B A doubt about tangential and normal aceleration, pilot g forces

[jbriggs444]

each astronaut momentum will be I=mr2=1*0.1*.1=0.01

that gives a total for the system of 6.62 m2/s2

Good. Now then, since we are agreed that angular momentum is conserved, this same angular momentum must still be present when the astronauts are at the ends of the tubes, about to drop out from the bottom. So at that point, the total angular momentum is 6.62 m²/s²

Back in post #33, @Dale suggested a next step (to which I raised an objection). I am going to suggest a different next step. This is almost certainly where he was heading in any case.

What is a formula for the total angular momentum of tube plus astronauts in terms of R (the astronauts' distance from center) and $\omega$ (the current rotation rate of the tube)?

 

[farolero]

you can not put total momentum in terms of R of the tube but in terms of the arm between the two trajectory vectors which are very different things and the formula is mvr or mw/r is constant but this wouldn't be valid after the intitial stage and sorry for saying it again but this is what 99% of people would fall for

 

[jbriggs444]

you can not put total momentum in terms of R of the tube but in terms of the arm between the two trajectory vectors which are very different things and the formula is mvr or mw/r is constant but this wouldn't be valid after the intitial stage and sorry for saying it again but this is what 99% of people would fall for

Read what I wrote. R is not the radius of the tube. R is the distance between one of the astronauts and the center of the tube.

Read what I wrote. I am not asking for momentum. I am asking for angular momentum.

Angular momentum can be computed based on R, $\omega$, the moment of inertia of the tube and the masses of the astronauts. You just finished doing it.

 

[farolero]

ok let's see where does this take us

L=Iw2+mvR=mr2/12*v2*R+mvR

 

[jbriggs444]

ok let's see where does this take us

L=Iw2=mr2/12*v2*R

That is not correct. It does not account for the angular momentum of the astronauts. Nor does it take the requested form. It is not a function of $\omega$, R, the moment of inertia of the tube and the mass of the astronauts.

In particular, the "v" and "r" terms are improper unless you can express them in terms of the given variables: $\omega$, R, the moment of inertia of the tube and the mass of the astronauts.

Edit: Note that above the top of the posting window, there is a bar of icons including one that looks like "x²". You can use it to generate superscripts. This would prevent the annoyance of meaning v² and writing v2. Also on this bar is an icon that looks like "Σ". You can use that to insert special characters such as ω so that you do not write w instead.

 

[farolero]

i don't know how to put the angular momentum of the tube as function of the radius of the astronauts i just know how to put the angular momentum of the astronauts as function of the radius of the astronauts and the angular momentum of the tube as a function of the radius of the tube

 

[jbriggs444]

i don't know how to put the angular momentum of the tube as function of the radius of the astronauts i just know how to put the angular momentum of the astronauts as function of the radius of the astronauts and the angular momentum of the tube as a function of the radius of the tube

I understand that the angular momentum of the tube alone depends only on the moment of inertia of the tube and its current rotation rate. That means that you could just write down:

$L_{tube}=I\omega$

That would be fine for that part of the formula.

What I am asking for is a single formula for the total angular momentum of the tube plus astronauts. That formula may use the following variables and only the following variables. If you do not need to use all of the variables, you are free not to use them all.

$I$: The moment of inertia of the tube. Do not express this as $\frac{mr^2}{12}$. Do not express it as 0.66 kg m^2. Yes, I know we already computed this value. Humor me. Just leave it as "I".

$R$: The current distance of one of the two astronauts from the center of the tube. R is not always 1/2 of the length of the tube (the ending position of the astronauts). It is not always 0.1 (the starting position of the astronauts). It is a variable.

$\omega$: The current rotation rate of the tube. $\omega$ is not always 10 radians per second (the starting angular velocity). It is not always [whatever the ending angular velocity turns out to be]. It is a variable.

$m$: The mass of each of the two astronauts. Do not express this as 1 kg. Yes, I know you said that their mass is 1 kg. Humor me. Just leave it as m.

The formula must be valid for starting conditions in which the tube has a different moment of inertia than you specified for this problem, or in which the angular rotation rate is different, the astronaut separation is different or in which the astronaut mass is different. [So it's no fair just writing down "6.62 kg m²/s²"]

 

[farolero]

oh thanks a lot for your help and time really appreciate it, you must be a vocational teacher :)

i think i figured it out:

I= Itube+Iastronaut=Itube+mRv=I+mR/w

but i would have some trouble using this formula if its what youre asking for

 

[jbriggs444]

oh thanks a lot for your help and time really appreciate it, you must be a vocational teacher :)

i think i figured it out:

I= Itube+Iastronaut=Itube+mRv=I+mR/w

but i would have some trouble using this formula if its what youre asking for

You used "v" in that formula. You are not allowed to use "v". You are only allowed to use I, R, m and $\omega$.

In any case, I was trying to get you to write down a formula for L -- angular momentum, not I -- moment of inertia. There is a reason for that. I had touched on that reason previously in this thread.

For a rigid object rotating in a plane about a fixed axis, angular momentum can be computed as moment of inertia (about the axis) times rotation rate (about the axis). But the complete system here is more than just a rigid object rotating around a fixed axis. So we cannot use that approach. (*)

The good news is that the tube alone is a rigid object rotating about a fixed axis. So its angular momentum can be computed as $L_{tube}=I \omega$.

[Another piece of good news is that angular momentum is additive. The angular momentum of a collection of objects is the sum of the angular momenta of each of the objects taken separately, provided that you use the same reference axis throughout].

But we still need to compute the angular momenta of the astronauts. Let's go back to fundamentals. The underlying definition of angular momentum is $L=m (\vec{v} \times \vec r)$ where m is the mass of a pointlike object, $\vec v$ is its velocity vector, $\vec r$ is its displacement from the chosen reference point and $\times$ denotes the vector cross product.

As you may already understand, the cross product of two vectors is a vector that is at right angles to both and whose magnitude is the product of the magnitudes of the two vectors times the cosine of the angle between them. $|\vec{a}\times\vec{b}| = |\vec{a}|\ |\vec{b}| \ cos(\theta)$. Alternately, you can think of this as multiplying $\vec{a}$ by the component of $\vec{b}$ that is perpendicular to $\vec{a}$

If we have an astronaut of mass m at a displacement R from the center moving along with a tube whose rotation rate is $\omega$, can you see how the above information allows you to compute to compute his angular momentum in terms of m, R and $\omega$?

Can you write down a formula for the angular momentum of one of the astronauts now?(*) As it turns out, computing the moment of inertia of rod plus astronauts and multiplying by the rotation rate of the rod would give the right answer in this case. But it would be more work to demonstrate that fact than to work the computation correctly instead.

 

[farolero]

im not very sure about this:

i know that I= w/arm between the two vectors

so the arm=w/I

also by pithagoras i know cos of the angle of the vector=arm/R and hence arm=cosvectorangle*R so i have:

cos vector angle*R=w/I

but how do i know the trajectory vector angle if initially is 90º from the radial direction and in infinity is 0º from radial direction?

 

[jbriggs444]

im not very sure about this:

i know that I= w/arm between the two vectors

We can stop right there. We are not trying to compute I (moment of inertia). We are trying to compute L (angular momentum). The formula that we are trying to use is one that does not even mention I (moment of inertia). So that first step will go nowhere.

The bit about "perpendicular component" was the hint that I was trying to get you to take. The displacement ($\vec{r}$) vector in this case is the displacement from the center of the tube to the astronaut. Its magnitude is given by R. Its direction is directly down the tube. We are trying to compute the cross product of this with the astronaut's velocity vector.

Let us make this a bit more clear by changing up the notation and using R instead of r.

We are trying to compute $\vec{R} \times \vec{v}$. As per #59 above, we can get the right result by multiplying the magnitude of $\vec{R}$ by the component of $\vec{v}$ that is at right angles to $\vec{R}$.

That component has another name: tangential velocity.

We have a formula for tangential velocity: $\omega R$

So the answer you are seeking is:

$L_{astronaut}=m R \omega R = m \omega R^2$

[Note that, as suggested previously, this matches what one would get by computing the moment of inertia of the astronaut ($I=mR^2$) and multiplying by $\omega$]

Now... can you put this together with the formula for the angular momentum of the tube to write down the formula for the total angular momentum of the tube plus astronauts?

 

[farolero]

but i don't agree with that:

the angular momentum of the astronaut would be:

Lastronaut=m*w*arm between the two vectors, hence:

Lastronaut=m*w*R2*cosalpha

where alpha is the angle of the trajectory with respect to the radial position

edit:

the formula L=m*w*R2 can only be applied to circular motion and this is SPIRAL non circular motion

that formula comes from L=m*v*R2 assuming v=wr and you got tired of me telling this assumption for this problem is only valid initially and otherwise false assumption

 

[jbriggs444]

but i don't agree with that:

the angular momentum of the astronaut would be:

Lastronaut=m*w*arm between the two vectors,

Do you know the definition of angular momentum? I believe that you do not. Hint: it does not involve rotation and was given in #59.

 

[farolero]

yes that's my whole point:

the product of the two vectors with the arm remain constant or that mvr remain constant

in circular motion you could extrapolate mw/r remain constant but you can NOT extrapolate it in SPIRAL motion cause then youre not accounting for the TRUE arm between the vectors and youre assuming than a spiral and a circle its the same and both have the same tangent

please if somebody sees what i say could you explain it here please, english is not my native language and I am rusty on physics

 

[jbriggs444]

yes that's my whole point:

in circular motion you could extrapolate mw/r remain constant

Angular momentum has nothing to do with extrapolation.

but you can NOT extrapolate it in SPIRAL motion

Angular momentum is independent of the history of a particle. That particle may be traversing a circular arc, a spiral arc, a parabolic arc or an unpatterned trajectory of any sort whatsoever. Its angular momentum at a particular time depends on only two things:

p: the momentum of the particle at the time
r: the displacement of the particle from the reference point at the time.

cause then youre not accounting for the TRUE arm between the vectors and youre assuming than a spiral and a circle its the same and both have the same tangent

The "tangent" that I am considering is the one to that is at right angles to the displacement vector pointing back to the center of the tube, NOT the one pointing parallel to the current trajectory. To put it another way, I am considering the "tangential" direction as the direction tangent to the trajectory of a point on the tube. The tube's velocity in this direction and the astronaut's velocity in this direction are necessarily equal as long as the astronaut remains in the tube. So the astronaut's "tangential" velocity (in this sense) is given by $\omega R$. The astronaut also has a component of velocity parallel to the tube which I will be referring to as his "radial velocity" when we get that far. [That's the next step when we get into energy considerations].

Using the term "tangential velocity" to refer to the component of velocity in the astronaut's current direction of travel would be a waste of a phrase and would put the notion of "radial velocity" on shaky ground. The velocity is always in the direction the velocity is in. We have a shorter word for that concept: "speed".

 

[farolero]

ok so you have two weights united by a string going at 1 m/s

you cut the string and when the distance is two meters then momentum has double since the radius of the weights have double if you apply I=mwR

this only would be true if the weights were going normal to radial direction between both weights but you have to account for the angle of the right tangent, actually 30º

so the right answer is I=mwRsenalpha so
I initially= 1*1*1
I finally=1*1*2*sen30

the same happens in this thread case

 

[jbriggs444]

ok so you have two weights united by a string going at 1 m/s

you cut the string and when the distance is two meters then momentum has double since the radius of the weights have double

Either you forgot the cosine term or you forgot that the rotation rate has reduced. Think again.

Note that it is "angular momentum", not "momentum".

 

[farolero]

ok i hope to make you see my point:

check where youre going wrong:

edit:

as a matter of fact if you solve the problem the way you want youll find out the astronaut decelerates and at infinity his velocity is zero

then where has energy gone?

 

[jbriggs444]

You are failing to understand what I am talking about and are arguing against things that I am not saying.

If you have two stones revolving around one another while connected by a string and you cut the string, the rotation rate of the imaginary line connecting the two stones (that is to say, $\omega$) will not remain constant.

Edit: I am trying to read the chicken scratchings that you call a diagram. You may be confused about the correct moment arm to use to calculate angular momentum, but I cannot tell because I cannot figure out which moment arm you think is right and what axis you are taking angular momentum about.

 

[farolero]

yes that was a bad example, never mind

do you see what i mean in my last diagram

do you see how you are taking as arm the radius of the astronauts along the tube to calculate angular momentum while that is not the right arm

im beginning to think I am not didactic enough

 

[jbriggs444]

The correct "arm" to use when calculating angular momentum is the one that runs from the selected reference axis to the object you are considering. That is, the vector that goes from the center of the tube to the object's current location. That is true by definition. It is not open for debate.

You can quickly google up a reference on, for instance, Wikipedia.

The relevant angle (for which the cosine will be taken) is the angle between this vector and the object's momentum vector.

It is easy to see that the component of v at right angles to r is given by $v \cos \theta$ where $\theta$ is that angle.

 

[farolero]

not the correct arm is the one normal to both vectors not an arbitrary one

maybe answering this question helps:

edit:

anyway if you really want to do the things your way you can do it and take the arm as the radius of the astronaut along the tube but then you must multiply the vector by the sin of the angle of the trajectory

and i would like to know how would you obtain this angle so far mine its been the only solution to the problem

 

[farolero]

it took me some time but i figured out where have you gone wrong:

check out this you said:

"That component has another name: tangential velocity.

We have a formula for tangential velocity: wR"

think what does tangential mean:

it means it has the same direction than the trajectory of the object

the wR figure its only valid FOR CIRCULAR MOTION

thats where you have gone wrong

for a spiral that cuadruples its radius each turn it would be wr2

 

[Vanadium 50]

check where youre going wrong

I think you misunderstood who is getting it wrong.

 

[farolero]

well the truth is not democratic

check out my first diagram on tangential and normal vector here:

you just can not take as tangent to the trajectory an arbitrary vector you feel like

 

[farolero]

so check out how v tangential has to follow the path of the trajectory:

of course if you take the v tangential of a circle where there's a spiral you get wrong conclusions

i like to make simple afirmations cause theyre easy to prove wrong unless right

whats wrong of that line affirmation id like to know

 

[jbriggs444]

well the truth is not democratic

Just because everyone disagrees with you, that does not prove that you are right and that everyone else is wrong.
A truth is that the word tangent can mean "tangent to the path being traced out by a point on the tube".
A truth is that the word tangent can mean "tangent to the path being traced out by the astronaut".
A truth is that the word tangent can mean "at right angles to the radial direction".

The meaning that I had in mind is the first one listed above. I clarified that meaning when you first indicated disagreement about it. If you cannot tolerate that word choice, pick another word or phrase to refer to that component of the astronaut's velocity which is at right angles to the tube's long axis and we can agree to use that instead.

 

[jbriggs444]

so check out how v tangential has to follow the path of the trajectory:

of course if you take the v tangential of a circle where there's a spiral you get wrong conclusions

i like to make simple afirmations cause theyre easy to prove wrong unless right

whats wrong of that line affirmation id like to know

That diagram is perfectly consistent with the meaning of tangential that I am using: "tangential to the path traced out by a point on the tube".

 

[farolero]

well then you could show to solution of the problen we could be discussing the meaning of words forever

any way be sure as soon i see your point ill admitt to it

edit:

i can not understand how that a point in the tube that actually makes a circle have any revelance for solving the problem

i don't think ill be able to sleep till i see your solution I am really expectant

 

[jbriggs444]

well then you could show to solution of the problen we could be discussing the meaning of words forever

You are the one who is insisting on discussing a particular word meaning to the point that you refuse to understand the analysis.

i can not understand how that a point in the tube that actually makes a circle have any revelance for solving the problem

I already explained that once. I will go over it again more slowly. In addition, you should realize that it is your job to solve the problem, not ours. The Physics Forum way is to guide people to the solutions they seek rather than to simply spit out numbers.

Let us review.

We are trying to find a formula for the angular momentum of the two astronauts plus tube in terms of the mass of each astronaut (m), the angular rotation rate of the tube ($\omega$), the moment of inertia of the tube (I) and the current distance of each astronaut from the center of the tube (R).

I have asked you for such a formula. You have been unable to provide it.

We have reached agreement (I believe) about the angular momentum of the tube alone:

$L_{tube}=I \omega$

We have failed to come to an understanding about the angular momentum of each astronaut. I have a correct formula in mind and have offered a justification for it. You have rejected that justification, apparently out of a disagreement on the meaning of the word "tangential". So it seems that we must avoid the use of that word.

Your current question is how the fact that a point in the tube makes a circle could possibly be relevant to determining the angular momentum of each astronaut.

The astronaut has a velocity. I think you would agree with this.

As viewed from any inertial frame, that velocity is not in a straight line away from the center of the tube. Possibly you will agree with this.

Instead, that velocity is at some unknown angle. Except at the start of the situation, it has a non-zero component in the direction away from the center. We can call this its "radial component". It also has a non-zero component at right angles to that. We can call this its "clockwise component".

At any time while an astronaut is still in the tube, the "clockwise component" of his velocity will match the "clockwise component" of the velocity of the spot on the tube wall that he is passing at that time. This is an assertion with which you may agree after some thought.

We have an easy formula that will give the "clockwise component" of the velocity of a spot on the tube wall.

The "clockwise component" of the astronaut's velocity is important because it is, by definition, perpendicular to the displacement of the astronaut from the axis of rotation.

That means that it is also equal in magnitude to $v \sin \theta$ where theta is the angle that the astronaut's velocity makes with his displacement from the center.

The angular momentum of the astronaut is the vector cross product $\vec{R} \times \vec{p}$ where R is the astronaut's displacement vector and p is his momentum.

Factoring out the m from p=mv, that is also equal to $m \vec{R} \times \vec{v}$

It is a basic property of vector cross products that this is equal in magnitude to $mRv \sin \theta$ where $\theta$ is the angle that the velocity makes with the displacement vector.

But $v \sin \theta$ is equal in magnitude to the component of the astronaut's velocity at right angles to his displacement.

And that is what we are calling the "clockwise component" of his velocity.

Which is, as above, equal to the "clockwise component" of the velocity of a spot on the wall of the tube at the astronaut's current position.

Which is given by $\omega R$.

Which is why the circular motion of a point on the tube is relevant.

Edit: $\sin \theta$, darnit.

 

[farolero]

yes i see where your going let's see if i can solve it the way you mean

initial angular momentum=6.6
initial energy=100
i know that clockwise speed=wR
L=mwR2

hence L=m*clockwisespeed*R3
hence final clock wise speed =L/mR3=6.6m/s
final w=6.6/1
I=ml2/16=0.25
energy of the tube in the final stage=1/2 Iw2=1/2*0.25*6.6*6.6=5.5
energy of the astronaut=initial total energy-final energy of the tube=100-5=95

so final speed of the astronaut would be 95=0.5*2*v2 then final v of the astronauts=9.7 m/s

would this be correct?

i hope you realize if its correct its wrong cause after one meter of artificial gravity free fall at 0.1 G speed must have gone from one to slightly less than two m/s

 

[jbriggs444]

yes i see where your going let's see if i can solve it the way you mean
initial angular momentum=6.6
initial energy=100

We went through this previously. 6.6 is good enough for the angular momentum (in kg m²/s²)
But the energy was 332 Joules.

i know that clockwise speed=wR
L=mwR2
hence L=m*clockwisespeed*R3 [...]

You've gone off the rails here, I am afraid.

You still have not provided the formula I asked for: total angular momentum of tube plus astronauts in terms of m, R, $\omega$ and I.

Please provide that formula and we can proceed.

Edit: note that @Dale asked you for [almost] this very thing back in #33. It is almost 50 posts later with no formula in sight. And still no proper equation formatting.

 

[farolero]

ok ill try again.

L=m*w2*R+I*w

anyway i already solved the problem you could tell me what's wrong about it:

youre at artificial gravity free fall for 1 m starting at 1 m/s if you take the centrifugal gravity as linear you get a final speed of 2 m/s if you take it as quadratic less than 2 that you would obtain integrating accounting for all energy in infinity has gone to the astronauts

i think a problem solved simply deserves an extra

also chek out what you say concerning the angular conservation of the astronauts and what i say concerning the angular momentum of the astronauts conservation

theyre very different things and it should be obvious at this stage whose right:)

what do you have to argue to this diagram:)

hence L=mwR2 if false while L=mwR2cosalpha is true

theyre very different things and its up to the people here to see what's the truth and what is not

edit.

if cos alpha is not equal to one and it equals 0.5 for example it can not be L=1 and L=0.5 and be both true at the same time

so either of us is wrong, it can not be we are both right which i seriously pondered

so please you or somebody else tell me what is conserved in the last diagram what you say or what i say

 

[jbriggs444]

ok ill try again.

L=m*w2*R+I*w

You switched from the correct formula for the moment of inertia of the astronauts to an incorrect formula and forgot to multiply by 2. Please try again.

anyway i already solved the problem you could tell me what's wrong about it:

youre at artificial gravity free fall for 1 m starting at 1 m/s if you take the centrifugal gravity as linear you get a final speed of 2 m/s if you take it as quadratic less than 2 that you would obtain integrating accounting for all energy in infinity has gone to the astronauts

If you do not know whether centrifugal gravity is constant, linear, quadratic or none of the above then you cannot use it to solve the problem. Further, if you are considering centrifugal force in a frame that is rotating at a non-constant rate then you also have to factor in the Euler force. And it would be good to quantify the rotation rate. The approach I am trying to lead you toward starts by quantifying the rotation rate. That's what we are busy doing with the angular momentum formula.

Let me give you a preview of where we will be going after you provide the correct formula for angular momentum as a function of m, I, $\omega$ and R...

We know the starting angular momentum. We know that angular momentum is conserved. So we know the ending angular momentum at the moment when the astronauts are about to drop out the bottom of the tube.

We also know I, m and R for the moment when the astronauts are about to drop out the bottom.

If we have a formula for angular momentum:"L = something $\omega$ something R something m something I" then we have one equation with one unknown: $\omega$.

So we solve for $\omega$.

With $\omega$ in hand, we can find the clockwise component of the velocity of the astronauts as they are about to drop out the bottom of the tube.

That leaves us not knowing the radial component of their velocity. But wait...

We also know the starting energy of the system. We know energy is conserved. We can figure out how much energy the tube has and how much energy the astronauts have due to their clockwise velocity component. What remains is the energy they have due to their radial velocity component.

So we solve for the radial velocity component.

With the radial and clockwise velocity components in hand, we can compute the total velocity (Pythagoras) and angle (arc tangent of the ratio of the components).

That's the plan. But we need to attack it step by step.

 

[Dale]

One has to be careful when trying to use $E=\frac{1}{2}I\omega ^2$ for a system which is not a rigid body.

So I got back to this today. You are correct, my expression for E is incorrect. The correct expression is:

$E=\frac{1}{24} L^2 M \dot{ \theta }{}^2+m \left(\dot{r}^2+r^2 \dot{\theta }{}^2\right)$ where L is the length of the tube, M is the mass of the tube and m is the mass of one astronaut

Since there is no potential energy involved, this problem is probably best solved using the Lagrangian method. In fact, the conservation of angular momentum falls out naturally and does not even need to be added by hand as I was considering originally.

 

[jbriggs444]

It is amazingly pleasurable to see an equation typeset in TeX and with a definition of the variables sitting there right next to the equation.

 

[farolero]

ok L of the astronauts=mvR=mwR2

l of the tube=0.25
L of the tube=0.25w

Ltotal=mwR2+0.25w=w(mR2+0.25)

i hope to have got it right now

edit:

advancing a bit into the future what will happen to the tube in infinity?

if it has some rotation according the formula I am using its angular momentum would be infinite

 

[farolero]

edit:

for the record you were right and i was wrong saying you were wrong on your afirmation angular momentum of the astronaut=mw2R, though i still wonder if there's some flaw in taking the center of the tube as center of reference as seen its wrong considering it as such initially in the thread concerning the calculation of the g forces felt by the astronaut

my apologies for taking so long to see it, its a question that the opposite momentums from the center of the tube nullify each other, just the clockwise ones count

but you said that in infinity the tube stil will have some rotation

then with the astronauts at infinite distance wouldn't angular momentum be infinite?

theres something wrong there

edit:

i also know centrifugal gravity is cuadratic for its paralelism with real gravity, as the distance to the center halfs its effect cuadruples also as expected from the cuadratic centrifugal force formula

edit:

i just realize of something extreamly puzzling i hope someone can solve it, ill elaborate:

problems like scape velocity are solved by imagining what happens after infinite time, specially those concerning cuadratic gravity

being my example of artificial GRAVITY let's imagine what happens in infinity:

the tube after infinity has some w:

this means angular momentum is infinite for r is infinite

the tube after infinity is still this means the energy of the astronauts is infinite cause w=0

:) anyone?

 

[Dale]

advancing a bit into the future what will happen to the tube in infinity?

Why don't you put some thought into it first. Are these equations valid at infinity, why or why not?

 

[farolero]

im not even sure if those equations are valid for a mid point i would have to see the result and compare it with the result you would obtain with an acceleration of approximately 1 m/s2

anyway i like better thinking on abstract than doing raw calculations where you lose sight of things

as i see it in the beginning there's the vector of the tube and the two vectors of the astronauts with the 0.2 m arm

and finally at infinity the tube vector has disappeared and there are two vectors of the astronauts in opposite sense bigger than initially, the module of the initial vector of the tube has gone to the astronauts

this is what i believe will happen, but being the tube straight and the vectors going along radially and the tube is still at infinity

this is a contradiction for angular momentum would be zero being the vectorial product of the astronauts vectors zero

edit:

the initial angular momentum is 6.6 and the initial energy is 100

i think angular momentum of the astronaut can be neglected:
I=mvR=2*1*0,1=0.2

0.2 can be neglected in front of 6.6 it would make things infinitally easier and see where taking 6.6=mwR2 take us instead of taking 6.6=mwR2sinalpha as i think it should be
maybe you should consider why its wrong to take 6.6=mwR2sinalpha for i don't think you can account for both things at the same time

 

[Dale]

@farolero this post is incomprehensible. Please use LaTeX for your equations, please use proper English grammar and capitalization and punctuation. Please put some thought into communicating your ideas as clearly as possible.

 

[farolero]

OK I will try to use latex though its quite difficult for a newbee, my apologies.

 

[Dale]

im not even sure if those equations are valid for a mid point i would have to see the result

So let's ignore the results for a moment, and instead let's think about the assumptions. Are the assumptions that guided the development of the equations applicable at infinity?

I will give you a hint. The conservation assumptions still apply. However, there are some additional assumptions made that don't apply. Consider the fact that two point particles and a rod together have 11 degrees of freedom, but our equations only use two. How did that happen, and does it apply at infinity?

 

[farolero]

I do not think they are aplicable to infinity for if there's a slight rotation of the tube angular momentum would be infinite as the arm is infinitely long, falling in a contradiction.

I think that going the astronauts in opposites senses limits the degrees of freedom of the system.

 

[Dale]

if there's a slight rotation of the tube angular momentum would be infinite as the arm is infinitely long

Are you talking about an infinitely long tube or about the behavior of the system a long time after the astronauts leave the end of the tube? I had thought that the length of the tube was fixed to a couple of meters and you were discussing the behavior of the system as the astronauts fly off to infinity

I think that going the astronauts in opposites senses limits the degrees of freedom of the system

Yes, but there are also other constraints that are important to reducing the number of variables.

 

[jbriggs444]

My suspicion is that @farolero is under the impression that angular momentum has something to do with an instantaneous center of rotation rather than with an arbitrary fixed reference point or axis. But, I am having a an extremely difficult time making sense of his claims.

 

[farolero]

I understand your point fully that L=6.6=mwR2 where m is the mass of the astronauts w is the rotational speed of the tube and R its the distance to the center of the tube of the astronauts but I do not see where it is wrong.

Neglecting the astronauts initial angular momentum applying that formula we have:

6.6=2*w*1 where w=3.3

This implies that just one of the components of the pythagoras triangle that will give the final speed,3,3 is already way bigger than the expected result of less than 2 m/s.

On the other hand i point that:
6.6=mwR2sinalpha

And i insist both can not be true and yours gives a contradictory result.

My apologies for not using latex yet but i am on learning it.

 

[farolero]

Are you talking about an infinitely long tube or about the behavior of the system a long time after the astronauts leave the end of the tube?

Well I imagine a 2 m long tube of 2kg and along it till infinity a massless force field, i hope this is allowed.

 

[jbriggs444]

I understand your point fully that L=6.6=mwR2 where m is the mass of the astronauts w is the rotational speed of the tube and R its the distance to the center of the tube of the astronauts but I do not see where it is wrong.

That is not a correct equation. The angular momentum of the system is not equal to $m \omega R^2$. The angular momentum of the system includes the angular momentum of the tube as well as the angular momentum of the astronauts.

You still have not bothered to write down the formula for the total angular momentum of the system in terms of m, R, I and $\omega$ that I asked for some 50 posts ago.

 

[jbriggs444]

Well I imagine a 2 m long tube of 2kg and along it till infinity a massless force field, i hope this is allowed.

That is allowed, certainly. But it is a different problem than you had proposed. In this modified setup, you are correct that the rotation rate of the tube must decrease toward zero as the astronauts recede toward infinity if angular momentum is to be conserved.