Rail gun - point of application of force?

[rcgldr]

TL;DR Summary

Rail gun - point of application of force?

A rail gun exerts an electromagnetic force on a projectile. The only moving object is the projectile, so is the point of application of force moving?

Update - For others reading this thread, after a re-think on this, for linear motion (as opposed to angular motion), the point of application of force is not a factor in calculating change in momentum or energy, although it might be interesting to some.

What started me on this was similar to jbriggs444's post#19: Although work is often defined in terms of the dot product of a force and the displacement of the point of application of said force, that definition is not, in fact, correct. More correct would be the dot product of the force and the displacement of the material at the point of application of said force.

However, for linear motion, it seems to me that the point of application of force doesn't matter when calculating change in momentum or energy for linear motion, only the requirement that the linear force be applied to the object somewhere, even if that point of application of force is moving with respect to the object. The example I used below is a vehicle accelerating on a flat surface with finite mass and free to move linearly, using the center of mass of vehicle and surface as an inertial frame of reference. The change in momentum of the flat surface is the {backwards force exerted by the tires of the vehicle} · time, and the change in energy is {backwards force exerted by the tires of the vehicle} · {distance the flat surface moves}, even though the point of application of force is moving with respect to the flat surface.

 

[A.T.]

A rail gun exerts an electromagnetic force on a projectile. The only moving object is the projectile, so is the point of application of force moving?

A projectile gains kinetic energy, so the gun is doing work on it, by applying a force over a distance. This is not specific to a rail gun.

 

[rcgldr]

Although not practical, what if the projectile was the source of energy, say a capacitor discharging to create the electromagnetic force within the projectile to propel the projectile? Assuming a lossless system, then the projectile would be converting internal potential energy into mechanical kinetic energy, so that the total energy of the projectile remains constant. In this case, could it be stated that no work is done, just a conversion of energy?

 

[russ_watters]

In this case, could it be stated that no work is done, just a conversion of energy?

Work is the expenditure energy via w=fd; the conversion from electrical to mechanical is doing work.

 

[rcgldr]

Work is the expenditure energy via w=fd; the conversion from electrical to mechanical is doing work.

But the only force in the direction of travel of the projectile is from the rail gun, which is not moving. In this case, the rail gun doesn't perform any work, the projectile's internal potential energy is the source of the work, but it's output is a backwards force on the rail gun (which isn't moving).

 

[A.T.]

Although not practical, what if the projectile was the source of energy, say a capacitor discharging to create the electromagnetic force within the projectile to propel the projectile? Assuming a lossless system, then the projectile would be converting internal potential energy into mechanical kinetic energy, so that the total energy of the projectile remains constant. In this case, could it be stated that no work is done, just a conversion of energy?

A car also uses internally stored energy to accelerate. In the reference frame of the road, no negative work is done on the road, but there is positive work being done on the car. This mismatch requires an energy input, that here comes from the car's internally stored energy.

 

[sophiecentaur]

Note that the OP uses the word "Force" but the replies have been using Work and Energy. It is this dichotomy that needs to be sorted out for the OP and not so much the details of who does what to whom. This is just another form of the same Old Chestnut that you can nearly always find in any list of recent posts.

The Energy can come from on board the projectile or from a device, bolted to the ground; it makes no essential difference because any Force on the projectile is the same as the Force back against the gun. The question is why does the projectile get 'all the energy'. Well it doesn't. It gets most of the energy because, although the Momentums of projectile and gun are equal and opposite, the Kinetic Energy if mostly with the (low mass) projectile and only a tiny proportion goes into the (high mass) gun + ground. Compare this situation with a rocket engine in space where the Energy from the fuel is split much more equally between rocket and ejected fuel.

This was a huge problem in Newton's time, until they realized there are two important quantities involved - Kinetic Energy AND Momentum.

 

[russ_watters]

But the only force in the direction of travel of the projectile is from the rail gun, which is not moving.

That isn't actually true, but it doesn't matter for analyzing the projectile. You can pretend it isn't. Work on the projectile is fd.

in this case, the rail gun doesn't perform any work...

As said above, it is Electrical energy that is converted to work. They don't have to exchange work if that's what you are thinking.

...the projectile's internal potential energy is the source of the work, but it's output is a backwards force on the rail gun (which isn't moving).

The output you are supposed to care about here is the forwards force on the projectile. It's what you are trying to move. Why do you think the work done to the gun matters?

 

[russ_watters]

Note that the OP uses the word "Force" but the replies have been using Work and Energy.

You're right; the answer to the question in the OP is yes, from the perspective of the gun the point of application of force is moving. But on the projectile, it is not. But [to the op]; so what? I think we've moved past that and are correctly addressing the real question, but we'll see...

 

[sophiecentaur]

Why do you think the work done to the gun matters?

Well, it would matter if the gun were not a lot more massive than the projectile. Just consider the massive artillery guns that were needed for throwing large shells over many km and it wasn't just barrel strength. The share of Energy is much greater for the less massive of the two. This is when the non-intuitive difference between Momentum and Kinetic Energy comes into the argument. It's strange, these days, how people are excited by the ideas in Modern Physics but they can be very uninterested in the basics of Mechanics and Electricity.

 

[russ_watters]

Well, it would matter if the gun were not a lot more massive than the projectile. Just consider the massive artillery guns that were needed for throwing large shells over many km and it wasn't just barrel strength. The share of Energy is much greater for the less massive of the two.

You're right, if there's recoil it matters. But from the OP's description I was assuming the gun is fixed to the ground (like a satellite launch gun).

 

[sophiecentaur]

Summary:: Rail gun - point of application of force?

A rail gun exerts an electromagnetic force on a projectile. The only moving object is the projectile, so is the point of application of force moving?

Can we look at this again? If something is being accelerated over a distance then you could often say that the Force is been applied at or from different places. Take a motor car as an example. In fact, it's only for a rocket propusion that the point of application doesn't change (relative to the rocket, that is). So what actually was your actual meaning?

 

[russ_watters]

Can we look at this again? If something is being accelerated over a distance then you could often say that the Force is been applied at or from different places. Take a motor car as an example. In fact, it's only for a rocket propusion that the point of application doesn't change (relative to the rocket, that is). So what actually was your actual meaning?

But that's just it; he's talking about the point of application in the ground/gun's rest frame. It isn't different from the rocket or car*. That's why I'm confused about what the OP's real question is.

*Caveat regarding the car's rolling wheels and stationary contact point is probably not relevant.

 

[rcgldr]

Clarifying, the rail gun is fixed, (assume attached to an object of infinite mass). The frame of reference is relative to the rail gun, so the rail gun never moves, only the projectile.

Looking at forces between rail gun and projectile, the projectile exerts a backwards force on the the rail gun, coexistent with the rail gun exerting a forward force on the projectile. The source of potential energy that is converted into kinetic energy of the projectile is internal to the projectile. The issue is that the only force in the direction of acceleration of the projectile is from the rail gun, which doesn't have an energy source, so it doesn't perform any work. So it would seem the point of application of force moves with the projectile.

I didn't want to get into the issues with a car, rolling wheels and contact points. However, even in this case, although there is no relative movement between the surfaces of tires and pavement at the points of contact, the point of application of force would seem to move with the car (due to the rolling motion of the tires).

 

[russ_watters]

The source of potential energy that is converted into kinetic energy of the projectile is internal to the projectile. The issue is that the only force in the direction of acceleration of the projectile is from the rail gun, which doesn't have an energy source, so it doesn't perform any work.

The location where the energy is originally stored is totally irrelevant, as are the nuts and bolts of how the force is generated. It could be an exterternal or internal source. What matters is it generates a force between the projectile and rail, and acceleration.

Think about a rocket vs a gun. Both have a propellant, but in one case it is internal and in the other, external. Doesn't matter and same here.

 

[rcgldr]

After a re-think on this, the point of application of force may not be relevant except for specific situations (such as a force applied off center to an object that results in angular as well as linear acceleration).

After a re-think on this, for linear motion (as opposed to angular motion), the point of application of force is not a factor in calculating change in momentum or energy, although it might be interesting to some.

Change the situation to where the rail gun is free to move and has a finite amount of mass. Assume an inertial frame of reference relative to the center of mass of the system of rail gun and projectile (and no external forces exerted from outside the system). Change in momentum of the rail gun = (backwards force from projectile) · time, change in momentum of the projectile = (forward force from rail gun) · time. Change in kinetic energy of the rail gun = (backwards force from projectile) · (distance rail guns moves). Change in kinetic energy of the projectile = (forward force from rail gun) · (distance projectile moves). Assuming a lossless situation, total increase in kinetic energy of the system == total decrease in potential energy.

The same logic could be applied to a vehicle on a flat surface of infinite mass. Although there is no relative motion between the vehicles tires and the flat surface at the point of contact it doesn't matter. Change in momentum would still == force · time, change in kinetic energy would still == force · distance vehicle moves == decrease in potential energy (in a lossless situation). (Note - this does not imply that the surface is performing any work, the work done is due to a decrease in potential energy, which in this case, is internal to the vehicle).

 

[sophiecentaur]

That's why I'm confused about what the OP's real question is.

I agree.

After a re-think on this, the point of application of force may not be relevant except for specific situations (such as a force applied off center to an object that results in angular as well as linear acceleration).

Perhaps that implies that the OP is getting there with considering the relevant factors. This sort of 'internal discussion' is, in many ways, best to be kept internal and not put up for discussion. The standard formulae and equations are the result of a lot of similar confusing 'public' thought in the past and the choices have already been made about what's really relevant.
I appreciate the nagging points that arise when thinking about a problem - we all get them - but following the standard approach (ignoring what can be ignored) tends to get the right answers and can be a quick route to understanding.
There's a lot to be said for doing a number of standard exercises (end of textbook chapter questions) to arrive at the best methods of sorlving more complicated practical problems. I know that doing sums can be a real pain though!

 

[rcgldr]

Perhaps that implies that the OP is getting there with considering the relevant factors.

I missed an edit, that should have been:

After a re-think on this, for linear motion (as opposed to angular motion), the point of application of force is not a factor in calculating change in momentum or energy, although it might be interesting to some.

Post #16 notes the factors in calculating change in momentum and energy to clarify things, but that wasn't the point of my original question, which was just asking about the motion of the point of application of force, not whether it was a factor in calculating change in momentum or energy. For the rail gun, the point of application force follows the projectile. For a vehicle and rolling motion of the tires (or treads), even though there's no relative motion between surfaces at the point of contact, the point of application of force follows the vehicle (since the center of the tires or treads move with the vehicle).

 

[jbriggs444]

You're right; the answer to the question in the OP is yes, from the perspective of the gun the point of application of force is moving. But on the projectile, it is not. But [to the op]; so what? I think we've moved past that and are correctly addressing the real question, but we'll see...

Although work is often defined in terms of the dot product of a force and the displacement of the point of application of said force, that definition is not, in fact, correct.

More correct would be the dot product of the force and the displacement of the material at the point of application of said force. [For a force applied at a fixed point on an object, there is no meaningful distinction, so one cannot really fault textbooks for ignoring the detail].

Consider, for example, a rocking chair sitting in the bed of a pick-up truck (think Granny in the opening sequence of the Beverly Hillbillies). As the truck accelerates forward, there is a forward force on the rocking chair, but the point of application of the forward force moves rearward along the rocker. The erroneous definition would hold that the work is negative -- the point of contact is moving rearward in the face of a forward force. The correct definition holds that the work is positive -- the rocker is moving forward in the face of a forward force.

Edit: A quick trip to Google reveals that this is not the first time I've jumped on this particular high horse in these forums.

 

[rcgldr]

Although work is often defined in terms of the dot product of a force and the displacement of the point of application of said force, that definition is not, in fact, correct. More correct would be the dot product of the force and the displacement of the material at the point of application of said force. ... rocking chair

My prior posts were about changes in kinetic energy related to force times distance, which is independent of the source of that energy that performs the work.

Using an inertial frame of reference, change in kinetic energy == (force exerted on object) times (distance object moves), or the integral equivalent if the force is not constant but varies with distance (such as a spring).

For example, a vehicle on a flat surface of finite mass, a constant force, using the center of mass of the system as an inertial frame of reference. The change in kinetic energy of the surface = {backwards force from vehicle} · {distance surface moves} (despite the fact that the point of application of force moves with the vehicle), and the change in kinetic energy of the vehicle = {forwards force from surface} · {distance vehicle moves}.

Again, in my original post, I was only interested in the point of application of force movement, but the thread evolved into changes in momentum and energy, in which case the point of application of force is not really a factor (as long as the force is somehow applied to the affected objects).

 

[russ_watters]

Again, in my original post, I was only interested in the point of application of force movement, but the thread evolved into changes in momentum and energy, in which case the point of application of force is not really a factor (as long as the force is somehow applied to the affected objects).

I guess why it moved fast -- and confused me -- is that original point is pretty self-evident. The purpose of the railgun is to fire the projectile, so of course the projectile (and therefore where the projectile is applying force to the barrel) is moving down the barrel. That's why I and I assume others quickly moved on to what I thought was the next point/question.

 

[rcgldr]

original point is pretty self-evident.

Originally I was wondering if it was similar to a maglev vehichle, where there is an eddy current in the projectile, but was also wondering if there was an eddy current in the "barrel" of the rail gun that follows the project, a field component of the rail gun that is moving, although the rail gun itself is not moving. (Somehow I missed including the part about eddy currents).