That is clever, and I did not see it so thanks for the solution. I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance. Although now that I think about it, I see how the ciphered solution answers the needle problem but not how it solves the disc problem, or rather not how it give the answer you says it gives. The disc problem has more symmetry so I can’t see why one orientation would be preferred. I’m probably missing something.
Imagine poking a needle through the middle of a disk, perpendicular to the disk. Then the needle is horizontal when the disk is vertical, and vertical when the disk is horizontal. So the fraction of vertical needles is going to be the same as the fraction of horizontal disks, and the fraction of horizontal needles is going to be the same as the fraction of vertical disks.
Yes, the way I finally understood this was imagine a horizontal disk whose circumference the equator. It can spin on one axis (through the center of the disk from north pole to south) while remaining horizontal. However a vertical disk oriented along a great circle can spin on two axes while remaining vertical, one from the north pole to the south pole crossing a diameter of the disk and one through the center of and perpendicular to the disk.
The key is that there are three different dimensions being collapsed into two “orientations”. With a needle, two dimensions are “horizontal”, and one is “vertical”. With a disk, two are “vertical”, and one is “horizontal”. Part of the issue with this problem is that people don’t generally have an explicit definition of “vertical” and “horizontal”. Ask them whether they know what they mean, and they’ll say “sure”, but ask them to give a definition, and they’ll flounder.
I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance
Vertical configurations are more stable, if I’m thinking about the problem correctly.
The question looks unclear to me because it’s not clear if they’re asking about statics or dynamics. If you look at the dynamics of the system, it seems clear that needles being vertical is an attractor, and we should expect randomly initiated particles to be more likely to be in the attractor as time goes on.
But suppose we neglect air resistance. Imparting a random force to the needles and a random attitude to the needles are different, because the first implies to me that most needles will have some angular momentum, and thus we need to look at paths on a sphere rather than points on a sphere, whereas the second makes shminux’s interpretation the obvious one.
“random force” is surely odd phrasing, but that’s what they used in their experiment. If the coins aren’t affected by air (which, oddly, they didn’t specify, but I suggested they shouldn’t be), and haven’t hit anything, then we can imagine most “random force” distributions to choose uniformly a random orientation (because the random distribution goes high, and enough time has passed to do so, even if the starting orientations are all the same). Most importantly, if we assume we can answer the question, then initial orientation can’t matter because it wasn’t specified.
That is clever, and I did not see it so thanks for the solution. I was going off on a tangent wondering if either vertical or horizontal configurations might be more or less stable due to air resistance. Although now that I think about it, I see how the ciphered solution answers the needle problem but not how it solves the disc problem, or rather not how it give the answer you says it gives. The disc problem has more symmetry so I can’t see why one orientation would be preferred. I’m probably missing something.
Imagine poking a needle through the middle of a disk, perpendicular to the disk. Then the needle is horizontal when the disk is vertical, and vertical when the disk is horizontal. So the fraction of vertical needles is going to be the same as the fraction of horizontal disks, and the fraction of horizontal needles is going to be the same as the fraction of vertical disks.
Yes, the way I finally understood this was imagine a horizontal disk whose circumference the equator. It can spin on one axis (through the center of the disk from north pole to south) while remaining horizontal. However a vertical disk oriented along a great circle can spin on two axes while remaining vertical, one from the north pole to the south pole crossing a diameter of the disk and one through the center of and perpendicular to the disk.
The key is that there are three different dimensions being collapsed into two “orientations”. With a needle, two dimensions are “horizontal”, and one is “vertical”. With a disk, two are “vertical”, and one is “horizontal”. Part of the issue with this problem is that people don’t generally have an explicit definition of “vertical” and “horizontal”. Ask them whether they know what they mean, and they’ll say “sure”, but ask them to give a definition, and they’ll flounder.
Vertical configurations are more stable, if I’m thinking about the problem correctly.
The question looks unclear to me because it’s not clear if they’re asking about statics or dynamics. If you look at the dynamics of the system, it seems clear that needles being vertical is an attractor, and we should expect randomly initiated particles to be more likely to be in the attractor as time goes on.
But suppose we neglect air resistance. Imparting a random force to the needles and a random attitude to the needles are different, because the first implies to me that most needles will have some angular momentum, and thus we need to look at paths on a sphere rather than points on a sphere, whereas the second makes shminux’s interpretation the obvious one.
“random force” is surely odd phrasing, but that’s what they used in their experiment. If the coins aren’t affected by air (which, oddly, they didn’t specify, but I suggested they shouldn’t be), and haven’t hit anything, then we can imagine most “random force” distributions to choose uniformly a random orientation (because the random distribution goes high, and enough time has passed to do so, even if the starting orientations are all the same). Most importantly, if we assume we can answer the question, then initial orientation can’t matter because it wasn’t specified.