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**[**initially at time ##t=0##

**]**moving towards right with a combination of:

(i) Angular speed of ##2 \pi## in clockwise direction (i.e. exactly one rotation per second).

(ii) Speed of the centre being ##\omega##.

The value ##x_0 \leq 0## is freely adjustable. Now suppose there are ##n## points on the circle separated by equal angles. That is, the angle between any two 'adjacent' points will be ##2 \pi/n##. Without loss of generality, let's assume that at time ##t=0##, one of these points is exactly on the top of the circle having coordinates ##(x_0,2)##.

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So, one natural question seems to be that for which values of ##n## a solution is possible by adjusting the value ##x_0##. By solution I mean an intial value of ##x_0## (at time ##t=0##) such that there is no intersection between any of the ##n## points and the line-segment/obstacle.

Via some plotting, it seems there is a solution for all values ##n=1## to ##n=5##. However, there didn't seem to be a solution for ##n=6##. (hopefully, I didn't make any mistake in the plots)

Plotting seems to suggest that one could try to prove this (more formally) using periodicity of curves traced by the points. I was thinking there would probably be a more elegant way to prove it (using reasoning that doesn't explicitly use periodicity).

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Of course, a more general version of the question would ask about values of ##n## serving as solutions ... for an obstacle of height ##0 \leq l \leq2##.