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 Re: Experiment at home
Сообщение03.09.2015, 18:52 
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03/09/13
85
France
Thanks

Цитата:
Can you split the total disk kinetic energy to parts?

Why not ? With a gyroscope when there is a precession, the sum of energy is always constant, the wheel of the gyroscope (spin + precession) has more energy but the torque necessary for the precession lost a potential energy (if it's springs for example or gravity) it's the nutation. Here, when the disk accelerates around itself the support must have a torque, no ?

Цитата:
Maybe the disk with jammed axle wil have more energy than the free-rotating disk?

I think but when the axle is jammed this is the support that give the energy.

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 Re: Experiment at home
Сообщение03.09.2015, 19:14 


10/02/11
6786

(Оффтоп)

Theoristos: why do not you write the equations of the motion or at least the solutions to these equations? Can you do that? :mrgreen:

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 Re: Experiment at home
Сообщение03.09.2015, 21:32 


24/01/09
1237
Украина, Днепр

(Оффтоп)

Oleg Zubelevich: looks like you miss something

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 Re: Experiment at home
Сообщение03.09.2015, 21:51 
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03/09/13
85
France
Oleg missed please I think

But it's not necessary to have all equations or solutions, it's possible to think and to understand with sentences.

Something I would like to know, if I accelerate the disk with an external spring (for example), does the support receives a negative torque ? Please ;)

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 Re: Experiment at home
Сообщение04.09.2015, 13:06 
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03/09/13
85
France
I'm just understood why the disk turns around itself. But I don't understand where is the torque on the support when the disk accelerates around itself. And I would like to understand if the support receives a torque if I accelerate the disk in rotation around itself. Maybe it's the same question ?

Maybe the disk moves and apply a torque but I can reduce the torque if I change the thickness of the disk at the contact of the red arm like that (cut view) :

Изображение

The torque:

Изображение

The support receives the torque like that ?

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 Re: Experiment at home
Сообщение05.09.2015, 18:32 
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03/09/13
85
France
Здравствуйте :)

I don't know if it's easy to do but in fact it could be very helpful to calculate the coordonates of a point on the disk in 3 axes (x,y,z), like that it could be possible to verify if I can use a spring for increase the kinetic energy of the disk and in the same time to increase the potential energy of the spring. I don't think the support will receive a torque if I add a spring. The spring must move in a plan in the same time its length increases so the (x,y,z) of a point is very good in fact. The simulation from Ansys shows the point move up but with equations it's perfect to demonstrate the sum of energy. Theoristos, maybe if you can do this it will be very friendly, пожалуйста

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 Re: Experiment at home
Сообщение06.09.2015, 12:35 
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03/09/13
85
France
Hello,

The trajectory of a point is not so difficult than that, it's an addition of a circle and an ellipse.

With:

$\alpha$ the angle of the axis of the orange disk from the vertical
$R_1$ the radius of the red arm
$R_2$ the radius of the orange disk
$\theta$ the angle of the rotation around the support
$\beta$ the angle of the rotation around the orange disk relative to the red arm
$\alpha=1.082$ in the Ansys simulation (62°)


I respected the direction of rotation of the simulation of Ansys (the video):

Around the red arm:
$x=R_1 cos(-\theta)$
$y=R_1 sin(-\theta)$

Around the orange disk:
$x=R_2 cos(\beta)$
$z=R_2 cos(\alpha) sin(\beta)$

The relation between $\theta$ and $\beta$ is:
$\beta=-cos(\alpha) \theta$

In 3d :
$x=R_1 cos(-\theta) + R_2 cos(\theta cos(\alpha)) cos(-\theta)$
$y=R_1 sin(-\theta) + R_2 cos(\theta cos(\alpha)) sin(-\theta)$
$z=R_2 cos(\alpha) sin((\theta cos(\alpha))$

I plotted it with Maxima:

plot3d([10*cos(-x)+2*cos(x*cos(1.082))*cos(-x),10*sin(-x)+2*cos(x*cos(1.082))*sin(-x),2*cos(1.082)*sin(x*cos(1.082))],[x,0,2*%pi],[y,-5,5], [grid,500,1], [gnuplot_pm3d,false]);

The plot3d:

Изображение

I'm not sure about my result, but the 3d curve seems pretty good with the simulation.

Now, I'm asking me about the spring and the energy, if you have any idea how the torque is returned to the support at start and with a spring it could be very helpful.

Have a good day :)

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 Re: Experiment at home
Сообщение13.09.2015, 15:42 
Аватара пользователя


03/09/13
85
France
I understood where is the torque on the support if I add a small mass on a point that moves up, it's because the force of gravity is to the bottom and a part of the force give a torque to the red arm.

Now, with the same device but I use a torus not a disk (orange). No friction. The torus turns at w around the support. The torus turns around itself at $-wcos(\alpha)$ in the arm reference. I add 2 grey torus with a mass, the grey torus turn at 0.99w around the support for example, just lower than the orange torus. But these grey torus can be cut at small part when I want (theory). I use the torus for collide the small grey part in a free 3d space.

The orange torus and 2 grey torus :

Изображение

I resumed my idea:

Изображение

I use like before the red arm that turns around the axis O at w. The disk is now the orange torus. I add 2 grey torus, I use these torus for decrease the angular velocity of the orange torus in the arm reference, so the angular velocity in the laboratory reference increases. The small parts of the grey torus are seperate just before to be collide from the orange torus, like that I'm sure the grey torus don't have any torque. If necessary I can collide 4 or 10 small parts of the two grey torus, each part of the grey torus can move in space with its linear velocity, I can recover in theory the kinetic energy.

The vectors of velocities are like that:

Изображение

The vectors of velocities are near perpendicular (in theory it can be perpendicular), here V2 is more to the left but V3 is more to the right. I think the orange torus has only a torque.

For me the energy won is:

$2*(-\frac{1}{2}*I*(w*(1-cos(\alpha)))^2) + (\frac{1}{2}*I*w^2)) $

$= 2*\frac{1}{2}*I*(-(-2w^2cos(\alpha) + w^2*cos^2(\alpha))) $

$= Iw^2 (2cos(\alpha)-cos^2(\alpha))$

with $I$ the inertia of the orange torus around itself and $\alpha$ the angle of the axis of the orange torus from the vertical.

If it's not clear enough I can explain or add another images.

Why I can't recover more energy than I gave at start in theory with this device, what prevents this ?

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 Re: Experiment at home
Сообщение23.09.2015, 23:17 
Аватара пользователя


03/09/13
85
France
There is no torque on the arm because the disk don't turn around itself in the laboratory reference when I turn the arm. So if I take this device:

изображение

Details:

изображение

The disks don't turn around themselves. So I can put inside a disk a gyroscope. The other disk don't have a gyroscope. I use 2 springs between the blue disk and the gyroscope for give a positive torque on the blue arm, the blue arm increases its angular velocity. The springs are not connected to the orange disk but to the gyroscope, so the orange arm don't receive a torque. The gyroscope will precession only. The lenght of the springs are always the same, so the springs never lost a potential energy. I need to have the forces like I drawn for 180° after I need to reverse the signs. I drawn the device at the angle 90°. Like that the energy of the device increases without an external energy so where I'm wrong ? I can't use the gyrsocope even the disk never turn around itself ?

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