Seemingly a contradiction of conservation of energy?

[Dale]

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?

You certainly can, but doing so produces electrical power.

 

[Phynn]

You certainly can, but doing so produces electrical power.

Is that necessarily true? If so, that would explain a lot.

 

[Dale]

Is that necessarily true? If so, that would explain a lot.

Yes, it is necessarily true.

It is just a question of what you do with that power. The electronics to use that power to charge a battery are more complicated than just dumping the power into a resistance. And also don’t forget that the motor/generator has its own internal resistance. The power is necessarily produced, but may be dissipated by the motor/generator internal resistance

 

[Lnewqban]

... 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?

See how the most common types of motors used in robotic arms work, and hold, and change angular positions:
https://robocraze.com/blogs/post/types-of-motors-used-in-robotics

I have edited post 23 above adding two diagrams for you.

 

[kered rettop]

I know nothing about electric motors.

Fortunately I do. Electric motors take more current when they exert more torque. That's all the OP needed to know.

 

[kered rettop]

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?

Without regenerative breaking, the weight just falls to the floor.

 

[kered rettop]

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.

Fair point. You can set the circuit up to dissipate the energy or to return some, ideally all, back to the power supply. Or a mixture, of course. As Dale explicitly stated in post #38.

 

[Nugatory]

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?

Yes, I was assuming regenerative braking because that’s how idealized frictionless and lossless electric motors behave. If we have a non-ideal motor there will be extra heat at the end of scenario 2, produced mostly by resistive heating as eddy currents flow through the machine.

 

[russ_watters]

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.

And elevators. My first thought of the OP's scenario was that it's a cumbersome elevator.

 

[kered rettop]

Yes, I was assuming regenerative braking because that’s how idealized frictionless and lossless electric motors behave. If we have a non-ideal motor there will be extra heat at the end of scenario 2, produced mostly by resistive heating as eddy currents flow through the machine.

Sorry to nit-pick, but eddy current losses should be quite small. The biggest resistive loss is usually in the windings. Same final effect of course, but what would a thread on Physics Forums be without some smarty-pants keeping a mentor on his toes?

 

[jbriggs444]

If we envision a non-ideal scenario such as a motor driving a worm gear, it may take electrical power both to drive the weight upward and to allow the weight to come back downward. Think about an old style screw jack lifting your car and then letting it back down.

Even though it takes non-zero work both ways, it takes more torque to advance the screw against a resisting weight and less torque to unscrew and let it back down. All other things being equal, you will consume more electrical power on the way up than on the way back down.

 

[Nugatory]

Sorry to nit-pick, but eddy current losses should be quite small. The biggest resistive loss is usually in the windings. Same final effect of course, but what would a thread on Physics Forums be without some smarty-pants keeping a mentor on his toes?

yep, you're right (or at least I was wrong with that bit)... typing quickly after I had made the main pint in the previous post.... And I did say the process was "error-prone", did I not?

 

[kered rettop]

yep, you're right (or at least I was wrong with that bit)... typing quickly after I had made the main pint in the previous post.... And I did say the process was "error-prone", did I not?

You may have revealed more than you intended with that typo!

 

[kuruman]

All this talk about ideal and non-ideal motors makes my head spin while regenerative braking makes me think "hold on a moment." How and where does the motor get the mechanical energy that is needed to raise the weight? I suggest scenario 3 in between the ones suggested by OP:

Scenario 3: The robotic arm is holding the box at fixed height for a length of time $t$. There is no raising and no lowering of the mass. Naturally, a feedback mechanism is needed to keep the mass fixed in position. If more mass is added to the box, something has to "work harder" to keep the increased mass in place. The input power to the motor has to increase for no apparent gain in mechanical energy.

My point is that, before one begins to argue where the mechanical energy goes when the robotic arm raises or lowers the mass, one has to understand where the mechanical energy comes from and where it goes when the mass is kept in place against gravity, ohmic losses notwithstanding.

A conceptually simple model to visualize electrical energy needed to keep a mass in place against gravity is a current-carrying wire segment of mass $m$ levitating in an external magnetic field. The magnetic force opposing gravity is $\mathbf{F_M}=I\mathbf{L}\times\mathbf{B}$. The current needed would be $I=\dfrac{mg}{BL}.$ An actual device that uses this idea is the current balance.

In this simple case, if the mass is doubled, twice as much current will be needed through the wire to keep it in place. That means 4 times as much $I^2R$ loss which is a separate issue from mechanical energy and does not address where (or in what form) the change in mechanical energy $mg\Delta h$ appears.

In very general terms, this conceptual electric motor takes in electrons at high (mechanical) potential energy and returns to the power company electrons at low (mechanical) potential energy while doing mechanical work and dissipating some of that mechanical energy as heat. The reference point of mechanical energy is the energy needed to keep the mass at fixed height for time interval $T$. Call it $E_0=N_0e\Delta V$ where $N_0$ is the number of charges returned to the power company at lower potential energy and $\Delta V$ the voltage drop across the levitating wire.

In Scenario 1 the current is adjusted in such a way as to raise the wire by $\Delta h$ in time $T$ and have it be at rest at the higher level. In that time interval the energy needed to do this is $E_{up}=N_{up}~e\Delta V$. Of course $N_{up} > N_0.$ The positive change in energy intake is due to the work $mgh$ that the electrical force does on the mass against gravity, the mechanical work needed to kill whatever kinetic energy the mass has acquired when it reaches the desired new height plus any new dissipative losses that are incurred as a result of the change in height. We can write $\Delta N =\dfrac{mgh}{e\Delta V}$ to represent the number of charges that contributed to convert electrical potential energy to gravitational potential energy. Raising the mass to a higher level requires $\Delta N$ more electrons to lose potential energy than just having the mass sit at fixed height in Scenario 3.

In Scenario 2 the current is adjusted in such a way as to lower the wire by $\Delta h$ in time $T$ and have it be at rest at the lower level. In that time interval the energy needed to do this is $E_{down}=N_{down}~e\Delta V$. The number of low potential energy electrons $N_{down}$ is less than $N_0$ by $\Delta N =\dfrac{mgh}{e\Delta V}$. In other words, the decrease in gravitational potential energy $\Delta U_g=-mgh$ "appears" as a decrease in electrical potential energy $\Delta U_g=-\Delta N~ e\Delta V$ in time $T$ at the source and a lower electricity bill. Using less power might be construed as "producing power".

This ended up being longer than planned, so I stop here. I don't know much about motors either.

 

[Vanadium 50]

You all got caught up in the Perpetual Motion Machine trap. "If I just add one more complication...." (This is one reason such threads usually get closed immediately).

Replace the setup with two weights on one pulley and zero morots. One goes up, one goes down. Ir should be obvious that the energy gained on one side comes from the other.

Now you can add all the bells and whistles,.

 

[jrmichler]

Where did the potential energy in scenario 2 go?

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.

This is about servomotors and perceptions. To oversimplify, there are two types of servomotors:
1) The little low voltage motors that the Arduino people play with,
and
2) Industrial servomotors.

Small low voltage motors tend to have low efficiency, and tend to be used in applications where friction is a large percentage of the total load. In these systems, it is possible that friction force could be larger than gravity force. There will be little, if any, power regenerated on a down move. In this case, the perception that electricity is used for both up and down moves is correct.

Industrial servomotors are different. They are more powerful, more efficient, and are used in applications where friction force is small compared to acceleration and gravity forces. When there is deceleration torque where the load is driving the servomotor, the motor acts as a generator and pushes energy back to the drive. Depending on the drive, that energy is either pumped back into the powerline, or absorbed by capacitors on the drive DC buss. This is well known to industrial servomotor manufacturers, and calculation of the maximum regeneration energy is an important part of system design. This is a case where perception is misleading, and it is critically important to do the calculations and trust the results.

I was part of an application that required moving a 900 lbs mass vertically about 36 inches in slightly over 500 milliseconds. The acceleration for both the up and down moves was 500 in/sec$^2$. The motion profile was a simple triangular velocity profile, where the acceleration was a constant 500 in/sec$^2$ for halfway, followed by equal acceleration in the opposite direction to bring it to a stop. Yes, it was being pushed down faster than gravity for the first half of the down move. The deceleration torque during the last half of the down move regenerated power into the drive. Since the regeneration energy was a known amount of energy, and would be followed by hard acceleration, it was only necessary to put enough capacitance on the drive DC buss to absorb that energy without exceeding the maximum DC buss voltage.

 

[kered rettop]

This is about servomotors and perceptions. ...

Yes to all of that. But the OP went to extreme lengths to describe an ideal (frictionless) case, not realising that they had overlooked a very fundamental part of the system. Which sent them off on a wild goose chase looking for the missing energy in the dynamics - acceleration etc - and unaccounted-for kinetic energy dissipated when the mass comes to rest. So I'd like to be sure they understand that motor losses and friction are actually just improvements on the basic model, after the fundamental flaw has been corrected. That's not a criticism of your post, of course, just putting it into the context of fine-tuning the model. I did like the examples, especially the reference to Arduino people. I immediately saw them as little elves posing on their microchips. That probably says more about me than the physics.

addendum:
I just looked up the meaning of "Arduino". Apparently, it means "little helper". My subconscious must be smarter than I thought!