# Rotational and inertial mechanics, overunity mechanism?

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17 minutes ago, J.C.MacSwell said:

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.

How and in which frame of reference?

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On 10/1/2019 at 9:59 PM, Ghideon said:

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.

You are correct, the principle here is the same as that of the mechanism in this thread. To make it more fully compatible, make the mass of the cart be zero and the pendulum to swing all the way around and make gravity be zero too.

2 hours ago, Ghideon said:

is the below a correct interpretation?

Yes this is also correct but just to be consistent with our mechanism I would have the rod rotating the other way, i.e. rotating clockwise. Also just to clear up any possible doubt about the initial input torque, it is input in a way which doesn't impart any translational force on the rod, it only imparts a turning force.

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5 minutes ago, studiot said:

How and in which frame of reference?

In this case, with the pivot point oscillating in the x direction while at the same time angular velocity remaining constant, it would vary in any inertial frame. (there is one exception based on that statement...but it isn't claimed or happening...and just adding noise to describe it)

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1 hour ago, J.C.MacSwell said:

In this case, with the pivot point oscillating in the x direction while at the same time angular velocity remaining constant, it would vary in any inertial frame. (there is one exception based on that statement...but it isn't claimed or happening...and just adding noise to describe it)

I did not say how the mass m1 would move, just that whatever energy was input by the initial impulse would remain and that the other parts of the mechanism would have zero KE at all times and that there were no dissipative or potential forces acting to change the PE of m1.

The question as to whether the parts of the mechanism would act in the way seanie describes has been challenged by all of us.

So I was hoping to dispel or remove the side issue of energy conservation.

By the way do you not find calling the ends of the rod m1 and m2 confusing?

I certainly do.

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5 minutes ago, studiot said:

I did not say how the mass m1 would move, just that whatever energy was input by the initial impulse would remain and that the other parts of the mechanism would have zero KE at all times and that there were no dissipative or potential forces acting to change the PE of m1.

That means m2 doesn't have any motion. In that case do you not find it hard to imagine how a rotating eccentric weight would not pull on the centre of its rotation? If it does pull on it and the centre can freely move then why would it not move? I'm hoping that we are at least seeing major problems with our analyses (including mine).

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Just now, Seanie said:

That means m2 doesn't have any motion. In that case do you not find it hard to imagine how a rotating eccentric weight would not pull on the centre of its rotation? If it does pull on it and the centre can freely move then why would it not move? I'm hoping that we are at least seeing major problems with our analyses (including mine).

Why is that ?

KE = 0.5 mv2

If m is zero then KE = 0, whatever v is.

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9 minutes ago, Seanie said:

That means m2 doesn't have any motion. In that case do you not find it hard to imagine how a rotating eccentric weight would not pull on the centre of its rotation? If it does pull on it and the centre can freely move then why would it not move? I'm hoping that we are at least seeing major problems with our analyses (including mine).

How do you find it easy to imagine it rotating when/while it is unable to pull on the centre?

Edited by J.C.MacSwell
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14 minutes ago, studiot said:

So I was hoping to dispel or remove the side issue of energy conservation.

We can remove that all we like but we are then stuck with a mechanism that doesn't make sense.

15 minutes ago, studiot said:

By the way do you not find calling the ends of the rod m1 and m2 confusing?

I agree. I had been working on similar mechanisms and for some reason had those labels and they just got carried over to this one. Then with all my notes being as they were I didn't want to come up with new names.

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1 hour ago, Seanie said:

Also just to clear up any possible doubt about the initial input torque, it is input in a way which doesn't impart any translational force on the rod, it only imparts a turning force.

I had hoped this meant you were beginning to listen and understand and could perhaps even answer my earlier hint question about the difference between a moment and a couple.

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6 minutes ago, studiot said:

Why is that ?

KE = 0.5 mv2

If m is zero then KE = 0, whatever v is.

because m2 can move even though its KE =0. Also if m2 is moving that means that m1 must have an extra x-direction motion and that requires energy.

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4 minutes ago, Seanie said:

because m2 can move even though its KE =0

the motion of m2 does not affect the KE of m1

4 minutes ago, Seanie said:

Also if m2 is moving that means that m1 must have an extra x-direction motion and that requires energy.

the motion of m1 is the motion of m1, whatever that is, and its KE is also given by my equation.

Not also that since the rod is free to pivot about m2 m1 has motion in both the x and y directions each with its corresponding KE, which both contribute to the total KE of m1.

Quote

that m1 must have an extra x-direction motion and that requires energy.

You appear to be trying to count something twice because you are jumping between frames again.

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6 minutes ago, studiot said:

I had hoped this meant you were beginning to listen and understand and could perhaps even answer my earlier hint question about the difference between a moment and a couple.

I'm not sure what you mean here, it certainly seems a bit derogatory, don't you think? You seem to not like the idea of the rod only being given an initial turning impulse (i.e. without translational force). Why is that? To keep the mechanism as simple as possible so as to focus on explaining what seems to be energy that is unaccounted for (an important matter surely), that's why I specified that the initial impulse only has a rotational effect on the rod about m2. Otherwise we would have to try to calculate the effect of translational force from the initial impulse which would unnecessarily complicate things. I suppose if the initial impulse was a moment that there would also be a translational force and if it were a couple there would not be. Are you clear now about the initial impulse or is there still something other than rotation of the rod that I need to consider (i.e. resulting from the initial impulse)?

25 minutes ago, J.C.MacSwell said:

How do you find it easy to imagine it rotating when it is unable to pull on the centre?

I'm not sure I get what you mean. We have a rotating mass because it has been given an initial impulse. Then because of its centre being free to move in a straight line the rotating mass will surely pull on its centre, will it not?

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9 hours ago, Seanie said:

I'm not sure I get what you mean. We have a rotating mass because it has been given an initial impulse. Then because of its centre being free to move in a straight line the rotating mass will surely pull on its centre, will it not?

These are contradictory. If you had a ball on a string and you spun it overhead in a circle you would feel a centripetal force...if you released it so that it was "free to move" it would fly away in a straight line...being free to move it would no longer travel in a circle...as it is unable to "pull on the centre" of that circle.

With no centripetal force their will be no centripetal acceleration...therefore no turning in a circle.

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6 minutes ago, J.C.MacSwell said:

These are contradictory. If you had a ball on a string and you spun it overhead in a circle you would feel a centripetal force...if you released it so that it was "free to move" it would fly away in a straight line...being free to move it would no longer travel in a circle...as it is unable to "pull on the centre" of that circle.

With no centripetal force their will be no centripetal acceleration...therefore no turning in a circle.

Ok, I get what you mean and that makes sense. The only thing is that I don't see how it relates to our mechanism here.

7 minutes ago, J.C.MacSwell said:

Then because of its centre being free to move in a straight line the rotating mass will surely pull on its centre, will it not?

To clarify my statement further: so if m1 is the rotating mass, rotating about the point m2 then m2 is the centre of the rotation, that's the "centre being free to move" that I referred to, it is free to move in a straight line in the x-direction. So when m1 is rotating (i.e. an unbalanced rotation) it will tend to pull m2, of course. How is that not the case? With your ball on a string its the same thing, your hand is at the m2 position (centre of rotation) and it will feel a pull towards the ball. Then since m2 if free to move in a straight line it will indeed move, how can it not? If you are saying it will not move that's what I don't understand. Thanks.

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10 hours ago, Seanie said:

because m2 can move even though its KE =0. Also if m2 is moving that means that m1 must have an extra x-direction motion and that requires energy.

It's not "extra" motion.

10 hours ago, Seanie said:

I specified that the initial impulse only has a rotational effect on the rod about m2.

Under the conditions you described, that may not be possible, and it's a difficulty if you have a set-up that isn't physically possible. There is no way for the mass to have a y component of velocity when the rod is aligned with the y axis. The only way to acquire one is if the pivot is in motion at the beginning, but that's not physically realizable, since any force on it will result in infinite acceleration. You have to get away from a description of forces causing the initial motion.

IOW, there are constraints on your setup, and you can be proposing something physically impossible when you make statements about the motion that are not the result of analysis.

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1 hour ago, Seanie said:

If you are saying it will not move that's what I don't understand. Thanks.

It will move if released...but it won't come back...m1 will continue on it's track...unless acted on by a force. Massless m2 will follow.

No force...no rotation of m1 about m2...no oscillation for m2.

Compare with two balls joined by a string: Grip one and spin the other overhead then release. Now they fly off but neither in a straight line as they also rotate about the centre of mass.

The string will be in tension keeping them together...supplying the necessary centripetal force for rotation.

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I have some heavy curtains, weighted down at the bottom.
They run on a very low friction track using brass bogie rollers.

Now if I grab the bottom an give an impulse all that happens is the curtains move bodily sideways.
They do not attempt rotation, even though the force of drawing the curtains has substantial mement about the rail.

If you do the same with your rod

ie grab m1 and give it a push the whole rod will simply slide sideways.

I am still waiting for you to tell me if you understand the difference between a couple and a moment (of a force)

A couple has the same moment about every point in the plane, including its point of application.

A moment of a force has a different moment about nearly every point in the plane and zero moment about its point of application.

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24 minutes ago, J.C.MacSwell said:

It will move if released...but it won't come back...m1 will continue on it's track...unless acted on by a force. Massless m2 will follow.

No force...no rotation of m1 about m2...no oscillation for m2.

Compare with two balls joined by a string: Grip one and spin the other overhead then release. Now they fly off but neither in a straight line as they also rotate about the centre of mass.

The string will be in tension keeping them together...supplying the necessary centripetal force for rotation.

Yes, that makes sense too and it is something I had overlooked, thank you.

Now I'm wondering what would happen if m2 had a real mass. I suppose m1 would rotate then as well as m2 oscillating. If so then we are back to the old question of where is the torque coming from that would oppose the rod's rotation? If there is none then how to account for the energy to oscillate m2?

1 hour ago, studiot said:

ie grab m1 and give it a push the whole rod will simply slide sideways.

Yes thanks for the example, I get the idea now.

1 hour ago, studiot said:

I am still waiting for you to tell me if you understand the difference between a couple and a moment (of a force)

Sorry I thought the answer was contained in what I said here:

14 hours ago, Seanie said:

I suppose if the initial impulse was a moment that there would also be a translational force and if it were a couple there would not be.

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28 minutes ago, Seanie said:

Now I'm wondering what would happen if m2 had a real mass. I suppose m1 would rotate then as well as m2 oscillating. If so then we are back to the old question of where is the torque coming from that would oppose the rod's rotation?

Why is a torque necessary?

28 minutes ago, Seanie said:

If there is none then how to account for the energy to oscillate m2?

It will come from m1, whose speed will change.

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10 minutes ago, swansont said:

It will come from m1, whose speed will change.

If the rod's angular speed changes (i.e. it must slow down) to supply energy to the oscillation then there must be a torque opposing its rotation because from Newton's 1st law it will otherwise continue at the same speed. Or to put it another way, if it slows there must be a force (i.e. torque) or some resistance to its rotation, to slow it. Is that correct?

To provide power to the oscillation it seems that m1's rotational energy is the only possible source of that power, yet how does this happen, I mean what are the mechanics of it exactly?

Because I don't see a way for there to be such an opposing torque, do you?

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6 minutes ago, Seanie said:

If the rod's angular speed changes (i.e. it must slow down) to supply energy to the oscillation then there must be a torque opposing its rotation because from Newton's 1st law it will otherwise continue at the same speed. Or to put it another way, if it slows there must be a force (i.e. torque) or some resistance to its rotation, to slow it. Is that correct?

If there is another mass internal to the system, it can gain the angular momentum the other mass has lost.

6 minutes ago, Seanie said:

To provide power to the oscillation it seems that m1's rotational energy is the only possible source of that power, yet how does this happen, I mean what are the mechanics of it exactly?

m1 and m2 will have energy. The sum must be constant, not the individual values.

6 minutes ago, Seanie said:

Because I don't see a way for there to be such an opposing torque, do you?

No external torque is necessary.

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So is the original arrangement being abandoned as unworkable?

1 hour ago, Seanie said:

Sorry I thought the answer was contained in what I said here:

14 hours ago, Seanie said:

I suppose if the initial impulse was a moment that there would also be a translational force and if it were a couple there would not be.

The point about a couple is that you can apply one to the rod to get the rotation started, but there is no guarantee it will continue for very long.

As a matter of interest, if you have a large m1 and a small m2 you have a situation that occurs in nature.

There are quite a few molecules that have a heavy atom at one end of the bond and a very light one at the other.

They perform a very peculiar sort of tumbling rotation.

I believe there are macro toys based on this as well.

Final comment.

When you have rotations about 3 axes (known as Euler axes) , two rotations are stable but the middle one is wildly unstable.
This can be demonstrated by trying to spin a matchbox/book/brick about each of the three axes and observing the behaviour.

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19 minutes ago, studiot said:

So is the original arrangement being abandoned as unworkable?

yes it seems so to me anyway.

40 minutes ago, swansont said:

If there is another mass internal to the system, it can gain the angular momentum the other mass has lost.

Ok but how does that happen here?

41 minutes ago, swansont said:

No external torque is necessary.

But to my understanding if a rotation (or anything else) changes speed its because some torque (or force) acted against its motion. Otherwise it would seem like a magical happening with no identifiable mechanics to explain it. By all means show me what I'm missing here.

30 minutes ago, studiot said:

but there is no guarantee it will continue for very long.

If you mean that the rotation may not continue for very long is that because of something about the couple itself? (if so I would love to understand that better), or is it because of friction or what?

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29 minutes ago, Seanie said:

yes it seems so to me anyway.

Ok but how does that happen here?

But to my understanding if a rotation (or anything else) changes speed its because some torque (or force) acted against its motion. Otherwise it would seem like a magical happening with no identifiable mechanics to explain it. By all means show me what I'm missing here.

The torque is internal to the system, exerted by the rod. Since it's internal, you can treat angular momentum as constant. You don't need to know the details of the torque the rod exerts, since it will exert an equal and opposite torque on the other mass. (similar to not needing to know the force of impact in a collision in order to apply conservation of linear momentum)

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51 minutes ago, Seanie said:

But to my understanding if a rotation (or anything else) changes speed its because some torque (or force) acted against its motion. Otherwise it would seem like a magical happening with no identifiable mechanics to explain it. By all means show me what I'm missing here

The classic example is the ice skater, spinning on the ice.

She speeds up or slows down her spin by drawing in or extending her limb(s).

53 minutes ago, Seanie said:

If you mean that the rotation may not continue for very long is that because of something about the couple itself? (if so I would love to understand that better), or is it because of friction or what?

Try this thought experiment.

Reach down give the rod a sharp twist with two fingers, like if you were spinning a coin on a table top.

Then let go.

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