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# Rotational and inertial mechanics, overunity mechanism?

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11 hours ago, Seanie said:

What do you think?

You have told me that the mass m1 will be whizzing round and round in circular motion.

You have told me that at certain points the force in the rod will be zero.

11 hours ago, Seanie said:

and so the tension will be zero.

Yet Newton's First Law demands that in the case where there is zero net force acting on a body it continues its motion in a straight line.

So why does m1 travel in circular motion?

Are you suggesting its circular motion speeds up and slows down in sympathy with your varying (centripetal) force in the rod?

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I think your difficulty in analysing the motion stems from jumping between different frames of reference.

m2 is only momentarily located at the origin of a rest frame (in this case the guide rail)

The rest of the time m2 is located elsewhere.

So you cannot use it as the fixed centre of a rotating vector to resolve into sin and cos components.

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13 hours ago, Seanie said:

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.

There is no conservation law that would make w1 remain constant if the pivot is free to move.

There is no torque on the system, since any force on the rod will be along its axis, and thus r x F will be zero. Angular momentum will be constant. But that's measured about some point/axis of rotation, which is not the physical point of rotation of this system, since the pivot point is free to move. You can't change the axis about which you are measuring the angular momentum in the middle of the problem.

edit: and I see studiot has made a similar observation while I was writing this

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11 hours ago, studiot said:

You have told me that the mass m1 will be whizzing round and round in circular motion.

Yes with respect to m2, but it apparently has another motion too which is extra motion in the x-direction, as revealed by the motion of m2 for example. The combination of these two motions produces an ellipse (i.e. m1's locus) and according to my calculations that ellipse is interesting for a few reasons one of which is that it has semi-major and semi-minor axes of phi and 1/phi respectively.

12 hours ago, studiot said:

Yet Newton's First Law demands that in the case where there is zero net force acting on a body it continues its motion in a straight line.

So why does m1 travel in circular motion?

Are you suggesting its circular motion speeds up and slows down in sympathy with your varying (centripetal) force in the rod?

Considering there to be two motions of m1 combined, i.e. its rotation about m2 as well as its extra x-direction motion (evidenced by m2's motion) then the "zero net force" you refer to in our case is the zero net centripetal force and indeed the extra x-direction speed is at its maximum when the centripetal force has reached zero, that is equal to m1's tangential velocity in its circular path about m2 and it is when theta = pi/2 radians. Of course the actual (net) motion of m1 is not in a straight line due to the 2nd of the combined motions which it undergoes. I believe what I have said here can be perceived to be true by thinking about it and looking at the diagram. I am now trying to add a better diagram to this post for our convenience, hopefully it is better than the last one I.

I still don't see how the circular motion can change its speed as the centripetal force will always act perpendicularly to the rotation. I think swansont made the same point in his recent post. I hope this addresses your points.

I'm still having the same problem as before uploading, the word "queued" continues to show all the time and file doesn't seem to upload. I will try putting the file where I put the last one and maybe someone here can add it here like before.

Diagram is at https://ibb.co/NjTmVb1

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42 minutes ago, Seanie said:

I believe what I have said here can be perceived to be true

As I implied it can only be true if it doesn't break Newton's Laws.

That means motion not in a straight line only occurs when a real force acts.

So this has not answered my point at all.

A further question.

You say there is tension in the rod.

How is this possible?
Tension requires two opposing forces.

45 minutes ago, Seanie said:

Considering there to be two motions of m1 combined, i.e. its rotation about m2 as well as its extra x-direction motion (evidenced by m2's motion) then the "zero net force" you refer to in our case is the zero net centripetal force and indeed the extra x-direction speed is at its maximum when the centripetal force has reached zero, that is equal to m1's tangential velocity in its circular path about m2 and it is when theta = pi/2 radians. Of course the actual (net) motion of m1 is not in a straight line due to the 2nd of the combined motions which it undergoes.

47 minutes ago, Seanie said:

Yes with respect to m2, but it apparently has another motion too which is extra motion in the x-direction, as revealed by the motion of m2 for example. The combination of these two motions produces an ellipse (i.e. m1's locus) and according to my calculations that ellipse is interesting for a few reasons one of which is that it has semi-major and semi-minor axes of phi and 1/phi respectively.

So calculating the force external and internal to the rod is the key to this.

The analysis of the scotch yoke is similar, which is why I suggested it.

A pity you rejected it.

Of course in the real world this includes friction, which is why the yoke requires constant drive.

Engineers use the "instantaneous centre concept" to analyse the motions of all the moving bits of machinery in a machine.

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19 minutes ago, studiot said:

That means motion not in a straight line only occurs when a real force acts.

yes but if one of the two combined motions produces motion in a straight line we may still not see a straight line motion because of the 2nd of the combined motion interfering with the first.

22 minutes ago, studiot said:

You say there is tension in the rod.

How is this possible?

For the same reason that tension would be in any rotating rod with a weight on the end.

I have described this mechanism and then tried to analyse it to clarify the physics to explain it. I have not succeeded in that and at this stage (long before starting this thread) am stumped by it and really don't know. The explanation seems elusive and beyond me. Finding fault with my analysis is a given. Perhaps you can come up with the correct explanation and be able to show it here. If classical physics is able to explain this then I have still not found anyone (including a number of physics experts) who can. They don't reply to me about it, or they are too busy or don't give it a proper look, or give a partial reply without dealing with the core aspects in need of explanation etc. I have searched exhaustively online for a long time to no avail, likewise in books but nothing. I have been referred to some online resource or book and in the end still find no explanation. Given the relative simplicity of the mechanism, basically just a rotation where the pivot is allowed to move in a line, I find it remarkable that I can not find an explanation. I wonder at what point is it reasonable to decide that classical physics does not have an explanation. If it does not then I'd say that should ring some alarm bells in physics circles. Thanks for bearing with me on this so far anyway and I will continue to try and to live in hope of getting to the bottom of it.

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1 hour ago, Seanie said:

For the same reason that tension would be in any rotating rod with a weight on the end.

When m2 is free to slide in the x direction...what supports the centripetal force when the rotating rod is aligned with the x axis?

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3 minutes ago, J.C.MacSwell said:

When m2 is free to slide in the x direction...what supports the centripetal force when the rotating rod is aligned with the x axis?

nothing supports it, that's why it is at full speed along the x axis at that point in the rotation.

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2 hours ago, Seanie said:

nothing supports it, that's why it is at full speed along the x axis at that point in the rotation.

What stops it?

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9 hours ago, Seanie said:

nothing supports it, that's why it is at full speed along the x axis at that point in the rotation.

Or it could be motionless. If your initial condition is that the rod is aligned with the x-axis, and your initial impulse is in the y direction. I think the resulting motion of the mass is only in the y direction, and the pivot is momentarily stopped when the rod is aligned with the x-axis.

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8 hours ago, J.C.MacSwell said:

What stops it?

In spite of their fictitious status I find it helpful nevertheless to use the terms centrifugal force and centrifugal acceleration to help understanding of the motion under investigation. Then the centrifugal acceleration of m1 in its "extra" x-direction motion is, understandably, what requires or implies m2's motion. So if m2 is at full speed when theta is pi/2 that's because the centrifugal acceleration is in the same direction that m2 can move when theta is pi/2. All of the centrifugal acceleration is in the positive x-direction at theta = pi/2. Therefore, when you imagine the rotating m1 (or rod) along with m2's freedom of motion in the x-direction you will see that m2 is bound to follow the "extra" x-direction motion of m1 and that since this must be an oscillating motion in the x-direction that means that m2 will have to slow down (theta between pi/2 and pi), stop (theta = pi) and reverse direction (theta >pi), because assuming constant angular speed of m1 for simplicity then theta determines the acceleration for the "extra" motion in the x-direction. Forgive me if I am overstating the explanation but I'm trying to make sure you get it. Another way of saying it is that m1 tends to pull m2 in the x-direction and since m1 is rotating then it constantly changes direction, thus m2 must stop and reverse at certain intervals.

1 hour ago, swansont said:

Or it could be motionless. If your initial condition is that the rod is aligned with the x-axis, and your initial impulse is in the y direction. I think the resulting motion of the mass is only in the y direction, and the pivot is momentarily stopped when the rod is aligned with the x-axis.

To clarify that, I am saying that the initial condition is that the rod is aligned with the positive Y-axis (theta=0). It will not be on top of the Y-axis but parallel to it and will lie along the line x=-r. That is at least if my understanding of m2's x-direction oscillation is correct. From that starting point it possesses its angular speed of w1 (omega 1) and m2 is at the point (-r, 0).

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43 minutes ago, Seanie said:

To clarify that, I am saying that the initial condition is that the rod is aligned with the positive Y-axis (theta=0). It will not be on top of the Y-axis but parallel to it and will lie along the line x=-r. That is at least if my understanding of m2's x-direction oscillation is correct. From that starting point it possesses its angular speed of w1 (omega 1) and m2 is at the point (-r, 0).

Then why would the motion in the x direction reverse itself? There's no force in that direction to change the x component of the momentum.

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16 hours ago, Seanie said:

If classical physics is able to explain this then I have still not found anyone (including a number of physics experts) who can. They don't reply to me about it, or they are too busy or don't give it a proper look, or give a partial reply without dealing with the core aspects in need of explanation etc. I have searched exhaustively online for a long time to no avail, likewise in books but nothing. I have been referred to some online resource or book and in the end still find no explanation. Given the relative simplicity of the mechanism, basically just a rotation where the pivot is allowed to move in a line, I find it remarkable that I can not find an explanation. I wonder at what point is it reasonable to decide that classical physics does not have an explanation. If it does not then I'd say that should ring some alarm bells in physics circles. Thanks for bearing with me on this so far anyway and I will continue to try and to live in hope of getting to the bottom of it.

I am not suprise some Physicists won't help you if you take that attitude with them, and given the peremptory way you have dismissed my attempts to help you.

4 hours ago, Seanie said:

To clarify that, I am saying that the initial condition is that the rod is aligned with the positive Y-axis (theta=0). It will not be on top of the Y-axis but parallel to it and will lie along the line x=-r. That is at least if my understanding of m2's x-direction oscillation is correct. From that starting point it possesses its angular speed of w1 (omega 1) and m2 is at the point (-r, 0).

Below I have set out a simple plan of the starting conditions in accordance with your instructions.

Now please tell me how you impart a  motion to your rod to set it rotating about m2.

Are you aware of the difference between a moment and a couple ?

Once you have demonstrated that it is possible to establish the rotating motion at all we can get down to the business of the geometry.

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4 hours ago, swansont said:

Then why would the motion in the x direction reverse itself? There's no force in that direction to change the x component of the momentum.

I'm not sure how else to show that it will. So if you imagine an unbalanced rotation (i.e. with an eccentric weight rotating) and you constrain the pivot (i.e. centre point of rotation) so it is free to move in a straight line then I suppose you would agree that the pivot will exhibit an oscillating motion on that line, yes? If so then what causes it to oscillate, i.e. to reverse its direction? Where is the force that does that? Well its the same idea with the mechanism I am discussing in this thread.

30 minutes ago, studiot said:

I am not suprise some Physicists won't help you if you take that attitude with them, and given the peremptory way you have dismissed my attempts to help you.

I believe you are mistaken and misjudging me here. I am not aware of a bad attitude with you or them, or of being peremptory. If you give me specific details of your basis for that then that would be fair to me so I can see it for myself it its there and if its not there then I can clarify the misunderstanding for you. Nevertheless I ask your forgiveness if I have offended you. Going forward though I fear that it may easily happen again since I don't know how it happened to begin with, hence my request for you to give details. Thanks in anticipation.

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57 minutes ago, Seanie said:

hence my request for you to give details.

And yet when I ask for details I get answer like

On 9/28/2019 at 11:20 PM, Seanie said:

It doesn't matter how.

On 9/29/2019 at 4:39 PM, Seanie said:

1. All of that about the position of the rail under or over the rod is irrelevant to the principle involved here.

On 9/29/2019 at 4:39 PM, Seanie said:

The mechanism under discussion here is fundamentally different from a scotch yoke.

On 9/29/2019 at 12:13 PM, studiot 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.

You can have no idea why I ask these questions, most have led to an improvement of your system description.

I have asked several times how you set the system going in the desired motion and other responders have queried whether it will ever reach that motion yet you seem to think there is nothing in the concerns of all these others.

Think about this very carefully.

If you simply apply a single temporary force to m1, you have specified zero friction on the x axis guide rail.

So why will the x component of that force not simply move the rod bodily along the guide rail when there is no resistance?
Real world mechanisms all have friction.
Applying a force with zero x component to the chosen starting configuration produces zero motion, but unstable equilibrium.

Edited by studiot
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38 minutes ago, studiot said:

Now please tell me how you impart a  motion to your rod to set it rotating about m2.

I am not being peremptory or dismissive here but I believe I have answered you on this already. Hoping for no bad feelings here so the answer is I don't have a method in mind because I don't see why it is relevant. If you have said something before re. why it is relevant then I'm sure I don't see it, perhaps you can clarify it and/or we can discuss it more if you like. Nevertheless since you are looking for something on this, here is a suggested way:

The rod is forcibly rotated anticlockwise against a spring which is not part of the mechanism and released so that the spring accelerates the rod to its initial angular speed. At the same time, to prevent any movement of m2, m2 is held at its starting position until the rod is at its starting angle (theta=0) at which point m2 is automatically released. The spring also does not exert any force on the rod from the moment m2 is released. In that case there should be no force which would cause a translational motion of any part of the rod. Maybe that's why you are asking, to see if the imparted torque imparts any other force. Maybe that's why you asked about the moment and the couple too.

Anyway regarding the mechanism the important matter to be dealt with is to explain the still unknown source of energy that powers the anticipated oscillation. If there is a way of inputting the initial torque which could account for the oscillation that still does not solve the apparent mystery of this mechanism, because then I would just change the way the torque is input so it doesn't account for it and we would still be left with an oscillation whose energy source needs to be explained. In this mechanism which is essentially a rotating eccentric weight in which the pivot is free to move in a straight line, then we do have an oscillation caused by that rotation do we not? And so far the explanation of how and where the energy comes from seems amazingly hard to find does it not? Well that's what I am interested in and looking to create awareness of this and to seek the answers.

11 minutes ago, studiot said:

So why will the x component of that force not simply move the rod bodily along the guide rail when there is no resistance?

because I believe I specified already that the initial torque is a torque about m2, that would mean there is no force to move the rod bodily as you put it.

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1 hour ago, Seanie said:

I'm not sure how else to show that it will. So if you imagine an unbalanced rotation (i.e. with an eccentric weight rotating) and you constrain the pivot (i.e. centre point of rotation) so it is free to move in a straight line

That's not what constrain means, in this context. You allow motion in the x direction.

Quote

then I suppose you would agree that the pivot will exhibit an oscillating motion on that line, yes? If so then what causes it to oscillate, i.e. to reverse its direction? Where is the force that does that? Well its the same idea with the mechanism I am discussing in this thread.

Indeed. What is the force that causes the system to reverse the motion? I don't see one. I don't agree it will behave as you claim.

This is a bit of a problem with how you have framed the problem. You have stated certain behavior will occur, but that's not a conclusion from some analysis, so it is very suspect. You say there is an angular speed, but you haven't shown there will be any circular motion. It is an unsupported assumption on your part.

The mass being at some y point, if it is given (or has) a velocity solely in the x direction, will simply move in the x direction. There is nothing in the problem to change that motion. The massless, frictionless arm will go along for the ride, as if it wasn't there.

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14 minutes ago, studiot said:
On 9/28/2019 at 11:20 PM, Seanie said:

It doesn't matter how.

On 9/29/2019 at 4:39 PM, Seanie said:

1. All of that about the position of the rail under or over the rod is irrelevant to the principle involved here.

On 9/29/2019 at 4:39 PM, Seanie said:

The mechanism under discussion here is fundamentally different from a scotch yoke.

On 9/29/2019 at 12:13 PM, studiot 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.

I was simply telling you what I understand to be true and still hold the same understanding since I haven't reason to change yet. In the case of the torque I said I don't have a particular method in mind, in that case why the need to think I have a bad attitude? And how am I supposed to tell you the method of inputting torque if I don't have a method in mind? I also said it doesn't matter how, because what I meant was that it is not something we should get hung up on given the main thing to be resolved with this mechanism. I think from my post just minutes ago that should be more clear. Then what is wrong with telling you what I said about the scotch yoke? I did give a reason why it is not the same. Do you still not see why they are not the same? Again its the same with the rail, as far as the main focus of explaining the energy source of the oscillation is concerned then it really doesn't matter whether it is above or below. Perhaps your concern for the rail position would be important if we were trying to build a physical model but my understanding of the matter of this thread is that it is about the physics principles involved. I even said it in a nice way. What am I supposed to do? Again my apologies for making you feel dismissed. When our communications are limited to typed text then there's plenty of room for misunderstandings, good if we were discussing this face to face, it should be much better then.

5 minutes ago, swansont said:

Indeed. What is the force that causes the system to reverse the motion? I don't see one. I don't agree it will behave as you claim.

If you agree that there will be a motion in the x-direction, i.e. the extra motion as I called it that is the motion of m2, then what force causes that? Do you agree that m2 will oscillate or what do you think will happen? Because whatever that force is will be the same force that causes the reverse. Also that's fine if you don't think it will behave as I suggest but will you kindly explain how it will move then?

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19 minutes ago, Seanie said:

If you agree that there will be a motion in the x-direction, i.e. the extra motion as I called it that is the motion of m2, then what force causes that?

That's a question studiot has put to you. To me, it seems to be an initial condition of the problem: the rod is aligned with the y axis and has a velocity in the x direction. I don't see how anything but linear motion follows from that initial condition.

19 minutes ago, Seanie said:

Do you agree that m2 will oscillate or what do you think will happen? Because whatever that force is will be the same force that causes the reverse. Also that's fine if you don't think it will behave as I suggest but will you kindly explain how it will move then?

I don't see why it would oscillate under the conditions I have described. There is no force, so motion continues in a straight line, in accordance with Newton's laws.

To get any motion in the y direction, it must have an initial velocity component in the ± y direction. Then the existence of the rod will come into play.

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I have a rough idea about how to analyse the device, I would like to have your opinion. But I'm not sure I've understood the device fully, it is quite possible the idea doesn't apply at all.

I think the device is similar to a pendulum on a cart where the cart can move horizontally. The pendulum mass is mounted on a rod instead of a string. The cart and pendulum is constructed so that the pendulum could swing full loops if given sufficient initial push. The equations of motion for a pendulum on a cart can be derived using Lagrangian mechanics and it would result in differential equations where gravitational constant g appears.

Below is a typical pendulum on a cart device. This particular drawing would need some adjustments since the pendulum can't swing full circles and rod l is not mounted symmetrically at m1. But I hope it shows the general principle.

Now the idea: If we the put the pendulum and the cart in free fall** so g=0, could the device behave as the device presented by OP?
If so: could we try some of the many available solutions* to typical pendulum on a cart problems and see what the math predicts about its movement when g=0?
But also: does the Lagrangian still work when potential energy is zero when g=0?

Since g=0 the initial movement must be generated by some means other than dropping the pendulum from some angle >0. The Lagrangian equations would need appropriate modifications.

Does the idea seem valid?

*) Probably someone with better skills in Lagrangian mechanics would just write down the equations, but I'm still learning the basics so I need to cheat. That's why I try to find a reasonably similar device that's easy to find solutions for and that possibly can be tweaked into describing the device in this thread.
**) It could also be tilted 90 degrees so rotation takes place in the horizontal plane.

Edited by Ghideon
grammar
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3 hours ago, Ghideon said:

could the device behave as the device presented by OP?

No. Setting aside the fact that the rod between the swinging mass (now m2, was m1) and the pivot point is now fixed in length, the fact that a second unfixed mass is in play will actually bring about some continued rotation, and x direction oscillation (or at least change in x component speed of the individual masses)

So it can behave in a manner similar to how Seanie believes his/her device would behave. Without of course, generating any extra energy.

Edited by J.C.MacSwell
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18 hours ago, J.C.MacSwell said:

No.

Thanks for the feedback! I'll start from scratch.

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On 9/29/2019 at 8:09 AM, J.C.MacSwell said:
On 9/28/2019 at 11:20 PM, 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?

It is worth reviewing the energy situation.

1) As with any isolated mechancial system the total mechanical energy is constant.

2) The total energy of the system comprises that input at first setting in motion (whatever that is) of the system.

3) Macswell suggests the KE is fluctuating, but is it?
The end m2 which is alleged to execute SHM has zero mass so zero KE at all times.

4) There are no dissipative forces as the system is frictionless and no energy is lost.

5) But no energy is produced either so there is no 'overunity'.

Does this sum it up?

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Other members seem to understand the device better than me, is the below a correct interpretation? I borrowed @studiot's picture.
m2 is free to move along the x axis. m1 may rotate any angle theta. The rod is solid so distance between m1 and m2 is constant. Note that I'm not stating if m2 will move and/or not or if m1 will rotate around pivot m2 or not (yet). I just want to be sure that x and theta is sufficient to describe the motion.

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1 hour ago, studiot said:

It is worth reviewing the energy situation.

1) As with any isolated mechancial system the total mechanical energy is constant.

2) The total energy of the system comprises that input at first setting in motion (whatever that is) of the system.

3) Macswell suggests the KE is fluctuating, but is it?
The end m2 which is alleged to execute SHM has zero mass so zero KE at all times.

4) There are no dissipative forces as the system is frictionless and no energy is lost.

5) But no energy is produced either so there is no 'overunity'.

Does this sum it up?

In Seanie's description of the movement (which does not properly describe how the device would move), m1(the only mass in the system) would necessarily vary in speed...thus vary in KE.

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