The gyroscope and its ability to avoid "falling down"

[Dadface]

If on the stone there was a net force that acts towards the center, the stone would move towards the center.

The stone does not move towards the centre but it accelerates towards the centre as given by Newton's second law. Remember that an acceleration is a change of velocity ( in other words a change of speed and or direction).

 

[jbriggs444]

The stone pulls this force out of its inertia which is "inert" only if no one stimulates it.

This is not correct. [Real] forces come in pairs. The force of the stone on the rope is part and parcel of the force of the rope on the stone. Given a particular tension on the rope, the force of the stone on the rope will match that force regardless of its mass.

If the rope (with real force) pulls the stone to one side, it is obvious that the stone responds by pulling the other side with its own inertia that stops being "inert" and becomes "real" force!

The inertia of a stone is never a real force.

 

[Dale]

If on the stone there was a net force that acts towards the center, the stone would move towards the center.

You again appear to be confusing velocity and acceleration. The stone does indeed accelerate towards the center due to the net force towards the center.

Whether or not it moves towards the center depends on the velocity. Newton’s 2nd law says that an object’s acceleration is proportional to the net force. It does not say that it moves in the direction of the net force.

Your confusion between velocity and acceleration is leading you to a faulty analysis of Newton’s laws.

Instead the stone does not approach and does not move away from the center of rotation, a clear sign that the two opposing forces (centripetal and centrifugal) cancel each other out.

Their vector sum is actually equal to zero.

As I have recommended several times, do the math*. You will see that this is false. The acceleration is nonzero, so by Newton’s 2nd law the vector sum is also nonzero.

*Write down the equation of a spiral. Take the second derivative. Note if it is zero and if it is non zero then note which direction it points.

 

[CWatters]

I think we are going in circles/repeating ourselves. Perhaps time to close this thread?

 

[Dale]

I think we are going in circles/repeating ourselves. Perhaps time to close this thread?

It is looking that way. If @Luigi Fortunati shows some effort to do the math then we can continue, otherwise I will post the math and close the thread.

 

[A.T.]

If on the stone there was a net force directed towards the outside, the stone would move away from the center.

That's not how acceleration works for moving objects.

 

[Mark44]

The stone is never pushed outwards. It’s acceleration is at all times approximately towards the center. It never accelerates outwards.

It never accelerates outwards?!?

Let's say that we have a stone on a string that is swung in a nearly horizontal circle above the person's head. If the string is released, does the stone then travel directly away from the person on a radial path?

The answer is "No." It travels on a straight line path that is tangential to the circle of revolution.

 

[Luigi Fortunati]

The stone does not move towards the centre but it accelerates towards the centre as given by Newton's second law. Remember that an acceleration is a change of velocity ( in other words a change of speed and or direction).

 

[Luigi Fortunati]

There is an obvious misunderstanding between me and my interlocutors. It is a misunderstanding that I want to clarify to myself and to others.

For everyone (and also for me until some time ago) it is peaceful to pass from strength to acceleration (and vice versa) as if they were (almost) the same thing.

It is true (obviously) that between them there is the relation F=ma of the second principle but they are not the same thing: the strength is autonomous with respect to the acceleration, so much so that if I push a table my strength certainly there is (in any case) but the acceleration of the table is only if the friction fails to oppose (otherwise the table remains stationary).

So we are forced to say that acceleration is not proportional to force in general but only to the "net" force.

So not all forces are proportional to acceleration but only a certain type of force.

And it does not end here.

Even in the rotating reference frame there is no precise correspondence between force and acceleration.

In the rotating reference "appears" a force that in the inertial one does not exist, and then we invent the "apparent" force, another sub-species of force, different from the "real" force.

But how many forces are there?

Let's say by whom the force exercises it and suffers it: the rope of the sling.

Imagine that the stone is not there and that the rope is formed by a series of elastic rings hooked to each other, in a row.

Then we grab one end of this sling with the hand and let it rotate: the rings lose their roundness and lengthen in the radial direction, on one side (the centripetal one) and on the other (the centrifugal one).

Here is the unequivocal "measure" of the two forces, the rings that lengthen!

It is an extraordinarily clear proof: the rings are longer (on one side and the other) because there is a force that pulls on one side and another force that pulls on the other!

And the elongation of these rings also has another peculiarity: they clearly demonstrate that the force that lengthens them does not depend at all on the reference because the elongation in the inertial system is identical to that in the rotating system.

I understand that all this reasoning will not be appreciated but I can not do anything about it.

It seems extremely reasonable to me, that's what I think and I say it.

I do not intend to suppress my ideas to adapt myself to the opinion of the majority, without first having discussed without prejudice to the subject.

I will listen to every polite protest but not the impositions.

As long as I will be allowed to attend the forum.

 

[berkeman]

So not all forces are proportional to acceleration but only a certain type of force.

And it does not end here.

Even in the rotating reference frame there is no precise correspondence between force and acceleration.

In the rotating reference "appears" a force that in the inertial one does not exist, and then we invent the "apparent" force, another sub-species of force, different from the "real" force.

But how many forces are there?

Let's say by whom the force exercises it and suffers it: the rope of the sling.

Are you familiar with the use of a Free Body Diagram (FBD) to work on such problems? The use of good FBDs should help to clear up your confusions, IMO.

 

[Dale]

So we are forced to say that acceleration is not proportional to force in general but only to the "net" force

Yes, that should be clear to anyone who has studied Newton's 2nd law. The f in f=ma is the net force.

So not all forces are proportional to acceleration but only a certain type of force.

The net force is not a type of force. It is the sum of all forces of any type acting on a given object.

Imagine that the stone is not there and that the rope is formed by a series of elastic rings hooked to each other, in a row.

Then we grab one end of this sling with the hand and let it rotate: the rings lose their roundness and lengthen in the radial direction, on one side (the centripetal one) and on the other (the centrifugal one).

Here is the unequivocal "measure" of the two forces, the rings that lengthen!

It is an extraordinarily clear proof: the rings are longer (on one side and the other) because there is a force that pulls on one side and another force that pulls on the other!

Note that the two forces you describe here act on the rope (or the rings), not the stone. This does not contradict anything that any of us said above regarding the forces on the stone, and I suspect that nobody would object to it.

I do not intend to suppress my ideas to adapt myself to the opinion of the majority

Stubbornly clinging to misconceptions is not a virtue, particularly when correct concepts have been taught to you. You are choosing ignorance instead of knowledge. It is not a teacher's job to exhaust themselves trying to shove knowledge into an unwilling recipient, and a student who demands that a teacher do that in order to teach them is simply being selfish and inconsiderate of other's efforts. You, in particular, have been taught correct principles here, but for whatever reason you are unwilling to learn them. That is fine, it is not mandatory for you to learn, but participation here is a privilege that this community reserves for those who are willing to learn from the community.

For future readers who may be interested, the math that I had requested @Luigi Fortunati to work out is fairly straightforward. It is easiest to work in polar coordinates, the acceleration is given by equation 4 at https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf

$\mathbf{a}=(\ddot{r}-r \dot{\theta}^2)\mathbf{e_r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\mathbf{e_{\theta}}$

For simplicity, if we let $\theta=\omega t$ and $r=r_0+b t$ then $\mathbf{a}=-r\omega^2 \mathbf{e_r} + 2 b \omega \mathbf{e_{\theta}}$. Since $-r\omega^2$ is always negative, the radial component of the acceleration is always directed inwards.