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Physics Help  
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Posted (10 years 9 months 5 days 5 hours ago) and read 8344 times:

If someone can help me with this and maybe some questions in the future, I'd really appreciate it. I know some of you out there love physics!

This has to do with a Rail Gun. I'm stumped.

This problem explores how a current-carrying wire can be accelerated by a magnetic field. You will use the ideas of magnetic flux and the EMF due to change of flux through a loop.

A conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength V in series with a resistor of resistance R makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left. The arrangement is shown in the figure.

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left.

What is the acceleration a(t)(sub r) of the rod? Take m to be the mass of the rod. Express your answer as a function of V, B, the velocity of the rod v(t)sub r, L, R, and the mass of the rod m.

I've tried fiddleling around with the units in the 6 variables but can't come up with acceleration. Thanks in advanced,

Matt







Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
14 replies: All unread, jump to last
 
User currently offlineJetService From United States of America, joined Feb 2000, 4798 posts, RR: 11
Reply 1, posted (10 years 9 months 5 days 4 hours ago) and read 8334 times:

David b. should be able to help you. He's an expert at dealing with friction from accelerating rods and fiddling with units.


"Shaddap you!"
User currently offlineDavid b. From United States of America, joined Jun 2001, 3148 posts, RR: 5
Reply 2, posted (10 years 9 months 5 days 4 hours ago) and read 8329 times:

David b. should be able to help you. He's an expert at dealing with friction from accelerating rods and fiddling with units



 Wow!  Wow!
 Wow!  Wow!

Do I know................well, there an other member that can help you more then I can. He/she has much more experience. Doesn't he/she JS?



Teenage-know-it-alls should be shot on sight
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 3, posted (10 years 9 months 5 days 3 hours ago) and read 8315 times:

Awesome, I needed that laugh  Smile


Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
User currently offlineDavid b. From United States of America, joined Jun 2001, 3148 posts, RR: 5
Reply 4, posted (10 years 9 months 5 days 3 hours ago) and read 8315 times:

744, do they let virgins fly for free? Laugh out loud


Teenage-know-it-alls should be shot on sight
User currently offlineMas a330 From , joined Dec 1969, posts, RR:
Reply 5, posted (10 years 9 months 4 days 19 hours ago) and read 8285 times:

I flunked my paper

User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 6, posted (10 years 9 months 4 days 16 hours ago) and read 8270 times:

Think of the railgun setup as an inductor, in which one element is free to move.

I'm not in the habit of giving people verbatim answers for their homework, and you haven't given all the information anyway, but we can do a walkthrough.  Big grin

You already know the current though the loop (Ohm's law). Use this to find the induced magnetic field. You could form a Biot-Savart integral in 4 parts (both rails, the rod, and the power connection) but this might be too deep. Whichever method you choose, be careful with vector directions.

In real life, resistance wouldn't be zero; you could apply Faraday's law to this function, and get the induced EMF. It's actually a very simple inductor, so the EMF effectively causes a "counter-current" such that the actual current flow through the rod is less than you would expect from V=IR.

Anyway, back to the question. You have a function for the induced magnetic field; add this to the existing ('background') magnetic field B to get a function for the total magnetic field. Use Lorentz to find the force that this field creates on the rod as current flows through it.

Once you have the force on the rod, the rest is basic calculus - just use F=ma &c. It's not linear because the force is a function of the rod's position.

If you want to be really thorough, go through the same solution relativistically - after all, the rails are infinitely long, the rod will eventually be moving pretty fast...



Cunning linguist
User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 7, posted (10 years 9 months 4 days 16 hours ago) and read 8263 times:

Oops. I included an unnecessary bit. Can you guess where it is?  Wink/being sarcastic
(it's only a few seconds extra work)

[Edited 2003-10-30 15:27:38]


Cunning linguist
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 8, posted (10 years 9 months 4 days 14 hours ago) and read 8248 times:

While I understand somewhat better now, no help.

I'm getting the mu_0 in my answer when applying the B field, and that is not in the answer.

If you don't mind, I'm passing the 12 hour mark, tell what you think the answer is, if it is correct I'd appreciate you explaining it to me in great detail, cause right now, I don't seem to be getting it.

Thanks

Matt



Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 9, posted (10 years 9 months 4 days ago) and read 8217 times:

If this helps anyone else trying to help:



Thanks



Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 10, posted (10 years 9 months 3 days 16 hours ago) and read 8209 times:

Argh...
I just spotted this as I was about to go out - don't have time for a detailed answer, sorry :-(

In the meantime, I spent a few seconds with Google, and found this:
http://www.public.iastate.edu/~hotdog/Projects/EE%20213%20Project.doc

There's a lot of railgun junk on the web, but I think this one is most similar to your example. Equation 34 is pretty close to what you need? It gives you a function for F, but you can change that to displacement [r] thanks to Newton; F=m(d²r/dt²).

Try that. If it still doesn't work, I'll be back in a few hours...



Cunning linguist
User currently offlineBobrayner From United Kingdom, joined Apr 2003, 2227 posts, RR: 6
Reply 11, posted (10 years 9 months 3 days 4 hours ago) and read 8191 times:

Is that OK? Or not?


Cunning linguist
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 12, posted (10 years 9 months 3 days ago) and read 8181 times:

When I try putting mu_0 in the answer, it says the solution is not dependant on mu_0.

I hate these stupid online homework assignments.

Thanks for that file though, I hadn't come across that one yet.



Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
User currently offlineJwenting From Netherlands, joined Apr 2001, 10213 posts, RR: 18
Reply 13, posted (10 years 9 months 2 days 22 hours ago) and read 8179 times:

darn it's been too long since I got my degree. Few years ago I could have solved this in my head without having to look up anything...


I wish I were flying
User currently offlineFlyVirgin744 From United States of America, joined Jul 1999, 1313 posts, RR: 1
Reply 14, posted (10 years 9 months 1 day 5 hours ago) and read 8158 times:

I finally was able to get the answer from a friend who gave up and pushed "Give answer."

Here it is for all of you interested:

(V-B*L*v_r(t))(R*B*L)/(R*m)


Thanks,

Matt



Sometimes I go about in pity for myself and all the while a great wind carries me across the sky.
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