Roll up a spring on a screw thread in a disk on rotation

[xyrth]

I try to find my mistake about that device:

After a while the device is like that:

 

My logic : the device without the spring conserved the sum of energy, so I add the spring and I watch what I recover and what I need, the sum must be at 0  too. But here I find a difference, I don't find my mistake.

[Strange]

You had another thread where you asked people to find your error. Is this the same thing?

Or, if you are claiming that energy is not conserved, you have posted this in the wrong place.

Which is it?

[xyrth]

Nothing relative to my old post. I try to find my mistake. It is a classical simple mechanics device. Maybe the rotational velocity of the helix ?

[John Cuthber]

Do you think that stretching and releasing a spring breaks the conservation of energy?

Do you think that a rotating disk breaks it?

Do you think that winding a string does so?

 

Do you think any component of the system can , on its own, break the law of conservation?

Do you think there is a way in which any of the components can "know about" the others so it can suddenly start breaking them?

Well, without that, there's no way that the ensemble can break the energy conservation law.

 

[Strange]

5 hours ago, xyrth said:

Maybe the rotational velocity of the helix ?

You could check this (and each other component of your calculation) by working backwards: you know that energy is conserved (I hope) so work out from that what the velocity should be and see if it is different from what you calculate. 

As you haven’t posted your calculations here (an illegible image doesn’t count) I don’t see how anyone can find your error for you. 

It looks like, once again, you have created something more complicated than you are capable of analysing. Why do you do that? And why do you expect others to find your errors for you?

 

[xyrth]

8 hours ago, Strange said:

(an illegible image doesn’t count)

The device is a disk in rotation around the orange axis. The disk rotates only, don't move in vertical translation. There is a hole inside the disk (hole at 30° relatively to the orange axis). In that hole I fixed a controllable nut. Controllable nut because I want to control the time the nut works, I can set OFF/ON the nut, it is simply walls that moves. I passed through the nut a screw thread. That's all. I drew 2 views, the left view is a side view and the right a top view. In the left view, I drew the thread by transparency to see the thread. Before start, the screw thread doens't rotate around itself. The device-nut is OFF, the spring is not attached on the thread. After, I rotate the disk at w=10rd/s for example, I need time for that. I set ON the device-nut and I attached in the same time the spring to the thread, and I count the energies.

It it easier to take a controllable nut, because I know the rotational velocity of the disk (it is not an acceleration). The controllable nut is only retractable walls, it is just to simplify the calculations.

 

 

 

 

[xyrth]

I redraw the start position without transparency and corrected the images:

 

 

 

 

 

[Strange]

23 minutes ago, xyrth said:

I redraw the start position without transparency and corrected the images:

Why?

[swansont]

21 hours ago, xyrth said:

Nothing relative to my old post. I try to find my mistake. It is a classical simple mechanics device. Maybe the rotational velocity of the helix ?

You don't really show your work in enough detail.  The diagram is also unclear to me — why is the turning of the bolt coupled to the rotation of the disk?

[xyrth]

23 minutes ago, swansont said:

why is the turning of the bolt coupled to the rotation of the disk?

Because there is a nut inside the disk. The disk rotates at 'w' when I start to count the energy. The bolt doesn't turn around itself when I set on the nut, the bolt is at -w*cos(30°) relatively to the disk when the spring is attached and the nut is set on.

 

28 minutes ago, swansont said:

You don't really show your work in enough detail.  

I give only the energies in comparison with the device without the spring (the energy is conserved with or without the spring). When I add the spring there is only the energy wins by the length of the spring and the energies lost by the torques from the force F on each axis 'x' and ''y'. What do you need ?

[swansont]

20 hours ago, xyrth said:

Because there is a nut inside the disk.

That illuminates nothing for me. 

20 hours ago, xyrth said:

The disk rotates at 'w' when I start to count the energy. The bolt doesn't turn around itself when I set on the nut, the bolt is at -w*cos(30°) relatively to the disk when the spring is attached and the nut is set on.

And from what I can glean from your diagram, the bolt and spring are rotating with the disk. There is no relative motion of anything that would affect the bolt.

20 hours ago, xyrth said:

I give only the energies in comparison with the device without the spring (the energy is conserved with or without the spring). When I add the spring there is only the energy wins by the length of the spring and the energies lost by the torques from the force F on each axis 'x' and ''y'. What do you need ?

The actual calculations, not the results, which you admit are wrong. And posted, rather than in a picture with small writing.

[xyrth]

On 9/24/2018 at 6:20 AM, swansont said:

There is no relative motion of anything that would affect the bolt.

The disk rotates but the bolt doesn't rotate around itself so if you are on the disk you see the bolt rotates counterclockwise. Without friction nor the spring it is always like that.

 

 

[Strange]

5 minutes ago, xyrth said:

The disk rotates but the bolt doesn't rotate around itself so if you are on the disk you see the bolt rotates counterclockwise. Without friction nor the spring it is always like that.

Are you not going to show your calculations?

How do you expect anyone to help you find your error if you won't show us what you have done so far?

You never now, by putting your detailed calculations here (in LaTex) you might even spot the error yourself. You would probably learn more from tat, than someone else pointing it out to you.

[swansont]

9 minutes ago, xyrth said:

The disk rotates but the bolt doesn't rotate around itself

OK

Quote

so if you are on the disk you see the bolt rotates counterclockwise. Without friction nor the spring it is always like that.

The only way this works is if the bolt is fixed with respect to some external coordinate system. But why would it remain fixed? Surely there is some friction that would keep it fixed with respect to the disk. And you have a spring attached to it, which would tend to keep it fixed with respect to the disk.

 

[xyrth]

35 minutes ago, swansont said:

And you have a spring attached to it, which would tend to keep it fixed with respect to the disk.

The spring will give a clockwise torque to the bolt and it will start to rotate. I gave the calculation for an angle of 2pi but it is possible to study for a small angle, in that case the thread don't have time to rotate (around itself).

 

35 minutes ago, swansont said:

But why would it remain fixed?

The bolt will rotate relatively to the disk and move. The center of gravity of the bolt will move more and more far away from the center of the disk. In my calculations I forgot the work lost because the center of gravity moves it is 2piRF  *  3/4*sqrt(3)/2, because in one turn of the disk, the bolt rotates only at w*cos(30°) and the angle of the force is at 30°. In one turn of the bolt the center of gravity moves of 2pi*R*sqrt(3)/2. Even, with that work, the sum is not at 0.

Edited September 27, 2018 by xyrth

[swansont]

42 minutes ago, xyrth said:

The spring will give a clockwise torque to the bolt and it will start to rotate. I gave the calculation for an angle of 2pi but it is possible to study for a small angle, in that case the thread don't have time to rotate (around itself).

Yes, if there is tension in the spring, it will rotate. The spring exerts a torque on the bolt. if you know how much frictional torque is present from the threads, you could solve for this motion.

It won't have time to rotate?

Quote

The bolt will rotate relatively to the disk and move.

Why will it do this? Are you claiming there is no friction between the threads, and the bolt is otherwise unimpeded? In some details you are treating this like a real device, and in others like it was an idealized system.

Quote

The center of gravity of the bolt will move more and more far away from the center of the disk. In my calculations I forgot the work lost because the center of gravity moves it is 2piRF  *  3/4*sqrt(3)/2, because in one turn of the disk, the bolt rotates only at w*cos(30°) and the angle of the force is at 30°. In one turn of the bolt the center of gravity moves of 2pi*R*sqrt(3)/2. Even, with that work, the sum is not at 0.

If it moves up or down you have gravitational potential energy to worry about, but I see no mention of that in the minuscule amount of work you've actually shown.

You might be better served starting with a simpler system and see if you can solve that.

[xyrth]

1 minute ago, swansont said:

It won't have time to rotate?

I mean around itself. For a small angle and because the bolt has a mass. The center of gravity of the bolt rotate like the disk but the bolt doesn't rotate around itself.

 

2 minutes ago, swansont said:

Why will it do this? Are you claiming there is no friction between the threads, and the bolt is otherwise unimpeded?  

The disk rotates around itself at w. The bolt rotates with the disk but the bolt doesn't rotate around itself at start. The bolt inside the disk is fixed to the disk, the disk rotates clockwise at w, so if you are on the disk you see the bolt rotates counterclockwise. No friction, correct.

 

6 minutes ago, swansont said:

You might be better served starting with a simpler system and see if you can solve that.

I added the consequences about the energy because I add the spring. Without the spring is conserved, I add the spring and I calculate the energies. I gave only the differences.

[swansont]

1 hour ago, xyrth said:

 I added the consequences about the energy because I add the spring. Without the spring is conserved, I add the spring and I calculate the energies. I gave only the differences.

Then it should be trivial to calculate, but you have to note that the bolt will not rotate through the same angle, nor with the same speed, with a torque being exerted on it.

Of course, nobody can confirm that this is the problem, because you haven't shown your work.

 

edit: you say energy is conserved in the first case, but it's not. The bolt rising adds potential energy, and the bolt CoM moving out raises the moment of inertia. Both of these would tend to slow the disk down, but you have it rotating at a constant speed. So you must have something doing work on the system. Again — there's no way to find your error.

[xyrth]

9 hours ago, swansont said:

The bolt rising adds potential energy,

Do you mean potential energy from gravity ? There is no external gravity.

The sum of energy of the closed device: disk + bolt is conserved (there is no friction). The center of mass of the bolt moves more and more far away from the center but this will change the angular velocity of the disk. It is possible to add a motor to the disk and keep constant the rotational velocity of the disk. The energy from the motor goes to the kinetic energy of the bolt. Again, the sum of energy is conserved in the close device (even composed of a motor if I take in account the energy I give to the motor).

 

My work, suppose the energy is conserved without the spring. And for a small angle of rotation I have the energies due to the presence of the spring:

 

a) The lead of the helix is 2πR√3 so the spring increases its length of 2*2πR in one turn of the helix, so the energy won by the spring is √3/2*2*δRF = + √3*δRF. I took the step of the helix like that to keep constant the angle Ω

b)/ The force F from the spring is applied in the axis 'x' with an angle of 30° on the disk, then the force that works is √3/2F, so in δ, the disk lost -√3/2F * δR = - √3/2*δRF.

c) The force F from the spring is applied in the axis 'y' with an angle of 60° so the force that works is F/2 (the helix applies a torque on the disk, the torque depends of F on 'y'), so in δ the disk lost F/2 * δR = - 1/2*δRF .

d)/ The center of mass of the helix changes its radius, the difference of length is δR*sqrt(3)/2  for a δ angle of the helix but the helix rotates at cos(30°) of the rotational velocity of the disk and the force is at 30° so the energy lost is - 3/4*sqrt(3)/2*δRF. I lost that energy because I can’t recover it with another device.

The sum of energy for an angle δ is (√3/2 - 1/2 -3/4*√3/2) * δRF = - 0.28 * δRF

 

Edited September 28, 2018 by xyrth

[swansont]

6 hours ago, xyrth said:

Do you mean potential energy from gravity ? There is no external gravity.

That's the sort of detail you should include when presenting a problem such as this.

6 hours ago, xyrth said:

The sum of energy of the closed device: disk + bolt is conserved (there is no friction). The center of mass of the bolt moves more and more far away from the center but this will change the angular velocity of the disk.

But you have presented the angular speed as constant.

6 hours ago, xyrth said:

It is possible to add a motor to the disk and keep constant the rotational velocity of the disk. The energy from the motor goes to the kinetic energy of the bolt. Again, the sum of energy is conserved in the close device (even composed of a motor if I take in account the energy I give to the motor).

 That's not what you said in the original setup. You need to actually determine the work done by the motor.

6 hours ago, xyrth said:

My work, suppose the energy is conserved without the spring.  

And I have explained why it's not. 

 

Quote

 

And for a small angle of rotation I have the energies due to the presence of the spring:

 

a) The lead of the helix is 2πR√3 so the spring increases its length of 2*2πR in one turn of the helix, so the energy won by the spring is √3/2*2*δRF = + √3*δRF. I took the step of the helix like that to keep constant the angle Ω

 

But the spring exerts a torque, so it will not rotate through this angle with the same rate as it did without the spring.

Quote

d)/ The center of mass of the helix changes its radius, the difference of length is δR*sqrt(3)/2  for a δ angle of the helix but the helix rotates at cos(30°) of the rotational velocity of the disk and the force is at 30° so the energy lost is - 3/4*sqrt(3)/2*δRF. I lost that energy because I can’t recover it with another device.

Some of the energy terms here have to be mass dependent, and I don't see any mass terms in your equations. The torque exerted by the spring will depend on the spring constant, and I don't see that referenced anywhere in your analysis.

[xyrth]

2 hours ago, swansont said:

That's the sort of detail you should include when presenting a problem such as this.

Of course, sorry for that miss.

 

2 hours ago, swansont said:

And I have explained why it's not. 

The angular velocity I supposed constant ?

 

2 hours ago, swansont said:

Some of the energy terms here have to be mass dependent, and I don't see any mass terms in your equations. The torque exerted by the spring will depend on the spring constant, and I don't see that referenced anywhere in your analysis.

I thought I could only do the sum of energy without the mass IF I compare with the same device but without the spring and for a small angle of study.

 

So, like I'm not able to study the transient sum of energy, I take a motor, I will count the energy I give to the motor to let constant 'w'. Before I attached the spring, I supposed all the parameters of the device like constant, I mean the bolt doesn't rotate around itself (no friction), the center of mass of the bolt rotates at 'w', the disk rotates at 'w' and the center of mass of the bolt is moving more and more far away from the center of the disk. The sum of energies of the closed device :{external power + motor + bolt + disk } is constant. The motor needs an energy but the center of mass will increase its kinetic energy.

 

At time=0, I add the spring:

The motor need to give an extra energy because there is the spring:

b) The force F from the spring is applied in the axis 'x' with an angle of 30° on the disk, then the force that works is √3/2F, so in δ, the disk lost -√3/2F * δR = - √3/2*δRF.

c) The force F from the spring is applied in the axis 'y' with an angle of 60° so the force that works is F/2 (the helix applies a torque on the disk, the torque depends of F on 'y'), so in δ the disk lost F/2 * δR = - 1/2*δRF

I win the potential energy from the length of the spring:

a) The lead of the helix is 2πR√3 so the spring increases its length of 2*2πR in one turn of the helix, so the energy won by the spring is √3/2*2*δRF = + √3*δRF. I took the step of the helix like that to keep constant the angle Ω

And I lost the energy that the spring recovers from the movement of the center of mass (farther from the center of the disk):

d)/ The center of mass of the helix changes its radius, the difference of length is δR*sqrt(3)/2  for a δ angle of the helix but the helix rotates at cos(30°) of the rotational velocity of the disk and the force is at 30° so the energy lost is - 3/4*sqrt(3)/2*δRF. I lost that energy because I can’t recover it with another device.

For a small angle of rotation, the bolt doesn't have time to rotate around itself because its angular velocity is 0 rd/s when I attach the spring. So the center of mass will move like no spring. I can take an angle of study of 1e-10 rd, with a mass of the bolt at 1000 kg and the force of the spring at 1 N.

 

 

 

Edited September 28, 2018 by xyrth

[swansont]

1 hour ago, xyrth said:

 So, like I'm not able to study the transient sum of energy, I take a motor, I will count the energy I give to the motor to let constant 'w'. Before I attached the spring, I supposed all the parameters of the device like constant, I mean the bolt doesn't rotate around itself (no friction), the center of mass of the bolt rotates at 'w', the disk rotates at 'w' and the center of mass of the bolt is moving more and more far away from the center of the disk. The sum of energies of the closed device :{external power + motor + bolt + disk } is constant. The motor needs an energy but the center of mass will increase its kinetic energy.

The motor won't add the same energy when you add the spring. And you aren't properly accounting for all the parameters in your idealized case, so there's no way to be sure if your assumptions are part of the reason you aren't getting the right answer.

Such as: how do you account for the change in the moment of inertia without the mass of the bolt being known?

[xyrth]

1 hour ago, swansont said:

how do you account for the change in the moment of inertia without the mass of the bolt being known?

For example, with a bolt of 1000 kg with a radius for the bolt at 0.2 m, the moment of inertia could be at I=20 (it is an example), a spring with a force of 1N, with w=10rd/s and a study for δ=1e-10 rd. The time of study is dt=δ/w=1e-11 s. The torque is T=0.2 Nm. The angular acceleration will be  T/I=0.01 rd/s². The angular velocity at the end of 1e-11s will be 1e-13 rd/s compared with w = 10 rd/s, I supposed it is constant or don't change the result enough to have the sum at 0.

[swansont]

22 minutes ago, xyrth said:

For example, with a bolt of 1000 kg with a radius for the bolt at 0.2 m, the moment of inertia could be at I=20 (it is an example), a spring with a force of 1N, with w=10rd/s and a study for δ=1e-10 rd. The time of study is dt=δ/w=1e-11 s. The torque is T=0.2 Nm. The angular acceleration will be  T/I=0.01 rd/s². The angular velocity at the end of 1e-11s will be 1e-13 rd/s compared with w = 10 rd/s, I supposed it is constant or don't change the result enough to have the sum at 0.

That's a really big bolt, and a really wimpy spring, and you were doing an analysis for a complete turn, not for a tenth of a nanoradian. And for that you can't assume the spring has a constant force. 

 

[xyrth]

16 minutes ago, swansont said:

And for that you can't assume the spring has a constant force. 

Why ? for a small angle of study δ=1e-10rd, the distance increased by the spring is sqrt(3)*δ*R, it is 3.46e-11 m, the force is constant for that small difference of length. And it is possible in theory to have a "device" like a spring but with a constant force.

22 minutes ago, swansont said:

and you were doing an analysis for a complete turn

No, for a small angle of rotation δ. And it is possible to have a device composed of several springs to have a constant force: