# Calculating the final velocity of a simple electric train

#### Himanshu Singh

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

#### Himanshu Singh

So lets 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

Homework Helper
Gold Member
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

Mentor
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.

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#### 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.
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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.

#### Tom.G

I just saw your response, and 8 amps for a 1.5v battery? Just double checking, thank you very much.
For a fresh, high quality, brand name, size AA, Alkaline battery; yes, up to 8 Amps short circuit current. Don't carry them in a pocket with loose change or keys, they get HOT. (Those are surprising little things!)

#### Himanshu Singh

okay haha, thanks.

#### Himanshu Singh

I have figured out which equation to use and finding an equation isn't the issue, I have another question about the force calculated.

Sorry for bringing this back up, but that initial equation I have used is working, I do get 0.35m/s (which is pretty fast) as a final result. Although, One major concern I still have is that I'm not sure whether the electromagnetic force would be doubled after the calculation? I'm assuming it would as there are two magnets, there are a total of 4 poles. The 2 n poles would cause the electromagnetic force and would be greater than the combined S forces (which are acting in the opposite direction). So would I be right in saying that the electromagnetic force would be doubled? I feel like they would have to be doubled as the magnets in this video seem to have a velocity of around 0.4m/s, the fastest ones.

I feel like this video explains it well, around 2 minutes and 45 seconds.

Thank you, and again sorry for bringing this back up.

Last edited:

Anyone?

#### Himanshu Singh

If anyone can help here, I'd greatly appreciate it. This question has been bugging me for the past few days.

#### Tom.G

I have figured out which equation to use and finding an equation isn't the issue, I have another question about the force calculated.
Although, One major concern I still have is that I'm not sure whether the electromagnetic force would be doubled after the calculation?
Since you did not post your calculations we have no way of knowing what you calculated, therefore your question cannot be answered.

#### Himanshu Singh

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.

I used this equation.
N was 18, the number of coils the magnet had "seen"
I was the current, around 3.8 amps with the magnets attached to them
A was 1.767x10^-4 (cross sectional area)
G was 0.013m (the width of magnets combined)

I then took this value and multiplied it by two. (this here is my concern)
After that I subbed it into w = f d (d being 0.4m, length of my track) and solved for w.

After this I subbed it into ke = mv^2/2 and I solved for v. The mass was somewhere around 50 grams (which i converted to kg)

I got somewhere around 0.3m/s.

So would I multiply the force calculated through FPull= ( n x I ) 2 μ0(A/(2g)^2) by 2?

#### CWatters

Homework Helper
Gold Member
Snip

After that I subbed it into w = f d (d being 0.4m, length of my track) and solved for w.

After this I subbed it into ke = mv^2/2 and I solved for v.
That's incorrect.

W is the work done by the force over one revolution. That's not equal to the KE of the train.

W is equal to the work done against friction and the _change_ in KE of the train over one revolution. If it moves at constant velocity the change in KE is zero and w is just equal to the work done against friction.

#### Himanshu Singh

That's incorrect.

W is the work done by the force over one revolution. That's not equal to the KE of the train.

W is equal to the work done against friction and the _change_ in KE of the train over one revolution. If it moves at constant velocity the change in KE is zero and w is just equal to the work done against friction.
So would I just use kinematics to solve for v2?

#### CWatters

Homework Helper
Gold Member
I don't think its possible to take this approach without knowing the friction.

I think this is one of those problems where there isn't enough information to solve it. The friction is unknown. The current and voltage when its moving are unknown.

#### Himanshu Singh

I'm calculating the velocity without any of the friction, since it is too complicated for me to calculate at the moment, the current also saying inbetween 3-3.5 amps while it is moving (I'm only taking the current given off by the battery and the magnets attached to it.
(at least for the double aa 1.2v battery). I just want the final velocity itself with no other factors affecting it.

#### CWatters

Homework Helper
Gold Member
I know that's what you want. However Newtons law F=ma tells you that a force produces an acceleration not a velocity. So you cannot calculate a velocity from a force.

If you assume no friction the net force is non zero so it should keep accelerating indefinitely. Others have already mentioned this.

#### Himanshu Singh

I know that's what you want. However Newtons law F=ma tells you that a force produces an acceleration not a velocity. So you cannot calculate a velocity from a force.

If you assume no friction the net force is non zero so it should keep accelerating indefinitely. Others have already mentioned this.
Do you know another method to solve for v then? As far as I know, using kinematic and work equations are the only way to get the velocity.

#### Tom.G

Here's a thought, perhaps not practical. See if the projectile can hold itself when trying to move vertically. This should greatly reduce the friction. If not, find the angle at which it will hold itself. Friction will be reduced by the cosine of the elevation angle, and motive force needed to stay still will be proportional to the sine of the elevation, once you account for friction.

For an approximation of the friction, try it with a completely dead battery. Tilt the track up until the projectile just starts to move.The cosine of the elevation angle gives you the coefficient of friction, at least for that orientation.

These are only approximations and the coil will have to be wound carefully so that each turn is in line with the next. Perhaps glue the coil to a ruler to help with this alignment.

Cheers,
Tom

#### Himanshu Singh

Here's a thought, perhaps not practical. See if the projectile can hold itself when trying to move vertically. This should greatly reduce the friction. If not, find the angle at which it will hold itself. Friction will be reduced by the cosine of the elevation angle, and motive force needed to stay still will be proportional to the sine of the elevation, once you account for friction.

For an approximation of the friction, try it with a completely dead battery. Tilt the track up until the projectile just starts to move.The cosine of the elevation angle gives you the coefficient of friction, at least for that orientation.

These are only approximations and the coil will have to be wound carefully so that each turn is in line with the next. Perhaps glue the coil to a ruler to help with this alignment.

Cheers,
Tom
Ok thank you.

"Calculating the final velocity of a simple electric train"

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