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

[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

 

[jbriggs444]

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.

You've not driven the approach that I have suggested to a solution yet.

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.

One can take components of a velocity in any arbitrary pair of orthogonal directions one pleases. The components need not be aligned with the trajectory.

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:

Stop right there.

Galilean invariance allows you to move the origin of your coordinate system from one place to another. For instance, you can place it at the tip of the diving board. Or you can place it at the surface of the water. Or you can place it at the bottom of the pool.

Galilean invariance allows you to rotate your coordinate system. For instance, you can have the x-axis horizontal and the y-axis vertical. Or you can have the x-axis angling down and the y-axis angling forward.

Galilean invariance allows you to give your coordinate system a uniform velocity. For instance, you can have the x-axis moving outward and upward at a 45 degree angle at 1 meter per second.

Galilean invariance does not allow you to give your coordinate system an acceleration. If the origin is accelerating downward at 9.8 meters per second per second then you will not observe a force from gravity.

Galilean invariance does not allow you to give your coordinate system a non-zero rotation rate. If the x-axis is spinning around the origin then you will observe a centrifugal force.

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

Yes, one can transform from an inertial reference frame to a non-inertial reference frame through the use of "magic forces". Though the usual term is "inertial forces" or "ficticious forces".

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

Certainly.

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.

You keep using the term "artificial gravity" and pretending that centrifugal force is the only inertial force that is required to make Newton's laws apply to a rotating frame of reference. Again, I refer you to Coriolis.

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.

We might agree. But let us stop and clarify what you mean by the "normal direction".

The correct meaning of the word "normal" is perpendicular. But perpendicular to what? You have not specified anything for this force to be perpendicular to. From context, I assume that by "normal" you actually mean "directly toward or away from the origin". With that understanding in mind, we can continue to use your unconventional definition.

If the astronaut is falling and is subject to no force in the direction of the origin then he does not feel any force in the direction of the origin. I can agree with that.

You seem to claim that the astronaut is falling in a direction directly away from the origin. That claim would be incorrect if you are actually making it. His velocity is not in such a direction. His acceleration under the combined influence of centrifugal, Coriolis and the contact force of the tube is also not purely away from the origin.

[Edit: You may be intending to use a coordinate system tied to the tube. It is correct that the astronauts are falling directly "downward" along a coordinate axis in such a frame. But that frame is not inertial and is not even rotating at a uniform rate. It would unwise to use that frame to analyze the problem -- as has been pointed out previously].

You claim that $\omega$ remains constant despite the fact that you agree that there is a "tangential" acceleration. Possibly we can focus on this because both @Dale and myself agree that $\omega$ is not constant. There is a force between tube and astronaut, perpendicular to the walls of the tube. This causes the astronaut to deflect from a straight-line trajectory and causes the tube to deviate from a constant rotation rate.

Edit: I must admit to some puzzlement. It is 121 posts into this thread and you do not yet understand that in your own scenario, the tube exerts force on the astronaut and the astronaut on the tube?

 

[farolero]

yes I've seen that since the begining, one of the walls of the tube is pushing the astronaut acelerating him, a work on almost the same direction the astronaut is going is being applied on him

you can take any x y coordinates you want to define the tangential velocity vector

but you can not say I=mwR2 for that would be assuming I=mvR and this is only valid for circular motion

for spiral motion as is the case you should say:

I=mvRsinalpha where alpha is the angle of the spiral from the radial direction

and hence I=mwR2sinalpha

 

[farolero]

Galilean invariance does not allow you to give your coordinate system a non-zero rotation rate. If the x-axis is spinning around the origin then you will observe a centrifugal force.

then think that by galileo invariance you took the end of the tube as the arbitrary Vx of the tangential velocity and as you pointed this is not allowed for being a rotational reference

 

[jbriggs444]

yes I've seen that since the begining, one of the walls of the tube is pushing the astronaut acelerating him, a work on almost the same direction the astronaut is going is being applied on him

you can take any x y coordinates you want to define the tangential velocity vector

but you can not say I=mwR2 for that would be assuming I=mvR and this is only valid for circular motion

You cannot even say "I" because we do not have a rigid body rotating in circular motion.

Fortunately, I have never said $I=m \omega R^2$

 

[jbriggs444]

then think that by galileo invariance you took the end of the tube as the arbitrary Vx of the tangential velocity and as you pointed this is not allowed for being a rotational reference

You are not making sense.

 

[farolero]

i meant L where i said I sorry

for what i understand you considered the clockwise component as the Vx of the tangential velocity

inititially you called it tangential with which i disagreed and you changed its name to clockwise component

the value of this component is measured in a rotational frame of reference, in this case the end of the tube

so youre measurig a value(the clock wise component of the velocity of the astronaut as you called it) with the rotational reference of the tube, and i understood from you you can not take rotating frame references

 

[jbriggs444]

i meant L where i said I sorry

for what i understand you consider the clockwise component as the Vx of the tangential velocity

I defined what I mean by "clockwise component" precisely so that I could avoid using the term "tangential velocity" that we did not agree upon.

inititially you called it tangential with which i disagreed and you changed its name to clockwise component

And in so doing, I made it clear that "clockwise component" was not the same thing as what you call "tangential velocity".

the value of this component is measured in a rotational frame of reference, in this case the end of the tube

No.

You do not get to make up new definitions for terms that I have already defined and used. When I use a term, it means what I define it to mean. It does not mean whatever you decide you want it to mean.

As I have said repeatedly, I am using the inertial frame of reference.

The value of "clockwise component" is measured in the internal frame of reference. I defined the term and I get to decide what frame of reference it is measured in.

 

[jbriggs444]

yes I've seen that since the begining, one of the walls of the tube is pushing the astronaut acelerating him, a work on almost the same direction the astronaut is going is being applied on him

Newton's third law applies. If the tube applies a force on the astronaut, the astronaut applies a force on the tube.

The force of astronaut(s) on tube amounts to a torque. It causes the rotation rate of the tube to decrease over time.

Edit: You might want to be careful about using the term "work" here. Depending on your choice of reference frame, the force of wall on astronaut may or may not do work. Recall that work done depends on two things. Force is only one of those two things. You should also be aware that work is a scalar, not a vector. It has no direction.

 

[farolero]

i see thanks for the lesson on reference frame but for today its been enough confusion maybe tomorrow ill research the wikipedia on frame reference

but id like to know if you see my point here and what's wrong about it:

you can not say L=mwR2 for that would be assuming L=mvR and this is only valid for circular motion

for spiral motion as is the case you should say:

L=mvRsinalpha where alpha is the angle of the spiral from the radial direction

and hence L=mwR2sinalpha

edit.

imagine two cannons aiming in opposite senses in each end of a beam and how everything changes weather your aim the cannons normal to the beam or 45º offset concerning the calculation of angular momentum

 

[jbriggs444]

i see thanks for the lesson on reference frame but for today its been enough confusion maybe tomorrow ill research the wikipedia on frame reference

but id like to know if you see my point here and what's wrong about it:

you can not say L=mwR2 for that would be assuming L=mvR

Stop right there.

An assertion that $L=m \omega R^2$ is not the same as an assertion that $L = mvR$.

See if you can figure out why.

Also, stop for a moment and think about how you are defining $\omega$ here. We do not have a rigid system rotating about a fixed axis. What exactly do you think that $\omega$ denotes?

 

[farolero]

this is precisely what has me all confused:

that assertion,L=mwR2 its true if its true L=mvR and its true v=wR in fact id say both are false for a spiral trajectoryim not even sure you can use w for an spiral that might work as well as using it for a straight trajectory

 

[jbriggs444]

this is precisely what has me all confused:

that assertion comes if its true L=mvR and its true v=wR in fact id say both are false

You are right. Both are false (or at least are not always true).

However $\vec{L} = \vec{R} \times \vec{mv}$ is always true.

The difference is between multiplying magnitudes and performing the proper vector cross product.

 

[farolero]

ive been podering the following:

L=mvRsinalpha
v=wR/sin alpha
hence
L=mwR2 which would make your assumption right

but i don't think v=wR/sin alpha

for this spiral is a varying openness spiral, its totally closed inititally and totally open at infinity

so you just in this spiral can not relate v,w and r in a simple manner as with a circle, youd need to integrate

that conservation of angular moment is a vectorial product i fully agree, that's why it can only be applied to instant velocity vectors and extrapolating the rotational formulas for it only works if its true that circular motion for which they were desiged for and not any different kind of motion

 

[jbriggs444]

ive been podering the following:

L=mvRsinalpha
v=wR/sin alpha

hence
L=mwR2 which would make your assumption right

but i don't think v=wR/sin alpha

Until you define $\omega$, we can never know. What is $\omega$ ?

 

[farolero]

w its the rotational speed, an universal term, in this case the rotational speed of the tube

i think v and w can be related for a spiral trajectory with the following formula:

v=wR2*constant

with which the equation you want to apply to solve the problem would look something like this:

L=m*w*R3*sin alpha : )=

a particle going in a straight line for example has a decreasing w which receeds to zero at infinity from an arbitrary center

 

[jbriggs444]

w its the rotational speed, an universal term, in this case the rotational speed of the tube

i think v and w can be related for a spiral trajectory with the following formula:

v=wR2*constant

If the angle for the velocity were constant then your thinking could be correct. But what if (as in the case at hand), the angle is not constant?

[Note that I am considering the angle that the velocity makes with the R vector]

 

[farolero]

i concluded this formula by paralelizing artificial and real gravity

as the radius doubles speed cuadruples hence you must fit the R2 in the formula, the varying angle makes it even more complex to relate v and w

 

[jbriggs444]

i concluded this formula by paralelizing artificial and real gravity

as the radius doubles speed cuadruples hence you must fit the R2 in the formula, the varying angle makes it even more complex to relate v and w

If I understand your reasoning properly, it contains two errors.

1. You believe that the speed associated with "artificial gravity" (aka Centrifugal Potential) scales as the square of the radius.

This is incorrect. It is correct that the centrifugal potential scales with the square of the radius. But that is a measure of energy. Speed scales as the square root of energy. So speed would scale directly with radius except that...

2. You appear believe that the local acceleration of "artificial gravity" in your scenario scales directly with radius according to a formula of $a=\omega ^2 R$ where $\omega$ remains constant.

But the tube is subject to an unbalanced net torque from the astronauts pushing on one wall. It will not rotate at a constant rate.I urge you to think before you post and to slow down so that you do not commit so many spelling errors. It is also a rule of English that the first letter in the first word in each sentence be in presented in upper case and that there be a punctuation mark (for instance, a period or a question mark) at the end of each sentence.

The spelling errors, I can forgive since you are not a native English speaker. The failure to use upper case and to terminate sentences properly are not as easily excusable.

The use of R2 instead of R², $R^2$ or R^2 is just plain rude. Stop that at once!

 

[farolero]

Ok my apologies I will try to correct that.

I think we can agree that L=mvR^2sinalpha, where alpha is the angle of the astronauts trajectory with the radial direction of the tube.

So to follow your approach what we need to do next is relate w,v and R and I am really lost on how to do this but i know its not v=wR.

I remember I asked a teacher once and his solution was to integrate.

 

[jbriggs444]

Ok my apologies I will try to correct that.

I think we can agree that L=mvR^2sinalpha, where alpha is the angle of the astronauts trajectory with the radial direction of the tube.

No, we do not agree about that. It is time to review some dimensional analysis.

Ordinary momentum has units of mass times distance divided by time. For instance, kg m/sec.
L (angular momentum) has units of mass times distance squared divided by time. For instance, kg m^2/sec.

You have suggested $mvR^2 \sin \alpha$ as a possible formula for angular momentum. Let us verify whether that is dimensionally consistent.

You have a factor of mass times velocity times distance squared. That's mass times distance cubed divided by time.

That's a non-starter. It cannot possibly be correct.

 

[jbriggs444]

@farolero, I have a couple of passages that may be of interest.

The OP asks a question and wants an answer. They need to be shown a reliable pathway to a solution. They have fixated on a direct and simple path to the answer and do not want to be lead off from what they see as that shortest obvious path. That fixation has blinded them and prevented them so far from finding an answer. We must break that fixation by education.

It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so.

@Baluncore and that Mark Twain fellow are smart guys.

 

[farolero]

Yes i know its not correct but i don't know where i did go wrong.

I consider that angular momentum is the vectorial product of the given vectors, in this case the astronauts velocities vectors.

Edit:

What if friction can not be neglected in this case? : )=

 

[jbriggs444]

Yes i know its not correct but i don't know where i did go wrong.

I consider that angular momentum is the vectorial product of the given vectors, in this case the astronauts velocities vectors.

The definition for the angular momentum of a point mass with momentum p at offset R from the reference point is $L=\vec{R} \times \vec{p}$

The definition for momentum is, as you well know, $\vec{p}=m\vec{v}$

One of the ways of defining the vector cross product is $|\vec{a} \times \vec{b}| = |a|\ |b|\ \sin \alpha$ (*)

If you put these together [and are casual about using the $|x|$ notation for "magnitude of" and $\vec{x}$ notation for vectors] then you get $L = mvR \sin \alpha$. The problem with your formula was the extra factor of R.

(*) I had carelessly written this same formula with cosine some time back. I apologize for that error.

 

[jbriggs444]

Edit:

What if friction can not be neglected in this case?

Before complicating things with friction, we need to finish solving the problem without it.

Off hand, I'd expect a differential equation. Possibly one without a closed form solution. But let's not get into a discussion about types of solutions to differential equations either.

 

[farolero]

It might turn out right your formula L=mwR^2 accounting as true L=mvRsinalpha.

All you would need to prove it, but i need to see it to believe is that v=wR/sinalpha.

But i insist you just can not take circular motion equations for an spiral motion just because.Friction might be vital actually to solve the problem.

If an spinning ice skater opens her arms with weights in her hands the weights will go slower so kinetic energy has been transformed into heat by friction.

You can not balance energy and momentum in the spining skater unless you account for the transformation of energy by friction.

 

[jbriggs444]

It might turn out right your formula L=mwR^2 accounting as true L=mvRsinalpha.

All you would need to prove it, but i need to see it to believe is that v=wR/sinalpha.

Let me be sure that I understand what you are want to see to believe.

You want a demonstration that $v=\frac{\omega R}{\sin \alpha}$? Is that what you are asking for?

But i insist you just can not take circular motion equations for an spiral motion just because.

It is not "just because".

 

[farolero]

Yes because if v=wR/sin alpha substituting v in L=mvRsinalpha, would give as result L=mwR^2 making your point right and we could go on with the problem.

But I think that to relate v,w and R in this case we would need to integrate as you need to integrate to obtain the perimeter of an spiral as I learned in a physics class.

Edit:

Applying circular motion equations to spiral motion is an arbitrarity that i don't think its allowed in physics.

For example you can not say v=wR for spiral motion, this would be false due to taking an arbitrary formula.

 

[jbriggs444]

Yes because if v=wR/sin alpha substituting v in L=mvRsinalpha, would give as result L=mwR^2 making your point right and we could go on with the problem.

But I think that to relate v,w and R in this case we would need to integrate as you need to integrate to obtain the perimeter of an spiral as I learned in a physics class.

No. One might need to integrate to obtain the total length of a spiral or the circumference of a circle. But velocity is a differential. It is the ratio between a tiny incremental bit of distance and the corresponding tiny incremental bit of time. Because the distance is tiny, it can be safely treated as a straight line. One can then use trigonometry to consider that line in terms of its components.

[If you will permit me, I will have the tube spinning counter-clockwise. The sign conventions will come out more easily that way]

Picture the tube aligned so that it is pointing straight right and left. It is rotating, at least momentarily, counter-clockwise with a rotation rate of $\omega$.

Picture a stationary grid of cartesian coordinates behind the tube. The x-axis extends out to the right and lines up with the tube. The y-axis extends upward and is at right angles with the tube The origin of the coordinate system is placed at the center of the tube.

Picture an astronaut within the tube. He is, at least momentarily, positioned at coordinate position (R,0). Since his x coordinate is R, he is R meters to the right of the origin. Since his y coordinate is 0, he is on the x axis.

Picture a spot on the wall of the tube where the astronaut is. The spot is also at coordinate position (R,0). It is on the x-axis and is R meters to the right of the origin.

The astronaut is moving with some unknown velocity. Call that velocity "v". We do not know what v is. But we do know that it is at angle $\alpha$ above the horizontal.

What is the y component of the velocity of the astronaut? (answer in terms of v and $\alpha$).
What is the y component of the velocity of the spot? (answer in terms of R and $\omega$).
Is the y component of the velocity of the astronaut the same as the y component of the velocity of the spot? (yes or no).

If you can answer these three questions, we can proceed. They are not trick questions. The answers are easy.

 

[farolero]

Edit:

Forgive me if I change the scenario again but I find it extreamly illustrating of the problem:

Theres an spining ice skater of two kg mass spinning at w=10, two spot weights are at 0.1 m of its center.

What speed will have the weights if the skater increases the radius of the arms to 1 m radius and releases the weights?

do you see how this problem is almost identical and can be only solved accounting kinetic energy has disappeared by friction?

 

[jbriggs444]

Theres an spining ice skater of two kg mass spinning at w=10, two spot weights are at 0.1 m of its center.

What speed will have the weights if the skater increases the radius of the arms to 1 m radius and releases the weights?

do you see how this problem is almost identical and can be only solved accounting kinetic energy has disappeared by friction?

I do see that this problem is almost identical and does require accounting for kinetic energy which may have been either removed or added by the skater. But the addition or subtraction of unknown amount of kinetic energy only adds complexity to the problem by injecting an unknown quantity.

I refuse to consider this scenario until you complete a solution for the problem we are already working on.

In the problem at hand there is no friction. The astronauts are free to slide up and down the tube without any resistance in the direction parallel to the tube.