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 The pencil (Posted on 2004-02-23)
I have a pencil that always rolls around on a slanted surface. One end is wider and heavier than the other. So whenever it rolls, it goes in a wide circle. Otherwise, the pencil is featureless, only becoming steadily wider towards one end.

The difference between the two diameters on the two ends is exactly 144 times smaller than the length of the pencil. If the pencil is pointing uphill on a slanted surface, how many times will it spin until it points downhill?

 See The Solution Submitted by Tristan Rating: 4.0000 (6 votes)

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 Trying to be exact. | Comment 1 of 15

First some assumptions:
By "the pencil is pointed uphill", in order to have a problem we must assume this means the large end is uphill of the small end, rather than the other way around, otherwise the pencil is already in its stable position.

Also, it is assumed that the distance from the top of the pencil measured to the bottom, along its central axis, is 144 times the difference in the two diameters.

As the difference determines the angle or slope of a ruling along the conical surface and that angle would be maintained all the way to the tip of a full cone (and since the problem didn't give us the actual top and bottom radii), we can substitute a full cone of height 288 and base diameter of 2.  The slant height of this cone is √(1+288²).

The semicircle traversed by the contact of the base with the slanted surface is then π√(1+288²).  The circumference of the base is 2π, so the ratio of the length of the semicircle to the circumference of the base is √(1+288²)/2, or about 144.0008680529392.  This is the number of times the pencil would spin on its axis while going from base uphill to base downhill.

 Posted by Charlie on 2004-02-23 15:29:48

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