A doubt about tangential and normal aceleration, pilot g forces

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Discussion Overview

The discussion revolves around the concepts of tangential and normal acceleration experienced by a pilot in a spinning tube, particularly focusing on the effects of varying radius and angular velocity. Participants explore the implications of these forces in different scenarios, including the pilot's motion along a spiral trajectory and the conditions under which the pilot may experience zero g forces.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that a pilot spinning in a circle at constant speed experiences g forces related to normal acceleration dependent on angular velocity and radius.
  • Another participant questions the definition of "tangential" and "normal" acceleration, seeking precise clarification.
  • A participant proposes a scenario where the pilot moves within a tube that allows for varying radius, leading to a discussion on the implications for normal and tangential acceleration.
  • Concerns are raised about the feasibility of programming the tube's motion to achieve specific conditions for the pilot's trajectory.
  • One participant asserts that if the tube exerts no force on the pilot, he would move in a straight line, challenging the initial assumptions about the pilot's motion.
  • Another participant argues that the proposed motion of the tube may not be possible, suggesting that assuming impossible conditions leads to contradictions.
  • A later reply indicates that the center of curvature of the pilot's trajectory must be considered to accurately calculate normal acceleration and g forces.
  • Discussion includes a hypothetical scenario involving two counter-rotating tubes and astronauts, raising questions about vector dynamics in an isolated system.
  • Participants express uncertainty about the implications of massless objects and their behavior under unbalanced forces.

Areas of Agreement / Disagreement

Participants express multiple competing views on the definitions and implications of tangential and normal acceleration, as well as the feasibility of the proposed scenarios. The discussion remains unresolved with no consensus reached on several key points.

Contextual Notes

Limitations include unresolved assumptions regarding the motion of the tube and the pilot, as well as the implications of massless objects in the proposed scenarios. The mathematical details of the spiral trajectory and the conditions for zero g forces are also not fully clarified.

  • #151
jbriggs444 said:
What is the y component of the velocity of the astronaut? (answer in terms of v and αα\alpha).
What is the y component of the velocity of the spot? (answer in terms of R and ωω\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).

vsinalpha.
wR.
yes.Edit:

Also I was wondering if your frame reference is rotating with the tube or not? I didnt understand that part.
 
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  • #152
farolero said:
vsinalpha
wR
yes
Good. Now that means that one can write:
$$v \sin \alpha = vy_{astronaut} = vy_{spot} = \omega R$$
Solving for v...
$$v = \frac{\omega R}{\sin \alpha}$$
 
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  • #153
Yes but is it the frame reference rotating with the tube? because otherwise I am not sure the aswer to the last question is yes.

From a fixed reference the Vy of the astronaut would be Vsin alpha plus the speed of the tube at that point.
 
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  • #154
farolero said:
Yes but is it the frame reference rotating with the tube? because otherwise I am not sure the aswer to the last question is yes.
*heavy sigh*
There is a reason that I said "stationary" when I asked you to imagine the cartesian coordinate grid.
From a fixed reference the Vy of the astronaut would be Vsin alpha plus the speed of the tube at that point.
That is not correct and does not match the answer you gave previously. You have already agreed that the Vy of the astronaut is equal to the Vy of the tube.

Unless the walls of the tube have holes in them, the two velocities must match.
 
  • #155
Ok i see you took an stationary reference i was not sure if it was stationary with the tube.

What you said seems only valid for that special position, if the tube is offset 45º from the x reference the Vy of the point is wrsen45 but the vY of the astronaut would actually be the net tangential velocity of the astronaut so Vy= Vt( supposing the trajetory of the astronaut is offset 45º from radially). And in this case the Vy of the point wouldn't be equal to the Vy of the astronaut.

So your main assumption to solve the problem though antiintuitive is not right.

Your assumption that the Vy of the point= Vy of the astronaut is only valid if you take a rotating reference with the tube, as a matter of fact you did so putting the x axe along the tube
 
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  • #156
farolero said:
Ok i see you took an stationary reference i was not sure if it was stationary with the tube.

What you said seems only valid for that special position, if the tube is offset 45º from the x reference the Vy of the point is wrsen45 but the vY of the astronaut would actually be the net tangential velocity of the astronaut so Vy= Vt( supposing the trajetory of the astronaut is offset 45º from radially). And in this case the Vy of the point wouldn't be equal to the Vy of the astronaut.
That is not correct. If the tube is offset by 45 degrees, then you just unstick the coordinate grid from the wall, rotate it by 45 degrees and stick it back down again.

Or you can do the analytical geometry to demonstrate that it works even if you do not unstick the grid.

Your basic problem is your unwavering conviction that you are right. Please review Mark Twain's admonition. You are so far wrong that it is not even entertaining.
 
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  • #157
jbriggs444 said:
That is not correct. If the tube is offset by 45 degrees, then you just unstick the coordinate grid from the wall, rotate it by 45 degrees and stick it back down again.

Then if you unstick and stick the x axe along the tube youre taking a rotational frame with the tube.

Ill demonstrate how the Vy of the point is not the Vy of the astronaut analitically:

When the tube has an angle of 45º with the x reference you stop it with an exterior torque, so the w is zero, then Vy of the spot=0 but Vy of the astronaut doesn't equal zero, so Vy of the astronaut is not equal to Vy of the spot bringing down your original assumption

I think mark twain quote is more appropiate for the teacher than for the student.
 
  • #158
I think it's high time to close this thread.

@farolero: Please review the entire thread, paying close attention to jbriggs444's very patient explanations.
 
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