A doubt about tangential and normal aceleration, pilot g forces

[jbriggs444]

ou can take the astronauts are on artificial gravity free fall and hence are acelerated radially

the magnitude of this centrifugal gravity aceleration with the given parametes would be 1 m/s2 or 0.1 G

so hence after 1 m fall both astronauts speeds will be aproximately 2 m/s

The initial outward acceleration will be 0.1 g, yes. [Edit: oops, no -- I get a different result, please show your work]. But as the astronauts move outward, the rotation rate cannot be maintained if angular momentum is to be conserved. The simple assumption of constant outward acceleration will not give a correct answer.

Also note that the tube is 1 meter long, so each astronaut has only 0.4 meters to fall before hitting the end. Also note that falling 1 meter at 1 meter per second² does not give you a velocity of 2 meters per second.

@Dale has suggested a good strategy. If angular momentum and energy are both conserved, one can write down equations that must always hold no matter how far the astronauts have fallen through the tube, including at the moment that they exit from the bottom end.

Your task is to write those equations down. See #33. What is the moment of inertia of the tube plus astronauts as a function of R, the distance of each astronaut from the center?

 

[farolero]

yes but what youre saying its taking the false assumption that when the astronauts leave the tube they do it perpendicular to its mouth while in truth they keep a much more radial vector along the tube

in fact the arm between these two vectors tends to decrease to keep conservation of momentum true as the vectors aim more and more radially

if you take the typical approach to solve the problem first question would be does the astronut accelearte or decelerate?

from conservation of momentum with the pointed false assumption youd expect it to decelerate which can not be

 

[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.

Oops, yes, you are completely correct.

 

[jbriggs444]

yes but what youre saying its taking the false assumption that when the astronauts leave the tube they do it perpendicular to its mouth while in truth they keep a much more radial vector along the tube

in fact the arm between these two vectors tends to decrease to keep conservation of momentum true as the vectors aim more and more radially

if you take the typical approach to solve the problem first question would be does the astronut accelearte or decelerate?

from conservation of momentum with the pointed false assumption youd expect it to decelerate which can not be

No. That is not what I am saying.

 

[farolero]

well what i did is calculate the energy of the rotating tube with e=w2*I taking as I=mr2
i obtained an energy of 200

the energy of the astronauts would be 0.5*m*v2=0.5
so the total energy is 200.5

i took that at infinite distance all energy will have gone to the astronauts so their speed will be 10

the initial momentum would be mvr=2*1*0.2 from the astronauts and L=i*w=mr2*w=2*1*10=200 from the beam

so total initial momentum =200,2

so the arm between the final vectors will be 200=v*r so r=200/10=20

seems i was wrong in the arm shrinking apparently id does very slightly grow

then just by thales as you can see in my diagram accounting for what i told i calculated gemoetrically the exact speed after advancing 1 meter through the tube that i correct now is 2 m long, i was just accounting for one side by the 1 m figure

i thought that if you start at 0 m/s and accelerate during 1 meter at 1 m/s2 you end at 1 m/s so if instead you start at 1 m/s you ennd up at 2 m/s

 

[jbriggs444]

well what i did is calculate the energy of the rotating tube with e=w2*I taking as I=mr2
i obtained an energy of 200

The formula for the moment of inertia of a 2 meter long tube (1 meter radius) rotating about its midpoint is not $I=mr^2$.

the energy of the astronauts would be 0.5*m*v2=0.5

The energy of each astronaut is $\frac{1}{2}mv^2$ = 0.5 Joules. Total 1 Joule.

i took that at infinite distance all energy will have gone to the astronauts so their speed will be 10

In the absence of friction, the tube spins forever. Its remaining energy is never imparted to the astronauts.

The hard part here is figuring out the rotational speed of the tube at the moment the astronauts drop out from the ends. Again, @Dale has given a strategy for computing that speed. You should consider pursuing that strategy.

 

[farolero]

i didnt fint the moment of inertia of a tube so i used mr2
the problem wih this problem is that v=wr only works in the initial stage so the easiest way to solve it is as potential artificial gravity that decreases to the distance to the center squared and in infinity is zero

edit:

another similitude with gravity is that if you treat the astronauts as spot masses and they pulled their cogs together the speed of the system would be infinite

so the tube potential artificial gravity with spot masses is infinite at the center and 0 at infinite distance

edit:
as v=wr is not valid for this problem for the astronaut doesn't leave the tube perpendiculary any ecuation based on this formula is not valid neither to solve it

edit:

again how i would solve the problem:

i calculate the energy of the whole system

i take that at infinity all energy will be on the astronauts and apply Ec=1/2v2 and obtain v which i obtained 10

then i draw a thales triangle rendering infinity and put the speeds i know:

0,1 m radius= 1 m/s
infinite radius= 10 m/s

i drew the triangle measure the proportion and apply the ratio that obtained a final speed at 1 m radius of 1,875

also what do you think of my method to instead of integrating using a thales triangle that renders infinity in perspective :)

 

[jbriggs444]

i didnt fint the moment of inertia of a tube so i used mr2

Using the wrong formula will almost always give the wrong answer. By contrast, using the right formula will give the right answer.

the problem wih this problem is that v=wr only works in the initial stage so the easiest way to solve it is as potential artificial gravity that decreases to the distance to the center squared and in infinity is zero

I agree that it might be possible to solve the problem using the notion of a centrifugal potential. However, that approach is more subtle than you may imagine. You have not given a formula for the relevant potential.

For the usual formulation of centrifugal potential, one is considering a frame of reference with a fixed rotation rate. That is to say, a fixed $\omega$. Since centrifugal force is $m \omega ^2 r$, the associated potential (the negation of the integral of the local force) is given by $- \frac{1}{2} m \omega^2 r^2$. If we fix potential 0 at the center then the potential is negative infinity at infinity. That makes it difficult to use this potential to reason about behavior for the astronauts at infinity.

In a frame of reference that rotates at a variable speed, one might pretend that this yields a coherent potential field. [It is not really a potential field because the path integral is depends on the path taken, but as long as one only cares about the two paths that are actually taken and since both yield the same result, one can ignore that fact and move on]. But in order to use this approach, one would need to tie the rotation rate to the current astronaut radius in order to obtain the correct centrifugal force at that radius. Part of that calculation would involve the moment of inertia of the tube. But you have refused to compute the moment of inertia of the tube.

The approach that has been suggested to you uses the non-rotating inertial frame.

another similitude with gravity is that if you treat the astronauts as spot masses and they pulled their cogs together the speed of the system would be infinite

I agree that conservation of angular momentum guarantees that the speed of the system would increase without bound. Whether that fact makes centrifugal force similar to gravity is not clear. However, we can all agree that both gravity and centrifugal forces are "inertial forces" and are similar for that reason.

as v=wr is not valid for this problem for the astronaut doesn't leave the tube perpendiculary any ecuation based on this formula is not valid neither to solve it

No one has posted a conclusion about the direction that the astronauts in this scenario will travel upon leaving the tube other than yourself. However, I will do so now.

It is quite clear that each astronaut will have a tangential velocity given by $v=\omega r$ at the instant he leaves the tube, assuming that by $\omega$ we mean the rotation rate of the tube at that time. A moment before leaving the tube, the astronaut must have this velocity in order to remain within the tube. There is no impulsive tangential force applied at the exit. So a moment after leaving the tube, the astronaut must still have this same tangential velocity. Given an absence of friction, it is clear that the astronaut will have a non-zero radial velocity as he leaves the tube. Accordingly, from the point of view of an observer on the end of the tube, the non-zero relative radial velocity together with the zero relative tangential velocity means that the astronaut will depart moving straight "down". His path (briefly) continues to follow the line of the tube.

From the point of view of an external observer, the same observations can be made, but from this point of view, the end of the tube has a non-zero tangential velocity. From the point of view of an inertial observer, the non-zero tangential velocity together with the non-zero radial velocity means that the astronaut will leave the tube moving in a diagonal direction, neither parallel to the tube nor perpendicular to it.

These two facts are both true and do not contradict each other. However, this does not help us calculate the radial or tangential velocity of the astronauts when they leave the tube. So...

What is the moment of inertia of a narrow uniform tube of total length 2 meters and mass 2 kg rotating end over end about its center?

 

[farolero]

well i found it but as if the problem wasnt complex enough i would have to account for inner and outer radius of the tube:

 

[jbriggs444]

well i found it but as if the problem wasnt complex enough i would have to account for inner and outer radius of the tube:

Narrow tube. Ignore the inner and outer diameter and consider it as a narrow rod.
https://en.wikipedia.org/wiki/List_of_moments_of_inertia

[Note that you get the same formula by approximating a and b as zero]

 

[farolero]

ok so moment of inertia of a rod I=ml2/12

so I of the tube=2*4/12=0,66
L=I*w2=0.66*100=66

Ec=1/2I*w2=330

all this energy at infinity is transferred to the astronauts so:
115=1/2mv2=0.5*1*v*v so v=15 m/s

if v at 0,1 m is 1 and at infinity is 15 then i just calculate by thales or integrating the exact result that will be slightly less than 2

beyond the initial stage using any formula that inludes v=wr is wrong including the rotational kinetic energy formula

edit:

i forgot the other thing you need to know is that the force of artificial gravity decreases to the distance to the center squared i forgot this when solving my thales triangle

btw the resemblances between inflating centrifugal gravity and normal gravity are amazing arent they?

 

[jbriggs444]

ok so moment of inertia of a rod I=ml2/12

$I=\frac{ml^2}{12}$ yes.

so I of the tube=2*4/12=0,66

0.66 kg m², yes

L=I*w2=0.66*100=66

L is angular momentum. That's just $I \omega$. So you should get 6.6 kg m²/sec²

Ec=1/2I*w2=330

The rotational kinetic energy of the tube along, $E_c=\frac{1}{2}I \omega ^2$ = 330 Joules, yes.

all this energy at infinity is transferred to the astronauts

At r=1m the tube is still rotating and the astronauts are no longer touching it. The tube rotates forever and never transmits all of its rotational kinetic energy to the astronauts.

But we are not there yet. You have not finished computing the initial kinetic energy yet.

Edit: for goodness sake, stop blabbering about $v=\omega r$ already. You are the only person even considering it.

 

[farolero]

well to that i sould add the astronauts intitial kinetic energy but is so small that i neglected it

Ec=1/2mv2=0.5*2*1=2

 

[jbriggs444]

well to that i sould add the astronauts intitial kinetic energy but is so small that i neglected it

Ec=1/2mv2=0.5*2*1=2

Fair enough. So the kinetic energy in the starting configuration is 330 Joules from the tube and 2 Joules from the astronauts for a total of 332 Joules.

And we have an initial angular momentum of 6.6 kg m²/sec² for the tube alone.
What is the initial angular momentum for each of the two astronauts?
What is the total initial angular momentum of the system?

 

[farolero]

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

 

[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