Frame of reference question: Car traveling at the equator

[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

 

[jbriggs444]

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

In the frame that rotates with the object, the object IS NOT MOVING! Its velocity is zero.

Centrifugal force is found in a rotating frame and depends on the rotation rate of the frame, not on the velocity of any object measured against that frame.

Scroll back up a few posts to see your question: You yourself have noted that the object has zero velocity and expressed bewilderment that despite this, it is subject to centrifugal force.

 

[John Mcrain]

In the frame that rotates with the object, the object IS NOT MOVING! Its velocity is zero.

Centrifugal force is found in a rotating frame and depends on the rotation rate of the frame, not on the velocity of any object measured against that frame.

Scroll back up a few posts to see your question: You yourself have noted that the object has zero velocity and expressed bewilderment that despite this, it is subject to centrifugal force.

Rotation rate of frame compare to what?

 

[jbriggs444]

Rotation rate of frame compare to what?

Absolute rotation rate. Measured, for instance, by determining centrifugal force. Or Coriolis. There are also fancier ways like ring laser gyros.

Alternately, one might reference the rotation to the fixed stars and call it a sidereal rotation rate. It turns out that a frame in which the stars are fixed is also one that has zero centrifugal, zero Coriolis and zero Euler forces. So we can treat such a frame as a standard of zero rotation.

 

[John Mcrain]

Can we solve any problem/task with inertial frame,if does why than useing non inertial frames?

 

[jbriggs444]

Can we solve any problem/task with inertial frame,if does why than useing non inertial frames?

Any problem can be solved in any frame. Non inertial frames sometimes simplify matters. Doing weather prediction in the inertial frame would be positively painful because all of the land would be moving at hundreds of miles per hour. If you are trying to predict weather for Chicago, it helps if Chicago stays put.

 

[John Mcrain]

Any problem can be solved in any frame. Non inertial frames sometimes simplify matters. Doing weather prediction in the inertial frame would be positively painful because all of the land would be moving at hundreds of miles per hour. If you are trying to predict weather for Chicago, it helps if Chicago stays put.

Any live creature feel fictitious forces because our own reference frame moving with us all the time.
So when drive car in turn , is better to say that centrifugal force push our body out of turn ,then car seat push our body into turn(centripetal).

 

[jbriggs444]

Any live creature feel fictitious forces because our own reference frame moving with us all the time.
So when drive car in turn , is better to say that centrifugal force push our body out of turn ,then car seat push our body into turn(centripetal).

No live creatures feel fictitious forces. Fictitious forces come with reference frames. Reference frames are mathematical constructs. They have no physical effects.

Both descriptions of the situation are equally valid. Whether it is centrifugal force which keeps us in place in the face of a real physical centripetal force or whether the real physical centripetal force has the effect of accelerating us around a corner in lock-step with the car does not change the nature of the real physical force of the seat on our body.

What we feel are the strains on our body as it adjusts itself to the real physical force. Those real physical strains exist regardless of the reference frame we might use to predict or explain them.

 

[John Mcrain]

No live creatures feel fictitious forces. Fictitious forces come with reference frames. Reference frames are mathematical constructs. They have no physical effects.

Both descriptions of the situation are equally valid. Whether it is centrifugal force which keeps us in place in the face of a real physical centripetal force or whether the real physical centripetal force has the effect of accelerating us around a corner in lock-step with the car does not change the nature of the real physical force of the seat on our body.

What we feel are the strains on our body as it adjusts itself to the real physical force. Those real physical strains exist regardless of the reference frame we might use to predict or explain them.

But why you call centripetal force real physical force and centrifugal not,for men point of view?
They are both equaly real for men in car,just depend what decription you decide to use..

 

[jbriggs444]

But why you call centripetal force real physical force and centrifugal not,for men point of view?
They are both equaly real for men in car,just depend what decription you decide to use..

The centripetal force is [the horizontal component of] the force of the seat on your body. It is real. It is physical. It has a third law partner force: the force of your body on the seat. When we speak of a "centripetal force", we are not speaking about some special type of force. Instead, we are talking about the net force [from whatever source] that acts toward the center of a curved path. The centripetal force may, for example, be the force of road on tires, of seat on body, of centrifuge on test tube or of sling on stone.

The centrifugal force is the fictitious force associated with the adoption of a rotating frame. Its existence or not depends on the coordinate system that you have decided to use. It is fictitious. It has no third law partner force.

If you have adopted an inertial frame then the centrifugal force does not exist. Your body is accelerating around the corner in response to the only two forces there are: gravity and the force of the seat on your body. The net force is inward toward the center of your curved path. Your body accelerates. Newton's second law is upheld.

If you have adopted the rotating frame then the centrifugal force does exist. Your body is at rest and remains at rest throughout the exercise. There are three forces in play: gravity, the force of the seat on your body and centrifugal force. The net of these three forces acting on your body is zero. Your body remains at rest in its seat. Newton's second law is upheld.

 

[John Mcrain]

It has a third law partner force: the force of your body on the seat.

This force is called reactive centrifugal force.

From your logic gravity is also fictitous force.Is Earth in same strange sense accelrate "upward" in same manner like when elevator accelrate up weight scale show increase in reading?

 

[jbriggs444]

This force is called reactive centrifugal force.

From your logic gravity is also fictitous force.Is Earth in same strange sense accelrate "upward" in same manner like when elevator accelrate up weight scale show increase in reading?

Gravity is indeed a fictitious force. But you'll have to wait for General Relativity to make sense of that idea. The fact that you can transform your weight away by adopting a freely falling frame of reference is a key motivator for that theory.

However, that is for another day. We are here discussing classical Newtonian physics. In the Newtonian model, gravity is a real, physical, inertial force. The third law partner of the force of Earth's gravity on your body is the force of gravity from your body on the Earth.

The reactive centrifugal force you mention is not the centrifugal force that I mentioned.

 

[John Mcrain]

Gravity is indeed a fictitious force. But you'll have to wait for General Relativity to make sense of that idea.

In the Newtonian model, gravity is an real, physical, inertial force. The third law partner of the force of Earth's gravity on your body is the force of gravity from your body on the Earth.

How do you mean wait?Is general relativity proved?

 

[jbriggs444]

How do you mean wait?Is general relativity proved?

It is correct. It is just that you are better served learning Newtonian mechanics first before we start complicating matters. It does no good discussing curvature in four dimensional pseudo-Riemannian manifolds when you are hazy on how centrifugal force works.

 

[haruspex]

No live creatures feel fictitious forces.

As I mentioned before, I think most people would say they 'feel' centrifugal force. Of course, what is felt directly is the normal force from the wall or side of the car, but since no acceleration is observed the brain infers a force pushing the body against the wall.

 

[John Mcrain]

As I mentioned before, I think most people would say they 'feel' centrifugal force.

Of course they feel centrifugal force.
In sharp left turn your head wonts to go the right,so your left muscle in neck must contract to keep head in stright position.
So from human perspective centrifugal force exist..

If you want imitate force on your head during left turn,you can put head strap which pull you to the right...

for same reason F1 drivers lean head into turn,so head gravity helps them fighting against centrifugal force..