't exist,so why then we agian see wind bend to the right which gives counterclok wise rotation at cyclon even in inertial frame?
If you look at missile form inertial frame ,missile is going in straight line ,so why wind will not do the same?
Why is my question so strange,isnt my question logic?

 

[PeterDonis]

Earth rotates clockwise

I think you mean counterclockwise, at least if you are looking down on the Earth from above the North Pole.

 

[jbriggs444]

In inertial frame coriolis effect don't exist,so why then we agian see wind bend to the right which gives counterclok wise rotation at cyclon even in inertial frame?
If you look at missile form inertial frame ,missile is going in straight line ,so why wind will not do the same?
Why is my question so strange,isnt my question logic?

The individual parcels of wind are doing the same. They move in straight lines unless deflected by a real force.

The individual parcels of wind start at rest relative to a rotating earth. They are already rotating about a common center. If all things were equal (uniform temperature, etc), this would be an equalibrium situation. No wind would be blowing anywhere. The air would stay in a stable rotation, at rest above the rotating Earth. The Earth itself would (and does) relax into an ellipsoidal shape so that the required centripetal acceleration is provided for.

But if you apply a real physical force to draw a air parcels in toward the center of rotation, that inward movement means that their original tangential velocity at a large radius is now a greater tangential velocity at a smaller radius. The rotation rate of all of the parcels about their common center has increased. The air is no longer at rest relative to the rotating Earth. Winds blow.

 

[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.

What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?

 

[jbriggs444]

What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?

Why does there have to be only one explanation for any given effect?

 

[haruspex]

What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?

It depends on your frame of reference. In the frame of reference rotating with the earth, the Eötvös effect (Coriolis) applies and you can consider that the reason. In an inertial frame there is no such effect, but the target is on a curved trajectory.

 

[John Mcrain]

It depends on your frame of reference. In the frame of reference rotating with the earth, the Eötvös effect (Coriolis) applies and you can consider that the reason. In an inertial frame there is no such effect, but the target is on a curved trajectory.

How can I set equation why bullet hit top of target using inertial frame?
I know that bullet absolute velocity is incresed (earth speed + bullet ground speed) so centripetal force is increased also,but centripetal force point down (not up), toward center

?

Why does there have to be only one explanation for any given effect?

Yes it can be more...

 

[haruspex]

How can I set equation why bullet hit top of target using inertial frame?

The point from which the bullet is fired has some instantaneous velocity.
The bullet is aimed as though the target had the same velocity, but it doesn't. If firing East, the target's velocity is, from the shooter's perspective, angled slightly down, so by the time the bullet arrives the target will have descended relative to the trajectory.
Conversely, if firing West, the target's velocity is tipped up, causing it to rise above the trajectory.

 

[jbriggs444]

I know that bullet absolute velocity is incresed (earth speed + bullet ground speed) so centripetal force is increased

Why do you say that centripetal force is increased? We need to examine that reasoning because it is faulty.

 

[haruspex]

so centripetal force is increased

Centripetal force is not an extra applied force. It is that component of the resultant of all the forces that are applied that is normal to the velocity. In this case, it is just gravity, and that is not about to change.
What you mean is that in order to hit the target on the anticipated trajectory the centripetal force would need to increase - but it doesn't.

 

[John Mcrain]

Centripetal force is not an extra applied force. It is that component of the resultant of all the forces that are applied that is normal to the velocity. In this case, it is just gravity, and that is not about to change.
What you mean is that in order to hit the target on the anticipated trajectory the centripetal force would need to increase - but it doesn't.

You proved in rotating frame how car/aircraft or bullet travel at east has reduced weight because of corioils net upward force.
In general I have trouble to prove reduction in weight of car/aircraft/bullet in inertial frame.

How can I set equations for this?
if car/aircraft/bullet travel et east with 463m/s (same velocity as Earth rotation )

Car: mg - N = mv2/r (where v is apsolute speed of car,v=car ground speed + Earth speed)
weight(mg) = mv2/r + N

Aircraft: mg - L = mv2/r (where v is apsolute speed of aicraft,v=aircraft ground speed + Earth speed)
weight(mg) = mv2/r + L

Bullet: mg = mv2/r (where v is apsolute speed of bullet,v=bullet ground speed + Earth speed)

in all three cases when v is increased,because of eastward travel(absolute speed increase) then weight(mg) is increased also,which is not consistent with results from rotating frame!

So what I am doing wrong?

 

[haruspex]

in all three cases when v is increased,because of eastward travel(absolute speed increase) then weight(mg) is increased

No, mg, the true weight, does not change.
How do you measure your weight?

 

[John Mcrain]

No, mg, the true weight, does not change.
How do you measure your weight?

I stay on weight scale.

Than what is changed in inertial frame?
Do both frames must get same final results?

 

[haruspex]

I stay on weight scale.

And what force does a scale actually measure?

 

[John Mcrain]

And what force does a scale actually measure?

it show normal force = mg - mv2/r (centrifugal force)

But can I call it "mv2/r" centrifugal force?

 

[haruspex]

it show normal force

Right, the apparent weight is the normal force, and what all of these questions concern is the change in apparent weight, not the actual weight.
Redo your work in post#61 on that understanding.

 

[John Mcrain]

Right, the apparent weight is the normal force, and what all of these questions concern is the change in apparent weight, not the actual weight.
Redo your work in post#61 on that understanding.

But is my "mv2/r " centifugal force or centripetal force? that confused me

in one way, must be centripetal because I am working in inertial frame
in other way, doesn't make sense is centripetal because it act toward center of Earth so must be add it to mg!
I am fighting with terminology

 

[John Mcrain]

Redo your work in post#61 on that understanding.

aircraft/bullet/car speed = 463m/s ,weight=1000kg
r=6371km
inertial frame;

Car: N = mg - mv2/r .
trevel EAST 1000kgx9.81 - 1000kg x (926m/s)2 / 6371000m...N=9675 Newtons
trevel WEST 1000kgx9.81 - 1000kg x (0m/s)2 / 6371000m...N=mg=9810 Newtons
136N/9810N= 1.38% Car apparent weight is 1.38% less when trevel EAST

aircraft : same calulation as car case, just normal force(N) I will call L (lift)
(neglect increased of radius becasue of fligt altitude)

bullet: again looking from inertial frame bullet trevel et east with 926m/s and to west 0m/s
but here is see only one force,gravity mg,
so mg=mv2/r ??
so if I put v for east and west ,what now?
has bullet decreased apparent weight like car and plane when trevel east??

 

[haruspex]

But is my "mv2/r " centifugal force or centripetal force? that confused me

in one way, must be centripetal because I am working in inertial frame
in other way, doesn't make sense is centripetal because it act toward center of Earth so must be add it to mg!
I am fighting with terminology

Centripetal force is a component of the net force. Specifically, it is the component which is normal to the velocity. That definition works in all circumstances, including non inertial frames.
This leads the object to travel in an arc around some point, at least transiently. The radius of that arc is given by $\frac{v_{tangential}^2}{a_{radial }}$.

Suppose there is a satellite circling at the equator at distance r from the Earth's centre, speed v, W to E. It's angular rate is v/r.
A ground based observer perceives the satellite's angular rate as v/r-Ω, so speed as |v-Ωr|, so computes the centripetal acceleration as (v-Ωr)²/r. The net force on it is calculated by taking mg and adjusting for centrifugal and Coriolis forces. If there is no fault in the math, this should equal m(v-Ωr)²/r.

In the inertial frame, the centripetal acceleration is calculated as v²/r, and no adjustment is made: the net force is just mg.

Now consider a geostationary satellite. v=Ωr. This makes the observed speed and acceleration zero in the non inertial frame, so no centripetal force and no Coriolis. But the centrifugal correction still applies, namely, mΩ²r. But that is the same as mv²/r.

In sum, centripetal force always equals mv²/r, but centrifugal force only equals that (with opposite sign) when the rotation of the non inertial frame makes the observed tangential velocity zero.

 

[John Mcrain]

In the inertial frame, the centripetal acceleration is calculated as v²/r, and no adjustment is made: the net force is just mg.

How then I write equation for bullet trevel at 463m/s(ground speed)for east and for west ,from inertial frame?
earth speed rotation is also 463m/s