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        riddles >> medium >> Rotating resistor wheel
        
(Message started by: Eigenray on May 27^th, 2008, 5:14am)

Title: Rotating resistor wheel
Post by Eigenray on May 27^th, 2008, 5:14am

A metal wheel, radius r, moment of inertia I, has n spokes, with n odd. Each spoke, and each arc between spokes, has a resistance R (imagine 2n resistors as shown in red). The wheel rotates through a fixed region of constant magnetic field B, which forms a sector with central angle 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n (shown in blue).

Find the angular velocity http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif(t) in terms of its initial speed http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif₀.

(From a US Physics Team exam. n may have been given.)

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 1^st, 2008, 7:28pm

Assuming B perpendicular torhe weel's plane, [hide] \omega = \omega_0 exp(-\lambda t) , where \lambda= B^2 r^4\divI(5+\sqrt5 +(2^n \sqrt5)/({3+\sqrt5}^{n-1}-2^{n-1}))R[/hide]

Sorry: I tried Latex, as in the [ FAQ : how to write (and view) math symbols] , but seems it doesn't work ( so the preview...)

Title: Re: Rotating resistor wheel
Post by towr on Jun 2^nd, 2008, 12:30am

on 06/01/08 at 19:28:34, Immanuel_Bonfils wrote:

Assuming B perpendicular torhe weel's plane, [hide]Sorry: I tried Latex, as in the [ FAQ : how to write (and view) math symbols] , but seems it doesn't work ( so the preview...)

You need to have a plugin and a userscript installed for it to work (which is explained in the faq here (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_suggestions;action=display;num=1171644142;start=1#1)).
It's not a full implementation of latex btw, it's only symbols, not other typographic functions. And the symbols aren't hidable either.

Title: Re: Rotating resistor wheel
Post by Eigenray on Jun 2^nd, 2008, 2:59am

Right, I forgot to say the field is perpendicular to the plane of the wheel.

on 06/01/08 at 19:28:34, Immanuel_Bonfils wrote:

[hide] \omega = \omega_0 exp(-\lambda t) , where \lambda= B^2 r^4\divI(5+\sqrt5 +(2^n \sqrt5)/({3+\sqrt5}^{n-1}-2^{n-1}))R[/hide]

Yep! But I think the answer looks nicer if you don't express it in closed form. In fact, if you solve the problem in two different ways, you get a physical interpretation of a certain identity.

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 2^nd, 2008, 9:37am

Thanks towr but I had it already installed, and used http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/beta.gif, for instance in an earlier post. The problem starts when I want powers or subscript like x^2 , x_0 in LaTex. Also I wonder if we can pull division like frac{x}{y} or {x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/o.gifver y} in tex... Could you, please, help me with that?

Title: Re: Rotating resistor wheel
Post by towr on Jun 2^nd, 2008, 10:16am

on 06/02/08 at 09:37:59, Immanuel_Bonfils wrote:

What the script does is replace codes for symbols with an image of the symbol; so anything else has to be done in another way.
To get a subscript or superscript, you can use [sub] or [sup] tags: x² and x₀

Quote:

Also I wonder if we can pull division like frac{x}{y} or {x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/o.gifver y} in tex... Could you, please, help me with that?

There really isn't a way to do that here, unfortunately. So you'd have to make do with the old fashioned x/y.

Title: Re: Rotating resistor wheel
Post by SMQ on Jun 2^nd, 2008, 10:28am

on 06/02/08 at 09:37:59, Immanuel_Bonfils wrote:

The problem starts when I want powers or subscript like x^2 , x_0 in LaTex.

True, my script doesn't understand any markup/layout; it only makes a (fairly robust) set of symbols available under the same names they're known by in LaTeX. Superscripts can be done either through ^, tags (e.g. x² gives x²), or most of the letter and number symbols have superscript equivalents (e.g. \x\sup2 gives http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/x.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup2.gif). The same is true for subscripts. Be aware, though, that there is only one "level" of superscripts or subscripts available -- there's no clean way to handle x^y^z, for instance.

Quote:

Also I wonder if we can pull division like frac{x}{y} or {x \over y} in tex... Could you, please, help me with that?

Because we don't have much in the way of layout options, I think usually around here people just put the terms in parentheses if necessary and use a slash, e.g. x/y or (3x+2) / (y-5). I think once or twice I've been known to resort to a table with lines drawn from dashes, but that gets "messy" very quickly.

Sorry we can't offer better formula capability at the moment, but it's still better than *most* furums where you're limited to text alone...

--SMQ

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 2^nd, 2008, 10:38am

Thanks People, you're quite cool!

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 3^rd, 2008, 11:10am

Eigenray
I'm not sure I've got your idea. Could you be more explicit?

Title: Re: Rotating resistor wheel
Post by Eigenray on Jun 3^rd, 2008, 12:08pm

Well, how did you solve it?

By the way, the case of n even is not really any harder, but the associated sequence is less well known.

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 3^rd, 2008, 3:40pm

OK
The electromotive force is http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif= Bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif₀^r {v dx}, where x localizes the generic spoke’s point P, relative the center, and v=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gifx is its velocity. So http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/epsilon.gif=0.5 Bhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif.r². We could get it, also from Faraday’s Law.
This “source” induces the current in the whole rod but the arc opposite the spoke considered above, ( that , by the way, is the only one immersed in the magnetic field in each moment), cause, by symmetry the current gets null and we can open the circuit there.
Now it's a matter of equivalent resistance in each half of the rod , associating R in series , parallel, series, parallel…(n-1)/2 times, and this leads to quotient of Fibonacci numbers (the source has an internal resistance R itself).

Getting the current , i = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gifB r²/(kR), where k, dependent of n, comes from the circuit analysis, we can calculate the elementary force of the field on the point P and its corresponding torque and integrate to get http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/tau.gif= (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gifB² r⁴)/(2kR).

The movement equation http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/tau.gif= - I (dhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif)/(dt), leads to the answer hidden in my first post, above.

I will think about the even case

Title: Re: Rotating resistor wheel
Post by Eigenray on Jun 4^th, 2008, 5:55am

I did it slightly differently, using symmetry and Faraday (and Kirchoff) to find the current through each resistor.

on 06/03/08 at 15:40:05, Immanuel_Bonfils wrote:

this leads to quotient of Fibonacci numbers

Yes that was what I was getting at. We can express it as

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/alpha.gif = -http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif* B²r⁴/(2RI) * F_n-1/(F_n-1+F_n+1),

which is prettier.

Now, what do you get if you add up all the power losses over each resistor in the circuit?

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 4^th, 2008, 6:55am

I see. By closed form you mean substituting Fibonacci numbers by their "numerical" expression? ( one could say this is a open form...).
And is always a good advise to make an "enrgetic check"(he he he), but seams we can't solve it this way... unfortunately.

// Excessive whitespace removed by Eigenray

Title: Re: Rotating resistor wheel
Post by Eigenray on Jun 4^th, 2008, 7:46am

I just mean that expressing it in terms of the Fibonacci numbers is more illuminating. We can always plug in the closed form for them, but it hides what's going on.

The currents through the resistors are

F₁x, F₂x, ..., and F_n-1x, each occuring twice; and 0 and 2F_n-1x, where

x = Br²http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif/[2R(F_n-1+F_n+1)].

If we add up all the power losses,

P = [ 2(F₁² + ... + F_n-1²) + 4F_n-1² ] * x² R,

we find that

P = B²r⁴http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif²/(2R) * F_n-1/(F_n-1+F_n+1)
= - d/dt [ Ihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif²/2 ].

Why is this?

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 4^th, 2008, 11:41am

Tested!
Thanks for fixing that excessive white space (rather black...). By the way, how can I use the Search tab? I just tried " Rotating resistor wheel" and it gives back "Sorry, no matches where found" !

Title: Re: Rotating resistor wheel
Post by Eigenray on Jun 4^th, 2008, 12:06pm

It works for me. You did remember to check the box for the medium forum, right? (I still forget to check them myself sometimes.)

And you can edit your own posts, by the way, using the 'Modify' link in the upper right of each post.

Title: Re: Rotating resistor wheel
Post by Immanuel_Bonfils on Jun 10^th, 2008, 3:41pm

I forgot about the even n.
A conversion of wye to delta in the node opposite to the “active” spoke, introduces three 3R resistor , but the “delta’s base” can be removed, since by symmetry the current is null. Like the odd case, successive series and parallel association leads to the Lucas Numbers, L[sub] j [\sub] instead of Fobnacci’s.