Frame of reference question: Car traveling at the equator

[John Mcrain]

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

When bullet fly to east his speed form inertial frame is 926m/s and when fly to west his speed is 0m/s,and if I put this numbers into centripetal formula mv2/r, I get higher centripetal force to east..

I know i am doing something wrong ,because my results are not same as in rotating frame

 

[jbriggs444]

When bullet fly to east his speed form inertial frame is 926m/s and when fly to west his speed is 0m/s,and if I put this numbers into centripetal formula mv2/r, I get higher centripetal force to east..

The "r" that you are putting into the formula. You are using the radius of the earth, right?

The "r" that belongs in the centripetal force formula is the radius of curvature of the projectile's path, right?

What makes you think that the two are identical?

 

[John Mcrain]

The "r" that you are putting into the formula. You are using the radius of the earth, right?

The "r" that belongs in the centripetal force formula is the radius of curvature of the projectile's path, right?

What makes you think that the two are identical?

Because I assume that bullet follow Earth radius for make calculation easier..
But I know that they are no identical..

So how to set equation for east and west bullet shoot for inertial frame,can you write?

 

[jbriggs444]

Because I assume that bullet follow Earth radius for make calculation easier..

Bullets do not follow Earth radius. Not even close. In order to follow Earth radius, a bullet would need to attain orbital velocity -- about 7.8 km/sec.

 

[John Mcrain]

Bullets do not follow Earth radius. Not even close. In order to follow Earth radius, a bullet would need to attain orbital velocity -- about 7.8 km/sec.

Yes I know that ,bullet radius is smaller then earth...

So how can I set equation for bullet?

 

[jbriggs444]

Yes I know that ,bullet radius is smaller then earth...

So how can I set equation for bullet?

It depends. What are you trying to calculate?

 

[John Mcrain]

It depends. What are you trying to calculate?

I want prove with equation why bullet hit east target high and west traget low using inertial frame.
in same maner like I do for aircraft and car case,in post #68

 

[jbriggs444]

I want prove with equation why bullet hit east target high and west traget low using inertial frame.
in same maner like I do for aircraft and car case,in post #68

I repeat: What is it that you want to calculate?

I am guessing that you want to know how far left and how much further up (or down) you will need to aim to hit your target.

So we have a target a distance d away on a flat and level plane. It is at the same height as the barrel of the rifle. Your muzzle velocity is v. The local acceleration of gravity is g. We start with a baseline situation of a stationary Earth. You may neglect air resistance. You may pretend that the Earth is flat.

How much above the horizontal do you need to aim?

When we get to the next step, by your own stipulation, you will be required to do all calculations in the inertial frame -- the gun will be moving, the target will be moving and the Earth will be spherical.

 

[John Mcrain]

I repeat: What is it that you want to calculate?

In short, is bullet apperent weight smaller when shoot east compare to west?(just like is case with plane or car which travel east,which I proved in post#68 )
I need equation that prove this in INERTIAL FRAME

 

[jbriggs444]

In short, is bullet apperent weight smaller when shoot east compare to west?(just like is case with plane or car which travel east,which I proved in post#68 )
I need equation that prove this in INERTIAL FRAME

You have forbidden us to use the rotating frame. That means that the concept of "apparent weight" is off the table. Not allowed by your own rules!

In the inertial frame, both shooter and target are moving. The moving target means that one must lead the target, aiming counter-clockwise (left of target). The moving shooter means that one must compensate for the shooter's movement as well. This is also a leftward correction.

If the target is eastward, there is also a downward correction required because the shooter is moving up and the target is moving down relative to a point at rest in the middle. If the target is westward, there is an upward correction for the same reaason.

If you want an equation, you have to put in the work. Let's see the equation for a stationary shot.

 

[John Mcrain]

You have forbidden us to use the rotating frame. That means that the concept of "apparent weight" is off the table. Not allowed by your own rules!

@ Haruspex call apparent weight the normal force .
normal force at plane is lift,for car is ground that push you up...
So what is normal force at bullet? don't exist or what?
I hate inertial frame grrrrrrrrrrrrrrrrrrrrrrrrrrrrr grrrrrrrrrrrrrr, I'm starting to go crazy

 

[jbriggs444]

@ Haruspex call apparent weight the normal force .
normal force at plane is lift,for car is ground that push you up...
So what is normal force at bullet? don't exist or what?

Correct. A bullet is ballistic. It is in a free fall trajectory. There is no normal force.

 

[John Mcrain]

Correct. A bullet is ballistic. It is in a free fall trajectory. There is no normal force.

So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?

 

[John Mcrain]

last 30 posts I ask for that simple equation...
there is no any " mv2/r" in bullet case

Fg = ma

 

[haruspex]

So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?

For a projectile, we can observe the trajectory and compare it with the known value for its gravitational acceleration. Doesn't matter whether it's parabolic (g effectively constant), or a satellite in orbit.

In an inertial frame there is no discrepancy, so no difference between true weight, mg or ##\frac{MmG}{r^2}##, and the apparent weight deduced from its trajectory.

In a rotating or accelerating frame, if we don't consider centrifugal and coriolis forces then we will deduce a different weight (i.e. gravitational force). This is the apparent weight in that reference frame. But if we do take those forces (and the Euler force if the rotation is not uniform) into account then once again we will deduce mg as the weight.

 

[John Mcrain]

For a projectile, we can observe the trajectory and compare it with the known value for its gravitational acceleration. Doesn't matter whether it's parabolic (g effectively constant), or a satellite in orbit.

In an inertial frame there is no discrepancy, so no difference between true weight, mg or ##\frac{MmG}{r^2}##, and the apparent weight deduced from its trajectory.

In a rotating or accelerating frame, if we don't consider centrifugal and coriolis forces then we will deduce a different weight (i.e. gravitational force). This is the apparent weight in that reference frame. But if we do take those forces (and the Euler force if the rotation is not uniform) into account then once again we will deduce mg as the weight.

So final conlusion is:

Bullet don't has any reduction in weight in any frame, just like car or plane have...

 

[jbriggs444]

So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?

WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?

 

[John Mcrain]

WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?

example:
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

so car normal force "weight" is reduced by 136N when trevel east,which is not case in bullet

proffesor Matt say centripetal force is not real force,he say it is sum of forces in radial direction at 6:37

 

[jbriggs444]

example:
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

so car normal force "weight" is reduced by 136N when trevel east,which is not case in bullet

proffesor Matt say centripetal force is not real force,he say it is sum of forces in radial direction at 6:37

I repeat: WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?

I am not asking for a formula. I am not asking for someone else's video. I am asking for a definition. What do you mean when you use the term? You have used the term. Surely that means that you know what it means when you use it.

 

[John Mcrain]

I am asking for a definition

I am confused with all this terminology and you ask me what is definition?
before this topic i didnt know that i am talking about normal force,because I didnt know what is normal force...I just understand that if airplane fly to east ,lift at wings is reduced compere to fly to west.That mean plane can fly with lower AoA,this reduce induced drag,lower fuel consumption etc etc

so "my aircraft logic" was;weight is reduced,because lift must be equal to weight

So my definition of weight was,it has opposite direction from lift and has equal magnitude as lift,to keep plane in level flight.

 

[PeterDonis]

if airplane fly to east ,lift at wings is reduced compere to fly to west

Why do you think this is true?

 

[jbriggs444]

I am confused with all this terminology and you ask me what is definition?
before this topic i didnt know that i am talking about normal force,because I didnt know what is normal force...I just understand that if airplane fly to east ,lift at wings is reduced compere to fly to west.That mean plane can fly with lower AoA,this reduce induced drag,lower fuel consumption etc etc

so "my aircraft logic" was;weight is reduced,because lift must be equal to weight

So my definition of weight was,it has opposite direction from lift and has equal magnitude as lift,to keep plane in level flight.

So for you, "weight" is the real, physical support force that would be required to keep the vertical component of an object's velocity constant.

Is that correct?

 

[haruspex]

So final conlusion is:

Bullet don't has any reduction in weight in any frame, just like car or plane have...

Your difficulty seems to be that you keep switching what you mean by weight.

On the one hand, there is mg (or more generally ##\frac{GMm}{r^2}##), the force of gravity. On the other, there is what an observer may measure as its value, perhaps not taking into account all the complications.
If you stand on the bathroom scales, what you directly measure is the normal force the scales exert. This is your apparent weight, and it may be less than mg because of Earth's rotation, i.e. centripetal force.
If you observe a ball or bullet in flight, you can deduce a value of g from its acceleration in your reference frame, but the value will be inaccurate if you do not adjust for Coriolis as well as centripetal force.

Bottom line, the true weight is always mg; the apparent weight differs only because the observer in a non inertial frame fails to make all the appropriate corrections. It makes no difference whether the object is a car, bullet, plane...

 

[John Mcrain]

So for you, "weight" is the real, physical support force that would be required to keep the vertical component of an object's velocity constant.

I don't understand well this combinations of words.I am not physicist.

I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...whatever that mean..
So this was my starting point when asked question...

 

[jbriggs444]

I don't understand well this combinations of words.I am not physicist.

I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...whatever that mean..
So this was my starting point when asked question...

If you are getting into fine details of centrifugal force and rotating frames of reference then fine details on the definition of weight matter.

You cannot have it both ways. You cannot ask for an understanding and then disavow any basis for understanding.

 

[haruspex]

I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...

Yes, but that's only to a first approximation.
Let's assume this refers to the true weight, i.e. mg.

In an inertial frame, the plane has to follow the curvature of the Earth, so it has a small downward acceleration. To provide the centripetal force necessary, the lift must be a little less than mg.

In the frame of reference of the plane, there is no acceleration, so no centripetal force. Instead, we should apply the centrifugal correction since we know we are in a rotating frame. This gives centrifugal (up) + lift = mg, so again, lift is a bit less than mg.

An observer on the ground observes a centripetal acceleration (but different from that measured in the inertial frame) and needs to apply both centrifugal and coriolis corrections. As in the other frames, lift will turn out to be a bit less than mg.

 

[John Mcrain]

In the frame of reference of the plane, there is no acceleration, so no centripetal force. Instead, we should apply the centrifugal correction since we know we are in a rotating frame. This gives centrifugal (up) + lift = mg, so again, lift is a bit less than mg.

Why in this frame don't exist coriolis force?

And what is difference with this frame and frame of observer from gorund?

 

[haruspex]

Why in this frame don't exist coriolis force?

In the plane's own reference frame it is stationary, so no acceleration (no centripetal) and zero velocity (no Coriolis).

And what is difference with this frame and frame of observer from gorund?

The observer on the ground sees the plane accelerating radially and as having a nonzero velocity.

 

[John Mcrain]

In the plane's own reference frame it is stationary, so no acceleration (no centripetal) and zero velocity (no Coriolis).

I allways think that centripetal force exist in every frame.
Isnt centirpetal force in plane own ref. frame gravity=mg?in this picture write it is present in every ref.frame ?

 

[haruspex]

I allways think that centripetal force exist in every frame.
Isnt centirpetal force in plane own ref. frame gravity=mg?in this picture write it is present in every ref.frame ?
View attachment 270511

It's wrong - where is it from?
Edit: Hi @A.T., I see this graphic is from an old post of yours.

In an observer's frame, the centripetal force is that component of the net force which accounts for an observed curved path. In the astronaut's own frame, the astronaut is stationary. No curved path to account for. Moreover, the centrifugal force exactly balances the normal force from the space station wall, so the net force is zero. This fits with the astronaut's view of being stationary.

 

[John Mcrain]

It's wrong - where is it from?

Picture is wrong?so centripetal force don't exist in every frame?

In an observer's frame, the centripetal force is that component of the net force which accounts for an observed curved path.

You mean on observer at Earth ground?

What is definition of centripetal force:
1)every force which act towards center of rotation ?
2) net force of all forceses which act in radial direction?
3)net force of all forces which act towards center of rotation

 

[haruspex]

Picture is wrong?so centripetal force don't exist in every frame?
You mean on observer at Earth ground?

What is definition of centripetal force:
1)every force which act towards center of rotation ?
2) net force of all forceses which act in radial direction?
3)net force of all forces which act towards center of rotation

The definition I use, and gave earlier, is that if the sum of the forces on the object is ##\vec F## and the velocity is ##\vec v## then the centripetal force is that component of ##\vec F## which is orthogonal to ##\vec v##.

The instantaneous centre of rotation can be deduced from the velocity and the centripetal acceleration, so I see no distinction between your options 2 and 3.

In a frame of reference in which the velocity has a constant direction the centripetal force must be zero. In particular, in the object's own frame there is necessarily no acceleration, so no centripetal force.

I think a lot of confusion arises because every student's introduction to centripetal force is in terms of objects constrained to circular paths by some tether. It conveys the notion that the radius is the given and the centripetal force is a consequence.

 

[John Mcrain]

In a frame of reference in which the velocity has a constant direction the centripetal force must be zero. In particular, in the object's own frame there is necessarily no acceleration, so no centripetal force.

In object own frame velocity is zero,so how than centrifugal force exist at object which is not moving at all?
How can we explain this contradiction?

 

[jbriggs444]

In object own frame velocity is zero,so how than centrifugal force exist at object which is not moving at all?
How can we explain this contradiction?

Centrifugal force does not depend on an object's velocity. It is a function of the frame's rotation rate and the object's distance from the center of rotation.$$F=m\omega^2r$$
The object's velocity (in the rotating frame) does not appear in that formula.

 

[John Mcrain]

Centrifugal force does not depend on an object's velocity. It is a function of the frame's rotation rate and the object's distance from the center of rotation.$$F=m\omega^2r$$
The object's velocity (in the rotating frame) does not appear in that formula.

How can you say that centrifugal force don't depend on velocity?

ω depend on velocity ,if you increase velocity ,you also increase ω..
ω =v/r

F=mω^2 r ... F=mv^2 /r ...these two formulas are same