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

[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.

 

[farolero]

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.

Think that in the beginning of the discussion taking arbitrarily the center of the system as center of reference failed to explain the relation between tangent and normal accelerations and the astronaut G forces that he feels and doing this lead Dale to consider it an imposible motion.

 

[jbriggs444]

Think that in the beginning of the discussion taking arbitrarily the center of the system as center of reference failed to explain the relation between tangent and normal accelerations and the astronaut G forces that he feels and doing this lead Dale to consider it an imposible motion.

You are mistaken. The center of the system (like any other point whatsoever) is perfectly valid to take as the reference point for calculation of angular momentum. Nothing impossible results. @Dale would certainly agree.

Nothing requires that the reference point for angular momentum be "tangent" or "normal" to any particular place on the trajectory of any particle in the system being considered. It can be chosen arbitrarily.

There is an obvious reference point to choose: the center of the tube. That choice is so obvious that we had assumed you would be using it. This impression was reinforced when you actually did use it (when you computed the moment of inertia of the tube).

Edit to add this note: You are free to use any point you like for the reference point for angular momentum. What you are not free to do is to change your mind in mid-calculation. If you change the choice of reference point, angular momentum can change. That means that you cannot then depend on the principle of angular momentum conservation to hold good. In addition, if you compute the angular momentum of one piece of the system using one reference point and another piece of the system using another reference point then you cannot simply add those two angular momenta together to obtain a useful result.

 

[farolero]

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.

It is not that I have not bother, it is that I have tried but I have not been able to, I just don't know how to put the moment of inertia of the tube in function of the radius of the astronaut, I took last physics five years ago, I am too rusty.

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.

If tubes rotation is zero then angular momentum is zero, either way you take it there's trouble.

And it is not a different problem is solving a centrifugal gravity problem similarly to real gravity problem

Edit:

In this problem you can NOT take an arbitrary center of reference as the center of the spiral for it would lead to contradictions:

If you did so given the intitial conditions of the starting of the thread you would conclude than tangential and normal acceleration equal zero which is blatantly false for the pilot does change direction constantly

 

[jbriggs444]

It is not that I have not bother, it is that I have tried but I have not been able to, I just don't know how to put the moment of inertia of the tube in function of the radius of the astronaut, I took last physics five years ago, I am too rusty.

As I had written previously:

Angular momentum is an additive property. The angular momentum of a system consisting of multiple objects is the sum of the angular momenta of the each of those objects. [With the requirement that you use the same reference point when calculating all of those angular momenta].

We have a formula for the angular momentum of the tube.
We have a formula for the angular momentum of each astronaut.
You have not bothered to put them together with a plus sign.

If you are rusty in physics then you have no excuse for your repetitive claims that the rest of us are wrong. Take the opportunity to learn rather than continuing to insist on the correctness of your misunderstandings.

If tubes rotation is zero then angular momentum is zero, either way you take it there's trouble.

And it is not a different problem is solving a centrifugal gravity problem similarly to real gravity problem

It is a different problem because you are considering what happens after the astronauts have receded toward infinity. If there is a force field extending to infinity, that ties the rotation of the tube to the motion of the astronauts. If there is no such field, the rotation of the tube is not tied to the motion of the astronauts after they drop out of the ends of the tube.

In this problem you can NOT take an arbitrary center of reference as the center of the spiral for it would lead to contradictions:

I did not say anything about the center of a spiral. I spoke of the reference point for calculation of angular momentum. You can place that reference point 1000 kilometers northeast of the center of the spiral if you please and analyze the problem accordingly. The laws of physics still work and produce no contradictions.

You might want to review: https://en.wikipedia.org/wiki/Galilean_invariance

 

[farolero]

I see thanks for pointing me to the Galilean Invariance and allow me to go back to the orginal question that seems to not have been cleared up yet:

A pilot is on artificial gravity free fall along a tube that decreases its rotational speed in proportion to the pilot radius, that is, when the pilot radius doubles w halfs.

Analizing this problem from the center of the spiral acounting for galileo invariance the normal acceleration is zero for he is in centrifugal gravity free fall.

and the tangential aceleration is zero as well because he is going at a constant tangential speed.

The g forces the pilot experiments go in function of tangential and normal acceleration, i hope we can agree on this.

Then by the Galileo Invariance I conclude the pilot changes course at zero g.

Whats wrong about that? The Galileo Invariance allows me to prescind of the instant center of the trajectory and use an arbitrary one as the center of the spiral.

 

[jbriggs444]

If tubes rotation is zero then angular momentum is zero, either way you take it there's trouble.

It seems that you have never taken your mathematics education as far as a course on real analysis. Or perhaps your last mathematics course was taken long ago.

There is a difference between "decrease toward zero" and "equal to zero". Similarly, there is a difference between "recedes toward infinity" and "is infinitely far away". Physicists are sometimes willing to play fast and loose with infinities. Playing fast and loose, one might want to say that the rotation rate of the tube (and the "clockwise" component of the velocities of the astronauts) becomes zero and that the positions of the astronauts become infinite.

If you had bothered to write down that formula for the angular momentum of the system as a function of R, I, m and $\omega$, then we could plug in infinity for R and see what result is obtained. You would get something along the lines of $0 \times \infty$. That's an undefined result.

As a mathematician, one could say "yes, that is undefined, but we can evaluate the limit of that formula as R increases without bound". That would produce a well defined result. Indeed, the law of conservation of angular momentum gives an effortless shortcut to evaluating that limit in the case of the problem that you pose.

 

[jbriggs444]

I see thanks for pointing me to the Galilean Invariance and allow me to go back to the orginal question that seems to not have been cleared up yet:

A pilot is on artificial gravity free fall along a tube that decreases its rotational speed in proportion to the pilot radius, that is, when the pilot radius doubles w halfs.

You assert that a proportion applies, but you continue to refuse to write down the formula that could confirm or deny that assertion. I will save you some time. The assertion is false. There is no inverse proportion.

Analizing this problem from the center of the spiral acounting for galileo invariance the normal acceleration is zero for he is in centrifugal gravity free fall.

That is meaningless gibberish. Further, the acceleration of the astronauts as they fall through the tube most definitely has a non-zero component normal to the walls of the tube.

[Note that Galilean invariance does not extend to rotating or accelerating frames of reference].

If you insist on adopting the rotating frame (as you must if you continue to invoke "centrifugal artificial gravity"), you should review: https://en.wikipedia.org/wiki/Coriolis_force

Note that I have been using the inertial frame for analyzing this problem. If you try to analyze it in a rotating frame then angular momentum will fail to be conserved unless you invoke the Coriolis force. If you are using a frame that is tied to the tube and rotating at a variable rate (which it will, given the effect of Coriolis on the astronauts and, therefore, on the tube) then you must also consider the Euler force. Far, far better to stick with the inertial frame.

 

[Dale]

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

I am not sure what you mean. If the tube is 2 m long then how can there be anything "along it till infinity". If the tube is 2 m long then anything that is along the tube can only be 2 m long also, otherwise it is beyond the tube.

Also, what field are you talking about? The tube and astronauts are uncharged, so there is no electric field, and the masses are so small that we can neglect gravity.

 

[jbriggs444]

I am not sure what you mean. If the tube is 2 m long then how can there be anything "along it till infinity".

I took this to mean that only the center 2m of the tube has mass. The rest of the tube is rigid but massless.

That is, despite being infinitely long, the tube has a finite moment of inertia. Although physically unrealizable, this understanding of the problem poses no particular problem for classical Newtonian physics.

 

[Dale]

I took this to mean that only the center 2m of the tube has mass. The rest of the tube is rigid but massless.

That is, despite being infinitely long, the tube has a finite moment of inertia. Depite being physically unrealizable, this understanding of the problem poses no particular problem for classical Newtonian physics.

Ah, yes, I agree. If that is what he meant then it is no particular problem to analyze. Indeed, it makes it easier than a finite tube.

 

[Dale]

allow me to go back to the orginal question that seems to not have been cleared up yet

It would greatly help if you would pose just one scenario and stick with that one scenario until you have finished it. In this thread I count at least 3 scenarios that you jump between with no warning and without finishing the previous one. It is very confusing for your respondents.

 

[farolero]

OK my apologies but I saw in the original problem the root of jbriggs444 consideration of L=mwR2 while I take L=mwR2sinalpha.

I think i have the formula you want:

L of the tube=Iw , L of the astronaut=mwR2

So Ltotal=lw+mwR2

From there i could calculate the w at R=1 and obtained clockwise speed as you kindly pointed to me.

 

[Charles Link]

To comment on the first post above, to get an object moving in a spiral path, you can apply a normal (perpendicular to direction of travel) force that increases linearly with time. In this case, the object will spiral inward and maintain a constant speed. Alternatively, to spiral outward at constant speed, you simply decrease the normal force at a linear (with time) rate.

 

[jbriggs444]

I think i have the formula you want:

L of the tube=Iw , L of the astronaut=mwR2

So Ltotal=lw+mwR2

Let me tweak that a bit to account for the fact that there are two astronauts.

$L_{total} = L_{tube} + 2L_{astronaut} = I \omega + 2m \omega R^2$

From there i could calculate the w at R=1 and obtained clockwise speed as you kindly pointed to me.

Excellent. I look forward to seeing that result.

[Note that I'd be interested in seeing how Dale's formula for energy plays out. I have never played with the Lagrangian formulation of classical mechanics.]

 

[Dale]

So Ltotal=lw+mwR2

Excellent. So, from this since $L=I \omega + 2 m \omega R^2$ and since L is conserved we can immediately see that if $R$ increases then $\omega$ must decrease. In fact, without even solving for $R(t)$ as a function of time we can still describe the behavior of $\omega$ as $R$ increases without bound.

This is one reason why conservation laws are so powerful in analyzing problems.

 

[Dale]

[Note that I'd be interested in seeing how Dale's formula for energy plays out. I have never played with the Lagrangian formulation of classical mechanics.]

I can post that later today when I return home where my notes are. I knew that the conservation of angular momentum was "supposed" to fall out automatically, and it was nice to see that it did.

 

[Charles Link]

To comment on the first post above, to get an object moving in a spiral path, you can apply a normal (perpendicular to direction of travel) force that increases linearly with time. In this case, the object will spiral inward and maintain a constant speed. Alternatively, to spiral outward at constant speed, you simply decrease the normal force at a linear (with time) rate.

See the "link" Problem 1: https://www.physicsforums.com/threads/micromass-big-october-challenge.887447/

 

[Dale]

[Note that I'd be interested in seeing how Dale's formula for energy plays out. I have never played with the Lagrangian formulation of classical mechanics.]

So, since there is no potential energy the Lagrangian is equal to the kinetic energy and can be written (trying to be consistent with @farolero's variables as much as possible):$$\mathcal L = \frac{1}{2} I \dot{\theta}{}^2+m(\dot{R}{}^2+R^2 \dot{\theta}{}^2)$$ where $I$ is the moment of inertia of the tube, $m$ is the mass of one astronaut, $R$ is the distance of the astronauts from the center, and $\omega=\dot{\theta}$ is the angular velocity of the system.

Then the Euler-Lagrange equation for $\theta$ is $$ \frac{d}{dt}\left( \frac{\partial \mathcal L}{\partial \dot{\theta}} \right) = \frac{\partial \mathcal L}{\partial \theta}$$ $$ \frac{d}{dt}\left( I \dot{\theta} + 2 m R^2 \dot{\theta} \right) = 0$$ $$ I \dot{\theta} + 2 m R^2 \dot{\theta} = L$$ which is our expression for the conserved angular momentum, $L$, but obtained automatically from the Lagrangian.

Then the Euler-Lagrange equation for $R$ is $$ \frac{d}{dt}\left( \frac{\partial \mathcal L}{\partial \dot{R}} \right) = \frac{\partial \mathcal L}{\partial R}$$ $$ 2m\ddot{R} = 2 m R \dot{\theta}{}^2$$ combining that with the expression for angular momentum and simplifying we get the equation of motion $$\ddot{R}=\frac{L^2 R}{(I+2mR^2)^2}$$

 

[Charles Link]

So, since there is no potential energy the Lagrangian is equal to the kinetic energy and can be written (trying to be consistent with farolero's variables as much as possible):$$\mathcal L = \frac{1}{2} I \dot{\theta}{}^2+m(\dot{R}{}^2+R^2 \dot{\theta}{}^2)$$ where $I$ is the moment of inertia of the tube, $m$ is the mass of one astronaut, $R$ is the distance of the astronauts from the center, and $\omega=\dot{\theta}$ is the angular velocity of the system.

Then the Euler-Lagrange equation for $\theta$ is $$ \frac{d}{dt}\left( \frac{\partial \mathcal L}{\partial \dot{\theta}} \right) = \frac{\partial \mathcal L}{\partial \theta}$$ $$ \frac{d}{dt}\left( I \dot{\theta} + 2 m R^2 \dot{\theta} \right) = 0$$ $$ I \dot{\theta} + 2 m R^2 \dot{\theta} = L$$ which is our expression for the conserved angular momentum, $L$, but obtained automatically from the Lagrangian.

Then the Euler-Lagrange equation for $R$ is $$ \frac{d}{dt}\left( \frac{\partial \mathcal L}{\partial \dot{R}} \right) = \frac{\partial \mathcal L}{\partial R}$$ $$ 2m\ddot{R} = 2 m R \dot{\theta}{}^2$$ combining that with the expression for angular momentum and simplifying we get the equation of motion $$\ddot{R}=\frac{L^2 R}{(I+2mR^2)^2}$$

@Dale I think you need a 1/2 factor on your $m ( \dot{R}^2+R^2 \dot{\theta}^2)$ term to get the kinetic energy for the Lagrangian. The term in parenthesis is $v^2$ in polar coordinates.$\\$ editing: Please ignore. You have it correct=I see there are two astronauts, each of mass $m$.

 

[farolero]

Thanks a lot for your time guys youre being hugely helpfull and kind :)

I don't know about Dale solution because I don't understand it but I am prety sure jbriggs444 solution is not correct.

The Galileo Invariance allows us to take any zero reference we want but doesnt allow us to take arbitrary tangential and normal components to the trajectory of the astronaut and that's what you did to solve he problem.

Let me elaborate, going back to the first case concerning the g forces the astronauts feels on artificial gravity free fall:

A diver is on a high platform about to jump to the water:

When the diver is in the platform he feels and acceleration of 9.8 in his body, 1 g, and when he is on free fall he feels zero g.

In other words when he is still he feels he is accelerating and when he is accelerating he feels he is still, this is pretty antiintuitive.

That is why is so tempting by the Galileo Invariance take as zero reference the diver:

When he is on free fall he is still feeling zero g and its the ground which its accelerating at 9.8 m/s2 with which gravity wouldn be a magic force any more just inertia

I hope we can agree that when youre at free fall you feel 0 g to go to next step.

In the case of the astronaut on artificial gravity free fall let's first do it wrong solving an spiral motion with circular motion equations:

The same happens in artificial gravity than in real gravity, when youre at artificial gravity free fall you feel 0 normal g though youre acelerating normally and when youre fixed to the tube though youre fixed radially you feel some g.

So we can agree though the astronaut is falling in the normal direction he feels zero normal g, but he does feel the tangential g for tangential velocity is increasing since tangentialv=wr and w remains constant but r grows so the tangential velocity grows.

But what would happen if as the radius of the astronaut along the tube doubles you half the w of the tube?

Then the tangential velocity would be constant and there would be not tangential acceleration so he doesn't feel any tangential acceleration and we had agreed that he doesn't feel neither any normal aceleration.

So we could conclude wrongly he is changing course without feeling any g, maybe some wild schauberger antigravity

But we solved the problem wrong, we took as tangential velocity the velocity if the astronaut was making a circle and we must take the tangential velocity accounting the astronaut is making an spiral, let's suppose that the spiral trajectory vector is offset 30º from radially

Then we would have that v=wr/sen30 and if i double the radius and half the w as i set initially there would be still tangetial acceleration the pilot would feel giving things back sense.

Again the Galileo Invariance allow us to take an arbitrary reference but doesn't allow us to take an arbitrary tangent component

edit:

first my apologies for not using capitals but my shift key is failing and it takes me up to 10 tries to write a capital.

what i think jbriggs444 did not right was taking an arbitrary tangent component of the trajectory.

this is not allowed in physics.

a trajectory is defined by the normal and tangent component

if the normal component is zero the trajectory is straight otherwise curved but the tangent MUST always be tangent to the trajectory

a different thing would be take the x and y components of the tangent velocity but momentum doesn't equal mVxR but at any case mVxsinalphaR=mVtR