MHB What is the critical cone angle for Cowboy Joe's lasso?

[Ackbach]

Many thanks to anemone for sending me some very helpful source links for problems. Here is this week's POTW:

-----

Cowboy Joe wants to climb an absolutely slippery glacier of the round conical form. He has a lasso (with a loop of a fixed size) which he can throw over the top and pull himself up - if only the loop won't slip off the vertex of the cone. It seems clear that when the cone is almost as flat as the plane, it will slip off, and when the cone is as sharp as a needle, the loop will hang. Thus there should exist some intermediate critical cone angle. What is it?

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

There were two valiant efforts this week towards a solution, but neither one was correct. I post a solution by Stuart Anderson:

Spoiler

First, assume that Joe is very small compared to the mountain so that the leader (section of the rope between him and the loop) is not held up off the surface of the mountain. In that case, the entire length of the rope, both the loop and the leader, hug the surface of the mountain, and can be treated like a curve lying on the surface of a cone. Second, assume that there is no friction at all between the surface and the rope, and that Joe's weight keeps the rope tight at every point.

A tight string lying along a surface will form a geodesic, a curve of (locally) minimal length, for the physically obvious reason that if the string could get any shorter, the tension ensures that it would. Now a cone has no intrinsic curvature and can be slit up the side and unrolled into a sector of a circle. Suppose we slit the cone from base to vertex along a straight line that passes through the location of the knot in the rope (cut-at-the-knot, rather than cut-the-knot).

After unrolling the cone into a circle sector, you will see that the point at the location of the knot will appear as two points, one on each of the radii that bound the sector. Of course they are equally distant from the vertex of the sector because they are images of the same original point. The path of the rope will be a line joining these points, and (this is the crucial part) it will be a straight line, because we already know that it is the shortest path joining these points.

If the sector is less than half a circle, then this straight line will fall on the sector, but if the sector is more than half a circle, the straight line will cut across the complementary sector instead. In that case, there is NO geodesic on the surface of the cone for the rope to follow. Any path that follows the cone surface must be longer than the straight path, and so the tension in the rope will cause it to move from any initial path towards the straight line path. But the cone vertex lies inside the closed path formed by the initial path and the straight path, and so when the rope moves to the straight path, it will inevitably cross the vertex of the cone.

Therefore, the rope slips off if and only if the cone is formed by rolling up a sector that exceeds half a circle. The boundary case is of course when the sector is exactly half a circle. In that case, the circumference of the cone base is exactly half the circumference of the circle that would be traced by the slant length(1). Therefore, the radius of the base is half the slant length, so the angle between the axis of the cone and any line on the cone surface must be 30°. The cone would therefore cast an equilateral triangular shadow at sunset. In the case where the rope does not slip off, the climber has a steeper than the 60° slope to contend with; no wonder he needs a rope!