Imagine a pole of unit radius stuck in the ground at the origin. A thin string is wrapped to the left many times around the pole (clockwise direction as seen from above) and you are holding it tightly against the pole at (1,0).
If you hold the string tightly, you can unwind it by walking in a spiral to the right (counterclockwise.)
Task 1: Find parametric equations for your path.
Task 2: If you adjust your speed so the string unwinds at a constant rate of 1 revolution every 2π seconds and graph your position over time, formulate the x- and y-intercepts of this graph.
Task 1:
The circumference of the pole is 2 pi. With theta in radians, the end of the string makes the figure in cylindrical coordinates:
r = 1 + theta
Thus, if you have walked around once, which is 2 pi, then 2 pi of string has been released and the end will be at
(r, theta ) = (1+2 pi, 2 pi) or
(x, y) = (1+ 2 pi, 0), etc.
This is an Archimedean spiral where all rays from the center cross the spiral at equal intervals.
The parametric form is:
x = (1+theta) sin(theta)
y = (1+theta) cos(theta)
Task 2
Since we are told each 2 pi unwinding takes 2 secs, the solution for the +x axis is easy: each 2 secs the end of the string will cross the +x axis at (1 + 2 pi, 0), (1 + 4 pi, 0), (1 + 6 pi, 0), ... at seconds 2, 4, 6, 8...
What happens at the other three axes (+y, -x, -y) is similar since the speed of the unwinding is keeping pace with the increasing radius of curvature. -x will be crossed at (-1-pi), (-1-3 pi), (-1-5 pi) etc. at seconds 1, 3, 5, 7...
+y will be crossed at (1 + pi/2), (1 + 5 pi/2), (1 + 9 pi/2), ...
at seconds; 1/2, 5/2, 9/2, ...
-y will be crossed at (-1 - 3 pi/2), (-1 - 7 pi/2), (-1 -11 pi/2), ...
at seconds: 3/2, 7/2, 11/2, ...
(I think... It all seems way too easy.)
There is one little hitch, I now see.... Consider the point where the string last touches the pole as it extends outward. At the start it is said "you are holding it tightly against the pole."
To keep it tight you must be applying force in the -y direction.
The point of contact stays not in a line between you fingers and the origin but at the vertex of a right angle, located on the pole, with one leg to the origin and the other to your fingers. The moving last point of contact with the pole will be located at a theta a bit less than pi/2 ahead of the end of the string. To get the string around once to the x-axis crossing, more than 2 pi of string must be released. A sufficient quantity is the hypotenuse of a right triangle with legs of 1 and 1+ 2 pi.
.....
I have thought of a perhaps simplifying way to solve this: rotate the rigid axes at a constant rate CW at 2pi per 2 sec about the origin. The string then extends not as a spiral but a straight line starting at (x, y)=(1,0) and extending in the -y direction at a rate 2pi per 2 sec as well. The endpoint of the string is crossed successively by the 4 axes at regular intervals (every pi/2 sec) like a 4 blade propeller. The only geometric complication is that the line of the string is displaced as a negative going vertical line starting at (0, 1) so it is not aligned with the origin.... TBC
Edited on July 24, 2021, 4:08 pm
Edited on July 25, 2021, 12:37 pm
soln
