Where do the input energy go?

In summary, this is a conversation about an experiment involving gyros on a wheel. The gyros, which resist any change in alignment, are connected to a gear mechanism that causes them to spin fast when the wheel is turned slowly. To maintain a constant RPM, energy must be constantly added, as the gyros will try to stop the wheel from turning. There is also friction present, which causes heat and dissipates energy into the environment. The energy spent to fight against the gyroscopic counter torque goes into increasing the kinetic energy of the gyros. The system is stable at one given velocity, so no additional energy is required to sustain the precession of the gyros. However, if the system slows down, there has been more energy applied
  • #36
Low-Q said:
What is the reason why the weight will move upwards?
Torque.

Low-Q said:
One would think at first glance that the vertical torque is solely caused by the weight and gravity.
No. Gravity cannot cause a vertical torque. Remember, torque is always perpendicular to the force, so torque from gravity is always horizontal.

Low-Q said:
Wether the precess wants to be clockwise or counterclockwise perpendicular to the guides wouldn't matter as there is no physical precess present due to the guides, right?
If horizontal precess really is an important factor of which vertical direction the weight will move in this system, I will accept that - but have trouble in understanding why.
Let's say that the system is in front of me with the gyroscope facing directly towards me and spinning counter-clockwise. This is an angular momentum pointing horizontally directly towards me.

The weight will cause a torque to the right so the gyro will want to precess to the right. Instead it will run into the guide which will prevent any precession to the right.

The force that the guide exerts will in turn make a torque vertically down. Since the guides allow motion down the gyroscope will precess downwards.

Once it reaches the bottom there will be no more torque due to gravity and therefore no more force from the guide and therefore no more torque from the guide and it will stop precessing.

Why don't you try the analysis if the gyro is spinning clockwise (angular momentum directly away), and see if it precesses up or down. I don't actually know.

Low-Q said:
instead of generating heat, the spinning weight must slow down. Energy must be conserved.
Yes.

Low-Q said:
So, then I am back to the initial question in this thread. Analyzed the above system, I now put a gear to the weight so the weight is spinning because I push it along the guides. Whould't that mean that I try to accelerate the RPM of the weight at the same time as the shift of the wheel force to stop it's spin? While doing this, the force from my finger is moving a given distance. This will appearently end up with an energy consume from my side that is not going anywhere - not heat, not increased kinetic energy ?
Obviously this is incorrect: energy is conserved. I have dealt with this question multiple times already and am not interested in repeating it yet again over and over. Please re-read my previous comments, if they are not sufficient then go get a textbook because apparently my communication style is not clear enough.
 
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  • #37
English isn't my native language, and I think I might have mistranslated some of the terminologies in your explanations, and probably therfor I have not fully understood what you explained. I did believe you posted contradictory explanations, until I actually found the correct translation of a few terminologies... I can now picture the outcome from the systems I have sketched.

Thanks for your patience :)
 
  • #38
You say that the weight possibly can move upwards. That will ofcourse depend on the RPMs of the weight. If it too slow, the weight will be too heavy for the precess to move it upwards.

You also agree that the spinning weight will slow down in order to conserve energy in the system when the weight reach the bottom or top, because this system does not generate heat from friction. After a second thought, something seems to be missing. I would believe that the spin slows down only during the motion along the circumference of the guides. As it approach a full stop, the spin will gradually increase to its originally spin again. This should be the only way the system can conserve its kinetic energy.

Because if the weight permanently spins slower after it has reached bottom or the top, one would have to apply energy to increase the spin to its originally spin. Which in turn means that there is in spite of no loss on the system necessary to apply energy into it to make it conserve the kinetic energy. That would be impossible.

I found an interesting thread from 2006 in this forum. It seems like this is about a similar case:
https://www.physicsforums.com/showthread.php?t=116579

Vidar
 
  • #39
Low-Q said:
You say that the weight possibly can move upwards.
Actually, that may have been a mistake. I encourage you to work it out for yourself the same way as I did above. I got that the weight goes down regardless of if the gyro is spinning CW or CCW, but it is complicated enough that I may have made a mistake somewhere.

Low-Q said:
You also agree that the spinning weight will slow down in order to conserve energy in the system when the weight reach the bottom or top, because this system does not generate heat from friction.
Its rotation would have to speed up as it reaches the bottom, not slow down. It is losing gravitational PE, so it must be gaining KE.
 
  • #40
DaleSpam said:
Its rotation would have to speed up as it reaches the bottom, not slow down. It is losing gravitational PE, so it must be gaining KE.
That PE was applied by the hand which put it up there in the first place, and should not affect the rotational speed.
 
  • #41
Low-Q said:
That PE was applied by the hand which put it up there in the first place, and should not affect the rotational speed.
It doesn't matter in the slightest what mechanism provided the PE. The fact is that it has higher PE when it is up and it loses PE when it goes down. Since it loses PE it must gain KE.
 
  • #42
You might have a point, but I don't think the PE in this case have anything to do with the rotation. This weight could likely be located at the pivot point where another force is pitching the spin. The viewing point can be anywhere so it is hard to determine wether the weight will spin up or slow down. I stop analyzing this and start to build this system. Then we will know what happens.
 
  • #43
Any system you build will slow down due to friction.
 
  • #44
Yes, but it is possible to compare one test with another. Then calculate the difference given that the friction is the same.
 
  • #45
OK, good luck. I hope you find it enjoyable.
 
  • #46
I had an old rotating mirror (Those in laser scanners for bar code reading). This wheel is made of aluminium and the diameter is approx 5cm, and 1cm thick.

I spun it up to about 20k RPM with high pressure air while I was holding the spinning wheel with a ball bearing. It took about 60 seconds for it to full stop. I repeated this several times - 60 seconds to full stop.

So I did several tests where I pitched the wheel back and forth. I felt resistance while trying to pitch the wheel perpendicular to the rotation (Whithout letting the wheel precess). The wheel spent shorter time to full stop - 50 sec in average.

These tests was very provisional, so I have no exact measurements. I used the sound as reference to determine the correct RPM. I have a good ear...

No conclusion yet.

Tomorrow I will test a 15" diameter, 1" thick MDF board as a spinning disc (Not at 20k RPM). I'll see what I can find out. I have good quality ball bearings for this experiment.

Vidar
 
  • #47
I think I have found an similar frictionless system that also require energy input to sustain motion. The system is commonly used in most engines: The piston.

If an engine have a very heavy piston, the engine vill be less efficient. But the lower efficiency is not due to more friction. The increased energy requirements goes to stop and accelerate the mass in the piston.

If an electric super efficient motor is powering a flywheel attached to a mass that is going back and forth like a piston, no matter how low the friction is, there is a constant energy input requirement, but without increasing the average KE in the system.

The gyroscope that is prevented from precess as force is pitching it will require similar energy input. The reason is that a given mass in the wheels circumference is accelerating as it moves from the "north pole" towards the equator, but deaccelerate from equator towards the "south pole". It is called the Coriolis effect. This effect will counterforce the change in pitch, and thus require energy input - even if KE does not change.

Where do the energy input go? Obviously in both systems, piston or gyro, there is spent energy on something that is not friction or increased KE.

So where do this energy go if it is not heat? As we cannot get back the same energy as we put in - even if friction is out of the equation!

Isn't this annoying?

Vidar
 
  • #48
@low-q

If a reciprocating mechanism has no friction and the reciprocation is sinusoidal with angle (ideal and with a crank) then there are no inherent energy losses because the KE of the piston will transfer backwards and forwards from the flywheel - remaining constant; hence the rotation of the flywheel will not be constant and the motion of the piston will not be sinusoidal with time. In an engine, there are losses because there is energy transfer as the engine does work and it's important not to confuse the two, I think. So I don't think you can come to conclusions about one case, based on the other (which is what you seem to be implying.) With a real engine, the non-uniform motion of the crank is probably not relevant because of resilience in the transmission but I know that shorter stroking engines can certainly rev higher.
Also, if there is no motion in the direction of the Coriolis force then no work is done either - so there will be no losses from that effect either.
 
  • #49
Low-Q said:
I think I have found an similar frictionless system that also require energy input to sustain motion. The system is commonly used in most engines: The piston.
So which of Newton's laws do you believe that a piston violates? Newton's laws imply energy conservation, so if energy is not conserved then at least one of Newton's laws must be violated. So which one?
 
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  • #50
I am not claiming any violations. I just ask where the input energy goes. For the piston, it could be thousands in order to reduce cogging, but still even without friction the rotational energy in the flywheel will get lost in the way those pistons are moving. So further input energy are necessary to sustain rotation of the flywheel. As there is no violations of any laws, this input energy, or work, must be conserved somewhere. Where?
 
  • #51
Low-Q said:
I am not claiming any violations. I just ask where the input energy goes. For the piston, it could be thousands in order to reduce cogging, but still even without friction the rotational energy in the flywheel will get lost in the way those pistons are moving. So further input energy are necessary to sustain rotation of the flywheel. As there is no violations of any laws, this input energy, or work, must be conserved somewhere. Where?

As my earlier post says, the energy is shared, alternately, with the flywheel, which varies in speed over the cycle.
 
  • #52
Low-Q said:
So further input energy are necessary to sustain rotation of the flywheel. As there is no violations of any laws, this input energy, or work, must be conserved somewhere. Where?
You are ASSUMING that input energy is necessary to maintain a constant KE for a frictionless piston which does no work. Just like you are ASSUMING that input energy is necessary to maintain a constant KE for a frictionless gyro which does no work. That ASSUMPTION leads to the conclusion that energy is not conserved. If energy is not conserved then Newton's laws must be violated. Therefore either the ASSUMPTION must be false or Newton's laws must be violated.

Personally, I think your assumptions are false. I.e. further input energy is not required to sustain rotation. You are simply asserting that assumption without any valid justification, and since that assumption leads to a contradiction it must be false.

Please post a detailed analysis using Newton's laws for a frictionless piston showing this continuous energy input requirement.
 
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  • #53
Low-Q said:
So further input energy are necessary to sustain rotation of the flywheel. As there is no violations of any laws, this input energy, or work, must be conserved somewhere. Where?

I think your problem is that you are trying to consider a model that is neither real nor ideal. If the piston is the only thing with mass or MI and everything is massless and rigid then its motion will be a 'sawtooth' in time, with uniform motion and a discontinuity when it changes direction - not a real situation: the Laws don't apply and Energy need not be conserved. You have to allow the flywheel to have some MI, or you can't have the piston moving properly with no discontinuity of motion. Under those circs, Energy can be conserved. (Or, I suppose, you could have resilient, ideal con-rods so the piston KE would be stored as elastic energy in the con rods.)
In a 'realy real', non-ideal mechanism, you have to allow the structure to be rigid (enough) to allow the loss-less transfer of KE between piston and flywheel. etc etc
 
  • #54
It does not really matter how heavy the flywheel is. I am sure the KE in that wheel is not changing anything regardin the pistons. Ofcourse, the heavier the flywheel is the longer it can transfer its KE to the pistons befor it stands still, if the motor which power it is released from it. I do question yor claim that this isn't a real thing. It is possible to build and test this in real life. So since conservation of energy must be conserved, there must be some place for this input energy to go - that isn't heat... Might the energy loss be "fed back" to the motor some how?
 
  • #55
Low-Q said:
Might the energy loss be "fed back" to the motor some how?
What energy loss? You haven't even demonstrated that there is any. You merely assume it, and that assumption is incompatible with Newtons laws.
 
  • #56
Low-Q said:
It does not really matter how heavy the flywheel is. I am sure the KE in that wheel is not changing anything regardin the pistons. Ofcourse, the heavier the flywheel is the longer it can transfer its KE to the pistons befor it stands still, if the motor which power it is released from it. I do question yor claim that this isn't a real thing. It is possible to build and test this in real life. So since conservation of energy must be conserved, there must be some place for this input energy to go - that isn't heat... Might the energy loss be "fed back" to the motor some how?

You have not understood what is responsible for the dynamics of the piston, then. Being "sure" and saying that an experiment could be done are not arguments to support what you say; you need to use some Physics.
We are assuming that the reciprocating mechanism is not powering anything or being driven - just running freely, with no added friction.
If the flywheel had no MI (mass) then the piston (in an open cylinder with zero mass con rod) would move at uniform speed along the length of the bore as the flywheel could not affect it in any way. How could it do otherwise (Newton 1), as there would be no forces on it? That would, as I have already said, result in a discontinuity in velocity at the ends of motion - the piston would either stop with a jolt at the extremity of its motion or be pulled back due to nothing more than resilience in the con rod etc. and that is not 'allowed' in a real model.

Also, the on-going mean KE in the flywheel is irrelevant in this argument as it doesn't change in a situation where there are no losses. If there is an inelastic collision at each end of piston travel then there is energy loss - of course. But there is a force on the con rod because of the presence of the flywheel which slows down and speeds up the piston and forces a (timewise) sinusoidal oscillation. That force times distance (integrated) represents work done to and by the flywheel. At the end of travel, the change of direction involves no motion so no work is done actually at that time. The energy is transferred from piston to flywheel and back again, leaving the mean KE of the flywheel unchanged.

Thinking about a situation like this, one must avoid being subjective about it. There is an understandable feeling that the 'jerk' at the ends of motion is necessarily a reason for loss of energy but, although the acceleration is at a maximum at the extremes of motion (as in all SHM) the motion is smooth and continuous so no energy is lost if the bearings are all tight and there is no slop.
 
  • #57
Low-Q said:
Might the energy loss be "fed back" to the motor some how?
The Energy that the piston loses (and gains) is exchanged with the flywheel on a periodic basis. You must stop thinking so subjectively if you want to understand it.
A mass bouncing on a spring 'loses' Kinetic energy to the spring and comes to a halt each end of its motion but gets it back on the way down and up.
 
  • #58
DaleSpam said:
What energy loss? You haven't even demonstrated that there is any. You merely assume it, and that assumption is incompatible with Newtons laws.
I agree that this is an assumption. However, as you probably have understood, that assumtion is based on the input energy requirements necessary to sustain the KE in the system (That is indeed a real thing).

Take a seesaw. Put a heavy weight on each side of the pivot, and power it up. How fast can this seesaw run with a given input work? I bet the seesaw will stop accelerating at a given frequency even - if the weights are perfectly balanced. The lighter the weights are, the higher the frequency will be with the same input work.

@sophiecentaur said, that shorter stroking engines can certainly rev higher, indicates that there is higher efficiency in such a system.
I want to add that a lighter piston, with same stroke as a heavy piston, performs better (Higher efficiency). I assume race cars use as light pistons as possible for the engine to perform as good as possible. So even if the friction would be the same in those two engines, the different KE in those pistons can't be neglectet. So I dare to claim that there is some wasted energy input we (Probably only myself) yet do not understand where is heading.

You don't need to comment this anymore if you feel that I'm not "getting it" (:confused:). I will try this in practice with some of the brushless RC motors I have available - running a piston, or similar, with a flywheel, and measure the energy consumtion with light and heavy pistons.

I will post my findings here when the temperature outside is acceptable (-20'C as I write)...

Vidar.
 
  • #59
Low-Q said:
However, as you probably have understood, that assumtion is based on the input energy requirements necessary to sustain the KE in the system (That is indeed a real thing).
No, it isn't a real thing. That is my point. If you had a frictionless system then it would not need any input energy to sustain the KE. You are simply assuming that based on your experience with systems with friction. It is a wrong assumption, not a real thing.

Low-Q said:
Take a seesaw. Put a heavy weight on each side of the pivot, and power it up. How fast can this seesaw run with a given input work? I bet the seesaw will stop accelerating at a given frequency even - if the weights are perfectly balanced. The lighter the weights are, the higher the frequency will be with the same input work.
And here you are making yet another unsupported assumption without any analysis.

Low-Q said:
So I dare to claim that there is some wasted energy input we (Probably only myself) yet do not understand where is heading.
Not if there are no friction/heating losses, and not if Newton's laws are obeyed.

Low-Q said:
I will try this in practice with some of the brushless RC motors I have available - running a piston, or similar, with a flywheel, and measure the energy consumtion with light and heavy pistons.
In practice you will always have friction. Your questions here have been about idealized lossless systems. Such ideal systems do not require input power to continue running.

I have already demonstrated the analysis for your early gyroscope, but you are coming up with far more new systems than I am willing to analyze in detail. So the burden of analysis falls to you. From first principles we know that energy is conserved, so we know any assumption to the contrary is wrong. The details are up to you to show.
 
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  • #60
Low-Q said:
I agree that this is an assumption. However, as you probably have understood, that assumtion is based on the input energy requirements necessary to sustain the KE in the system (That is indeed a real thing).

Take a seesaw. Put a heavy weight on each side of the pivot, and power it up. How fast can this seesaw run with a given input work? I bet the seesaw will stop accelerating at a given frequency even - if the weights are perfectly balanced. The lighter the weights are, the higher the frequency will be with the same input work.

@sophiecentaur said, that shorter stroking engines can certainly rev higher, indicates that there is higher efficiency in such a system.
I want to add that a lighter piston, with same stroke as a heavy piston, performs better (Higher efficiency). I assume race cars use as light pistons as possible for the engine to perform as good as possible. So even if the friction would be the same in those two engines, the different KE in those pistons can't be neglectet. So I dare to claim that there is some wasted energy input we (Probably only myself) yet do not understand where is heading.

You don't need to comment this anymore if you feel that I'm not "getting it" (:confused:). I will try this in practice with some of the brushless RC motors I have available - running a piston, or similar, with a flywheel, and measure the energy consumtion with light and heavy pistons.

I will post my findings here when the temperature outside is acceptable (-20'C as I write)...

Vidar.

The reason that a short stroking engine can run faster is because the reciprocating forces are lower on bearings. But that has nothing to do with your ideas about the theory.
If you try an experiment, the results will be meaningless because they will be swamped by practical errors.

Just think about the following, ignoring the slight errors in geometry when a short con rod is used. Read it through with your mind switched to 'look and learn' mode.
A piston being driven by a simple crank will be oscillating with simple harmonic motion (sinusoidal oscillation) and so will the same piston, bouncing up and down on a spring, which has been chosen to produce the same period of oscillation. The two graphs of displacement in time will be precisely the same. There is no loss of energy in the mass on spring so where will the energy loss come from when precisely the same motion is produced by another method? (The crank can be chosen to be as long as you like, to suit this argument - it's assumed to be mass-less).
This is theory, which your argument is supposed to be based on. But your argument is based on a subjective view of the situation - which ain't Physics.
 
  • #61
I have tried to analyze:

I use a clock as reference. At 12 o'clock (Of the flywheel) the piston starts to accelerate. At 3 o'clock the piston has the highest possible velocity and KE.
This KE is transferred back to the flywheel between 3 o'clock and 6 o'clock. The cycle repeats with acceleration from 6 o'clock to 9 o'clock. Maximum KE at 9 o'clock. KE is transferred back to the flywheel between 9 o'clock and 12 o'clock.
The net energy is zero - zero loss. Can't believe I have not seen that. The similar will apply to the seesaw, and probably the gyro as well...
 
  • #62
Well done. You stuck at it and got there. :smile:
 
  • #63
I found an interesting and well explaining article about gyros:

"A gyro will resist any force that attempts to change the direction of its spin axis. However, it will move (precess) in response to such force; NOT in the direction of the applied force, but at right angles to it. The direction a gyro will precess also depends on the direction the gyro is spinning. Precession is actually the result of two forces: angular momentum (spinning force) and the applied force (torque). The direction of precession is always offset from the direction of the applied force. The offset is always in the direction of rotor spin. For example, when a force is applied upward on the inner gimbal, as shown in figure 3-8, the force may be visualized as applied in an arc about axis Y-Y. This applied force is opposed by the resistance of gyroscopic inertia, preventing the gyro from rotating about axis Y-Y. With the rotor spinning clockwise, the precession will take place 90º clockwise from the point of applied force. The gyro precesses about axis Z-Z in the direction of the arrow "P"."

14187_137_1.jpg


Vidar
 
<h2>What is input energy?</h2><p>Input energy refers to the energy that is put into a system or process to make it work. It can come in various forms such as heat, electricity, or mechanical energy.</p><h2>Where does the input energy go?</h2><p>The input energy can be converted into different forms of energy and used for various purposes. It can be transformed into kinetic energy, potential energy, or stored as chemical energy.</p><h2>What factors determine where the input energy goes?</h2><p>The distribution of input energy depends on the type of system and the laws of thermodynamics. Other factors such as the efficiency of the system and external forces also play a role.</p><h2>Can input energy be lost?</h2><p>According to the law of conservation of energy, energy cannot be created or destroyed, only transferred or converted. Therefore, input energy cannot be lost, but it can be transformed into other forms of energy.</p><h2>How is input energy measured?</h2><p>Input energy can be measured in different units depending on the type of energy. For example, heat energy is measured in joules, while electrical energy is measured in watts. The amount of input energy can be calculated using specific equations and formulas.</p>

What is input energy?

Input energy refers to the energy that is put into a system or process to make it work. It can come in various forms such as heat, electricity, or mechanical energy.

Where does the input energy go?

The input energy can be converted into different forms of energy and used for various purposes. It can be transformed into kinetic energy, potential energy, or stored as chemical energy.

What factors determine where the input energy goes?

The distribution of input energy depends on the type of system and the laws of thermodynamics. Other factors such as the efficiency of the system and external forces also play a role.

Can input energy be lost?

According to the law of conservation of energy, energy cannot be created or destroyed, only transferred or converted. Therefore, input energy cannot be lost, but it can be transformed into other forms of energy.

How is input energy measured?

Input energy can be measured in different units depending on the type of energy. For example, heat energy is measured in joules, while electrical energy is measured in watts. The amount of input energy can be calculated using specific equations and formulas.

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