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, . . . (N1)g, Ng?
(In reply to
Solution by rohit)
Rohit, I'm a little confused about your sixpoint 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 distancetoaxis has shifted
to a unique extent. Within a 90degree arc, [sin(x)sin(x1)] 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 20051019 18:57:47 