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

[farolero]

so a pilot is spining in a circle at a constant speed, the g forces he will feel will be the same than normal acceleration dependant on wr

but now you put him inside a tube where he can move along varying the radius of the circle

now the normal acceleration will be zero because he is in the equivalent of centrifugal gravity free fall

but he will feel as g forces the tangential acceleration as tangential velocity is function of wr and r is increasing as the pilot moves along the tube

but now you propel the spining tube in such a way than as the pilot doubles radius the w velocity of the tube halves, on this way the tangential velocity is constant and hence tangential acceleration is zero

but how can the pilot be in this last case in zero tangential acceleration and zero normal acceleration which implies he endures zero g forces if he is making a spiral trajectory?

 

[A.T.]

Define "tangential" and "normal" precisely.

 

[farolero]

i thought they were universal terms:

normal or tangential to the pilot trajectory

 

[Dale]

but now you propel the spining tube in such a way than as the pilot doubles radius the w velocity of the tube halves,

Exactly how would you do that? What motion of the tube would make this happen?

 

[farolero]

well you put a precise gps on the pilot and connect the data with the engine of the tube

alternatively you could calculate an spiral tube in such a way that the spiral of the tube made the w half as the radius doubles with the right shape of the spiral tube

 

[Dale]

well you put a precise gps on the pilot and connect the data with the engine of the tube

And what function would you program into the GPS? What motion of the tube could you program which would result in the desired motion of the pilot?

alternatively you could calculate an spiral tube in such a way that the spiral of the tube made the w half as the radius doubles with the right shape of the spiral tube

What shape spiral is the right shape?

 

[farolero]

well i would take as the center of reference the center of spin of the tube and from this reference the pilot will make a perfect spiral for which i don't know the formula of position but i guess someone does

 

[A.T.]

so a pilot is spining in a circle at a constant speed, the g forces he will feel will be the same than normal acceleration dependant on wr

but now you put him inside a tube where he can move along varying the radius of the circle

now the normal acceleration will be zero ...

Only initially. It doesn't stay zero if angular velocity of the tube is constant.

but how can the pilot be in this last case in zero tangential acceleration and zero normal acceleration which implies he endures zero g forces if he is making a spiral trajectory?

If you spin the tube such that it exerts no force on the pilot, then he moves on a straight line.

 

[Dale]

well i would take as the center of reference the center of spin of the tube and from this reference the pilot will make a perfect spiral for which i don't know the formula of position but i guess someone does

The requirement that you proposed in your original post is for the motion of the tube that would produce the specified motion of the pilot. I don't think that the motion you proposed is even possible, and it seems that you are simply assuming that it is possible without any knowledge of the details.

If you assume that something impossible is possible then clearly you will wind up with contradictions.

 

[farolero]

i think i figure it out with your help guys, thanks

the pilot makes a spiral trajectory and this is the key

but the center of curvature in a given instant is not the center of the spiral but its somewhere offset from the center

so you just have to calculate this center and obtain the normal aceleration with this new instant temporal center that will be the g forces the pilot will experiment as he makes an spiral trajectory

 

[Nidum]

(1) Draw us a picture .

(2) Aircraft and pilot g forces have been well studied over many years . Have you done a search for any existing information about your problem ?

 

[farolero]

g forces a pilot feels for what i know go in function of At and An

as a curiosity on free fall on vacuum you feel zero g

edit:

where i wrote At i meant An

 

[Cutter Ketch]

... but now you propel the spining tube in such a way than as the pilot doubles radius the w velocity of the tube halves, on this way the tangential velocity is constant and hence tangential acceleration is zero

but how can the pilot be in this last case in zero tangential acceleration and zero normal acceleration which implies he endures zero g forces if he is making a spiral trajectory?

Your analysis is incorrect. Increasing the tube speed proportional to radius is NOT the condition that keeps the tube from touching him.

Start with the idea that with no acceleration he must move in a straight line. For there to be no acceleration the tube must be positioned around him at all times but not touch him

One might question how it came about that the pilot started moving in a straight line in the first place, but I didn't make up the problem. You did.

So if the pilot moves parallel to a radial, the initial tube velocity is 0 and dividing by the pilots radial position still gives 0. The tube does not move.

Now say the pilot moves at a constant velocity v starting at t0 on a tangent at radius r0. The pilots position at time t is angle = atan(v t / r0) and radius = sqrt((v t)^2 + r0 ^2)
For small angles atan(v t / r0) ~= v t /r0 and the angular speed increases linearly with time, but not linearly with the pilots radius. In fact at small times the sqrt approximates to r0 + (v t)^2 / (2 r0). For the tube not to touch him and keep the acceleration at zero the tube moves proportional to the square root of the radial position. This is because in addition to the radius becoming larger the fraction of the velocity which is tangential becomes less and less. This is where your logic failed. Any other motion of the tube, say angular velocity proportional to radius as you propose WILL accelerate the pilot.

 

[farolero]

well obviously as soon as the wall of the tube pushes the pilot he will feel g forces and if he goes in a staright line the tube is not spining which is a condition

what i initially failed to see is that the center of the trajectory at a given instant is not the center of the spiral but the center of curvature of that portion of the spiral where the pilot is at the given instant

if instead you wrongly take the center of curvature in the center of the spiral that's when the contradiction appears

 

[farolero]

i still have some trouble with this case problem:

imagine the case there are two counterrotating tubes in space:

the tubes are massless and frictionless, i can make this assumption for being them counterrotaing they still can apply a torque

there are two astronauts in the center of each tube and exactly at the same moment they start a centrifugal gravity free fall

we measure their trajectory exactly in the moment both astronauts intersect

the trajectories will be V which added will be a I aiming up

if initially the astronauts vectors in an isolated system was zero, how can it the isolated system go to a vector I aiming up without counterpart since the tubes are massless and hence can not have a vector

i hope you don't mind me asking questions of things i don't see and setting my self physics problems in order to solve them for entertainment

 

[A.T.]

the tubes are massless and hence can not have a vector

If they were massless they would have an infinite acceleration from any unbalanced force, so you already found the flaw in your assumptions.

 

[farolero]

i think what would be going on is that actually none of the astronauts would be moving forward or backwards just lateraly, it would be the massles platform which would be rocking back and fordward :)

 

[farolero]

now a last case that will help me understand vectors and their arms:

now there's a single masless frictionless tube in space and two astronauts that jump to free fall artificial centrifugal gravity in opposite sense

initially there's a vector aiming upwards from the surface of the paper where the drawing is

how did it mathematically go to those two opposite vectors and the growing arm?

 

[Dale]

What you describe cannot happen without some external torque.

 

[farolero]

well then make the tube have the same mass than the both astronauts and be 10 m long, then the torque will be provided by the tubes inertia to keep spinning

consider than the tube with the astronauts inside its spinning at w=10

the initial separation between the astronauts cog would be 10 cm

 

[Dale]

then the torque will be provided by the tubes inertia to keep spinning

You really need to describe your scenarios more carefully. In this case, is the tube initially rotating?

Assuming so then the angular momentum of the tube + astronaut system is L. This can be written as $L= I\omega$, which is constant. As the astronauts go down the tube $I$ increases and therefore $\omega$ decreases.

 

[farolero]

i see thanks yes the tube is spinning together with the astronauts at w=10

if theyre initially in the center of the spinning tube separated by 10 cm could anybody help me calculate the speed they will have at 1 m centrifugal artificial gravity free fall?

im not very sure how should i solve it

 

[Dale]

Forget solving it for now. Do you know how to set up the problem? What equations would you use?

I am not sure what you have learned so far. Have you studied Newton's laws, free body diagrams, rotational motion, Lagrangian mechanics, ...?

 

[farolero]

well it would be easier for me to solve it accounting for a massless tube

i think the main equations i should use are conservation of momentum equations, the angular momentum must remain constant since its an isolated system

would you please allow me to solve it accounting for a very insignificant low weight tube?

 

[Dale]

Ok, so first step is to specify the problem completely.

Describe the geometry and the masses involved. How much mass are the astronauts, are they considered point masses, if not then what shape, the mass of the tube, any friction or other forces, any external forces, etc.

Next, describe the initial conditions and any events that set up the problem. Like initial rotation, astronauts letting go of each other, etc.

 

[farolero]

well first correct me if I am wrong please:

you were right a massless tube wouldn't result in an spiral motion but both astronauts going in an straight line

but if i take a very low neglectable mass tube they will go in a very open spiral, if increase the mass of the tube the spiral motion of the astronauts will be less open

the initial conditions are this:

theres a spining tube in space with two astronauts inside both separated by 0.1 m who are holding each other and suddenly let go on artificial gravity free fall(the astronauts for simplification are considered point masses and for simplifictaion reason friction along the tube is minimum and its neglected)

the spinning tube with the astronauts its rotating at w=10

the mass of each astronaut is 1kg

the mass of the tube can be consider and I am no very sure of this as 2 kg or neglectable

the tube is one meter long

what speed will have the astronauts when leaving the tube?

if there's any teacher here they could put this problem to their students to see who can face unseen problems and whose just been memorizing standard cases

edit:

i realize now it would be wrong to neglect the tube mass so please if you can take it as two kg it would help me a lot

 

[Dale]

OK, so you have a scenario in mind. The next thing to do is to start selecting the laws that you want to use.

At your level, the standard laws to use are Newton's laws. Those laws apply whenever there are forces acting and you can neglect relativity (which is fine here). So you can use Newton's laws by calculating the forces acting onnthe tube and on each astronaut.

However, it is usually easier to apply conservation laws. So think about your system. Which conservation laws apply and why do they apply.

 

[farolero]

well in this case above all i would check conservation of momentum so the mvr of the astronauts remain constant

they would apply for being practically an isolated system

 

[Dale]

Ok, so conservation of angular momentum. That is a good start.

Are there any other conservation laws that would apply?

 

[farolero]

yeah of course conservation of energy

so let's see the astronauts go initially at 1 m/s(0.1*10)(wr) so their kinetic energy is 0.5*2*1=1

the momentum is initially 2*0.1*1(mvr)=0.2

but i don't remember how to fit the tube in the equation, i don't remember very well the rotational equations

i know rotational momentum is w/r and rotational energy is 0.5*w*I2 but what's the moment of inertia of a tube?

 

[jbriggs444]

yeah of course conservation of energy

so let's see the astronauts go initially at 1 m/s(0.1*10)(wr) so their kinetic energy is 0.5*2*1=1

the momentum is initially 2*0.1*1(mvr)=0.2

You specified the astronauts had 0.1 meter separation. Is that from each other or from the center point?

 

[farolero]

its a separation from each astronaut cog and as well from the center of the tube as set initially, i would take the astronauts as spot masses since this problem is already enough confusing to me but certainly interesting to me

 

[Dale]

Yes, since the tube is frictionless the mechanical energy is also conserved. There is no potential energy involved, so all of the mechanical energy is kinetic energy. So your two conservation equations are:

$E=\frac{1}{2}I \omega^2$
$L=I\omega$
With $dE/dt=dL/dt=0$

Once you have the equations in mind then you need to think of how to solve them. In this case, the first step is to think of a good way to parameterize the equations. There is only really one internal degree of freedom, and that is the distance of the astronauts from the center, so let's call that $R$.

So the next step is to write $I$ as a function of $R$

 

[jbriggs444]

Yes, since the tube is frictionless the mechanical energy is also conserved. There is no potential energy involved, so all of the mechanical energy is kinetic energy. So your two conservation equations are:

$E=\frac{1}{2}I \omega^2$

Assuming that "I" includes the contribution from the mass of the astronauts (which I expect is what you intend), that will pick up the contribution from astronauts' tangential velocities, but I do not see an accounting for their eventual radial velocities.

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.

 

[farolero]

well here my answer after much thought about it:

if you were a teacher and you put this unexpected problem as real problems are let's see if I've found the right solution and i pass or not and i fail :)

the problem is pretty simple to solve accounting for one thing:

you 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

conservation of energy is acomplished for the increase of energy of the astronauts comes with a decrease of energy of the spinning tube

conservation of momentum keeps true as well for the spiral trajectory they make is a kind of spiral not fractal but that opens more and more and hence the arm between each astronaut trajectory vectors reduces as the atronauts advance

in infinity the tube will remain still and all energy of the system will have gone to the astronauts having both an speed of aproximately 10 m/s, you could calculate the arm between these two vectors accounting for the initial momentum that wouldn't have changed

as the artificial gravity force reduces each moment the final result of the problem is not 2 m/s but slightly less what I am trying to figure out now

edit:

well the eacts answer would be an speed of 1.875 m/s after one meter free fall

as i don't remember very well how to integrate i solved it by thales:

 

[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