Electric motor problem -- Power to accelerate a toy train

[ChiralSuperfields]

Homework Statement

Pls see below

Relevant Equations

Pls see below

For part (b),

The solution is,

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]

 

[hmmm27]

If the acceleration is not constant, then what does that tell you about the applied power ?

Also, please find better translation software.

 

[ChiralSuperfields]

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!

 

[hutchphd]

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?

 

[berkeman]

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):

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?

 

[ChiralSuperfields]

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!

 

[ChiralSuperfields]

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 !

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!

 

[hutchphd]

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.

 

[ChiralSuperfields]

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!

 

[hutchphd]

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?

 

[ChiralSuperfields]

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!

 

[hmmm27]

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.

 

[ChiralSuperfields]

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!

 

[hmmm27]

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.

 

[ChiralSuperfields]

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,

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

Many thanks!

 

[haruspex]

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 ?

 

[hutchphd]

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

 

[hmmm27]

I was hoping they were asking for minimum value of "peak trainpower rating" which could accomplish this feat. Slightly more interesting

The whatnow ?

 

[hutchphd]

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.

 

[ChiralSuperfields]

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!

 

[ChiralSuperfields]

I was hoping they were asking for minimum value of "peak trainpower rating" which could accomplish this feat. Slightly more interesting

The whatnow ?

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 !

 

[haruspex]

How can it be answered as zero?

Because it could have been no power for some of the time and lots at other times.

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.

 

[ChiralSuperfields]

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?

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!

 

[haruspex]

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.

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?

 

[erobz]

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.

 

[haruspex]

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 $E_K/t$.

 

[ChiralSuperfields]

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!

 

[ChiralSuperfields]

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 $E_K/t$.

Thank you for you replies @erobz and @haruspex !

 

[erobz]

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!