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?
The pencil will spin exactly 144 times.
Explanation:
Let L be the length of the pencil, d be the diameter of the thin end, and D be the diameter of the thick end.
144(D-d)=L
D-d=L/144
D-(L/144)=d
After a length L, the diameter of the pencil declines by L/144.
So if we imagine a long imaginary pencil beginning at the thick end and extending past the thin end for a total length of 144D, the thickness will decline to the very end:
D-(144D/144)=0.
When the pencil spins, it is as if this long imaginary pencil were rotating about that fixed imaginary point of zero thickness, 144D away from the thick end.
When the pencil makes a complete rotation, the thick end of the pencil describes a large circle of radius (not diameter) 144D and circumference 2*pi*(144D)=288*pi*D
Half this rotation will give us the situation in this puzzle: 144*D*pi.
Every time the pencil spins just one time, the surface at the large end covers pi*D.
So the pencil will spin exactly 144 times before the pencil is pointing from up to down.
Edited on February 24, 2004, 9:43 am
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Posted by Penny
on 2004-02-23 17:15:04 |
Trying to be even more exact
