How Does Angular Momentum Affect Gyroscope Movements and Moon Orbits?

[bobie]

We read everywhere that a spinning wheel, a gyroscope, offers resistence to change the plane of rotation.
Suppose a gyroscope has angular momentum L = 10 J*s:

What force F must we apply, what work W is needed to rotate the spinning plane by 2π?
- do we need to know only L or do we need to know r?
I suppose also:
- the work required is the same in every direction, and that
- the amount of work don is proportional to the angle of rotation, 90° = W/4
Is that correct?

Thanks for your help

 

[Simon Bridge]

Hint: Conservation of angular momentum.

Note: if the system ends up in the same state it started in, without friction or other losses, then no net work was done on the system.

It is not as simple as applying a force to rotate the gyroscope - you need a constant torque to maintain the gyroscope at the new angle. Remove the force and the gyroscope falls back to it's original orientation.
So I think you need to do more reading on how rotating systems work so that you can refine your questions.
http://www.cleonis.nl/physics/phys256/gyroscope_physics.php
More generally:
http://faculty.ifmo.ru/butikov/Applets/Precession.html

 

[bobie]

Hint: Conservation of angular momentum...
It is not as simple as applying a force to rotate the gyroscope - you need a constant torque to maintain the gyroscope at the new angle.

I couldn't watch the applet because I do not have Java, it seems really interesting, but,
Before I start getting deeper into the issue, please clarify these basic concepts:

- why is conservation relevant to this issue? we are not applying any torque to the spin , not changing the radius of the spinning wheel, right?, we are only changing the orientation of the spinning plane.
If we push the gyroscope in any direction (change its position) we must do the usual work (Ke) we would do if it were not spinning Ke = mv²/m
- if we tilt the gyroscope (changing its orientation) when it is not spinning, we must do some work w , if we repeat that while it is spinning, even if there were no extra resistence we must still do the work w, plus (if any) some extra resistence w₁ caused offered by angular momentum L, for a total (w+w₁) = W.
The fact that the gyroscope returns to the original position (that's amazing, is that in the applet ordo you know any video showing this phenomenon?) shouldn't affect that, the same as when we pull a spring, we do some work, even if the spring returns to the original position.
What is wrong in these beliefs?
Thanks a lot, Simon

 

[BruceW]

- why is conservation belong to this issue? we are not applying any torque to the spin, not changing the radius of the spinning wheel, right?, we are only changing the orientation of the spinning plane.

we are not applying torque to change the magnitude of angular momentum. But we are changing the direction of angular momentum. This still requires a torque. Angular momentum is a conserved vector, just like linear momentum. So it is not enough to just consider the magnitude of these vectors.

As an analogy to conservation of linear momentum, if we have an object moving in a circle at constant speed, the magnitude of linear momentum doesn't change, but the direction of linear momentum does change, so we still must supply a centripetal force to keep the object moving in a circle.

 

[bobie]

This still requires a torque... Angular momentum is a conserved vector, ... so we still must supply a centripetal force to keep the object moving in a circle.

So , I thought, a force/some work is needed to change the direction of the spinning plane. How do I calculate this force/work?

 

[jbriggs444]

Hint: Conservation of angular momentum.

Note: if the system ends up in the same state it started in, without friction or other losses, then no net work was done on the system.

It is not as simple as applying a force to rotate the gyroscope - you need a constant torque to maintain the gyroscope at the new angle. Remove the force and the gyroscope falls back to it's original orientation.

It takes no torque to maintain a gyroscope at a particular angle. It will not "fall back" to its original orientation. Conservation of angular momentum assures us that this is true.

The force applied to cause a gyroscope to precess to a new rotation plane is at right angles to the resulting deflection at the point of application. It requires no work.

 

[bobie]

The force applied to cause a gyroscope to precess .

Why precess?
Suppose we have a flywheel weighing 1 ton with two . If I grab the axle and rotate it by 90° do I do work? If the wheel is spinning do I do the same amount of work or more? does the wheel precess?does it bounce back?

Thanks

 

[jbriggs444]

Why precess?
Suppose we have a flywheel weighing 1 ton with two . If I grab the axle and rotate it by 90° do I do work? If the wheel is spinning do I do the same amount of work or more? does the wheel precess?does it bounce back?
Thanks

It is easier to demonstrate with a bicycle wheel rather than with a 1 ton flywheel.

Take an ordinary bicycle wheel, removed from the bicycle and give it a spin. Hold it by the axle, one end in your right hand and one in your left (with the treads of the tire spinning downward in the rough neighborhood of your nose).

Now push up with your left hand and down with your right. This applies a clockwise torque to the bicycle wheel. Over time, this will naturally impart clockwise angular momentum. In order to have a clockwise angular momentum, the spinning bicycle wheel will twist in your hands. The axle end in your left hand will twist toward you. The axle end in your right hand will twist away from you. The treads will be coming down somewhat to the right of your nose and you will be looking at a bicycle wheel which has turned so that its rotation is slightly clockwise.

This effect is called precession.

Note that you applied an upward force with your left hand. But the resulting movement was at right angles -- toward you. You applied a downward force with your right hand. But the resulting movement was at right angles -- away from you. In both cases, the force and the displacement were at right angles from one another. That means that no work was done.

If you stop applying the torque with your two hands, the bicycle wheel will keep spinning on its new axis. It will not "remember" its original orientation. It will not try to twist back to that orientation.

 

[A.T.]

If I grab the axle and rotate it by 90° do I do work?

To rotate the spin-axis (S) by 90° around an axis (R) you have to apply a net external torque around an axis (T), which is perpendicular to both S & R. The instantaneous net angular velocity vector (ω) is somewhere in the plane spanned by S & R. So the applied torque T and ω are perpendicular, and no work is done.

 

[bobie]

. In order to have a clockwise angular momentum, the spinning bicycle wheel will twist in your hands. The axle end in your left hand will twist toward you.

That is what I am trying to say: I resist the twist and keep my left hand up and right hand down allowing the wheel to change only in that direction. Is it clear now?
What is the work I have done? I surely applied a force, didn't I?
The wheel will stay that way or if I release my force it will move in any direction?

 

[bobie]

To rotate the spin-axis (S) by 90° around an axis (R) you have to apply a net external torque around an axis (T), which is perpendicular to both S & R. The instantaneous net (around S & R) angular velocity vector (ω) is somewhere in the plane spanned by S & R. So the applied torque T and ω are perpendicular, and no work is done.

You mean that if the wheel weighs 10 tons I can rotate it with my little finger? and the wheel (if there's no friction) will rotate at any speed with no effort on my side? I can make it rotate at 100 rps?

 

[jbriggs444]

That is what I am trying to say: I resist the twist and keep my left hand up and right hand down allowing the wheel to change only in that direction. Is it clear now?
What is the work I have done? I surely applied a force, didn't I?
The wheel will stay that way or if I release my force it will move in any direction?

If you resist the twist then you are now pushing away with your left hand and pulling toward you with your right. This is a clockwise torque as viewed from above. The wheel will twist against this torque and will precess to obtain clockwise angular momentum as viewed from above.

The result is that the wheel will now precess in the direction you originally tried to get it to move. Your left hand will rise, your right hand will fall and the treads coming down near your nose will be moving diagonally downward from your upper right to your lower left.

But you did not need to lift up with your left hand and down with your right to make this happen. That force was irrelevant. What was relevant was the outward push from your left hand and the inward pull from your right. But those forces were at right angles to the upward movement of the axle end in your left hand and the downward movement of the axle end in your right. So no work was done.

 

[A.T.]

no effort on my side?

Your effort has little to do with mechanical work done. Holding up a 100kg weight does no work, but it is quite an effort for a human.

 

[jbriggs444]

You mean that if the wheel weighs 10 tons I can rotate it with my little finger? and the wheel (if there's no friction) will rotate at any speed with no effort on my side? I can make it rotate at 100 rps?

You can rotate it through any angle you like with as little torque as you care to apply. Though it may take a long time.

The less torque you apply, the slower the rate of precession. If you want to precess a 10 ton wheel moving at 10,000 rps so that the axis of rotation itself rotates at 100 rps, that will take vastly more torque than you can apply with your little finger.

 

[bobie]

Your effort has little to do with mechanical work done. Holding up a 100kg weight does no work, but it is quite an effort for a human.

I might do no work, but I am applying a force equal and opposite to gravity, right?
That's why I enquired aboutF/W, force/work.
If there is a 10 ton flywheel must I apply a force to make it rotate at 1 rps? and if I want it to rotate at 10 rps? Do you think I can make it rotate at 100 rps?
Now, whatever your answers, if that wheel is spinning does it make any difference?
That is what I would like to know
Thanks, everybody

 

[bobie]

This is a clockwise torque as viewed from above. The wheel will twist against this torque and will precess

I haven't been clear enough:
I lift my left hand, lower my right and resist any twist, precession, force ... etc...that oppose my move, in any direction, and leave it like that.
Even if you say there is no work done, what force do I exert? no force at all?

 

[Nugatory]

I might do no work, but I am applying a force equal and opposite to gravity, right?
That's why I enquired about force/work.

If you actually change the orientation of the spin, you're doing work because you're applying torque across an angle, and that does work just as applying a force across a distance does. Indeed, it's fairly easy to derive the formula for the work done by a torque from the the old classic $W=Fd$ (and I strongly suggest that you try doing that as an exercise).

But first you have to understand how torque relates to changes of the direction of spin as well as the rate of spin. That came up in your other angular momentum thread: https://www.physicsforums.com/showpost.php?p=4728955&postcount=14

 

[jbriggs444]

I might do no work, but I am applying a force equal and opposite to gravity, right?

Right.

That's why I enquired about force/work.

But force and work are different things. Work is defined (in freshman physics) as the product of force applied time distance moved in the direction of the applied force.

If there is a 10 ton flywheel must I apply a force to make it rotate at 1 rps? and if I want it to rotate at 10 rps? Do you think I can make it rotate at 100 rps?
Now, whatever your answers, if that wheel is spinning does it make any difference?

Yes, whether it is spinning makes a difference.

If the flywheel is not spinning and if it has frictionless bearings, you can make it spin at any rate you like with as little force as you like. You just have to apply that force for a long enough time. It is actually a "torque" that you need to apply. Any force that is not applied right at the axis of rotation will have an associated torque.

If the flywheel is spinning and you want to change its axis of rotation then the situation is different. If the flywheel spin rate is much faster than the desired precession rate then the rules of gyroscopes in freshman physics apply. To a first approximation, the precession rate that you get is proportional to the applied torque and inversely proportional to the angular momentum that the gyroscope possesses by virtue of its spin rate.

The area that freshman physics does not cover is when the desired precession rate is not very small compared to the existing spin rate. As I understand it, you need tensors to deal properly with rotational mechanics in that regime.

 

[jbriggs444]

If you actually change the orientation of the spin, you're doing work because you're applying torque across an angle, and that does work just as applying a force across a distance does.

To a first approximation in the case of high spin rates and low precession rates, if you apply a torque, the resulting precession is at right angles to the applied torque. No work is done.

If the applied torque changes the rotation rate of the spin... then work is done.

 

[bobie]

If the flywheel is not spinning and if it has frictionless bearings, you can make it spin at any rate you like with as little force as you like. You just have to apply that force for a long enough time. .

Now, that really baffles me:
suppose we have a wheel (m, r)on frictionless gimbals (http://en.wikipedia.org/wiki/File:Gyroscope_operation.gif)
If we want to spin the wheel on its plane xy, we must apply a force/ give it momentum/Ke/work which depends on the speed of rotation: rps/v/ω. If instead of the wheel we have two balls (m= .5 kg) at the ends of a diameter (1 m) , and v is 10 m/sec, then we know we must give the 'wheel' 50J of Ke and L is 25J*s
You are saying that I can rotate the same wheel on a normal plane , give it same speed and a roughly similar Ke, with almost no force applied?
Now suppose the wheel is spinning on its plane at 10/π rps, and we make it rotate at the same speed on a normal plane, what force have we applied, what is the Ke in that direction?

Now, coming back to my first question, suppose I grab the 2 golden axles (in the .gif) in my hands, rotate it by 90° in any direction and keep it like that,
- what work have I done?
- what happens if, after a while, I let go of them?

Thanks a lot for your Kind help

 

[Nugatory]

- what work have I done?

Take the torque that you applied, multiply by the angle (be careful about the units here - the angle is in radians), you'll have the work. This is just $W=\tau\theta$ where $\tau$ is the torque and $\theta$ is the angle it's applied over.

- what happens if, after a while, I let go of them?

Nothing. It just sits in its new position until some other torque is applied (which may happen pretty quickly if the supporting gimbals allow for a net torque from the force of gravity).

 

[Nugatory]

You are saying that I can rotate the same wheel on a normal plane , give it same speed and a roughly similar Ke, with almost no force applied?

A small force will accelerate an object to just as great a speed and KE as a larger force - it just takes longer. Look at $F=ma$ for a moment; no matter how small $F$ is, $a$ is non-zero. Now speed $v=at$, so as long as $a$ is non-zero, there's some value of $t$ that will give you any value of $v$ that you please.

the same thing works with torques and rotational velocities.

 

[jbriggs444]

Now, that really baffles me:
suppose we have a wheel (m, r)on frictionless gimbals (http://en.wikipedia.org/wiki/File:Gyroscope_operation.gif)
If we want to spin the wheel on its plane xy, we must apply a force/ give it momentum/Ke/work which depends on the speed of rotation: rps/v/ω. If instead of the wheel we have two balls (m= .5 kg) at the ends of a diameter (1 m) , and v is 10 m/sec, then we know we must give the 'wheel' 50J of Ke and L is 25J*s
You are saying that I can rotate the same wheel on a normal plane , give it same speed and a roughly similar Ke, with almost no force applied?

I am saying that if you start with wheel that is already spun up to an L of 25J*s rotating on one plane that you can then get it rotating with an L of 25J*s in a new plane with almost no additional force applied and almost no additional work done.

It will take time. The lower the force, the longer the required time. But it will not take work.

Now suppose the wheel is spinning on its plane at 10/π rps, and we make it rotate at the same speed on a normal plane, what force have we applied, what is the Ke in that direction?

What is the angular momentum in the first plane of rotation? What is the angular momentum in the new plane of rotation? Angular momentum is a vector quantity. What is the vector difference between the two angular momenta?

How much time will you allow for the change to take place? How long is the moment arm for the force that you are applying?

Divide the change in angular momentum by the time and you have the required torque. Divide the torque by the length of the moment arm and you have the required force.

Now, coming back to my first question, suppose I grab the 2 golden axles (in the .gif) in my hands, rotate it by 90° in any direction and keep it like that,
- what work have I done?
- what happens if, after a while, I let go of them?

You will have done no work. The force that you will have exerted will turn out to have been at right angles to the resulting movement of the axles. The torque that you will have applied will turn out to have been at right angles to the resulting precession of the rotation axis.

Work computed as the dot product of force times distance ends up being zero. Work computed as the dot product of torque times rotation also ends up being zero. Work computed as the change in kinetic energy of the rotating wheel also ends up being zero. The books balance.

If you let go of the axles, the gyroscope will retain its new orientation indefinitely.

 

[bobie]

I am saying that if you start with wheel that is already spun up to an L of 25J*s rotating on one plane that you can then get it rotating with an L of 25J*s in a new plane with almost no additional force applied and almost no additional work done.
.

I do not understand that, jbriggs: either you do work or not.
Please correct me if I am wrong, but if a body moves, acquires v and therefore KE, a definite amount work must have been done. And we should be able by a formula to calculate it.

Suppose the flywheel has m = 10 tons, if I spin it at k rps on the wheel plane (xy) I know exacttly the energy required/spent ( a huge amount of work)
If it is not spinning, and I spin it on the zx plane I know exactly the energy required/ spent (which is still huge)
If it is already spinning on the xy plane and I spin it in the zx plane the wheel acquires v and therefore Ke also in that plane, I should be able to calculate exactly , and it cannot be almost zero.
Isn't there a formula for that? am I the first in the world to ask this question?

It seems rather unrealistic that I can give a 10-ton-wheel a jerk and set it spinning at any speed ad lib with almost no work done, do you agree or is this practical aprroach wrong?

 

[Nugatory]

Isn't there a formula for that?

yes, there is a formula. Post #21 in this thread has it.

But before you just plug some numbers into it (correctly applying it to a gyroscope can be tricky) try considering how the problem changes when the spin rate is small and when it is very large.

 

[jbriggs444]

I do not understand that, jbriggs: either you do work or not.
Please correct me if I am wrong, but if a body moves, acquires v and therefore KE, a definite amount work must have been done. And we should be able by a formula to calculate it.

Yes, absolutely.

Suppose the flywheel has m = 10 tons, if I spin it at k rps on the wheel plane (xy) I know exacttly the energy required/spent ( a huge amount of work)

Yes, absolutely.

If it is not spinning, and I spin it on the zx plane I know exactly the energy required/ spent (which is still huge)

Yes, absolutely.

If it is already spinning on the xy plane and I spin it in the zx plane the wheel acquires v and therefore Ke also in that plane, I should be able to calculate exactly , and it cannot be almost zero.

Note the phrase that you have emphasised: "Ke also in that plane". KE does not occur in planes. It is a scalar quantity. By spinning in the xy plane, the flywheel already has KE.

If you grasp the golden axles (in your previous .gif) and apply a torque, always at right angles to the rotation of the flywheel it will precess from rotation on the xy plane to rotation on the xz plane with no expenditure of energy

Isn't there a formula for that? am I the first in the world to ask this question?

Yes, there is. The formula is "0".

This result appears at first glance to contradict the formula in post 21 that Nugatory references above.However, as Nugatory points out, the problem changes when spin rate is small compared to the precession rate you are attempting to impose and when the spin rate is large compared to the precession rate.

Work = torque times rotation is a correct formula. But in three dimensions it is a vector formula. It is the vector dot product of torque times rotation (more generally it is the integral of the vector dot product of instantaneous torque times incremental rotation)."0" is the limit for low precession and high spin because the applied torque in such a case is always at right angles to the resulting precession and the dot product is always zero.

It seems rather unrealistic that I can give a 10-ton-wheel a jerk and set it spinning at any speed ad lib with almost no work done, do you agree or is this practical aprroach wrong?

For a non-spinning 10-ton wheel, you cannot "set it spinning" with no work done. No one has said that you can.

For an already-spinning 10-ton wheel, I suspect that you will have problems changing the spin axis with no work done using a jerk. Rather than imposing a spin on a new axis you will have introduced (I expect) a three-dimensional wobble. This is a case of "high precession rate, low spin rate".

But... for an already-spinning 10-ton wheel, you can apply an arbitrarily small force for a correspondingly long time and do arbitrarily little total work to change the spin to a new axis.

 

[A.T.]

Note the phrase that you have emphasised: "Ke also in that plane". KE does not occur in planes.

Bobie do you understand the above for linear dynamics? If you apply a force perpendicular to velocity, you are doing no work. You are not adding any "KE along some direction" additionally to the "KE along the old direction". You are just changing the direction, while preserving KE.

The analogy for rotational dynamics is: If you apply a torque perpendicular to angular velocity, you are doing no work. You are not adding any "KE in some plane" additionally to the "KE in the old plane". You are just changing the rotation axis, while preserving KE.

 

[bobie]

... do you understand ...
If you apply a force perpendicular to velocity, you are doing no work. You are not adding any "KE along some direction" additionally to the "KE along the old direction". You are just changing the direction, while preserving KE..

I understand that this contradicts what I have studied in other occasions. Do you remember that Bohr model was judged wrong as a body changing direction accelerates? (.. and therefore should radiate and fall down).
I have studied that if you change direction to a vector you are always giving KE in the normal direction: (http://en.wikipedia.org/wiki/File:Normal_and_tangent_illustration.png ).
- if you hit a speeding ball on its center-of-mass you give it the same amount of KE you'd give if it were at rest:
- you change the direction of the vector and increase it. I cannot think of a real case when you can change direction doing no work, can you?

KE does not occur in planes.

As to planes and rotational dynamics, a flying (foot/base)ball can have both KE in direction of the goal and rotational KE on the normal plane, if you do not throw the ball straight but give it also a spin you add rotational KE and therefore you must do work.
Isn't it so? why should it be different with a gyroscope?

Thanks, everybody for your time and efforts! (you are doing hard work with me)

 

[jbriggs444]

I understand that this contradicts what I have studied in other occasions.

References, please.

Do you remember that Bohr model was judged wrong as a body changing direction accelerates? (.. and therefore should radiate and fall down).

A body changing direction accelerates, yes. An electric charge that accelerates radiates, yes. Neither fact has anything to do with the situation at hand.

I have studied that if you change direction to a vector you are always giving KE in the normal direction: (http://en.wikipedia.org/wiki/File:Normal_and_tangent_illustration.png ).

That's a diagram, not an explanation.

If you change direction of a vector you are NOT always giving KE. You are never giving "KE in the normal direction". KE is a scalar. It has no direction.

The moon orbits the Earth in a [nearly] circular orbit. It is constantly changing direction under a force at right angles to its motion. This force does not change the speed of the moon. Accordingly, it does not change the kinetic energy of the moon.

- if you hit a speeding ball on its center-of-mass you give it the same amount of KE you'd give if it were at rest:

I can hit a speeding ball to slow it down, speed it up or change its direction while leaving its speed unchanged. The angle of the hit controls whether the hit removes KE, adds KE or leaves KE unchanged.

By contrast, if I hit a ball at rest, I can only increase its KE.

- you change the direction of the vector and increase it. I cannot think of a real case when you can change direction doing no work, can you?

The work that you need to supply can be arbitrarily small. A trampoline is one example. The moon orbiting the Earth is another. A car being steered through a turn is another.

As to planes and rotational dynamics, a flying (foot/base)ball can have both KE in direction of the goal and rotational KE on the normal plane, if you do not throw the ball straight but give it also a spin you add rotational KE and therefore you must do work.

A football does not have KE in the direction of the goal. KE is a scalar. It has no direction. A football has KE by virtue of its linear motion toward the goal. But that KE has no direction.

A football does not have rotational KE on the normal plane. KE is a scalar. It has no direction. A football has rotational KE by virtue of its rotational motion. But that KE has no plane.

Yes, if you were to start with a moving but non-rotating football and give it a rotation while preserving its original state of linear motion, this would increase its energy and therefore require work. No one here has said anything different.

Isn't it so? why should it be different with a gyroscope?

In the case of a gyroscope you are not starting with a non-spinning football. You are starting with a football spinning on one axis and ending with a football spinning on a different axis. It has the same KE before and after. So no work is done.

 

[A.T.]

Bobie do you understand the above for linear dynamics? If you apply a force perpendicular to velocity, you are doing no work. You are not adding any "KE along some direction" additionally to the "KE along the old direction". You are just changing the direction, while preserving KE.

Do you remember that Bohr model was judged wrong as a body changing direction accelerates? (.. and therefore should radiate and fall down).

That is EM radiation, which occurs only for charged bodies. In General Relativity masses radiate gravitational waves when accelerated. But that is a negligible effect for any real world human made gyroscope. It has nothing to do with classical Newtonian Mechanics, which we use to analyzes such devices.

I cannot think of a real case when you can change direction doing no work, can you?

Yes, uniform circular motion:
http://en.wikipedia.org/wiki/Work_(physics)#Constraint_forces

For example, the centripetal force exerted inwards by a string on a ball in uniform circular motion sideways constrains the ball to circular motion restricting its movement away from the center of the circle. This force does zero work because it is perpendicular to the velocity of the ball.

 

[bobie]

the centripetal force exerted inwards...restricting its movement away from the center of the circle. This force does zero work because it is perpendicular to the velocity of the ball.

The force does zero work not because it is perpendicular but because the ball cannot move, the same as when you are holding up a weight. But, if you lower your arm and let the weight drop a little, G does some work.
In our case the wheel moves, gets velocity, no matter in what direction:
if we push the spinning wheel on the normal plane zx or zy it will start to spin also in that direction, if you think that is not possible, there are videos on the web.

 

[A.T.]

The force does zero work not beacase it is perpendicular but because the ball cannot move,

What are you talking about? The ball IS MOVING. That's why it is called uniform circular MOTION.

 

[Doc Al]

The force does zero work not because it is perpendicular but because the ball cannot move, the same as where you are holding up a weight.

The ball's in uniform circular motion. It's moving.

 

[bobie]

In the case of a gyroscope you are not starting with a non-spinning football. You are starting with a football spinning on one axis and ending with a football spinning on a different axis. It has the same KE before and after. So no work is done.

Thanks for your explanations, jbriggs, I hope you forgive my inaccurate wording, of course Ke is a scalar, I used it instead of velocity.
Before I comment your statements, please let me know if you understood the problem I am posing and if you, too, are excluding that a gyroscope can spin at the same time on two planes/ axes.

For an already-spinning 10-ton wheel, I suspect that you will have problems changing the spin axis

The wheel is spinning (on xy), and we shoot a bullet to a point on its circunference (on y or x), the wheel starts spinning also on the normal plane (zy or zx) is this a real case?

 

[bobie]

What are you talking about? The ball IS MOVING. That's why it is called uniform circular MOTION.

The ball's in uniform circular motion. It's moving.

The ball cannot move perpendicularly, it is of course moving tangentially, you can hold up a weight and run.
The ball is not allowed to fly off at a tangent by the spring, the same as you are not allowing the weight to fall to the ground. Is there any difference?