Electric motor problem -- Power to accelerate a toy train

In summary, the conversation discusses the relationship between acceleration, power, and time in regards to a train's speed and energy usage. The experts conclude that if the acceleration is not constant, the power required will be larger. They also mention the importance of finding the most efficient scheme for transmitting energy in a fixed amount of time.
  • #1
ChiralSuperfields
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Homework Statement
Pls see below
Relevant Equations
Pls see below
For part (b),
1677189254531.png

The solution is,
1677189269208.png


However, if the acceleration was not constant during the 21 ms, then would the power required be larger?

I believe the average power required would be larger because if the train started off at a lower speed and then speed up very rapidly towards the end reaching the same speed, a very large force would have to be applied.

Many thanks!

[post edited to fix some typos]
 
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  • #2
If the acceleration is not constant, then what does that tell you about the applied power ?

Also, please find better translation software.
 
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  • #3
hmmm27 said:
If the acceleration is not constant, then what does that tell you about the applied power ?

Also, please find better translation software.
Thank you for your reply @hmmm27!

If the acceleration is not constant then the power will be a function a time.

Also what do you mean translation software?

Many thanks!
 
  • #4
One can easilly find the minimum energy supplied over the 21.0ms: that is ths final KE of the train (which you can calculate). Any inefficiencies will waste energy.
The question then becomes what scheme will transmit that energy at minimum (max power usage) over a fixed time. It should not take long to figure that out. What is your answer?
 
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  • #5
Callumnc1 said:
Also what do you mean translation software?
Your post that he was replying to has a bunch of typos in it. He may have mistaken those for poor translation, but I'm guessing maybe you didn't proofread it when posting? See (emphasis mine):
Callumnc1 said:
However, could if the acceleration was not constant then the power required would be larger? For example, since the if it started slowly then speed up very rapidly to reach the same speed in the same time it would require a very large force?
 
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  • #6
berkeman said:
Your post that he was replying to has a bunch of typos in it. He may have mistaken those for poor translation, but I'm guessing maybe you didn't proofread it when posting? See (emphasis mine):
Thank you for your reply @berkeman ! Sorry for those typos, I will fix them.

Many thanks!
 
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  • #7
hutchphd said:
One can easilly find the minimum energy supplied over the 21.0ms: that is ths final KE of the train (which you can calculate). Any inefficiencies will waste energy.
Thank you for your reply @hutchphd !
hutchphd said:
The question then becomes what scheme will transmit that energy at minimum (max power usage) over a fixed time. It should not take long to figure that out. What is your answer?
Sorry I don't quite understand what you mean what scheme will transmit energy at minimum over a fixed time.

Also please see my post #1 as my question may be clearer now after fixing some typos.

Many thanks!
 
  • #8
I think I understand your first post (I don't know what the issues were) The question I asked was "if you wish to transmit a bolus of energy in a fixed time limiting maximum power what is the profile in time of the power. I believe I am overthinking an obvious result (turn it up to the value required for the interval), but what is "obvious" is idiosynchratic.
 
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  • #9
hutchphd said:
I think I understand your first post (I don't know what the issues were) The question I asked was "if you wish to transmit a bolus of energy in a fixed time limiting maximum power what is the profile in time of the power. I believe I am overthinking an obvious result (turn it up to the value required for the interval), but what is "obvious" is idiosynchratic.
Thank you for your reply @hutchphd !

Sorry, what is a bolus of energy? Is it just a like a small amount?

Also what is what dose 'fixed time limiting max power' mean sorry?

Many thanks!
 
  • #10
Sorry I was being pedantic. "bolus" just means a known amount in a finite time. If you wish to deliver that at minimun peak power it is not difficult to see that a constant value for power is the answer. So are you good?
 
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  • #11
hutchphd said:
Sorry I was being pedantic. "bolus" just means a known amount in a finite time. If you wish to deliver that at minimun peak power it is not difficult to see that a constant value for power is the answer. So are you good?
Thank you for your reply @hutchphd ! No worries!

I am still a bit confused.

However, if the acceleration was not constant during the 21 ms, then would the power required to reach the same speed be larger?

Many thanks!
 
  • #12
As you know, varying the power varies the acceleration. You could apply more power, less power, even negative power variously over the time period and end up with the same, greater, or lesser end velocity.
 
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  • #13
hmmm27 said:
As you know, varying the power varies the acceleration. You could apply more power, less power, even negative power variously over the time period and end up with the same, greater, or lesser end velocity.
Thank you for your reply @hmmm27!

Are you saying that it dose not matter whether there is acceleration during the time interval, only the initial and finial velocities determine the power. I guess the power equation dose not depend on acceleration.

##P = F \cdot v##

Many thanks!
 
  • #14
Can you show how the back-of-book solution got the answer ?

After that, explain how you think that power equation relates to the information given in the problem.
 
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  • #15
hmmm27 said:
Can you show how the back-of-book solution got the answer ?

After that, explain how you think that power equation relates to the information given in the problem.
Thank you @hmmm27 ! Here it is,
1677214262016.png

They just used the work energy theorem to find the average power.

Many thanks!
 
  • #16
Callumnc1 said:
I believe the average power required would be larger because if the train started off at a lower speed and then speed up very rapidly towards the end reaching the same speed, a very large force would have to be applied.
The question as posed can be answered as "zero". What they should have asked for is the minimum value of the average power. Since ##P_{avg}=\frac{\Delta E}{\Delta t}##, it does not then matter how the power and acceleration vary over the interval.

So why might the average power be a bit more ?
 
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  • #17
haruspex said:
What they should have asked for is the minimum value of the average power.
I was hoping they were asking for minimum value of "peak trainpower rating" which could accomplish this feat. Slightly more interesting
 
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  • #18
hutchphd said:
I was hoping they were asking for minimum value of "peak trainpower rating" which could accomplish this feat. Slightly more interesting
The whatnow ?
 
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  • #19
What is the minimally powered engine (peak power) that could have done this. Same answer but you have to figure out the power is constant.
 
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  • #20
haruspex said:
The question as posed can be answered as "zero". What they should have asked for is the minimum value of the average power. Since ##P_{avg}=\frac{\Delta E}{\Delta t}##, it does not then matter how the power and acceleration vary over the interval.

So why might the average power be a bit more ?
Thank you for your reply @haruspex !

I agree that they should have been clearer and said the minimum average power.

How can it be answered as zero?

I think average power could be a bit more due to not all the work going into kinetic energy due to non-conservative forces acting.

Many thanks!
 
  • #21
hutchphd said:
I was hoping they were asking for minimum value of "peak trainpower rating" which could accomplish this feat. Slightly more interesting
hmmm27 said:
The whatnow ?
hutchphd said:
What is the minimally powered engine (peak power) that could have done this. Same answer but you have to figure out the power is constant.
Thank you for your replies @hutchphd and @hmmm27 !
 
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  • #22
Callumnc1 said:
How can it be answered as zero?
Because it could have been no power for some of the time and lots at other times.
Callumnc1 said:
if the acceleration was not constant during the 21 ms, then would the power required be larger?
As has been noted, the average power does not depend on the pattern of acceleration, only on the speed reached and the time taken. But for minimum peak power, as @hutchphd notes, you certainly do not want constant acceleration. Because P=Fv=mav, as it gains speed you need more power for the same acceleration.
 
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  • #23
haruspex said:
Because it could have been no power for some of the time and lots at other times.
If finial KE = inital KE?, how would that be possible?
haruspex said:
As has been noted, the average power does not depend on the pattern of acceleration, only on the speed reached and the time taken. But for minimum peak power, as @hutchphd notes, you certainly do not want constant acceleration. Because P=Fv=mav, as it gains speed you need more power for the same acceleration.
Thank you for your reply @haruspex !

Sorry, what is minimum peak power?

Many thanks!
 
  • #24
Callumnc1 said:
If finial KE = inital KE?, how would that be possible?
No, initial and final KE as before, but all the work being done in less than the allowed time.
Do nothing for 1ms, then go from rest to 0.620m/s in 20ms.
Callumnc1 said:
what is minimum peak power?
If the maximum power the engine produces during the acceleration is ##P_{peak}##, what is the minimum possible value of that to achieve the given speed in the given time?
 
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  • #25
haruspex said:
If the maximum power the engine produces during the acceleration is ##P_{peak}##, what is the minimum possible value of that to achieve the given speed in the given time?
Is that a calculus of variations problem? It seems like you would have to find some ##P(t)## with a bounded area = ##\frac{1}{2}mv^2##, and then attempt to optimize its peak? Quite the problem indeed @hutchphd.
 
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  • #26
erobz said:
Is that a calculus of variations problem? It seems like you would have to find some ##P(t)## with a bounded area = ##\frac{1}{2}mv^2##, and then attempt to optimize its peak? Quite the problem indeed @hutchphd.
No, nothing so advanced. If the peak power available is ##P_{peak}## then the most energy you can get out of it in time t is ##P_{peak}t##, and you can only get that by operating constantly at that power. So for a given KE achieved, the minimum ##P_{peak}## is ##E_K/t##.
 
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  • #27
haruspex said:
No, initial and final KE as before, but all the work being done in less than the allowed time.
Do nothing for 1ms, then go from rest to 0.620m/s in 20ms.

If the maximum power the engine produces during the acceleration is ##P_{peak}##, what is the minimum possible value of that to achieve the given speed in the given time?
Thank you for your reply @haruspex ! I will reply to the thread when I have thought about it a bit more.

Many thanks!
 
  • #28
erobz said:
Is that a calculus of variations problem? It seems like you would have to find some ##P(t)## with a bounded area = ##\frac{1}{2}mv^2##, and then attempt to optimize its peak? Quite the problem indeed @hutchphd.
haruspex said:
No, nothing so advanced. If the peak power available is ##P_{peak}## then the most energy you can get out of it in time t is ##P_{peak}t##, and you can only get that by operating constantly at that power. So for a given KE achieved, the minimum ##P_{peak}## is ##E_K/t##.
Thank you for you replies @erobz and @haruspex !
 
  • #29
haruspex said:
No, nothing so advanced. If the peak power available is ##P_{peak}## then the most energy you can get out of it in time t is ##P_{peak}t##, and you can only get that by operating constantly at that power. So for a given KE achieved, the minimum ##P_{peak}## is ##E_K/t##.
Ahh, so "peak power" is a bit misleading. It's actually just constant power which gives you this. Thanks!
 
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