All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Centrifugal Balance II (Posted on 2005-10-18)
A circular centrifuge has 30 slots spaced evenly around its circumference. Thirty samples need to be spun in the centrifuge, their masses being 1g, 2g, 3g, . . . 29g, 30g. How can all the samples be placed in the centrifuge at once while keeping it balanced properly?

For what other values of N is it possible to balance an N slot centrifuge with samples weighing 1g, 2g, 3g, . . . (N-1)g, Ng?

 No Solution Yet Submitted by Brian Smith Rating: 4.2857 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(8): Simplification | Comment 23 of 30 |
(In reply to re(7): Sums of irrationals by Leonidas)

The previous post might be clearer if you arrange the moment equation this way:

(A-D)(sin x)+(B-E)(sin x+60)+(C-F)(sin x+120)=0

In the general case, each of the three sines is a different irrational number with an integer coefficient, which means that you can't balance the equation unless you make the coefficients zero.

You can only make this equation true with non-zero coefficients by picking special values of x.  The easy way is to make x=0, so that sin x = 0 and sin x+60 = sin x+120.  Then you can balance the wheel if (B-E)+(C-F) equals zero.  It may also be possible if x=30, though I doubt it.  But for any other value of x, you've just got three different irrational numbers, and no clever arrangement of integer coefficients (other than zeroes) is going to make them total zero.

 Posted by Leonidas on 2005-11-01 18:19:55

 Search: Search body:
Forums (0)