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Rotational and inertial mechanics, overunity mechanism?


Seanie
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Please see the attached "concept" pdf file. If the mechanism does not generate energy then there should be physics which can clearly show that without it being assumed in the method of analysis used. Included in the analysis should be clearly shown details of a torque opposing the rod's rotation since the absence of such a torque would mean that the mechanism continues to operate indefinitely and in doing so continues to power the motion of m2 indefinitely. 
If it does produce energy then energy is not conserved.
More could be said about this interesting mechanism e.g. acceleration of m1 in the x direction without a force being applied to it in that direction, the force on m2 in the x direction without there being a reaction force, how rotation produces inertia, implications for Newton's laws of motion and so on. Thanks for your interest.

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44 minutes ago, Seanie said:

Please see the attached "concept" pdf file. If the mechanism does not generate energy then there should be physics which can clearly show that without it being assumed in the method of analysis used. Included in the analysis should be clearly shown details of a torque opposing the rod's rotation since the absence of such a torque would mean that the mechanism continues to operate indefinitely and in doing so continues to power the motion of m2 indefinitely. 
If it does produce energy then energy is not conserved.
More could be said about this interesting mechanism e.g. acceleration of m1 in the x direction without a force being applied to it in that direction, the force on m2 in the x direction without there being a reaction force, how rotation produces inertia, implications for Newton's laws of motion and so on. Thanks for your interest.

 

I really can't say if the opening post is any more clear than the non existant pdf.

If it doesn't generate energy then..

If it does then...

 

Surely your analysis should be able to predict which?

Edited by studiot
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1 hour ago, Seanie said:

Please see the attached "concept" pdf file

No file is available. I'll comment on what is available at this time. 

Quote

Rotational and inertial mechanics, overunity mechanism?

The header seems to imply a perpetual motion device, seems confirmed by: 

1 hour ago, Seanie said:

the absence of such a torque would mean that the mechanism continues to operate indefinitely and in doing so continues to power the motion of m2 indefinitely. 

Then no analysis is really needed. Applying currently available physical theories will result in confirmation that the device is not an over unity device or perpetual motion device. That is already known. But it could be interesting to analyse the device just to show what makes this specific device fail to run forever and what mistakes that may have caused such claims. So the question, in case the non existent file claims perpetual motion, is:
What new physics beyond Newton, Einstein and others have been discovered, that allows construction of a perpetual motion / over unit device? 

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!

Moderator Note

The OP assures me that they are totally committed to mainstream physics and want to understand the mechanics of this mechanism as well as to find out what is wrong with their analysis. On that basis, I am moving it back to Physics.

Any suggestion that the OP is promoting an over unity mechanism will lead to the thread being closed.

 
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4 minutes ago, Seanie said:

For some reason the file with the concept details will not upload to my post so you can view it at https://ibb.co/mHPKmRK instead.

!

Moderator Note

The still require you to describe it here. With, references to offsite material for support if necessary.

 
!

Moderator Note

That text is completely illegible. Post the material here or this thread will be closed.

 
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Iwould say that what stands out immediately is the phrase at the beginning of line 5

"assume there is no friction"

There appears to be no reason for the rotation either.

How does a centrifugal acceleration act on the centre of rotation ?

Surely this is zero at the centre?

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Description of the set-up as given in the uploaded file: Consider a rod of length r, viewed from above, one end of which is called m2 and the other end is called m1. m2 is pivoted in a linear bearing so it can freely slide in a straight line in both directions as well as being free to rotate either clockwise or anticlockwise about m2 as the centre of rotation. The linear bearing lies along the x axis. At m1 there is a point mass of m1 kilograms. The rod is given an initial momentary torque to cause it to rotate clockwise about m2.

Assumptions: Other than the force which produced the initial torque there is no subsequent force or torque given to the system from outside. The whole assembly operates in a horizontal plane so is unaffected by gravity. There is no friction. Everything has zero mass except for m1. The linear bearing is long enough to accommodate any range of motion which m2 may undergo.

My analysis: Due to m1's rotation m2 should exhibit simple harmonic motion symmetrically about a point on the x-axis, let this be the point (0,0) in the Cartesian plane. According to Newton's first law of motion the rod should continue to rotate at it's initial angular speed of w1 (omega-1) since (as far as I can tell) there is no torque opposing its rotation. Let theta be the angle the rod makes with the y-axis so that when the rod is in line with the positive x-axis theta = pi/2 radians.

m2's motion: It seems to me that the position of m2 is given by x=rcos(theta). Therefore m2's speed, v2, is the rate of change of x with time, which is -rw1sin(theta). Therefore the magnitude of m2's acceleration, a2, is the rate of change of v2, which is -r(w1)^2cos(theta) (i.e. minus r omega1 squared cos theta). Therefore the magnitude of the force, F2, acting on m2 is m1a2. Therefore the power, P2, in m2's motion is F2v2. The kinetic energy in m2's motion, E2, can be calculated from the kinetic energy formula of 1/2 mv^2 which gives the same result as getting the integral of P2.

Discussion & request for help: If the analysis given above were correct that would mean energy is not conserved in this system since there is a quantifiable power output even if it is constantly changing in direction and magnitude, even though power is not being input to the system. Assuming therefore that the physics used here is flawed, then why and how exactly is it flawed since it is apparently the standard Newtonian mechanics that I know? Also exactly what is the correct physics for this mechanism? The answer to that must among other things be able to show exactly how torque arises which opposes the rod's rotation since it seems clear that as long as the rod rotates then it will provide power to m2's motion. Thanks in anticipation. 

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1 hour ago, Seanie said:

The rod is given an initial momentary torque to cause it to rotate clockwise about m2.

How do you give a torque to the rod?

1 hour ago, Seanie said:

The whole assembly operates in a horizontal plane

I am guessing that you mean the x axis guide rail and the mass m1 are supported on a frictionless horizontal table ?

How can the rod and m1 pass the guide rail to complete a circle of motion?

Edited by studiot
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1 hour ago, Seanie said:

 

My analysis: Due to m1's rotation m2 should exhibit simple harmonic motion symmetrically about a point on the x-axis, let this be the point (0,0) in the Cartesian plane. According to Newton's first law of motion the rod should continue to rotate at it's initial angular speed of w1 (omega-1) since (as far as I can tell) there is no torque opposing its rotation. Let theta be the angle the rod makes with the y-axis so that when the rod is in line with the positive x-axis theta = pi/2 radians.

 

I'm not sure I follow your description, but if m2 is exhibiting simple harmonic motion, how is angular speed constant?

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because the SHM applies to m2's linear oscillating motion along the x-axis and the constant angular speed applies to the rod's rotation about m2.

 

18 minutes ago, studiot said:

I am guessing that you mean the x axis guide rail and the mass m1 are supported on a frictionless horizontal table ? yes.

19 minutes ago, studiot said:

How can the rod and m1 pass the guide rail to complete a circle of motion? because they are raised slightly above it. (ok so its not exactly all in the same plane then)

20 minutes ago, studiot said:

How do you give a torque to the rod?

It doesn't matter how.

 

 

 

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8 minutes ago, Seanie said:

It doesn't matter how.

Of course it matters.

Are you telling me that a rigid, solid rod (weightless or not) can pass through a rigid solid rail?

Surely if they are in the same plane the rod must strike the rail and then what?

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1 minute ago, studiot said:

Surely if they are in the same plane the rod must strike the rail and then what?

Sorry for not clarifying that, they are not all in the same plane. let the rod including m2 be in the same plane as each other and m2 is mounted in the rail just beneath it. So rod rotates just above the rail. Let me know if that is still not clear.

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14 minutes ago, Seanie said:

Sorry for not clarifying that, they are not all in the same plane. let the rod including m2 be in the same plane as each other and m2 is mounted in the rail just beneath it. So rod rotates just above the rail. Let me know if that is still not clear.

It is perfectly clear, it just won't work.

If you mount the rod beneath an inverted channel chaped rail then the pivot through m2 can project upwards into the guide channel.
Then the rod can describe a complete circle with m1 at the other end sitting on the horizontal table.

Now please explain the method of applying the initial torque.

If you consider the rod as a free body you will see that you need to include reaction components at the m2  end of the rod.

Edited by studiot
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3 minutes ago, studiot said:

If you mount the rod beneath an inverted channel chaped rail then the pivot through m2 can project upwards into the guide channel.
Then the rod can describe a complete circle with m1 at the other end sitting on the horizontal table.

Actually I was thinking of the rail being beneath the rod but I don't think it makes any difference whether it is above or below.

4 minutes ago, studiot said:

Now please explain the method of applying the initial torque.

I don't have a method in mind. All we need to know is that the rod is rotating at some arbitrary angular speed.

6 minutes ago, studiot said:

If you consider the rod as a free body you will see that you need to include reaction components at the m2  end of the rod.

I'm not sure what you mean. The only reaction forces I see at m2 are the forces against the sides of the rail which constrain it. m2's motion in the x direction is caused by the centrifugal force acting on it from m1's rotation, or if we consider centrifugal force not to be real then it is the inertia of m1 in the x direction which is transferred to m2 to make it move along the rail. The rod's rotation gives rise to inertia of m1, hence its motion in the x direction (& thus to m2 in the x direction also) and so the whole rod is in effect moving in the x direction as it rotates. As I understand it this motion in the x direction is not caused by a real force, so there is no reaction force to it. Perhaps you can show me where I am wrong here.

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8 hours ago, Seanie said:

because the SHM applies to m2's linear oscillating motion along the x-axis and the constant angular speed applies to the rod's rotation about m2.

 

 

Then the kinetic energy of the system is fluctuating. Where is the excess energy stored when the kinetic energy drops?

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49 minutes ago, J.C.MacSwell said:

Where is the excess energy stored when the kinetic energy drops?

Good question. IF I were unaware of Newton's laws of motion then I may suggest that kinetic energy of m2 is caused by m1 taking on inertia (& therefore acceleration) in the x direction and that this inertia is given to m1 by virtue of the rod's rotation and since this happens without the need of extra torque being given to the rod then the inertia, acceleration and kinetic energy arise from the rotation itself. But since I am aware of said laws of motion then I can't seriously propose that.

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You described the m2 end as massless, so there is never kinetic energy at m2 in this idealization. m1 has the only mass.

Unless you have energy gain, dissipation, or storage other than the kinetic energy, m1 must be at constant speed. That doesn't appear to be consistent with your description.of constant angular velocity of the rod while m2 is oscillating linearly.

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4 hours ago, J.C.MacSwell said:

Then the kinetic energy of the system is fluctuating. Where is the excess energy stored when the kinetic energy drops?

A very good insight question. +1

 

12 hours ago, Seanie said:

Actually I was thinking of the rail being beneath the rod but I don't think it makes any difference whether it is above or below.

 

 

Yes I know you were, but I have already put quite a lot of effort into helping you propose a viable mechanism, resulting in you modifying some of your original propositions.

You cannot analyse the mechanics until you have a viable mechanism.
Just stating things are in a plane or that gravity doen't matter doesn't make it so.

I suggested the underslung rotator for the reasons shown in the diagram

seanie2.thumb.jpg.cdb157b6dc71fe9969a723f931b10b95.jpg

 

If the rod is mounted above the rail then the weight of m1 will exert a moment on the mounting point m2.

That is why I said to draw some free body diagrams including all forces and moments.
You need to think in three dimensions here.
 

In turn this moment will be applied to the rail by the pivot stick
So don't tell me there are no reaction forces.
Moments generate reactions as well as normal forces.

However if the rod is underslung the weight of m1 can be supported on the frictionless table, so there is no moment applied to m2.

Of course, if the clearance is too small the real mass m1 will strike the rail.

 

The reciprocating mechanism you seem to be trying to describe is called a Scotch Yoke.
This is described and analysed in standard books on Mechanics of Machines.

The crank wheel corresponds to your rod and mass,
The yoke to your guide rail.

But the SY requires a power drive, and has stops at the ends of the yoke which can provide reactions.

There is no overunity or magic power generated.

https://www.google.co.uk/search?source=hp&ei=ypGQXcujHOaflwTh3b3IBg&q=scotch+yoke&oq=scotch+yoke&gs_l=psy-ab.3..0l10.534.5406..5720...1.0..0.1140.5876.1j1j5-3j2j2......0....1..gws-wiz.......0i131j0i13j0i13i30.tlnpwS--xe0&ved=0ahUKEwjLsN6k8_XkAhXmz4UKHeFuD2kQ4dUDCAc&uact=5

 

Edited by studiot
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I am grateful for your efforts to explain the mechanism under discussion. However I would like to point out a couple of things:

1. All of that about the position of the rail under or over the rod is irrelevant to the principle involved here. That's because as long as the rod moves as I described then there is the need to explain the apparent power in m2's motion in spite of an absence of power being input to the rod's rotation. That is the issue. 

2. The mechanism under discussion here is fundamentally different from a scotch yoke. In the mechanism here the power in m2's motion is not being taken from the rotational kinetic energy of the rod, at least not that has been shown yet. In the scotch yoke the yoke's motion is clearly being powered by the rotational kinetic energy of the rod or crank. The crux of the matter here is to explain where the power comes from which causes m1 to have a constantly changing acceleration in the x direction. 

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2 hours ago, Seanie said:

That's because as long as the rod moves as I described

You have yet to prove that the rod moves in the way described.

2 hours ago, Seanie said:

apparent power in m2's motion

You have mentioned Newton several times.

The terms Energy and Power were not available in Newton's day and those concepts were not formalised.

He did understand central forces however and of course we have his famous 3 Laws of Motion.

 

You have yet to answer the question

What forces do you think act on and in the rod after the initial push has set it in motion?

Do you think it is in tension or compression or what?

Edited by studiot
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3 hours ago, studiot said:

You have yet to prove that the rod moves in the way described.

I may not be able to prove that it moves exactly the way described. So let's see what aspects of its motion could reasonably be doubted. With a given rotation of the rod m2 will surely oscillate about the origin, that seems self-evident, is it not? Then consider two cases:

1. If w1 remains constant: Then m2's oscillation will surely be simple harmonic about the origin, will it not? If so then it moves as described. If not then how does m2 move? Also, regardless of whether the oscillation is simple harmonic there will surely be an oscillation which at least approximates simple harmonic will there not? Then even with any acceleration of m2 there must be energy to make that happen and apparently the only place that energy can come from is from the rotational kinetic energy of m1 which means w1 must be reduced but that is not an option here because in this paragraph it is assumed that w1 is constant. Therefore, logically, m2 cannot move at all and it seems self-evident that this cannot be the case. So then if m2 moves and the rod has a constant w1, where is the energy coming from?

2. If w1 changes: If m2 were fixed at the origin then clearly by Newton's first law w1 will remain constant. Therefore with m2 moving and since w1 changes (assumed in this paragraph) and such change must be a reduction, that means the motion of m2 must somehow be able to effect this change in w1. In this case the motion will differ from what I assumed it to be only in the reduction of w1 until the rod stops rotating. In that case also it must be that the energy for m2's motion comes from the rotational kinetic energy of m1, and with a diminishing w1 that seems plausible. Still the physics of how that happens needs to be worked out. 

4 hours ago, studiot said:

What forces do you think act on and in the rod after the initial push has set it in motion?

It seems to me that the rod will experience a tension which varies from zero to a maximum according to theta. So when theta = 0 (i.e. rod extending from the origin along the positive y-axis) the rod will momentarily be rotating about what is essentially a fixed point (m2 at the origin), therefore the standard formula for centripetal force applies. As the rod rotates past this point (i.e. theta > 0) m2 will begin to move in the positive x-direction and thereby increasingly reduce the tension in the rod. By the time theta reaches pi/2 radians the rod will not offer any resistance to m1's (or m2's) motion in the positive x-direction and so the tension will be zero. For theta > pi/2 the tension will rise again until it reaches a maximum at theta = pi radians, and so on. I don't see how there will be any compressive force on the rod.

What do you think?

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