Frame of reference question: Car traveling at the equator

[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

 

[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.