A doubt about tangential and normal aceleration, pilot g forces

[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.

 

[farolero]

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.

 

[jbriggs444]

vsinalpha
wR
yes

Good. Now that means that one can write:
$$v \sin \alpha = vy_{astronaut} = vy_{spot} = \omega R$$
Solving for v...
$$v = \frac{\omega R}{\sin \alpha}$$

 

[farolero]

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.

 

[jbriggs444]

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.

 

[farolero]

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

 

[jbriggs444]

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.

 

[farolero]

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.

 

[Doc Al]

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.