Experiment at home

[m441]

Hello :)

I done this experiment at home:

A support (grey color) turns at $\omega_ 0$. The axis of the purple disk is fixed on the support. The purple disk can turn around itself but at start the angular velocity is 0 (laboratory reference). Like that :



The point A is fixed on the purple disk. I'm looking for the trajectory of the point A in the side view:



The point A moves up/down.

Is it logical for you ?

I tested with a low friction for the rolling-element bearing but it's possible to test with a wheel of a bike. Take the wheel in your hands and turn your body like the support can do. It's possible to see the point A moves up/down (altitude).

[Pphantom]

Тема перемещена из форума «Физика» в форум «Помогите решить / разобраться (Ф)»

[Theoristos]

Hello.
Is the purple disk well balanced? When grey disk is still, the A point is always up at the end?

If it is not balanced, after gray disk $\omega_0$ is enough high, purple disk would turn upside down due to centrifugal force.
Also, when purple is not balanced, gray disk angular acceleration will lead to some displacement of A.

I still can't figure out the idea of this machine.

[m441]

Hello,

Yes, the purple disk is well balanced (I tested with 3 disks and 2 wheels to look if there is a difference). I balanced the grey disk with a weight. The point A can't be at top always because the purple disk don't turn in the laboratory reference. If the point A is at top always this could say the purple disk turns at the same angular velocity than the support but at start the purple disk don't turn. And the support can't turn the wheel I think.

I don't understand this: " When grey disk is still, the A point is always up at the end?"

It's just a test, an experiment.

[Theoristos]

m441 в сообщении #1049308 писал(а):

I don't understand this: " When grey disk is still, the A point is always up at the end?"

It concerns to balancing, the position of A when all disks are still.

m441 в сообщении #1049308 писал(а):

The point A can't be at top always because the purple disk don't turn in the laboratory reference.

Well. Am I right:
1. Initially the both disks are still.
2. Then you give gray some angular acceleration till it angular speed reaches w0 and look to purple.

One can project the vector of gray disk angular speed $\vec \omega_0$ (as angular acceleration too) to purple disk coordinate system. The $\vec \omega_0$ component along the purple axis is $\omega_0 \cdot \cos a$.
The component perpendicular to the axis is $\omega_0 \cdot \sin a$. This one gives us nothing.
But, if friction if purple disk axle would be zero, the frist component gives pruple disk the rotation relative to gray disk with angular velocity $\omega_1 = - \omega_0 \cdot \cos a$. Due to friction momentum, this rotation will cease down to zero. If fruction is quite big, the total relative rotation angle will be quite small. Exact value depends on values of friction momentum, $a$, $\omega_0$ and spinup angular acceleration $\varepsilon$ and some other subtle things.

[m441]

Цитата:

Well. Am I right:
1. Initially the both disks are still.
2. Then you give gray some angular acceleration till it angular speed reaches w0 and look to purple.

Right, I do that. And I'm interesting at the altitude of the point A when the support accelerate too.

I don't understood all your message. Do you said the point A move up/down only when the support accelerates ?

[Theoristos]

No.
When gray disk accelerate, the purple one accelerate too, but "backwards" and with lower rate due to angle a and friction if purple disk axle.
Thus, when gray reaches $\omega_0$, the purple one will reach (relative to grey disk) the maximum by module $|\omega_1|=\omega_0 \cdot \cos a$ if there are no friction at all, and some less $|\omega|<|\omega_1|$ if there are.
Then, when gray continues rotation at constant w0, the relative rotation speed of the purple disk will decrease due friction in it axle, down to zero.
So, the total rotation angle of A is highly depends on friction. If friction is low and a is not far from 0 (it lowers friction moments too), A will manage make even more then one turns relative to grey disk.
If friction is high - it will be only some degrees (or even zero due to static friction).
Higher angular acceleration of grey disk gives higher (relative to friction momentum) inertial "momentum", and angle of A rotation will be higher.

If you, after all, stop the gray disk, the purple will take similar turn in opposite direction.

[m441]

Ok, I understood better your message, thanks. I would like to study only when there is no friction (in theory).

If the angle is at $0^\circ$ the purple disk can't accelerate around itself, ok ? And your formula seems to say $\omega_1=\omega_0$ but it's not the rotation of the disk around itself it's the angular velocity of the axis of the purple disk no ?

Without friction the point A moves up/down always ?

[m441]

I think the expression of the altitude of the point A is :

$R\,(1-\cos \alpha) \cdot \cos \alpha \cdot \cos \omega_0 t$

With $R$ the radius of the purple disk.

Is it correct ?

[m441]

Maybe the device is easier to understand with these images (click on the image to increase the size):



The black axis is fixed. The red arm can turn only around the black axis. The orange disk can only turn around itself.

At start, the orange disk don't turn around itself. I launch the red arm and I'm looking at the point A to look if the point A moves up/down:



But I'm not sure. A friend tested and find the same result. But I would like to be sure maybe there is a problem from the friction.

[Sender]

Are you sure that disk's axis support (red arm at the last picture) is rigid enough to ensure the absence of its oscillation during rotation? Maybe it is woth considering to add some additional edges in order to increase rigidity of the construction?

[m441]

I tested with the disks in the first message, I gave these last images for show how the device works. There is no oscillation.

Take the first message, take the angle $\alpha=0$ (2 axes vertical), you're ok that like the purple disk don't turn in the lab reference, the positions of the point A will be like that at different times:



Now, set $\alpha=5$ degrees (or 1 degree), why the point A don't be like before ? And like there is an angle for the axis of the purple disk this could say the point A moves up/down.

If $\alpha=0$ degree the point A never moves up/down
If $\alpha=90$ degrees the point A never moves up/down

But with intermediates value from 0 to 90° I think it's logical the point A moves up/down. And my tests confirm that. I'm waiting for the 3d simulation without friction. I don't how to calculate this device. If the point A if at top at start, if the point A don't moves up/down, this could say the purple disk turn around itself, and how the axis could do this, and this need an energy for turn the purple disk around itself.

[Theoristos]

m441 в сообщении #1049374 писал(а):

I think the expression of the altitude of the point A is :

$R\,(1-\cos \alpha) \cdot \cos \alpha \cdot \cos \omega_0 t$

Is it correct ?

$R \sin \alpha \cdot \cos(\cos\alpha \cdot \omega_0 t)$,

but only for a very low friction

[m441]

Thanks :)

Is it:

$R \sin \alpha \cdot \cos(\cos(\alpha) \cdot \omega_0 t)$

or

$R \sin \alpha \cdot \cos(\cos(\alpha \cdot \omega_0 t))$

For $\alpha=\pi/2$ the point A must be always at the same altitude, so it's the first expression ?

[Theoristos]

m441 в сообщении #1049546 писал(а):

Is it:
$R \sin \alpha \cdot \cos(\cos(\alpha) \cdot \omega_0 t)$

It is.

m441 в сообщении #1049546 писал(а):

For $\alpha=\pi/2$ the point A must be always at the same altitude, so it's the first expression ?

You are right.