184_projects:railgun_sol

Dr. McPaddel woke up to a great surprise this morning when she walked outside to admire the junkyard and noticed a mysterious glowing creature stuck to her magnetic crane. Apparently, in all the excitement of showing the contraption to her neighbors, she had forgotten to turn off the machine. The creature had wandered into the scrapyard over night looking for some metal to eat, and when it wandered near the crane, the metal became magnetized, and pulled the creature up into the air, where it remained stuck until Dr. McPaddel discovered it this morning. She alerted the authorities, but it was the Hawkion researchers who arrived before anyone else, to transport the creature to a research facility to learn more about it. They are thinking these animals might be responsible for the disappearance of fish from the lake and green-bellied canaries from the cliffside.

By the time you learn of all this, the creature is already at the research facility. You arrive and the scientists fill you in. They have acquired this beast to study it, but by now they have figured out Dr. McPaddel's crane snagged a young version of the monster. Because of the apparent resemblance to both boars and tigers and their electromagnetic glow the scientists at the compound have taken to calling the monsters EM-boar tigers. So you have captured an EM-boar tiger cub. It turns out that EM boar tigers are super protective of their young and they have organized to attack the Lakeview compound. Surprisingly, even though it is a government facility, there are no guns at the compound. But you are scientists and engineers godammit! You decide to construct makeshift launcher that is using magnetic force to launch projectiles so that they will leave the launcher at a speed of $300m/s$. You must construct your launcher on the floor of the laboratory and position it at an angle of 63 degrees so that it is firing out of the window of the laboratory.

At your disposal to construct your launcher, you have access to two copper rails that can be cut to your specifications. You also have perspex that you can cut to any dimensions and chunks of aluminum that can be cut to any shape. You also have access to multiple variable power supplies that can be set to a requested current and a bunch of wires that can be cut to any length and some switches. You must outline your design, explaining the physics behind the design of the magnetic force launcher and indicating the current needed to reach an exit velocity from the launcher of $300m/s$.

caged.jpg

Learning Goals

  • Understand how superposition of magnetic fields works and practice adding magnetic fields together
  • Relate magnetic fields to forces and use those forces to calculate motion
  • Review some basic kinematics concepts
  • Practice building complex models using assumptions and approximations to simplify

Learning Issues

  • Students typically have a lot of fun with this problem, but there are a lot of assumptions which go into making this problem do-able. Make sure that students are documenting all of the assumptions that they are making and being very explicit about the limitations of their model, the things that their model cannot do, etc.
  • If students don't know what they are building and are getting stuck, then suggest they start looking up a rail gun. There are a lot of cool videos on the internet that they can look at to help them get ideas.
  • There is often a lot of variations in the ways that students envision making this rail gun. Some of these models are not feasible. The best argument to use when helping students reflect on whether or not the model is going to work is to use the right hand rule and discuss forces with them. There needs to be some mechanism for there to be a force exerted on the projectile. If there is no force, there will be no motion. If the force doesn't point in the correct direction, then the projectile won't properly launch.
  • There can be a tendency to use the formula for a moving point charge in a magnetic field to calculate forces. The argument for using this model is flawed in that it's not very easy to make a charged block of metal stay charged and shoot it. The better model is the force on current carrying wires. There are flaws to this model, too, though, so discussions about limitations are again going to be important.
  • Expect a lot of trouble with this problem. Conceptually, this is very challenging, but it can be a lot of fun.
  • People don't remember kinematics that well, so just be aware of that. You may have to remind students about kinematics equations and how they work.

Timing Issues

  • If students don't know that they're building a rail gun by 30 minutes in, clue them in on this.
  • After 60 minutes or so, if they don't have a similar model as shown in this solution, start talking with them to get them moving in the right direction.
  • Once they have the correct model, the math and logic are going to take a while to work through, so get them trucking on working with force equations and kinematics with no less than 45 minutes remaining before you want to start asking tutor questions.

Diagram 1

Diagram 3

Diagram 2

NOTE TO TUTORS: This problem is very complex and you have to make many approximations and assumptions to make the math do-able. Make sure you help your students make these assumptions and evaluate the limitations of their model.

Assumptions:

  • Magnetic field between two rails has a constant value. This is not completely accurate since the magnetic field should get weaker further away from the rails, but this becomes a better approximation if the rails are long compared to the length of the projectile and if the projectile is started in the middle of the rails rather than near the ends.
  • The magnetic field in the center of the two rails can be calculated using the formula for the magnetic field due to a long wire. Again, this is a rough approximation. The “long wire” formula assumes you are in the middle of a long wire. In this case, there is no electric current in the rail past the cross over the wire.
  • The force that the projectile feels will not be based on where in the launcher it is (again this is a better approximation if the projectile starts in the middle of the rails rather than near the ends).
  • The current provided by the power source is constant.
  • You will have to choose values for Mass, Length, and Distance between rails (R).

There are potentially many more assumptions to be made. Make sure that all of the assumptions the students are making are documented and be sure to discuss each of them in detail. The real value of this problem is in making assumptions to turn a complex problem into a much less complex model.

Step 0: Realize that this problem is about building a rail gun. From there, we need to have a “correct” picture of the rail gun in question. This is the diagram shown above. The rail gun consists of two parallel copper rails hooked up to the power supply. The projectile is a chunk of metal which bridges the two railings. The current flows from one rail, through the projectile, into the next rail. The current flowing through the copper rails creates a magnetic field, which causes a force to act on the current carrying metal projectile. We can verify that this is correct by using the right-hand rule. We also need to make sure that the projectile goes the correct place. There are a lot of ways that students could set-up the gun so that it fires out the window. The easiest is to just build it so that the barrel is directly up against the window and fires directly out towards the incoming boar tigers. This step is all concerned with making the correct mental model and setting the problem up correctly.

There are other models that students might try, but this one is the “only one” (there could be more of course) that has the behavior that we are looking for. The common issues are trying to have the metal be a moving charge and using that magnetic force equation, $F = q v \times B$, but there are issues to that such as the metal wouldn't remain charged very well and we would have to throw it in to the magnetic field or something to get it moving. The other common issue is the direction of the forces are incorrect. We need the force directed along the rail gun so that it fires out the window. Students like to use solenoids for this because the magnetic field points directly along the solenoid, but the right hand rule shows that can't be possible because the force can't be parallel to the magnetic field. There is some ambiguity here, but that is mostly because it's hard to predict what students will do. Relying on the right-hand rule here will quickly evaluate many of the potential models that students come up with.

Step 1: Realize that this is a force problem and write an equation for net force. $F_b$ is the force provided by the magnetic field. $F_g$ is the gravitational force.

$$\vec{F_b} - \vec{F_g} = \vec{F_{net}}$$

$$\vec{F_b} - \vec{F_g} = m_{projectile} \vec{a}$$

Step 2: Setup our kinematics equation so that we can solve for acceleration so and use it to substitute in. We need to choose a kinematics equation which we know enough of the variables for. We know the speed we want to launch the rail gun at. We don't want to be messing around with time here because that variable is going to be harder to solve for. The easier thing to use as a “known variable” is the length of the rail gun because we can control for that. This also represents the length over which the projectile will be accelerated over, so it is easier to control for.

$${V_f}^2 = {V_i}^2 + 2aL$$

$$a = \frac{{V_f}^2 - {V_i}^2}{2L}$$

Step 3: Write out an equation for $F_g$ (even though it will be negligibly small in comparison to the magnetic force) which points along the railing. Gravity provides a force in two components, but what we are really interested in is the amount of the force that is going to oppose the accelerating projectile as it moves along the railing and is accelerated by the magnetic force.

$$F_g = mgsin\theta$$

Our force equation along the rail gun rails (because these are forces, we have to be very diligent with noting direction) is as follows

$$F_b - mg \sin\theta = ma$$

These are all along the direction that the projectile will travel, so we drop our vector signs.

Step 4: Find the magnetic field at the location of the projectile. To do this we make two major assumptions: The magnetic field between two rails has a constant value and the magnetic field in the center of the two rails can be calculated using the formula for the magnetic field due to a long wire. In this case, the best approximation would be to say that the magnetic field at the center of the rails is due to two halves of a long wire - because there is not current flowing in the rails beyond the projectile (shown in the figure).

halveslongwire.jpg

So we would say the total magnetic field is given by:

$$\vec{B} = 1/2 \vec{B}_{rail1}+ 1/2 \vec{B}_{rail2}$$

Using the right hand rule, we know that in the center of the rails these field should add together and point into the page, so approximating each rail as a long line of current gives a total B-field in the middle of: $$\vec{B}= - \frac{\mu_0 I}{2\pi R} \hat{z}$$

In this case, it's easier to make this calculation if we assume that the force on the middle of the projectile is the same everywhere along the projectile. The middle of the projectile is a convenient choice because it's halfway between the two railings. So, $R$ in this case is not the distance between the railings, but the distance from each railing to the middle of the bar, so it is half of the distance between the two railings.

These are some pretty rash assumptions which don't really hold up in the real world, but it allows us to model the force, so we stick with it. We also, in this step, are assuming that as the projectile moves up the railings and current flows through more of the railing, our magnetic field doesn't change. That way, our equation holds constant for the entire time the projectile is being launched. This assumption holds if we assume our power supply can be programmed to output a variable current, which we can describe exactly for this problem. Or we assume that it doesn't matter and if the magnetic field changes, it only gets stronger (magnetic field flowing in more of the railing could make for a larger B-field) which just means we shoot the projectile faster. There are, again, a lot of assumptions going on in this problem, so make sure to not glaze over them.

Step 5: Calculate the force on the current in the projectile from the magnetic field (from the rails). If we treat the projectile like a current carrying wire, then we can use the following equation with the magnetic field we just calculated.

$$\text{d}\vec{F}= I \text{d}\vec{l} \times \vec{B}$$

We integrate this force over the length of the projectile (our current-carrying wire)

$$\vec{F}_{B} = \int_0^l I \text{d}\vec{l}_2 \times \vec{B}_1$$ $$\vec{F}_{B} = \int_0^l \frac{\mu_0 I^2}{2 \pi R}\text{d}l \hat{x}$$

$$\vec{F}_{B} = \frac{\mu_0 I^2 l}{2 \pi R} \hat{x}$$

Since we are working in the direction along the railing, then we can ignore the vectors and carry them implicitly, that is, we know the direction we are working in, so we don't have to write it down mathematically, so long as we keep things consistent. Plugging this back in for $F_b$ in our original equation.

$$\frac{\mu_0 I^2 l}{2 \pi R} - mgsin\theta = m(\frac{{V_f}^2 - {V_i}^2}{2L})$$

If we assume the projectile is shot from rest, then the final equation becomes the following

$$\frac{\mu_0 I^2 l}{2 \pi R} - mgsin\theta = m(\frac{{V_f}^2}{2L})$$

where I is the current provided by the source, R is half of the distance between the rails, $\theta$ is the launch angle, $L$ is the length of the railings, $l$ is the length of the projectile we are firing, and $V_f$ is how fast we want the projectile to launch.

Step 6: At this point you have to choose a mass, length and a R which is the horizontal distance between the rails. Solve for $I$ once you have chosen reasonable values.

Tutor Questions
  • Question: What forces are acting in this case? Draw a force diagram for your projectile
  • Expected Answer: They should include the magnetic force from the rails on the projectile, the gravitational force, and the normal force from the rails at the very least. (This is a good place to remind them that the physics from mechanics is still true).
  • Question: How do you know what direction the magnetic force on the projectile should be?
  • Expected Answer: Use the right hand rule! (This is another chance to go over how the right hand rule works.)
  • Question: Does the amount of current that is running through the projectile make sense?
  • Expected Answer: The current might be so high that it would melt the aluminum.
  • Question: How does the current relate to the mass of the projectile (draw a graph of this relationship)?
  • Expected Answer: It should be a quadratic equation.
  • Question: What are the practical concerns for a gun designed in this way?
  • Expected Answer: Resistive heating (gun would be intensely hot), repulsion (The current in each rail of a rail gun runs in opposite directions. This creates a repulsive force, proportional to the current, that attempts to push the rails apart. Because the currents in a rail gun are so large, the repulsion between the two rails is significant) and as mentioned above melting is also a problem.
  • Question: Are the values you chose as your approximations reasonable?
  • Expected Answer: Depends on their choices, evaluate based on realistic values for rails in a building and how mass is related to velocity required. R's resonableness ties into the size of the particle that is being projected.
  • Question: What assumptions and approximations did you make? What are limitations of the model you made?
  • Expected Answer: See the giant list at the beginning…
  • Question: Any lingering questions?
  • 184_projects/railgun_sol.txt
  • Last modified: 2021/08/19 19:05
  • by dmcpadden