Calculating the final velocity of a simple electric train

[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.

 

[Himanshu Singh]

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]

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]

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]

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.