The lines L and M are horizontal and intersect at O. A sphere rolls along supported by L and M. What is the locus of its center?
Solution
Easy, except for one trap, and one ambiguous point.
Let the angle between the two lines be 2θ. Evidently the center rolls above the angle bisector. Let its projection onto the angle bisector have length x and let it be a height y above the angle bisector. Suppose the sphere has radius r. Then (x sin θ)2 + y2 = r2. So the locus is an ellipse with semi-axes r and r/sin θ.
If we assume the rolling is under gravity, then the sphere falls off when it reaches x = r/sin θ, so the locus is just the upper half of the ellipse. If we assume the sphere is somehow kept in contact with the lines, then it can roll back underneath and so the locus is the whole ellipse. That is the ambiguity.
The trap is that it can also roll the other way, over the other angle bisector, giving another ellipse (or upper half) with semi-axes r and r/cos θ.
© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999