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Centrifugal Balance II (Posted on 2005-10-18) Difficulty: 5 of 5
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)

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re: Solution | Comment 6 of 30 |
(In reply to Solution by rohit)

Rohit, I'm a little confused about your six-point circle.  You're right that all the diagonals balance, but only if you draw them through the weights themselves.  What happens if you draw the diagonals through points halfway between the weights?

My trigonometry is a little rusty, but I'm pretty sure that you calculate the moment as each point's distance from the axis, multiplied by the weight.  The angle between the weight and the axis is 36 degrees, so the distance to axis is sin(36)=.5877, if we say that the circle as a radius of one.

Let's consider an axis drawn perpendicular to the line between three and four.  On the three side, we have moment of (3x1)+(6x.5877)+(2x.5877)=7.7016.  On the four side, we have moment of (4x1)+(1x.5877)+(5x.5877)=7.5235.

More generally, I don't see how any circle like this could truly be balanced along every axis, no matter how many weights there are.  Once you start rotating that axis, the moment generated by each weight starts moving in its own direction.  And the change in moment doesn't change in a nice linear function; it changes on a sine function. 

If you had identical weights on opposite sides of the circle, then of course those sine functions would always cancel each other out.  But I doubt that it's possible for any other arrangement to stay balanced along the axis at every possible angle.

Or think of it this way: You have some arrangement of weights that's balanced along a given axis, and then you tilt the axis one degree.  On each side, each weight's distance-to-axis has shifted to a unique extent.  Within a 90-degree arc, [sin(x)-sin(x-1)] is different for every value of x.  So the only way to change the other side's moment to the same extent is to have exactly the same calculation on the other side.

  Posted by Leonidas on 2005-10-19 18:57:47

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