The gyroscope and its ability to avoid "falling down"

[Drakkith]

It is certainly true that the inertia of the stone pulls outwards.

But what is this thing that we call "inertia"? It is the property of the mass not to exert force if it does not undergo it (first principle) but to react with real force when it is subjected (according to the principle F = ma: the accelerated mass reacts with the force F).

The inertia of the stone that "pulls" towards the outside is a force that is anything but "apparent", because it is a mass that moves with *accelerated* moviment.

The stone obviously exerts a force on the rope that points directly opposite of the tension. But we're not talking about the force on the rope, we're talking about the force on, and motion of, the stone. And the only acceleration going on here is the stone's acceleration towards the center or, if in the process of increasing its angular velocity, its acceleration tangentially to its motion. Not outwards.

When the angular velocity increases, the stone goes to cover a larger circumference outside the smaller one.

No bigger circumference can stay inside a smaller one!

Geometrically.

Of course. No one is arguing otherwise. But this movement isn't an acceleration outwards. No force is ever exerted on the stone in the outward radial direction so no acceleration can take place.

Before replying, please do the following: draw a picture of the stone and the rope. Label the real forces (not fictitious) on the stone. You should only have 2 forces, neither of which are capable of accelerating the stone outwards.

 

[Dale]

When the angular velocity increases, the stone goes to cover a larger circumference outside the smaller one.

No bigger circumference can stay inside a smaller one!

Geometrically.

Yes, clearly.

You appear to be confused about the distinction between velocity and acceleration. The velocity indeed has an outward component, but the acceleration only has an inward component.

Please work out the math for yourself to convince yourself of this fact: Take a spiral path and calculate the first derivative (velocity) and the second derivative (acceleration). What is the sign of the radial component of each?

If you get stuck, post your work and we can help.

 

[CWatters]

When the angular velocity increases, the stone goes to cover a larger circumference outside the smaller one.

No bigger circumference can stay inside a smaller one!

Geometrically.

Earlier you were talking about the angular velocity increasing and we were answering accordingly.

Now you seem to be talking about the rope getting longer. That's the only way the circumference can increase.

 

[Dale]

Earlier you were talking about the angular velocity increasing and we were answering accordingly.

Now you seem to be talking about the rope getting longer. That's the only way the circumference can increase.

No, he is correct on this point. Suppose that you have a rope of 50 cm length. If you spin very slowly then you might have a 30 cm radius with the rope hanging down 40 cm. If you spin a little faster then the rope will hang down only 30 cm and the radius will be 40 cm. In both cases the rope is 50 cm.

 

[Luigi Fortunati]

Earlier you were talking about the angular velocity increasing and we were answering accordingly.

Now you seem to be talking about the rope getting longer. That's the only way the circumference can increase.

I have never talked about the lengthening of the rope but the lengthening of the circumference traveled by the stone as the angular velocity increases.

 

[CWatters]

Ok I meant increasing the radius.

 

[Luigi Fortunati]

The stone obviously exerts a force on the rope that points directly opposite of the tension. But we're not talking about the force on the rope, we're talking about the force on, and motion of, the stone.

The two things are absolutely connected to each other: it is obvious that if the stone exerts a force on the rope, this force must have it.

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

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!

And the only acceleration going on here is the stone's acceleration towards the center or, if in the process of increasing its angular velocity, its acceleration tangentially to its motion. Not outwards.

The tangential velocity is directed towards the outside even if not radially.

And it is precisely the increase of this speed that makes the circumference of the stone path widen.

Of course. No one is arguing otherwise. But this movement isn't an acceleration outwards. No force is ever exerted on the stone in the outward radial direction so no acceleration can take place..

The force that radially accelerates the stone outwards is that of its own inertia which is activated when the rope pulls inward.

It is exactly the same and opposite force provided by the third principle which arises spontaneously when the horizontal component of the tension of the rope acts on the stone.

Before replying, please do the following: draw a picture of the stone and the rope. Label the real forces (not fictitious) on the stone. You should only have 2 forces, neither of which are capable of accelerating the stone outwards.

This is precisely what should make you think!

If the forces were only 2 the stone could not accelerate outwards and could not extend its rotation radius!

The forces are 3 (can not be 2!).

 

[Dale]

The force that radially accelerates the stone outwards is that of its own inertia which is activated when the rope pulls inward.

I am getting tired of repeating this. Please do the math. The stone does not accelerate outwards. At all times the acceleration is inwards, so there is no force outwards.

The tangential velocity is directed towards the outside even if not radially.

Velocity, yes. Acceleration, no.

If the forces were only 2 the stone could not accelerate outwards

It doesn’t accelerate outwards.

and could not extend its rotation radius!

This does not follow. Even while it extends the rotation radius, the acceleration is still inwards. Do the math.

The forces are 3 (can not be 2!).

There are only two forces acting on the stone: tension and gravity.

 

[Drakkith]

@Luigi Fortunati I'm done. It's pretty obvious you have no wish to actually learn what's going on here, so I'll just bow out of this thread. Have a nice day.

 

[A.T.]

The forces are 3 (can not be 2!).

In the inertial frame there are 2 forces. In some non-inertial frame might be more.

If the forces were only 2 the stone could not accelerate outwards...

It doesn't accelerate outwards in the inertial frame.

...and could not extend its rotation radius!

It can extend it's radius by accelerating inwards less than required for a constant radius.

 

[rcgldr]

Maybe this will help, at the connection between rope and stone, in a steady state, the rope exerts a tension force on the stone, and the stone exerts an equal and opposite force on the rope. The rope's horizontal force on the stone is a centripetal force, the stones horizontal force on the rope is an outwards reaction force, sometimes called a centrifugal reaction force (not to be confused with the fictitious centrifugal force of a rotating frame of reference). The rope also exerts an upwards vertical force onto the stone, and the stone exerts a downwards vertical force on the string. Gravity exerts a downwards force on the stone, which opposes the rope's upwards vertical force on the stone, and in a steady state, the net vertical force on the stone is zero, so the only net force on the stone is a centripetal force causing the stone to follow a circular path (again assuming a steady state).

 

[rcgldr]

I assume the gyroscope issue is how a gyroscope can precess and remain horizontal (or nearly so) while only supported at one end of the gyroscope. In this case, the support exerts an upwards force onto the end of the gyroscope, and the gyroscope exerts a downwards force onto the support. Gravity exerts a downwards force on the gyroscope that opposes the upwards force that the support exerts on the gyroscope so it has zero net vertical force (in an ideal state). The upwards force at the support and the downwards force at the gyroscope's center of mass produces a torque that would tend to lower the gyroscope. The precession generates a torque that would tend to raise the gyroscope. In an ideal state, the torques cancel and the gyroscope remains horizontal while precessing.

If the support is a string suspended from above, then ideally, the gyroscopes center of mass moves about a horizontal plane as the support point of the string moves inwards and outwards in a spiraling pattern. For an example of this, go to video #9 (top link is a .wmv file) on this web page. Note that the professor starts the gyro tilted upwards a bit to induce the spiral like pattern, and later he's not quite happy when the gryo support point goes through one outwards and inwards pattern, then tends to stay outwards for a while and he interferes with it. I'm wondering if the support path pattern is really supposed to cycle or instead tend to approach some constant radius circular path. I'm not sure of the dampening factors or the issue of the gyroscope slowing down over time.

http://www.gyroscopes.org/1974lecture.asp

 

[CWatters]

The two things are absolutely connected to each other: it is obvious that if the stone exerts a force on the rope, this force must have it.

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

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!

If the centripetal force and the centrifugal force are both "real" and equal magnitude then their vector sum would be zero (because they act in opposite directions) and the stone would travel in a straight line. If the stone moves in a circle there must be a net force acting on it towards the centre.

 

[Luigi Fortunati]

If the centripetal force and the centrifugal force are both "real" and equal magnitude then their vector sum would be zero (because they act in opposite directions) and the stone would travel in a straight line. If the stone moves in a circle there must be a net force acting on it towards the centre.

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

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

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.

But then, why is the motion not straight in the inertial reference?

Because the two forces (centripetal and centrifugal) also rotate and cancel each other at different points in the circumference.

And in fact, even in the inertial reference, the stone does not approach and does not move away from the center of rotation.

 

[weirdoguy]

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

No it wouldn't because it's velocity is perpendicular to net force. Direction of the net force alone does not tell you how the stone will move. You need "initial conditions", and in this case, the initial velocity is the most importnat (it's direction).

Their vector sum is actually equal to zero. But then, why is the motion not straight in the inertial reference?

Because the net force is not zero. You should really start reading what other people write to you, because otherwise this discussion is pointless. You have some BIG misunderstandings about inertial forces and Newton's second law and you really don't want to give up on them.

 

[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.