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?
Let's name L = length of the pencial, D = wider diameter and d = smaller diameter.
If we say, for example, that L = 144, D = 1, d will be 0 and the pencil will rotate over its extremity with d = 0 and it will have to rotate half of the perimeter of a circle with radius = 144. So, the distance to cover will be 2 x Pi x 144 / 2. As the perimeter of the extremity that will be rolling is given by 2 x PI x (1/2), we conclude that it will be necessary exactly 144 rotations.
My thought
