[Jaaanosik (OP)]

Halc,
here is a wheel, top view and the side view:



The wheel and the brakes calipers have friction less bearings.
The wheel and the calipers can move up along the axles without any obstruction.
The setup is here on the Earth, the axles point up, the wheel is horizontal to the ground.
The wheel is rotating clock wise, omega points down.
Let us apply an instantaneous breaking to the wheel.
The angular acceleration (actually deceleration) is in upward direction.
Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?
Jano

[Halc]

Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?

You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?

For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.

[Jaaanosik (OP)]

Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?

You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?

For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.

Halc,
I assure you, this is not trolling.
Please, have a look:



https://en.wikipedia.org/wiki/Torque

The figure shows the setup from my question.
If we take a realistic time, let's say dt = 0.1s for the braking.
When Torque - T, Omega - w, Inertia - I, Angular acceleration - a

T = I (dw/dt) + (dI/dt) w
We know that dI/dt = 0 during the breaking, so the torque
T = I (dw/dt)
or
T = I a

Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?
Jano

[Bobolink]

Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?

I assure you, this is not trolling.

It is trolling, because regardless of ANY information you're given you will continue to maintain that the conservation of momentum does not hold.

[Jaaanosik (OP)]

Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?

I assure you, this is not trolling.

It is trolling, because regardless of ANY information you're given you will continue to maintain that the conservation of momentum does not hold.

Bobolink,
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,
Jano

[Kryptid]

Bobolink,
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,

LOL!

[Jaaanosik (OP)]

Bobolink,
on the contrary, if you say that the wheel is not going to jump up then you are defending position that the conservation of the momentum does not hold,

LOL!

Kryptid,

2:50min into the video, how is it possible that the cube jumps?
The axles in my setup keep the reaction wheel on a straight up jump.
As I said the wheel and the breaks have bearings in my example and they allow the motion along the axles.
The cubes in the video jump 'randomly' but they use inertial actuators - reaction wheels, the same principle,
Jano

[Jaaanosik (OP)]

15 seconds into the video and 2:00min,
Jano

[Kryptid]

It amazes me that even after so much explanation, you still just don't understand angular momentum.

[Jaaanosik (OP)]

It amazes me that even after so much explanation, you still just don't understand angular momentum.

Really?
Please, explain how the cube can fly above the table?
What caused the jump?
How is it going to be different from my explanation?
Jano

[Kryptid]

Really?
Please, explain how the cube can fly above the table?
What caused the jump?
How is it going to be different from my explanation?
Jano

If you haven't understood it by now, nothing I can say will fix that problem.

[Bobolink]

2:50min into the video, how is it possible that the cube jumps?

Hint:  It wouldn't jump anywhere in a freefall reference frame.

[Jaaanosik (OP)]

2:50min into the video, how is it possible that the cube jumps?

Hint:  It wouldn't jump anywhere in a freefall reference frame.

Bobolink,
here is the spaceship, the axles are attached to the spaceship frame.



Do you see how easy it is to generate counter torque by the CMGs?
The wheel will move in the direction of the yellow angular acceleration and the spaceship has no reason to move.
The sum of all the CMGs torques is going to compensate for the disturbance,
Jano

[Jaaanosik (OP)]

Is the wheel going to jump up due to the angular acceleration/deceleration, assuming angular acceleration is bigger than gravitational acceleration?

You know it will not.
The fact that you got the α and ω vectors correct suggests you're not completely uneducated, and yet you ask a question like that. Now why is that?

For the third time, you are adding vectors of different units (bold above), meaning you're ignoring everything I say and intend to continue to do so. Hence I see no reason to continue this troll discussion.

Halc,
I assure you, this is not trolling.
Please, have a look:



https://en.wikipedia.org/wiki/Torque

The figure shows the setup from my question.
If we take a realistic time, let's say dt = 0.1s for the braking.
When Torque - T, Omega - w, Inertia - I, Angular acceleration - a

T = I (dw/dt) + (dI/dt) w
We know that dI/dt = 0 during the breaking, so the torque
T = I (dw/dt)
or
T = I a

Please, tell me a good reason why the wheel is not going to jump up if the angular acceleration a is bigger than gravitational g?
Jano

Halc,
my friend, look above and here:



The dark blue force F in wiki diagram - the braking force is not torque.
If two calipers/pads on the wheel are 1m from the axis and the tangential braking force is 1N then we have 2Nm of torque.
If two calipers/pads on the wheel are 0.5m from the axis then we need 2N of the tangential braking force in order to have 2Nm of the torque.
It is important to establish 'the flow', the braking force is the cause the torque is the effect.

The simple F=ma, right, it is simple.
Well, there could be lots of stuff happening to generate force F, the cause, and how it affects the acceleration, the effect.
There is going to be difference in acceleration a if I push a wheelbarrow not lifted and lifted.
The same force will give us different acceleration.
The same goes for the torque. If the brakes on the wheel can supply only 1N then the torque is different based on the position of the brakes, 0.5m gives us smaller torque compared to 1m for the same braking force 1N.
Jano

[pzkpfw]

It seems to me you've misunderstood L (angular momentum).

The right hand rule as applied to rotation is essentially a mathematical artifact of a cross product.
That the pseudovector L appears to have a "direction" does not imply a force in that direction, in the way you're thinking.
Why would L be a force up or down, due to the disk spinning?
The pseudovector L (being a vector) is more about resistance to a change in the orientation of that vector. e.g. like a gyro wanting to keep spinning in its plane.

When the brakes are applied to your spinning disk, it won't move up (or down).


(Edit: removed section too easily misconstrued.)

[Jaaanosik (OP)]

It seems to me you've misunderstood L (angular momentum).

The right hand rule as applied to rotation is essentially a mathematical artifact of a cross product.
That the pseudovector L appears to have a "direction" does not imply a force in that direction, in the way you're thinking.
Why would L be a force up or down, due to the disk spinning?
The pseudovector L (being a vector) is more about resistance to a change in the orientation of that vector. e.g. like a gyro wanting to keep spinning in its plane.

When the brakes are applied to your spinning disk, it won't move up (or down).


(Edit: removed section too easily misconstrued.)

Hi,
No, to the pseudovector (pseudo force?).
The torque is a real force during the braking of the wheel otherwise we would not have the gyro precession.
Here  are some analysis:



https:  //www.youtube.com/watch?v=bYF0PGsF92k




https:  //www.youtube.com/watch?v=qS_dcNqs3d4

Please, notice the different position of the torque vectors in those two images.




The top view of the wheel. The darker gray balls represent two centers of mass of two halves of the wheel.
The orange vectors are torques positioned at the same position as in the second video.
It does not matter because they can be summed up and put at the total center of mass, the yellow vector, when angular acceleration is multiplied by the rotational inertia,
Jano

[Kryptid]

If you're so certain this would work, then you should invest some time and money into building a prototype. If it did work, you'd become quite famous.

[pzkpfw]

A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.

[Halc]

A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.

He hasn't explicitly mentioned that since post 252.

Still trolling however. Most recent post had at least four errors, one of which was the lack of direct relevance of the videos to the situation being discussed.
I find it best to just not feed the trolls. He's not here to learn anything, only to push our collective buttons and get his jollies from those who continue to respond.

I try to respond to intelligent questions, but the last several post have neither asked questions nor displayed any deliberate intelligence.

[Jaaanosik (OP)]

A bicycle and a spinning office chair would be enough to put aside this silly idea of a braked wheel experiencing a force in the direction of L.

Hi,
Please, provide a diagram and math to support your claim,
Jano