Why is the work done double its expected value? (conveyer belt)

In summary: I think that is a little clearer. The displacement of the box on the belt is doubled. This occurs due to the inelastic collision between belt and box. The amount of energy dissipated (in friction between box and belt) is fixed regardless of the acceleration profile.
  • #36
jbriggs444 said:
The top of the belt may or may not accelerate with ##F=ma## (or ##F=\frac{dp}{dt}##). You get an effective mass and an effective momentum regardless.

In particular, for a free-wheeling belt (no motor), the effective mass will be the mass of the belt plus a mass contribution from the angular momentum of the rollers if they are not ideal. If you are being picky, the belt material that rolls around the end has an increased effective mass due to rotation as well. For a thin, continuously flexible belt the rotational contribution from the belt material at the ends is negligible. For a belt constructed of thick plates (like a Caterpillar tractor), it may not be negligible.

If it does not accelerate at all, that's an effective infinite mass and infinite momentum. You may need to exercise due care cancelling infinities in the math.

If it does accelerate, that's a finite effective mass.

You can use this same analytical trick when considering the motion of two weights hung by a string over a pulley. You have the net force of gravity ##g(m_1 - m_2)## and the total mass of the pair ##m_1 + m_2##. The effective mass in this case is equal to the total mass. Assuming an ideal pulley and ideal string.

If you are adventurous, you can use the trick with more complex pulley arrangements, multiplying or dividing the effective mass of each weight based on the pulley setup.

Edit: Let's try explaining things another way

Pretend that we have a long flat plate of mass ##M## moving at some speed ##v##. As long as we do not run out of plate, the relevant behavior of this long flat plate will be identical to the top surface of a conveyor belt of total mass ##M## whose surfaces are both moving at speed ##v##.

Since the relevant behavior of the two is identical, we can reason about the behavior of the flat plate even though the reality is a looping belt. Yes, that does mean that when we speak about the "momentum of the belt", we are jumping scenarios and talking about the equivalent situation rather than the literal physical truth.
It didn't dawn on me until you were talking about a sphere in the other post. Aren't we really just conserving angular momentum?

Maybe I’m behind the 8 ball in this…
 
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  • #37
erobz said:
Aren't we really just conserving angular momentum?
That could be the better way to look at it. Typically, the belt consists mostly of two equal masses, M/2, one above the rollers and one below. If the separation is 2s, we can take angular momentum about the midpoint to get Mvs. This has the advantage that the unbalanced horizontal forces exerted by the end rollers have no torque. The landing crate has to acquire angular momentum mvs.
 
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  • #38
haruspex said:
I cannot work out what you mean by that. By "rack and pinion" I thought you meant like gear teeth on the base of the crate engaging similar on the belt.

1684375727122.png
Thanks Britanica. https://cdn.britannica.com/77/3677-004-BCF27DE8/Rack-pinion-Gear-wheel-cogwheel.jpg?s=1500x700&q=85 By long movable rack and fixed axis pinion which is separable/attachable to rack, I would discuss on linear motion of rack and crete on it with only one axis pinion motor to drive.
 
  • #39
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  • #40
jbriggs444 said:
It is quite generic. The nature of the propulsion of the target does not enter in, except to the extent that the target might slow down temporarily as a result of the interaction. Such a slow down would reduce the energy loss. [See the analysis in #23]
Thanks. Now I perceive in my #30 that process of sharing a same velocity or adhesion of crete with belt ineivitably dissipates kinetic enegy of maximum 1/2 m v^2 where v is speed of belt before the adhesion.
 
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  • #41
In assembling charges the work done by external force is calculated equal to the work done against repulsive force in bringing it from infinity to required distance as asked and answered in this question .
Work Done In Assembling Equal Charges At Corners Of An Equilateral Triangle
(//I didn't ask that question)
 However considering the fact that electrostatic force is an internal force and in the situation behaves the same as friction, Why is the work done not double the calculated value (as there will be heat loss)?
 
  • #42
Aurelius120 said:
 However considering the fact that electrostatic force is an internal force and in the situation behaves the same as friction, Why is the work done not double the calculated value (as there will be heat loss)?
I do not see any analogy to be made between the two situations. There is no loss of energy into heat in the case of assembling three like charges.

However, squinting my eyes just right...

Suppose that we have two of the three charges already pre-assembled and nailed into place so that they cannot fly apart. We figure out how much energy will be required for the third charge to be hurled from infinity and come to rest precisely at its prescribed location. We give the third charge that much velocity and send it on its way.

But then we change our mind about what reference frame to use. We firmly attach the frame with the two charges to a railroad flat car and send the flat car hurtling toward the third charge which is now at rest.

We put a pusher engine behind the flat car and give the engineer instructions to maintain a constant speed, come hell or high water.

The flat car arrives at the location of the third charge and the three charges come to relative rest in the equilateral arrangement. We ask the engineer how much energy he has used. It turns out to be twice the kinetic energy that the third charge would have needed to start with in the originally chosen frame.

You ask how much work was done in assembling the charges. That is equal to the kinetic energy that the third charge would have needed to start with in the originally chosen frame.

The extra energy is required to get the train back up to speed after having been decellerated during the collision.

We were much better off sticking with the original frame where the flat car remains at rest. The analysis is dead simple there. No doubling to worry about. No huge before and after energies to cancel out with each other while trying to calculate the miniscule delta.
 
  • #43
jbriggs444 said:
I do not see any analogy to be made between the two situations. There is no loss of energy into heat in the case of assembling three like charges.

However, squinting my eyes just right...

Suppose that we have two of the three charges already pre-assembled and nailed into place so that they cannot fly apart. We figure out how much energy will be required for the third charge to be hurled from infinity and come to rest precisely at its prescribed location. We give the third charge that much velocity and send it on its way.

But then we change our mind about what reference frame to use. We firmly attach the frame with the two charges to a railroad flat car and send the flat car hurtling toward the third charge which is now at rest.

We put a pusher engine behind the flat car and give the engineer instructions to maintain a constant speed, come hell or high water.

The flat car arrives at the location of the third charge and the three charges come to relative rest in the equilateral arrangement. We ask the engineer how much energy he has used. It turns out to be twice the kinetic energy that the third charge would have needed to start with in the originally chosen frame.

You ask how much work was done in assembling the charges. That is equal to the kinetic energy that the third charge would have needed to start with in the originally chosen frame.

The extra energy is required to get the train back up to speed after having been decellerated during the collision.

We were much better off sticking with the original frame where the flat car remains at rest. The analysis is dead simple there. No doubling to worry about. No huge before and after energies to cancel out with each other while trying to calculate the miniscule delta.
Get it. In this case we are ignoring the work needed to keep the two charges fixed and only considering the energy needed to bring the third there.
 
  • #44
Aurelius120 said:
Get it. In this case we are ignoring the work needed to keep the two charges fixed and only considering the energy needed to bring the third there.
Zero work is required to maintain two charges at a fixed separation or in a fixed location.

The mental image in my head is that we nail the two charges to a piece of plywood and glue the piece of plywood to the floor. As long as the floor stays in place and the plywood does not tear apart, neither the nails nor the glue are transmitting any energy.
 
  • #45
Consider the maximal power transfer theorem (MPT) although it applies Electrical Engineering, it also applies to all physics where power is transferred at maximal power (AFAIK I haven't proven this yet). The requirements are perfectly matched impedance and conjugate reactance if there any i.e. if a spring , there must also be equal damping just as L and C must be equal. This results in an ideal 50% loss in order to transfer the maximum power available. The same applies to mechanical systems where the impedance is friction and both convert the loss to heat. The power in this case is the constant acceleration force due to friction in a bounceless displacement drop to the conveyor. The impedance between is just the shared friction, so there is no mismatch defined.Thus the work or power*time done is twice the output kinetic energy regardless of the shared coefficent of friction ##work ~done=E~=~2P*t~=~ mv^2 ~[J]##

How about optical to electrical energy transfer?
This MPT theorem is also used in photo-voltaic (PV) or Solar panels from current source (high Z) diode arrays into low-Z voltage source batteries or converters and regulated at an operating "Point" in the VI power plot that gives MPT so they might call it the MPPT regulator. Although PV's it may use one of many a max power measure and hunting algorithm with MPT end up having the incremental impedance dV/dI equal to the matched impedance of the PV array and the input impedance of some switched mode converter. If that incremental slope R= dV/dI it not also equal to R= Voc/Isc (open cct/short cct) at that solar level, it is not putting out maximum power. So where is the 50% work lost?How about Nuclear to electrical? .. If it is maximal power transfer it must be only 50% efficient. They don't want to waste heat, but they have to in order to generate as much as possible. i.e. a 1 GW generator output also loses 1 GW into heating, massive amounts of water only a few degrees to a lake or evaporates waste energy into steam. (Well it might not be wasted if heat can be utilized elsewhere) But Candu Reactors just dump heat into Lake Ontario with a controlled conservative temp rise like 10'C.

mass to energy ? ##E= mc^2## Was there a 0.5 lost in this process? ;)
 
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  • #46
TonyStewart said:
mass to energy ? ##E= mc^2## Was there a 0.5 lost in this process? ;)
No. That E is total energy, so there can be nothing lost.
 
  • #47
TonyStewart said:
The same applies to mechanical systems where the impedance is friction and both convert the loss to heat. The power in this case is the constant acceleration force due to friction in a bounceless displacement drop to the conveyor.
You seem to be arguing that the 50% work loss is inevitable when dropping a crate onto a conveyor belt if the time to bring the crate up to speed is to be minimsed.
TonyStewart said:
Consider the maximal power transfer theorem (MPT) although it applies Electrical Engineering, it also applies to all physics where power is transferred at maximal power
I fail to see the analogy. The MPT is about tuning the load to match resistance in the source in order to maximise power in the load. There is no requirement to get the crate up to speed in the minimum time.
To reduce loss, we need some elasticity in the belt. That could avoid any sliding or impact losses. The challenge is to convert the stored elastic energy into crate KE without too much in the way of residual horizontal oscillations.
 
  • #48
haruspex said:
You seem to be arguing that the 50% work loss is inevitable when dropping a crate onto a conveyor belt if the time to bring the crate up to speed is to be minimsed.
Exactly . Just like a drop in charge voltage from one Cap switched to anothe Cap.
The resistance (like friction) only affects the rate of change of a) current for cap or b) acceleration of object on conveyor.
haruspex said:
) I fail to see the analogy. The MPT is about tuning the load to match resistance in the source in order to maximise power in the load. There is no requirement to get the crate up to speed in the minimum time.
Let me clarify.
Matched impedances (source to load) in all circuits that use the Maximum Power Transfer Theorem whether 50 ohms or any resistance will automatically lose 50% of the power generated from source. The shared or matched impedance between objects with different energy available when switched together will lose 50% due to the resistance.

So when transferring power over any time through any lossy friction or resistive channel into a stored energy like kinetic or charged battery a potential energy source like a capacitor voltage or gravitational will automatically lose 50% due to the MPT theorem if viewed as 1 path of resistance.

Overcoming these losses (not part of this question) relies on lossless acceleration to translate from vertical to horizontal such as a frictionless curve slide with the height matched to the land at the conveyor H velocity.

Of course in theory with two capacitors switched together at different voltages AND zero resistance the current rises to infinity as a singularity but as previously shown only half the total energy remains.

Does that make it any better to consider the analogy for MPT in lossy energy transfers?

Of course the loss can be less or more by changing reaction to the vertical stop forces and the change in frictional forces, but in this question the friction was defined as constant.

Other info

Fortunately we do not match impedances for audio power or grid power and the % losses are equal to the ratio of the source to load resistance so the efficiency can be high. In other words Load regulation (loss) is equal to the source/total load impedance ratio and when equal the efficiency is 50%.
 
  • #49
TonyStewart said:
So…
There's no "so" about it. You have not established that the MPT applies in anything like such a general context. Shouting (bold text) does not suffice.
TonyStewart said:
when transferring power over any time through any lossy friction or resistive channel into a stored energy like kinetic or charged battery a potential energy source like a capacitor voltage or gravitational will automatically lose 50% due to the MPT theorem
No, the MPT does not say that. It says that if you wish to maximise the power in the load then you must lose 50%. You can avoid losing so much by making the load resistance, r, greater than the inherent source resistance, R. The fraction lost is then ##\frac R{R+r}<\frac 12##, but the power then the load is then less than it could have been.
In the present context, there is no requirement to maximise useful power (i.e. minimise the time to bring the crate up to speed) or anything else.
TonyStewart said:
Overcoming these losses (not part of this question) relies on lossless acceleration to translate from vertical to horizontal
The issue in this thread has nothing to do with KE of vertical motion. The crate could have been dropped from 1mm.

All that said, it may be true that there is no way to avoid the 50% loss here with a single crate. E.g. if the crate were to land on a pad that slides smoothly on the belt, but which is held in place by an elastic string, the belt would do work ##mv^2## to bring the crate up to speed, with half of that stored in the elastic. If there are more crates, the trick then would be to use that stored energy to help bring the next crate up to speed…
 
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  • #50
@haruspex Thankyou for your indulgence.

We both understand MPT and are saying the same thing. If ##\frac R{R+r}<\frac 12 or >\frac 12## then efficiency changes and power transferred is reduced from maximum.

The drop height and cap resistance are irrelevant in both cases and MPT is still ##\frac 12## of supplied energy. Time, resistance and friction are irrelevant in both cases.

Agreed?

The implications are significant.
Whenever lossy processes exist in converting any energy type to any other energy type at maximal power or energy being transferred, the loss is 50% . (at least)

other:

Many molecular reactions are known to be lossless while others being lossy with endo or exothermic.

BTW All caps is shouting, Bold is just for TL;DR highlights. (if I may be so bold to presume)
 
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  • #51
TonyStewart said:
Whenever lossy processes exist in converting any energy type to any other energy type at maximal power or energy being transferred, the loss is 50% . (at least)
Say what? Motor generator set efficiencies cap out at near 100%, not near 25%.

Or are you just saying that if one stresses a motor generator set well past design limits in order to maximize transmitted power that the efficiency is sure to take a 75% hit?
 
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  • #52
jbriggs444 said:
Say what? Motor generator set efficiencies cap out at near 100%, not near 25%.

Or are you just saying that if one stresses a motor generator set well past design limits in order to maximize transmitted power that the efficiency is sure to take a 75% hit?
Motor generator sets if rigidly coupled are lossless coupling and not lossy. MPT is when impedances are matched and Motor generators are usually much lower impedance (<1%) than their loads. This is performed due to negative feedback which reduces output impedance from loop gain.

Load regulation##= \frac {R_L} {R_L+R_S}= \frac 12 \text{ if } R_S=R_L ##

Commercial power Audio amplifiers are usually 0.1% which means Rs= 0.1% of 8 ohm load for example anf that means Damping factor = 1000.

Grid load regulation is about 5 to 10% max of load then they use tap changers to tune to meet regional specs. ( far-eastern global infrastructures e.g. India =do not meet this spec of 10% V error at service entry)
 
  • #53
TonyStewart said:
Motor generator sets if rigidly coupled are lossless coupling and not lossy.
A motor generator set converts electrical power to mechanical power. That is lossy.
A motor generator set converts mechanical power to electrical power. That is lossy.
The mechanical connection between motor and generator is not the point.
 
  • #54
jbriggs444 said:
A motor generator set converts electrical power to mechanical power. That is lossy.
A motor generator set converts mechanical power to electrical power. That is lossy.
The mechanical connection between motor and generator is not the point.
{ Motor nor Gens nor Motor-Gens } are never operated at MPT or max. power transfer for operation with the load impedance matched to the generator source.

That would be matching the DCR of the coil and would be equivalent to an electronic brake for generators and a mechanical brake for motors to measure brake Hp at any RPM..
 
  • #55
TonyStewart said:
We both understand MPT and are saying the same thing.
Except, that is not what you wrote in post #48.
If you wish to apply the MPT result in this dropped crate scenario you need to explain the analogy. What here corresponds to the maximum power requirement?

Consider this arrangement:
The belt and crate have the same mass. The crate slides smoothly on the belt, but after being dropped it is initially stationary in the ground frame, while the belt continues at speed v. A vertical buffer fixed to the belt strikes the crate perfectly elastically. The crate acquires speed v and the belt comes to a halt. The belt's motor then restores the belt to speed v.
Unless you can find a reason that 50% of the motor's work is wasted, this seems to avoid the losses.
 
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  • #56
haruspex said:
A vertical buffer fixed to the belt strikes the crate perfectly elastically. The crate acquires speed v and the belt comes to a halt. The belt's motor then restores the belt to speed v.
Sorry but I do not follow this. Can you elaborate? How does this buffer work? (If the belt is elastic it will oscillate horizontally)
 
  • #57
hutchphd said:
Sorry but I do not follow this. Can you elaborate? How does this buffer work? (If the belt is elastic it will oscillate horizontally)
au contraire, perfect vertical elasticity creates a vertical oscillator or trampoline.
This modulated loss of traction from friction being on/off vertically only affects time to stop skidding which we have stated is lossless and irrelevant given vertical momentum is assumed lossless.
 
  • #58
TonyStewart said:
au contraire, perfect vertical elasticity creates a vertical oscillator or trampoline.
This modulated loss of traction from friction being on/off vertically only affects time to stop skidding which we have stated is lossless and irrelevant given vertical momentum is assumed lossless.
This does not seem relevant to either the scenario proposed by @haruspex or the incomprehension expressed by @hutchphd.

@haruspex proposes a collision in the horizontal dimension which is perfectly elastic. The box lands on the equally massive belt. We ignore the vertical component of the impact and any vertical oscillation which may ensue. It is irrelevant to the horizontal momentum balance and to the portion of energy which can be attributed to the horizontal component of motion.

The box now sits [horizontally] stationary on the frictionless surface of the moving belt. It collides elastically with a tab standing up vertically from the surface the moving belt. The belt comes to a stop. The box rebounds at the original speed of the belt and slides forward on the frictionless surface.

The motor speeds the belt back up to match the speed of the box.

I agree with @haruspex that this arrangement contrives to avoid all frictional losses.

An elastic collision does not involve an ensuing oscillation. Indeed, no new vibration can result from the collision. In an elastic collision, the total bulk kinetic energy (##\frac{1}{2}m_\text{tot}v_\text{com}^2## for each colliding object) summed after the collision is the same as that summed prior. If any vibration also results, that would mean an increase in total energy. But we do not consider a collision "elastic" if it involves an increase in total energy.
 
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  • #59
Thanks to @jbriggs444 for the description, I now understand the proposal.

I propose a simpler equivalent method:
  1. Have the motor slow down the belt to zero (motor does negative work ##W##)
  2. Drop the box
  3. Speed up the belt and box to v. Motor does positive work ##W+ \frac {mv^2} 2##
  4. Net work by motor is ## \frac {mv^2} 2##
So the motor control is the real difference. Of course one can charge a capacitor without loss using active control as well. Maxwell Demon? '
 
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