I'm Looking For Special Research Group

Hey

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I didn't get your respond back by email.
I'm still looking for group, who can repeat or similar experiment and can produce good science report.

 

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Given the flawed nature of your entire approach that is impossible, no matter how competent the person writing it.

 

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What do you mean?
The science report is showing just result of experiment. It doesn't mean report should imagine some hypothesis.
Just experiment description, series of experiments and results with standard statistics deviation. That's all.
Base on this result could possible say something about hypothesis.

NOBODY CAN DO IT?:bawl:

 

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Let's say hypothesis is very simple.

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I think the repulsive objects should use this way for calculation.
http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18#
Need experiment to test it

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And you're unable to this because....? Basic experimental skills and hypothesis testing are high school level things, can't you manage them yourself?

Those who can likely see that their time is better spent not working with you. Besides, the applicability of energy and momentum conservations are tested every day by people all over the world in a myriad of ways, many far more precise and accurate than anything you could manage. Rockets rely on an accurate understanding of momentum and thrust. Satellite trajectories depend on an understanding of angular momentum, gravity and energy conservation.

There's a firm theoretical reason why linear momentum, angular momentum and energy are all separately conserved, they all relate to different continuous symmetries in nature (spatial translation, rotation and time translation respectively) via Noether's theorem. Their conservation form the pillars on which much of physics and engineering is built, they are put into practical use all the time. If they weren't extremely close to being independently conserved it would have been noticed by now. They are used to simplify fluid mechanics which is used to design planes. They are used to simplify general relativity equations to design the GPS network. They are used to design spinning projectiles ie artillery shells and the military takes accurate weaponry very seriously. There isn't an area of physics which doesn't make use of them in some way and thus pretty much every application of physics to the real world tests them and yet they remain as pillars of physics.

Yes, its entirely possible they aren't perfectly conserved individually but the level of accuracy required far outstrips anything someone like yourself could examine, you'd need serious specialist equipment (like the LHC or Gravity Probe B) to examine them. Plus you'd also need to know basic physics to determine whether or not an experimental result implies conversation is violated or not and since you don't even if someone gave you all the experimental data you could ever wish for you'd be unable to analyse it.
The kinematics of a spinning projectile

 

At home?

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I did. How big precision of this experiment you may see it on my website.

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I putted about 20% deviation and I got 25% difference of translational velocity. It's sign to repeat experiment in Lab.

I don't think so

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Actually, I graduated rocket/aviation scientists university. So what? It's a technology based on existing physics laws.

How easy hide everything behind theorems. But this theorem says nothing about motion. It's condition what we include there. Base on these conditions we are getting result.

The problems, what usually solved, have a local results which assuming the base laws works ever. However, the wrong assumptions about Newtonian frame of reference insert errors into problem solutions.

I assume the law of momentum conservation for Newtonian mechanics works ever. However, I think, we got bad assumption in rotational and translational motion description. If describe rotational and translational motion as a product of sum of two simple motions then need a little rule which says: each simple motion should follow it's own law of momentum conservation from force which inducted this certain motion. The net force of these motion equal to whole force which applied to this object.
Unfortunately, the modern classical mechanics doesn't have this rule for rotational and translational motion in free move rather than rolling objects on surface divide net force between translational and rotational motions.

 

? what

"if the velocity is parallel to the rotation axis, the acceleration is zero.
if the velocity is straight inward to the axis, the acceleration is in the direction of local rotation.
if the velocity is straight outward from the axis, the acceleration is against the direction of local rotation.
if the velocity is in the direction of local rotation, the acceleration is outward from the axis.
if the velocity is against the direction of local rotation, the acceleration is inward to the axis."(wiki

 

I understood what was happened. Modern physics use static model for forces which applied to an object. This is the same model if calculate loads in civil engineering.
If you this logic for physics problem "a rolling body along incline" then object has two reaction forces. In static mode each of this forces equal to gravity force projection on incline. However, solution of this problem use dynamic model where rotational and translational motion is a sum of two motion and sum of forces which conducted certain type of motion equal to net force(i.e. gravity force projection on incline)
The physics problem for rotational and translational motion in free move doesn't use dynamic model and substitute physics process by static model where torque as add-on(not part of it) for force in linear direction. This is mistake. The model should be the same as modern physics use for rolling objects on surface

 

Oh please. Civil engineers use static forces if they are studying static systems. Even civil engineers use dynamics to study systems with dynamics.

 

OK.
Rotational and translational motion can be described as a sum of 2 simple motions, which conducted in two separate events.

Then for translational motion equation is:
\(F_1=ma\)
Work for this type of motion which equal objects kinetic energy translational part is
\(E_k_t=m\frac{v^2}{2}\)
Base on third Newtons law symmetric force \(-F_1\) is present for this event

Same for rotational motion with fixed axis equation is
\(F_2R=I\alpha\)
Work for this type of motion which equal objects kinetic energy rotational part is
\(E_k_r=I\frac{\omega^2}{2}\)
Base on third Newtons law symmetric force \(-F_2\) is present for this event

====

Base on sum of two motions the full kinetic energy which is work of 2 forces is equal:
\(E_k_f=m\frac{v^2}{2}+I\frac{\omega^2}{2}\)
Therefore sum of two symmetric forces for each simple motion necessary to initiate rotational and translational motion:
\(-F=(-F_1)+(-F_2)\)

===

However, the modern classical mechanics says if force applied away from center mass of object then just force \(F_1\) with symmetrical force \(-F_1\) is enough to initiate rotational and translational motion with energy:
\(E_k_f=m\frac{v^2}{2}+I\frac{\omega^2}{2}\)
How this part of force do this part of work \(I\frac{\omega^2}{2}\) - mystery.

This paradox base on suggestion what law of momentum conservation works always and adding symmetric force \(-F_2\) is crashing this law.
If use little rule where each motion law of momentum conservation should adequate force which induct this certain type of motion and all law of momentum conservation works locally, then not necessary hide force \(-F_2\).
Center mass of isolated system is not the same as center of mass of objects and doesn't follow same rule.

Dark matter is not a mystery anymore. Just a miscalculation

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http://knol.google.com/k/paradox-of-classical-mechanics-2
http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18

 

The third Newton’s law and law of momentum conservation.

The third Newton’s law and law of momentum conservation.

The third Newton’s law declares what each reaction of forces should be symmetrical.
-F=F
where: F - reaction forces of objects

The consequence of this is law of momentum conservation.

-Fdt=Fdt
-mv=mv
-P=P

where: F - reaction forces of objects, m - mass of objects, v - velocity of objects, dt - time trame of action, P - momentums of objects.

This consequence should be used for case where objects induce same identical motions and this does not cover the case where objects induce different type of motions. In experiment 2 thin cylinders induce different type of motions and this simple consequence should not cover this objects repulcing action. Therefore, for correct explanation of experiment 2 needs to use prime third Newton’s law and identify all reaction forces by this law. The consequence law of momentum conservation where objects have symmetrical behavior should not be used for this objects repulsing case.

 

http://knol.google.com/k/paradox-of-classical-mechanics-2#
For example, the rotated object gets more kinetic energy on experiment 2 rather than this non rotated object on experiment 1. It means, for this rotated object, like on experiment 1 one force is conducting work for translational motion only and another force(torque) is conducting work for rotational motion only. Base on third Newton’s law, these forces must have symmetrical reaction forces. These symmetrical reaction forces are applying to other non rotated object which conducts translational motion only. Unfortunately, the modern classical mechanics use trivial consequence from third Newton’s law where objects have symmetrical behavior. Base on this model, look through center of mass of isolated system these two objects are getting symmetrical linear and angular momentums with symmetrical behavior on repulsing action. However, different objects alignment brings asymmetrical behavior of these objects. The non rotated object does not conduct rotational motion around center of mass of isolated system. Therefore, these objects asymmetrical behavior cannot be described by trivial consequence of third Newton’s law.

 

For crying out loud, Alex! Of course when you do things wrong you will get wrong answers. This statement from your document is wrong:

Base on third Newton’s law the interaction must be symmetrical. It means, both cylinders should have same translational and angular momentums.​

All of your problems, and they are your problems, stem from your refusal to learn physics.

 

So let's do your experiment 2 correctly. One thing to notice is that as soon as object 2 starts rotating away from the vertical is that the forces will no longer be purely horizontal. Both objects 1 and 2 will end up with non-zero vertical components in their velocity vectors. So right off the bat experiment 2 is going to differ from experiment 1. However, this effect will be very small if the extended lengths of the springs are much smaller than the length of object 2.

I'm going to generalize experiment 2 so that it covers experiment 1 as a special case. Rather than applying the force from the end of object 2, let's put the spring on object 2 at some distance r from object 2's center of mass. Experiment 1 results when r=0 while experiment 2 results when r=L/2. I'm also going to make the masses of objects 1 and 2, m[sub]1[/sub] and m[sub]2[/sub], not necessarily be the same. Your experiments 1 and 2 arise as special cases where m[sub]1[/sub]=m[sub]2[/sub].

One way to look at the result of this generalized experiment is from the perspective of the conservation laws. In the case where the spring lengths are very small compared to l[sub]2[/sub], the length of the second rod, we have object 1 moving with a velocity v[sub]1[/sub] in the -x direction, object 2 moving with a velocity v[sub]2[/sub], and object 2 rotating about the -z axis, with x,y, and z forming a right-handed system. (Aside: you have a left-handed coordinate system, Alex. The z-axis should be coming out of the page.)

For convenience, I am going to use a coordinate system in which the origin is at the initial position of object 2's center of mass. The initial position of object 1's center of mass in this system is \(-x_0\hat x + r\hat y\), where x[sub]0[/sub] is the sum of the lengths of the compressed springs and r is the distance between the center of mass of object 2 and the location of the spring on object 2. Ignoring the small \(\hat y\) velocity components (which we can make very small by making the springs very small), the final configuration will have

• Object 1 moving at a velocity of \(-v_1\hat x\) along the line \(y=r\),
• Object 2 moving at a velocity of \(v_2\hat x\) along the x-axis, and
• Object 2 rotating with an angular velocity \(-\omega\hat z\).

Note that I have defined the scalars \(v_1\) and \(\omega\) so that they will have positive values.

Conservation of linear and angular momentum apply assuming no other forces act on the objects. Assuming that all of the potential energy in the compressed springs went into kinetic energy. So what do the conservation laws say about this final state?


Conservation of linear momentum. Since both objects were initially at rest, the initial linear momentum of the system was zero. Conservation of linear momentum says

\(m_1\vec v_1 + m_2 \vec v_2 = 0\) or

\(m_1v_1 = m_2 v_2\)


Conservation of angular momentum. Since both objects were initially at rest, the initial angular momentum of the system was zero. Conservation of angular momentum says the total angular momentum of the system will also be zero. In the final state, object 1 has angular momentum about the origin due to its translational motion along a line that does not intersect the origin, \(\vec L_1=r\hat y \times (-m_1 v_1 \hat x) = rm_1v_1\hat z\). Object 2 is moving along the line y=0, so it has no angular momentum due to translational motion, but it does have angular momentum due to rotation about its center of mass: \(\vec L_2 = \mathbf{I}_2 \vec{\omega}_2 = -\frac{m_2{l_2}^2}{12}\omega \hat z\). Applying conservation of angular momentum yields

\(rm_1v_1 = \frac{m_2{l_2}^2}{12}\omega\)

Combining the results of the conservation of momentum laws lets us eliminate all but one of \(v_1\), \(v_2\), and \(\omega\). For example, in terms of \(v_2\),

\(v_1 = \frac{m_2}{m_1} v_2[tex] [tex]\omega = \frac{12\,r}{{l_2}^2}v_2\)


Conservation of energy. The initial energy of the system is in the form of potential energy in the compressed springs. The final energy is in the form of kinetic energy. Applying conservation of energy yields

\(\frac 1 2\left(m_1 {v_1}^2 + m_2 {v_2}^2 + I \omega^2\right) = \frac 1 2 \left(1+\left(\frac {m_2}{m_1}\right)^2 + 12\left(\frac {r}{l_2}\right)^2\right)m_2v_2^2 = E_{\text{springs}}\)


In experiment 1, m[sub]1[/sub]=m[sub]2[/sub]=m and r=0. Thus in this case the two objects will have equal but opposite velocity vectors with magnitude given by

\(v_1 = v_2 = \sqrt{\frac{E_{\text{springs}}}{2m}}\)

In experiment 2, m[sub]1[/sub]=m[sub]2[/sub]=m and r=l[sub]2[/sub]/2. Thus in this case the two objects will have equal but opposite velocity vectors with magnitude given by

\(v_1 = v_2 = \sqrt{\frac{E_{\text{springs}}}{5m}}\)

Note that in this case, most (3/5) of the energy is in the form of object 2's rotation.

 

DH. Thank you for perfect explanation.

Here is another one.
I added more realistic model where it show how rolling objects with same mass and different moment of inertia may have different translational acceleration during repulsing action.
http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18#

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I understand your sceptic point of view on this problem. However, you have a good physics education and could count better then me this.
The third newtons law of symmetrical interaction must works alway. Base on this law the net forces of both sides must be identical. The more realistic digram doesn't let platforms conduct rotational and translational motion. Just translational motion only. Base on kinematic equations for rolling bodies on surface, the net force of each side will be the sum of force which applied for translational motion of rolling objects plus torque which apllied for rotation motion divided by radius of rolling bodies. It shows on my solution.
Base on this equation the rolling bodies with same mass and different moment of inertia have a different translational and rotational accelerations. For instance, classic problem for rolling bodies on incline shows, the rolling bodies with same mass and different moment of inertia have a different translational acceleration on the end of incline. Here's same situation. The rolling bodies with same mass and different moment of inertia have different translational acceleration during repulsing action.
This problem shows paradox of classical mechanics where third newton's law of symmetry of interactions does not follow into consequence of simple law of momentum conservation. The modern law of momentum conservation in simple form works just for trivial cases where objects have symmetrical motion behavior. On other objects asymmetrical behavior during interaction, the law of momentum conservation has a complex form.
http://knol.google.com/k/paradox-of-classical-mechanics-2#

 

The mistake of modern classical mechanic is a third newtons law consequence of modern classical mechanics where translational momentum must be equal. For simple form where objects have same translational motion - yes.
However in our case it doesn't work.
Ok. Let's imagine simples case where objects are rolling on one side only. Let's assume the translational momentums for our case equal. Then, forces which induce translational motion must be equal too. But which force induce rotational motion? Yes, the objects rotation in both direction are equal. But what force? No any spring inside subset of rolling objects. Just weightless platform between them.

 

Have you read anything I've written?

And for crying out loud, Alex, stop with the stupid Rube Goldberg devices.

 

Yes. Perfect modern classical mechanics explanation.

Did you read my previous post for my "the wheels" problem?
http://knol.google.com/k/alex-belov/the-wheels/1xmqm1l0s4ys/18#

What is asymmetrical force which allow rolling bodies rotate on one side?
The assumption where law of momentum conservation works always in simple form is incorrect.
I hope you agree with third newtons law for symmetrical interaction where \(\vec F=-\vec F\)

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Your equations of motion are incorrect (which unfortunately is very typical for you).

Alex, while relativity and quantum mechanics do show that Newton's laws ultimately are incorrect, Newtonian mechanics most certainly are valid in the realm in which you are using them. You simply are not going to find a flaw in Newtonian mechanics in this realm of slow moving, somewhat large objects (objects a lot more massive than an atom, a lot less massive than a star, and with relative velocities much less than the speed of light).

Stop tilting at windmills and learn some physics.

 

What do you mean? You never see kinematic equations for rolling objects on surface?

It's your personal opinion about opponent. offtop

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Again. Back to my question.

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Ok. Let's imagine simples case where objects are rolling on one side only. Let's assume the translational momentums for our case equal. Then, forces which induce translational motion must be equal too. But what force induce rotational motion for rolling objects on one platform? Yes, the rolling objects rotation are equal in both direction. But what force is inducing rotational motion? No any spring inside subset of rolling objects. Just weightless platform between them.

Speed of light?
This may be interesting for you.
http://knol.google.com/k/alex-belov/electrostatic-propulsion-system-concept/1xmqm1l0s4ys/15

I have a question, I need an answer

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