Seemingly a contradiction of conservation of energy?

[Phynn]

I've had this question for a while now and I wonder if anyone can make sense of it. It's about two scenarios where the difference between them seems to contradict conservation of energy:

Scenario 1: In a vacuum chamber, there is a robotic arm, a box, a lower platform and a higher platform. At the beginning of the scenario, the box lies on the lower platform. The robotic arm lifts the box from the lower platform onto the higher platform, and that is the end of the scenario. Some electrical energy was turned into heat and some electrical energy was turned into potential energy.

Scenario 2: In the same vacuum chamber, the same robotic arm, box, and platforms are present, but this scenario starts with the box on the higher platform. With the exact same but reversed motion as in scenario 1, the robotic arm lowers the box from the higher platform onto the lower platform. In this scenario, the same amount of electrical energy was used, since the motion of the arm was exactly the same but reversed, but some potential energy was lost instead of gained.

Where did the potential energy in scenario 2 go? The only thing I can imagine is that is was turned into heat, but that would mean there is more total heat at the end of scenario 2. How can that happen if the motion of the robotic arm (the thing that generates heat) is exactly the same? It shouldn't matter that the motion is reversed, because the motion would still require the exact same amount of acceleration in every direction.

What am I missing?

Edit: To clarify: I am not arguing that conservation of energy is actually being broken here. I just want to find out how my reasoning is incorrect/incomplete.

 

[Dale]

In this scenario, the same amount of electrical energy was used, since the motion of the arm was exactly the same but reversed

This is not correct. It takes more electrical energy for a robotic arm to lift a weight than it does to lower the weight.

 

[sophiecentaur]

Where did the potential energy in scenario 2 go?

The best way to go with this is
A. Realise that energy is conserved. You've gotta believe this.
B. Draw a diagram with some arrows and stuff to help the brain.
C. Consider all the energy transfers involved - even the subtle ones. All the motions and the Force times Distances. What's the difference between the Potential Energyies at the start of 1 and stat of 2.? The heat flow is always away from where work is done (unless you have some sort of heat engine in there). Assume that the electric motor is 100% efficient so any heat energy will be generated from within the system.

 

[Ibix]

Think of it this way: when lowering the mass you could not bother with a motor and just use brakes on the joints. Where does the energy go in this case? Put the motor back in and take out the brakes. What's the motor doing now? And where does the energy go?

 

[Phynn]

Thanks for the replies! If I understand correctly, you are saying that there should be more energy consumed in the first scenario, or more heat released in the second scenario. Making the assumption that the motor is 100% efficient is a good suggestion so I will do that. That would mean that no heat is released at all and that more electrical energy should be consumed in the first scenario.

But how is that possible? The accelerations are exactly the same in both motions (only ordered differently) and the motions take the same amount of time to complete which means that they spend the same amount of energy to compensate for gravity.

Edit: I see now that it has to be true that the upward motion consumes more energy, but I still don't understand how that's possible for the reasons above.

 

[sophiecentaur]

Making the assumption that the motor is 100% efficient is a good suggestion so I will do that. That would mean that no heat is released at all and that more electrical energy should be consumed in the first scenario.

You've spotted a problem with my ideas because we are (i think) assuming that the system is stationary at the start and at the end. That implies careful speed control when raising by motor and some braking at the end of the fall. Fact is that your systems need to be specified more accurately if you are going to answer your question.

There are a million similar questions in which there are initial assumptions that render the problem insoluble. In real life, friction often comes to the rescue in practical situations and deals with these things.

The accelerations are exactly the same in both motions

I see you are using velocities and forces in your argument, which is a common approach. However, it's often much more convenient to talk in terms of energy and the transitions between Kinetic and Potential because the time taken is often not important - certainly not for initial ideas.

 

[Phynn]

You've spotted a problem with my ideas because we are (i think) assuming that the system is stationary at the start and at the end. That implies careful speed control when raising by motor and some braking at the end of the fall. Fact is that your systems need to be specified more accurately if you are going to answer your question.

There are a million similar questions in which there are initial assumptions that render the problem insoluble. In real life, friction often comes to the rescue in practical situations and deals with these things.

I see you are using velocities and forces in your argument, which is a common approach. However, it's often much more convenient to talk in terms of energy and the transitions between Kinetic and Potential because the time taken is often not important - certainly not for initial ideas.

You're right, I was assuming that the system is stationary at the beginning and end. I'm not sure what kind of specifications for the system you mean, but I will do my best here:

There is no friction. There is a perfect vacuum. The electric motors are perfectly efficient. The acceleration is perfectly constant, so that there are 3 perfectly horizontal lines in the graph of the acceleration. The platforms can phase in and out of existence at our will so that the motion of the box is only vertical and the robotic arm can just be a piston moving the box up and down. The motion of the box does not overshoot the platforms, so that it is contained in the exact same distance as the distance between the platforms. Again, the motions follow the exact same path but in opposite direction in each scenario. And lastly, no brakes are used: to slow down the electric motors will just accelerate in the direction opposite to the direction of the motion.

Are these the specifications you mean? I realize that this system is not possible in real life, but its consistency with conservation of energy shouldn't be affected (right?).

Lastly, the reason I approach this problem with forces and velocities instead of energy transitions is because that is the approach that makes it look like conservation of energy is being broken here, and I want to resolve that.

 

[AZRN]

I think it still relates to differences in gravitational force. They both initially start with different forces applied to them.The overall net energy would be the same if gravity was not involved.

 

[PeroK]

Thanks for the replies! If I understand correctly, you are saying that there should be more energy consumed in the first scenario, or more heat released in the second scenario.

If you do weights at the gym, then you'll know that it's a lot harder to lift a weight than to lower it. And that you're muscles get tired from the lifting and hot from the lowering.

You could forget your hypothetical robot arm and do some bench presses!

 

[AZRN]

This is not correct. It takes more electrical energy for a robotic arm to lift a weight than it does to lower the weight.

But he still has to lift the object in order to move it down. And also since it is a controlled mechanical move, there is extra energy spent resisting gravity. I think this resistance adds also adds more energy spent to the equation?

 

[PeroK]

But he still has to lift the object in order to move it down. And also since it is a controlled mechanical move, there is extra energy spent resisting gravity. I think this resistance adds also adds more energy spent to the equation?

Perhaps you need to try some bench presses.

Technically, when lowering a weight under control you are doing negative work on the weight and are absorbing energy. If your muscles were spring-like, you could store energy from lowering a weight and use that to lift it again.

 

[Phynn]

Perhaps you need to try some bench presses.

Technically, when lowering a weight under control you are doing negative work on the weight and are absorbing energy. If your muscles were spring-like, you could store energy from lowering a weight and use that to lift it again.

If I understand electric motors correctly, it works different for an electric motor: acceleration (or starting to move the weight upward) consists of magnetically pushing and pulling wires in the electric motor, and deceleration (or stopping the downward movement) works exactly the same. The only difference with deceleration is that the wires are pulled and pushed at opposite moments (so pushed and pulled instead of pulled and pushed, so to say). Because the electric motor is essentially doing the same while accelerating and decelerating, it shouldn't require different amounts of energy.

This is also why I chose electric motors in the scenario instead of muscles or springs or something.

 

[PeroK]

I know nothing about electric motors, but work done is force times displacement, where force and displacement are vectors. Technically, for more than one dimension it's the scalar product.

The motor is certainly not doing the same work in each case. There is no issue with energy conservation from the perspective of forces, displacements and mechanical energy.

How that relates to energy consumption by the motor is a different matter. Any paradox here, I suggest, is a result of your not understanding how a motor works.

 

[A.T.]

Making the assumption that the motor is 100% efficient is a good suggestion

No, that is not a good assumption at all here.

If a motor is used to lower a weight slower than free fall, it is acting as a brake. So it is doing negative work on the weight, while also consuming electric energy, thus it has less than 0% efficiency. It is dissipating the potential energy of the weight and the supplied electric energy as heat.

 

[Dale]

But he still has to lift the object in order to move it down. And also since it is a controlled mechanical move, there is extra energy spent resisting gravity. I think this resistance adds also adds more energy spent to the equation?

No. In fact, it is the opposite. Let’s look at the math.

For an ideal motor the torque is proportional to the current and the angular velocity is proportional to the voltage. For simplicity we can approximate that as the force is proportional to the current and the velocity is proportional to the voltage.

When you reverse the motion the velocity reverses, but the gravitational force does not reverse. For the motion to reverse, the voltage must reverse $$V\rightarrow -V$$ But since the gravitational force does not reverse, the current to overcome gravity does not reverse $$I \rightarrow I$$

Now, electrical power is current times velocity so $$P=IV \rightarrow I(-V)=-P$$ So the part of the power due to the load from gravity changes sign, when the path is reversed. When the arm lowers the object it is acting as a generator, not as a motor. A motor and a generator are the same device, just run in reverse. A motor takes electrical power and converts it into mechanical power, and a generator takes mechanical power and converts it into electrical power. But it is the same device, a stator, a rotor, a magnet, and a coil.

 

[Dale]

But he still has to lift the object in order to move it down. And also since it is a controlled mechanical move, there is extra energy spent resisting gravity.

No. None of this is correct. Think of a hydroelectric power plant. It literally produces electrical power to run a city precisely by “resisting gravity”. It does not cost electrical power, it produces electrical power.

 

[A.T.]

When the arm lowers the object it is acting as a generator, not as a motor.

That is possible, but I don't think all robotic arms use regenerative braking by default. It depends on the technical details.

 

[Dale]

That is possible, but I don't think all robotic arms use regenerative braking by default. It depends on the technical details.

The assumption was an ideal motor, which does indeed act as an ideal generator when run in reverse.

 

[A.T.]

The assumption was an ideal motor, which does indeed act as a generator when run in reverse.

True. But is this assumption realistic for how motors usually act when lowering a weight? For example a crane cable motor. Or do they rather act as described in post #14?

 

[Gordianus]

We have millions of electric cars that rely on regenerative braking, just the topic of the thread.

 

[AZRN]

Perhaps you need to try some bench presses.

Technically, when lowering a weight under control you are doing negative work on the weight and are absorbing energy. If your muscles were spring-like, you could store energy from lowering a weight and use that to lift it again.

That energy transfers back into the machine/arm. The original question was ""where is the extra potential energy going during scenario 2".

 

[A.T.]

We have millions of electric cars that rely on regenerative braking, just the topic of the thread.

True, but you don't need to invoke regenerative braking to explain where the energy went to. It can be simply dissipated into heat, not just by frictional brakes, but also by a motor itself.

 

[Lnewqban]

If I understand electric motors correctly, it works different for an electric motor: acceleration (or starting to move the weight upward) consists of magnetically pushing and pulling wires in the electric motor, and deceleration (or stopping the downward movement) works exactly the same. The only difference with deceleration is that the wires are pulled and pushed at opposite moments (so pushed and pulled instead of pulled and pushed, so to say). Because the electric motor is essentially doing the same while accelerating and decelerating, it shouldn't require different amounts of energy.

This is also why I chose electric motors in the scenario instead of muscles or springs or something.

Welcome!

In this case, you are using the magnetic field in the electric motor as a brake, just to slow down the free fall of the box from the higher platform onto the lower one.

That brake effect could transform most, but not all (due to real life inefficiencies, like Eddy currents), of the potential energy of the box into heat or into chemical reaction inside a battery.

At most, the robotic arm is only needed to produce the horizontal displacement of the box out and onto the platforms.

A regular crane uses its engine to rise a load, and only brakes to lower it.

I would repeat your experiment using a horizontal cylinder with the box fixed to its wall and rotating about a fixed axis or shaft.

I would need to input mechanical energy into the shaft in order to move the box from the lowest natural location to the zenith.

I would only need to let go the shaft for the gained potential energy of the box to relocate it to the lowest point again.

I fully agree with post 13 above.

 

[Dale]

But is this assumption realistic for how motors usually act when lowering a weight?

It is realistic. The regenerative braking system in any of the millions of hybrid vehicles on the road uses the same device to act as a motor when accelerating a car and as a generator when decelerating the car. When a hybrid car is driven up hill the battery discharges, reverse the motion and the battery charges. Anyone with a hybrid in a hilly region can run this experiment and confirm.

Motors that do not behave this way are not ideal

 

[Gordianus]

I thought dissipative processes were ruled out.

 

[Dale]

I thought dissipative processes were ruled out.

They were. This is an idealized scenario.

There is no friction. There is a perfect vacuum. The electric motors are perfectly efficient.

 

[A.T.]

It is realistic. The regenerative braking system in any of the millions of hybrid vehicles...

I was asking in the sense of how common is it for robotic arms to have regenerative braking, not whether it is possible to have it.

 

[Dale]

I was asking in the sense of how common is it for robotic arms to have regenerative braking, not whether it is possible to have it.

It doesn’t matter how common it is. The OP specified an idealized scenario.

The OP’s simplification is reasonable and didactically helpful. The idealized scenario they suggest can be well approximated by existing devices, and making that simplification helps the OP focus on what is confusing them.

 

[A.T.]

The only thing I can imagine is that is was turned into heat, but that would mean there is more total heat at the end of scenario 2.

Given all the other constraints in your original post (same electric energy input), that is the only possible answer.

How can that happen if the motion of the robotic arm (the thing that generates heat) is exactly the same? It shouldn't matter that the motion is reversed, because the motion would still require the exact same amount of acceleration in every direction.

Work is a signed quantity so the direction of motion vs direction of applied force does matter. When raising the weight the arm is doing positive work on the weight. When lowering it slower than free fall, it is doing negative work on the weight, so it is absorbing energy.

If that energy is not recovered (due to your same electric energy input constraint) it must be dissipated as heat, along with all the electric energy.

 

[A.T.]

The OP specified an idealized scenario.

The original post didn't assume ideal motors, but rather same electric energy input. One cannot have both those constraints at the same time. But it's worth to consider the consequences of both when applied separately.

 

[Nugatory]

Lastly, the reason I approach this problem with forces and velocities instead of energy transitions is because that is the approach that makes it look like conservation of energy is being broken here, and I want to resolve that.

This approach is generally discouraged when we just want an answer because it’s a lot more work than just using conservation of energy - we have to understand and properly analyze every detail of the machinery we’re using, and that’s hard and error prone. (Indeed, this is why we value and teach energy conservation - it makes so many problems easy).

However, here you don’t “just want an answer”, you are trying to see how and why the painfully detailed analysis ends up producing results that obey the energy conservation law. An electrical motor is especially challenging because the energy transfers are not between between stationary windings and rotating armature, they’re between armature and time-varying magnetic field and between time-varying magnetic field and stationary windings. These are the forces we have to consider and doing it right through entire range of motion from rest at bottom to rest at top and then back to rest at bottom is non-trivial.

In fact, it’s so non-trivial that I’m not going to try to do it…. Instead I’m going to wave my hands and tell you that the energy transfer is from the magnetic field to the armature when the armature is moving in the direction that the field is pushing it, and from the armature to the field (and thence back to the battery) when the armature is moving against the field. This is the basic asymmetry that requires that we add energy to lift the weight but take energy out when we’re lowering it.

 

[Phynn]

This approach is generally discouraged when we just want an answer because it’s a lot more work than just using conservation of energy - we have to understand and properly analyze every detail of the machinery we’re using, and that’s hard and error prone. (Indeed, this is why we value and teach energy conservation - it makes so many problems easy).

However, here you don’t “just want an answer”, you are trying to see how and why the painfully detailed analysis ends up producing results that obey the energy conservation law. An electrical motor is especially challenging because the energy transfers are not between between stationary windings and rotating armature, they’re between armature and time-varying magnetic field and between time-varying magnetic field and stationary windings. These are the forces we have to consider and doing it right through entire range of motion from rest at bottom to rest at top and then back to rest at bottom is non-trivial.

In fact, it’s so non-trivial that I’m not going to try to do it…. Instead I’m going to wave my hands and tell you that the energy transfer is from the magnetic field when the armature is moving in the direction that the field is pushing it, and from the armature to the field (and thence back to the battery) when the armature is moving against the field. This is the basic asymmetry that requires that we add energy to lift the weight but take energy out when we’re lowering it.

Thank you! You get me completely and your answer really helped me understand.

I do have one question: it sounds like you are assuming regenerative braking (because of the "and thence back to the battery") but I meant the scenario without regenerative braking. Without regenerative braking, would there be extra heat at the end of scenario 2? And if so, through what mechanism would that extra heat have been generated?

 

[Dale]

The original post didn't assume ideal motors, but rather same electric energy input. One cannot have both those constraints at the same time. But it's worth to consider the consequences of both when applied separately.

Which is why I didn’t feel the need to correct your approach and object to the principles you were trying to teach. I wish you had shown me the same courtesy.

 

[Dale]

I meant the scenario without regenerative braking. Without regenerative braking, would there be extra heat at the end of scenario 2? And if so, through what mechanism would that extra heat have been generated?

Then your scenario is no longer the idealized one you specified in post 7. The mechanism for generating the extra heat depends on the specific non-ideal scenario you have in mind.

You could dissipate heat in the mechanics (e.g. with a friction brake) or in the electronics (e.g. with resistors)

 

[Phynn]

Then your scenario is no longer the idealized one you specified in post 7. The mechanism for generating the extra heat depends on the specific non-ideal scenario you have in mind.

You could dissipate heat in the mechanics (e.g. with a friction brake) or in the electronics (e.g. with resistors)

If you were to dissipate in the electronics, would that mean that you are using resistors to slow down? And does that imply that you can't simply slow down by generating magnetic fields that repel and attract the wires in the electric motors at the right moments to slow down it's rotation?

Or does this question not make any sense? At this point it might be over my head.