Linear Accelerator (Gauss Rilfe)

I want to know if you think my Gauss Rilfe is gonna work? Please tell me how far?

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3 vote(s)
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4. I'll be able to lay seige to the next town!

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1. Nov 27, 2003

dagleykb

This is a little bit of a mouthful but any help would be greatly appreciated.
I am building a linear accelerator similar to the one found on this site: http://www.scitoys.com/scitoys/scitoys/magnets/gauss.html This is for a physics 12 project with my friends, but only it is on a larger scale. The project involves launching a regular tennis ball as far as possible without using combustion or fuels. We are constructing it as follows: A copper pipe is cut in half to create the "grooved" barrel were the steel balls will roll. (We are using copper because it has a low coefficient for friction.) At one end is a compressed spring which will release the first steel ball towards the first magnet. (Magnets are 1.0" x .125" neodymium-iron-boron (NdFeB) rare-earth permanent magnets fixed between two washers for safety and enhanced strength.) There will be 5-6 magnets with an undetermined spacing between them. We are using 10-12 stainless steel balls with a diameter of 1.0" weighing at approximately 66.4g or 0.0664kg we think. This entire barrel will be placed in an aluminum pipe, also cut in half, so that the copper barrel will allow the last steel ball to preform an inelastic collision against the tennis ball directly (thus an oblige collision is avoided). Because of projectile motion we will be attempting to fire this at an angle (probably 45 degrees, or a little less if wind resistance occurs) to achieve a maximum distance.

My questions are as follows: 1)How can we find the resultant velocity of the tennis ball(Please supply me with formulas I won't bother you with all the exact measurements, and besides I don't have all of them.)?
2)How can we calculate the pulling force of our magnets fields (in newtons) at a distance in meters?

For question 1:
A)I know that we need to find the spring constant (k) in (newton meters)of our spring using Hook's Law (F=kx) which is easily done.
B)Then we use (K) to find the Elastic Potential Energy for our spring (EPE=1/2kx^2) were (x) is the springs compression in meters, EPE is measured in joules.
C)Then we have to deal with Static Friction. (I need help here!)
D)Followed by drawing an incline plane problem at 45 degrees, where the Elastic Potential Energy is converted into Kinetic Energy (KE)(Both in joules). The kinetic energy and the attraction force of the magnet (this is where question 2 comes in) are pulling the steel ball up the slope after the spring system is released. The kinetic friction and acceleration due to gravity are pulling the ball down the slope and into the slope resulting in a loss of velocity. [Normal force is mgcos(45) {Where m is mass (kg) and g is gravity (9.8m/s)}]
E)When the ball does hit the magnet most of the kinetic energy is transferred to the furthest ball on the other side of the magnet, where it in turn overcomes static friction. (More static friction help needed!)
F)Another incline plan problem with KE and the 2nd magnets attraction force are pulling uphill. The acceleration due to gravity, kinetic friction, and 1st magnets attraction force slowing the velocity.
G)Parts E and F repeat until the inelastic collision of the last steel ball against the tennis ball where we dip into momentum. p=mv where p is momentum in (kgm/s), m is mass in (kg) and v is velocity (m/s). Because we have to objects (one at rest) we use the equation p=(m1)(v1)+(m2)(v2) were mass 2 is at rest (the tennis ball). Using momentum we then find the resulting velocity of the tennis ball!

Other formula: Ffr=(mu)N were Ffr is force of friction in newtons, (mu) is coefficient of friction and (N) is Normal force.
Ffr=(mu)mg involving force of friction, coefficient of friction, mass and acceleration due to gravity. (But someone told me this is for objects kept horizontal?)

That really was a mouthful, it covers a lot of different physics topics too! Please inform me if I'm even approaching this right? Thanks in advance.

Kristan

2. Nov 27, 2003

Sniper__1

what is the power in newtons of the magnets

3. Nov 27, 2003

dagleykb

That's one of the things I need to find a formula for to find out, sadly enough. If someone could tell me a way to figure that out I would be greatly appreciated as it's a major problem in my efforts to find or resulting velocity!

4. Nov 27, 2003

Sniper__1

i unfortunatley dont know that formula but im sure you could find it on any mathematical physics sight.

5. Dec 1, 2003

HallsofIvy

If you are using permanent magnets they are going to contribute nothing. The first magnet will accelerate the ball until it is past the magnet, then decelerate it.

A "Gauss rifle" (linear magnetic induction) uses electro-magnets carefully controlled to turn off as soon as the object has passed each magnet.

You are not, by the way, going to find a "formula" that will tell you the strength of a magnet. You would have to measure that experimentally.

6. Dec 2, 2003

GENIERE

HallsofIvy – I believe the poster is describing a rail gun wherein the steel bearings actually impact the magnets. I built one of these about 25 years ago for my kids. In my case, I used a wooden ruler of the type that has a small grove running lengthwise down the ruler. About every 2-3inches, I scotch-taped a magnet. In front of each magnet I placed 2 steel ball bearings in the grove (balls placed on muzzle side of magnets). The magnet holds the steel balls in place. The gismo is placed horizontally on a table. A ball is started rolling slowly towards rear-most magnet. It will accelerate and impact the magnet. The lead ball of the pair in front of the magnet is released and accelerated toward the second magnet. After the third (and last) collision, the last ball ejects at considerable speed. Mine went about 5-6 feet until it hit the kitchen floor when shot from table top. The speed of the initial ball was just enough to allow it to be attracted to the magnet.

Last edited by a moderator: Dec 2, 2003
7. Dec 3, 2003

Integral

Staff Emeritus
I would stay away from stainless steel for this purpose, many alloys of stainless are not magnetic. Get yourself some regular steel bearings.

8. Dec 8, 2003

pallidin

I am very familiar with the device and concept you are wanting to construct. A great project! And it does work!
Keep in mind that the magnets might shatter from the shock wave created during momentum transfer, especially in the faster "stages" Neodymium magnets(a rare earth magnet), which I assume you are using for their exceptionally high field strength, are also very brittle.
Also, the use of washers, though not a bad idea, should be thought through carefully. That is, if the washers are on both sides of the magnet facing the balls, and the washers are magnetic, say, steel, the magntic field of the magnet itself will not go very far at all beyond the washers, as the washers channel the field. Therefore, magnetic attraction on an approaching ball is very, very low.
So, if you use washers, do not use any material that is magnetically attracted(again, that would interfere grossly with the attraction of the steel balls) A hard plastic washer might be a good choice.
Second, remember that this effect is propagated through momentum transfer due to direct impact. A washer protecting the magnet will still transfer a shock wave through the magnet, through the other washer, and onto the next ball. So, washers might help a little in protecting the magnet from a sudden shock, but not by much.
A functional workaround to this problem is this: Have each magnet attract the balls, but, instead of transfering the momentum through the magnets, transfer the momentum around the magnets. This type of protection can preserve your magnets for many, many runs, but comes at the cost of a slight reduction in acceleration.
Anyway, just some thoughts to add, and maybe none apply in your case. Have fun in your project! Oh, and use caution.

Pallidin