Calculating the final velocity of a simple electric train

[Himanshu Singh]

I am currently in grade 12, and for a physics project (a huge project, which gets assigned to you at the start of the semester and needs to be completed just before exams. You think up your project, make a proposal for it and you prove your hypothesis to be right after a presentation and lab report.) I need help with calculating the velocity of a simple electric train. Currently, we are learning gravitational fields and have not learned anything about magnetic fields and what not (we will be in about a week) , so many of the equations and especially variables seem foreign to me.

Here is the video of the simple electric train:
(I need to re-do this but with different variables such as different types of batteries, different gauges of copper wire, more mass etc etc.)

I went through my textbook and had seen where they had made qvb equal to centripetal acceleration and had solved for velocity which resulted in v=qbr/m. Although I am certain that this is incorrect.

I was thinking something along the lines of using w=fd, subbing the force of the magnetic force into f (I do not know what that equation is, to solve for the magnetic force) and then taking the result in joules and subbing it into (result) = mv^2/2 and solving for v. But once again I am not sure if this is right or not.

If anyone can help me with this it would be greatly appreciated, thank you.

Edit: I had calculated the velocity using v=qbr/m using test values (except b, I had measured the mt earlier today)

 

[Himanshu Singh]

So I have found this equation to calculate an electromagnetic force.
FPull= ( n x I ) 2 μ0(A/(2g)^2)
F : Force.
n : Number of turns.
I : Current.
μ0 : permeability of air.
A: Area in m2
g : the gap that is separating the electromagnet and the object.
Once again, if someone can help me with my process to finding the final velocity, it would be very helpful.
Ok, here goes:
So, assuming that I know how to calculate this (or somehow calculate the electromagnetic force), my plan is to (with the current knowledge I know) multiply this force by 2 (since there is one force attracting and the other force repelling) which causes a net force to be directed in one direction (am I wrong here?). Once this is done, I will plug this force into w=fd. The distance would be the total distance that the battery will travel (0.3m-0.6m, i am still thinking how long my "track" would be) and get the work done in joules. After this subtract the value I got in joules by thermal energy (still need to know how to calculate this, as I do not know the coefficient of friction between my two materials). Once I get this net energy, I will substitute it into the kinetic energy equation (mv^2/2). The mass would be the total mass of the battery and I will solve for v. Does this make sense at all, and will it work? Or do I have to take another approach to this? If anyone could help me with this it would be very helpful, as I am stumped and am having a very difficult time trying to solve for the final velocity. Thank you.

I have also noticed that I do not have any B (magnetic field) value within my plan to calculate the final velocity, and I am guessing that I need one.

If anyone can help out it would be greatly appreciated, thank you once again.

 

[haruspex]

using w=fd, subbing the force of the magnetic force into f (I do not know what that equation is, to solve for the magnetic force) and then taking the result in joules and subbing it into (result) = mv^2/2

But what d would you use? The problem is that the electricity continues to do work on the train. Why does it not just go faster and faster?

 

[Himanshu Singh]

But what d would you use? The problem is that the electricity continues to do work on the train. Why does it not just go faster and faster?

The d would be the distance of the entire solenoid from one end to the other

 

[Himanshu Singh]

But what d would you use? The problem is that the electricity continues to do work on the train. Why does it not just go faster and faster?

Although it seems as if the speed of the battery seemed to remain constant, don’t resistive forces cancel it out?

 

[Himanshu Singh]

Anyone?

 

[CWatters]

Yes. If the velocity is constant the acceleration is zero so the net force is zero. That means the propulsion force equals the friction force.

The problem will be calculating the friction force in this set up with any hope of accuracy. The train won't be sliding smoothly but rattling off individual turns of wire.

 

[haruspex]

The d would be the distance of the entire solenoid from one end to the other

Then what you calculate would be the KE gained in one pass through the solenoid. I thought you wanted the steady state speed.

Although it seems as if the speed of the battery seemed to remain constant, don’t resistive forces cancel it out?

Yes, that was my point, and as CWatters writes that means you need to know what the frictional force is.

 

[Himanshu Singh]

Hm, say that I calculate this without any friction, I get the value of the actual train using v =d/t during the experiment, and I use both velocities to calculate the force of friction itself.

Also I used this equation : v = qBr/m to calculate for the velocity (this was given to me by my teacher) and when I plugged all the value's in (q constant) (8.355 mt) (0.075)/(0.35) -> (the mass was a guess, whereas the B value was measured, and the radius is an approximate value, not exact). I ended up getting 2.68m/s. Although when I actually constructed the project and had seen it travel across a very short distance, it seemed to be going much much slower (somewhere around 0.2 m/s to 0.4 m/s, I am assuming this is due to the frictional force correct?) Although I was not using the exact mass or radius. (mass of battery + magnets and the radius of the solenoid).

FPull= ( n x I )^ 2 * μ0(A/(2g)^2)

Also I have a few doubts about the variables of this equation:
n, are these the total amount of coils done, or just the amount of coils from one end of the battery the other.
I, how would this be measured, just with the battery alone or with the battery IN the solenoid
A , Would this be the cross sectional area for the neodyium magnets?
g, the gap that is separating the electromagnet and the object. <-- although as you can tell that there is NO gap whatsoever separating the electromagnet from the solenoid. (would the gap be the distance between the two magnets, or the width of them?)

Again, thank you so much for the responses, it helps me greatly.

 

[CWatters]

Can you quote the source for that equation? Its probably intended to calculate the max spring tension in a relay? In your application g=0 so that's not going to work.

Regarding the equation for the velocity your teacher gave you. I don't recognise that equation but what is r? You appear to have set it to 75mm?

 

[CWatters]

Thinking aloud here...

Basically what you have is a motor. Normally simple DC motors accelerate until the back emf is roughly the same as the applied voltage. _If_ that works for this circuit then we could look to equations like this for the back emf...

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

Vemf= -N dBA/dt

Where N is the number of turns between the magnets. B the magnetic field, A the cross sectional area of the coil.

I'm thinking that N/dt is related to the velocity eg number of turns passed per second?

What do others think of this approach?

Edit: It suggests that spacing out the turns would speed it up and bunching them up would slow it down?

Edit2: The voltage may well not be anything like 1.5V as the battery is virtually shorted.

 

[Tom.G]

Speed from the video with the long coil is [/i]about[/i] 3.5sec for one revolution. It will probably be faster if the coil is in a straight line.

For finding the coil current you don't need it to be coiled, you just need the same electrical length as when it is coiled. If you have access to an Ammeter that should be easy. Without an Ammeter you will need to measure the battery voltage with it connected to the wire. Then you can either measure the resistance if you have an Ohmmeter that reads that low (remember to acocunt for the resistance of the meter leads), or knowing the wire diameter, look up the resistance in a wire table. Hint it is less than 10 Ohms per 1000 ft (305m).

Cheers,
Tom

 

[Himanshu Singh]

Can you quote the source for that equation? Its probably intended to calculate the max spring tension in a relay? In your application g=0 so that's not going to work.

Regarding the equation for the velocity your teacher gave you. I don't recognise that equation but what is r? You appear to have set it to 75mm?

http://info.ee.surrey.ac.uk/Workshop/advice/coils/force.html#sxs

Here is where I had gotten that equation from, as well as a few other posts for forums had mentioned this.

This is one
https://physics.stackexchange.com/questions/53680/force-of-electromagnet-on-piece-of-iron

Also, what is the d supposed to be, and for N/dt I am assuming that I have to physically see it and calculate it when the experiment is on-going?
Vemf= -N dBA/dt (is that d before the variable B a typo?)

And I had asked this question on another forum, they had mentioned that the gap would be approximately the width of one magnet ( or the distance separating the electromagnet from the two other magnets).

Here is the link to that:
https://physics.stackexchange.com/questions/444177/calculating-the-final-velocity-of-a-simple-electric-train-cont/444342#444342

To get the equation my teacher gave me, all we did was take f = qvb and the centripetal force eqn and made them equal to each other. the v's canceled out (leaving only one) and we just isolated for v.

qvb = mv^2/r
qb=mv/r
v = qbr/m

Thank you for your reply

 

[Himanshu Singh]

Speed from the video with the long coil is [/i]about[/i] 3.5sec for one revolution. It will probably be faster if the coil is in a straight line.

For finding the coil current you don't need it to be coiled, you just need the same electrical length as when it is coiled. If you have access to an Ammeter that should be easy. Without an Ammeter you will need to measure the battery voltage with it connected to the wire. Then you can either measure the resistance if you have an Ohmmeter that reads that low (remember to acocunt for the resistance of the meter leads), or knowing the wire diameter, look up the resistance in a wire table. Hint it is less than 10 Ohms per 1000 ft (305m).

Cheers,
Tom

So, I would just attach the copper wire (the length being the amount used in the coil) and connect both sides to the ends of a 1.5 v battery, essentially making a circuit and then cutting in between that circuit and then taking an ammeter to measure the current yes?

like this video.



Thank you for your reply.

 

[Windadct]

I just scanned through this - but where is the energy coming from? It is an unlimited source? What determines the current "I" in the equations above.

Also - the "train" is not on a frictionless track - correct? How could you experimentally estimate the friction it sees?

 

[Himanshu Singh]

I just scanned through this - but where is the energy coming from? It is an unlimited source? What determines the current "I" in the equations above.

Also - the "train" is not on a frictionless track - correct? How could you experimentally estimate the friction it sees?

The energy being calculated it coming from the equation FPull= ( n x I )^ 2 * μ0(A/(2g)^2). (Which I still need to know how to use) The Energy is coming from the electromagnet.

The current would be the current flowing through the solenoid.

I am aware the train is not on a frictionless track and I do not know how to calculate it, I will just calculate the actual velocity when the experiment is done (using v = d/t) and then I will just compare the velocity calculated and work backwards to find the thermal energy.

 

[Windadct]

Really ? The electromagnet is the source of the ENERGY?

 

[Himanshu Singh]

Speed from the video with the long coil is [/i]about[/i] 3.5sec for one revolution. It will probably be faster if the coil is in a straight line.

For finding the coil current you don't need it to be coiled, you just need the same electrical length as when it is coiled. If you have access to an Ammeter that should be easy. Without an Ammeter you will need to measure the battery voltage with it connected to the wire. Then you can either measure the resistance if you have an Ohmmeter that reads that low (remember to acocunt for the resistance of the meter leads), or knowing the wire diameter, look up the resistance in a wire table. Hint it is less than 10 Ohms per 1000 ft (305m).

Cheers,
Tom

Another question, I am assuming that the circuit for the experiment would be a series circuit. Doesn't current stay the same throughout the circuit? Or am I wrong here and will have to measure it by using the amount of copper wire used?

 

[Himanshu Singh]

Really ? The electromagnet is the source of the ENERGY?

As far as I know, yes. I've never really worked with energy in this situation (have only worked with kinetic, potential, elastic etc.) , so I am assuming it is because of the equation I found.

 

[Windadct]

Make a diagram of the system... show the coil, magnets, etc... what does each element do. Other than a magnet / electromagnet - what else is going on?

 

[Himanshu Singh]

I made this a while back.

 

[CWatters]

http://info.ee.surrey.ac.uk/Workshop/advice/coils/force.html#sxs

Here is where I had gotten that equation from, as well as a few other posts for forums had mentioned this.

This is one
https://physics.stackexchange.com/questions/53680/force-of-electromagnet-on-piece-of-iron

Perhaps try measuring the force using a thread, pulley and weights to see if it gives similar results?

Also, what is the d supposed to be, and for N/dt I am assuming that I have to physically see it and calculate it when the experiment is on-going?
Vemf= -N dBA/dt (is that d before the variable B a typo?)

d in the equation I posted is meant to be delta. See the link I posted. I couldn't enter the symbol on this tablet. The one before the b isn't a typo. Its just i didn't finish off my train of thought..

I'm hoping that for this case we can modify the equation from...

Vemf=-NdBA/dt

to..

Vemf =-BAdN/dt

Because in this case BA is constant.

So now let's say the coil has K turns per Meter and V is the velocity of the train, then dN/dt= K*V so we can substitute that back in an we get..

Vemf=BAKV

Then rearrange to give the velocity of the train as..

V= Vemf/BAK

You would replace Vemf with the actual battery voltage on load.

However I'm not sure if we can make the step from dB/dt to dN/dt. I could use some input from others to confirm this is OK.

To get the equation my teacher gave me, all we did was take f = qvb and the centripetal force eqn and made them equal to each other. the v's canceled out (leaving only one) and we just isolated for v.

Perhaps I'm rusty but I don't thing the two v's are the same.

In f=qvb I think v is the velocity of a charged particle which would be the electrons in the wire not the velocity of the train?

And doesn't this assume the "train track" is in a circle? What if its in a straight line?

 

[Himanshu Singh]

Perhaps try measuring the force using a thread, pulley and weights to see if it gives similar results?
d in the equation I posted is meant to be delta. See the link I posted. I couldn't enter the symbol on this tablet. The one before the b isn't a typo. Its just i didn't finish off my train of thought..

I'm hoping that for this case we can modify the equation from...

Vemf=-NdBA/dt

to..

Vemf =-BAdN/dt

Because in this case BA is constant.

So now let's say the coil has K turns per Meter and V is the velocity of the train, then dN/dt= K*V so we can substitute that back in an we get..

Vemf=BAKV

Then rearrange to give the velocity of the train as..

V= Vemf/BAK

You would replace Vemf with the actual battery voltage on load.

However I'm not sure if we can make the step from dB/dt to dN/dt. I could use some input from others to confirm this is OK.
Perhaps I'm rusty but I don't thing the two v's are the same.

In f=qvb I think v is the velocity of a charged particle which would be the electrons in the wire not the velocity of the train?

And doesn't this assume the "train track" is in a circle? What if its in a straight line?

Yes, it is assuming the train track is in a circle, but my teacher had just given this to me when I asked.

V= Vemf/BAK

Vemf is the voltage on the train correct? B is the magnetic field in mt, A is the cross sectional area of the magnet and I still do not know what K is. what do you mean by the coil has K turns per Meter?

FPull= ( n x I )^ 2 * μ0(A/(2g)^2)
and would this equation still work? considering that the gap is the distance between the two magnets (or the thickness of one)Thank you for your reply.

 

[CWatters]

The track is a coil of wire. Let's say it has 400 turns of wire and is 2 meter in length. Then K is 400/2 = 200 turns per meter.

 

[Himanshu Singh]

Perhaps try measuring the force using a thread, pulley and weights to see if it gives similar results?
d in the equation I posted is meant to be delta. See the link I posted. I couldn't enter the symbol on this tablet. The one before the b isn't a typo. Its just i didn't finish off my train of thought..

I'm hoping that for this case we can modify the equation from...

Vemf=-NdBA/dt

to..

Vemf =-BAdN/dt

Because in this case BA is constant.

So now let's say the coil has K turns per Meter and V is the velocity of the train, then dN/dt= K*V so we can substitute that back in an we get..

Vemf=BAKV

Then rearrange to give the velocity of the train as..

V= Vemf/BAK

You would replace Vemf with the actual battery voltage on load.

However I'm not sure if we can make the step from dB/dt to dN/dt. I could use some input from others to confirm this is OK.
Perhaps I'm rusty but I don't thing the two v's are the same.

In f=qvb I think v is the velocity of a charged particle which would be the electrons in the wire not the velocity of the train?

And doesn't this assume the "train track" is in a circle? What if its in a straight line?

OK, so i just subbed in 1.5 volts, an 8.355 mt b value, 5.02x10^-3 cross sectional area value (in m) and 100 k value ( I still don't understand this, are these the coils in total?) and I got 0.357m/s which is surprisingly accurate..

 

[Himanshu Singh]

I just read that. The post where you answered what k value was

 

[Himanshu Singh]

So let's say I have 200 coils for a 0.4m track, my k value would be 200/0.4 correct?

 

[Himanshu Singh]

IS there any way to take mass into consideration while calculating for the velocity using this equation?

 

[CWatters]

Normally the mass or the armature in a motor doesn't feature in basic motor equations. It might affect acceleration if it's moment of inertia is significant.

I think in your case mass would only effect friction force.

 

[gneill]

It seems to me that the only part of the coil that is energized by the battery lies between the magnets at either end of the battery. You might consider improving the efficiency by insulating all but the end magnets from the coil so that the inward magnets aren't shorting out the coil turns that they span. This might involve making the end magnets slightly larger in diameter (maybe wrap them with a conductive sleeve of some kind.

The length of wire you should consider for finding the resistance that the battery "sees" is the length of the coil comprising the turns that are contributing to the magnetic field. Measuring the resistance of such a short piece of copper wire is probably more trouble than it's worth. Look up the resistivity of gauge of wire you['re using and plug in the length of wire in that many turns of the coil.

 

[Tom.G]

So, I would just attach the copper wire (the length being the amount used in the coil) and connect both sides to the ends of a 1.5 v battery, essentially making a circuit and then cutting in between that circuit and then taking an ammeter to measure the current yes?

Another question, I am assuming that the circuit for the experiment would be a series circuit. Doesn't current stay the same throughout the circuit? Or am I wrong here and will have to measure it by using the amount of copper wire used?

@gneill said it well in post #30. Use the amount of wire that the current would be flowing thru when in actual use. That is, the Blue colored turns in the photo gneill posted.

You do not have to cut the wire for this measurement, just make the connections for current measurement across the same wire length.
And Yes, it is a series circuit as in your photo of measuring current. The length of wire replaces the light bulb and resistor, and your small battery with magnets replaces that big 6V battery.

Cheers,
Tom

 

[Himanshu Singh]

@gneill said it well in post #30. Use the amount of wire that the current would be flowing thru when in actual use. That is, the Blue colored turns in the photo gneill posted.

You do not have to cut the wire for this measurement, just make the connections for current measurement across the same wire length.
And Yes, it is a series circuit as in your photo of measuring current. The length of wire replaces the light bulb and resistor, and your small battery with magnets replaces that big 6V battery.

Cheers,
Tom

It seems to me that the only part of the coil that is energized by the battery lies between the magnets at either end of the battery. You might consider improving the efficiency by insulating all but the end magnets from the coil so that the inward magnets aren't shorting out the coil turns that they span. This might involve making the end magnets slightly larger in diameter (maybe wrap them with a conductive sleeve of some kind.

View attachment 235035

The length of wire you should consider for finding the resistance that the battery "sees" is the length of the coil comprising the turns that are contributing to the magnetic field. Measuring the resistance of such a short piece of copper wire is probably more trouble than it's worth. Look up the resistivity of gauge of wire you['re using and plug in the length of wire in that many turns of the coil.

Oh, so the amount of current measured would be in a circuit that had a copper wire as long as the number of coils in between the two magnets?

 

[Tom.G]

Oh, so the amount of current measured would be in a circuit that had a copper wire as long as the number of coils in between the two magnets?

Yes!
Expect around 3 to 8 Amps for a new battery. (that's just a guess )

 

[Himanshu Singh]

Ok, say this distance is like 20-25 cm of copper wire. What would the current be? any estimate?

FPull= ( n x I )^ 2 * μ0(A/(2g)^2)

n would be the number of copper wire coils in between the two magnets

i would be measured using the distance of the coils in between the two magnets

A would be the cross sectional area of the magnets

And g would be the thickness of the magnets.
-----------------------------------------------------------------------------------------------------------------------------------------
I actually calculated earlier today by using the total number of coils (throughout the solenoid), say like 150. (in meters)

I estimated I to be 0.1amps (although i feel like this is too high for a 1.5v battery, I googled the maximum amount of current that a 1.5v battery would be able to provide)

u0 constant

the cross sectional area of the neodymium magnets was pi(0.015/2)^2

the gap was the thickness of magnet (around 1cm)

(the amps in this test calculation were a very rough estimate)

the mass I used for f = ma was 50-55 grams (as a triple a alkaline battery weighs around 11.5 grams)

I then calculated for the final velocity using f=ma and kinematics equations and got around 0.22 m/s. This is also without any friction. this value seems to be realistic.

Although I still have yet to use actual values.

Also, if I use the amount of coils in between the magnets rather than 150, the current would obviously increase due to how little the measurement of the coils of wire in between the two magnets would be compared to a total of 150 coils, would it "scale down" and would I still be able to get the same values?

I had also used the equation CWatters had provided me with and I got accurate values as well.

Thank you everyone for your responses.

 

[Himanshu Singh]

Yes!
Expect around 3 to 8 Amps for a new battery. (that's just a guess )

I just saw your response, and 8 amps for a 1.5v battery? Just double checking, thank you very much.