# Points on a Circle (moving through an obstacle)

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SSequence
This is a problem in two dimensions. Consider an obstacle (in the form of a line segment) placed at ##x=1##, ##0 \leq y \leq 1##. Now consider the circle of radius ##1## and center at ##(x_0,1)## [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##.

Homework Helper
Gold Member
Let the angular velocity of the circle be ##\omega## clockwise and the rightward velocity of its centre be ##v##. Then the path of a point ##P## on its circumference that is at ##(x_0,0)## (lowest point of trajectory) at time 0 is given by:
\begin{align*}
x &= x_0 + vt - \sin\omega t\\
x &= x_0 + vt - \sqrt{1 - (1-y)^2}
\\& = x_0 + vt - \sqrt{y(2-y)}
\end{align*}

First consider the case where ##v\ge \omega##, so that no point on the circumference ever goes leftwards (backwards). That makes the trajectory a function on the x-y plane, ie for every x coordinate there is only one possible y coordinate.

From symmetry arguments we see that this curve intersects that of a curve that has a phase difference of ##-2\pi/n## when ##x## is half of the value of ##x## at which the delayed-phase curve touches the ##x## axis. Since the distance between touches of a single point is ##2\pi v/ \omega## and the points are equally spaced, the delayed curve must touch the axis at ##x=2\pi v/\omega n##. So we need to work out the value of ##y## when ##x=\pi v/\omega n##.

Using the above equation and taking ##x_0=0##, that gives:
\begin{align*}
\pi v/\omega n &= v t + \sqrt{y(2-y)}
\\&= v t - \sqrt{(1 - \cos\omega t)(1 + \cos\omega t)}
\\&= v t - \sqrt{(1 - \cos^2\omega t)}
\\&= v t - \sin\omega t
\end{align*}

We solve this for ##t##. We have to solve it numerically, given numeric values of ##\omega, v, n##, as it has no general analytic solution.

The trajectory will clear the obstacle iff ##y=1-\cos\omega t>1##, which will occur iff ##t\in 2\pi k/\omega + (\pi/2\omega,3\pi/2\omega )## for some integer ##k##.

Below is some output that shows that, for ##v\ge \omega##, the maximum possible ##n## is 5, as you found.

But if we allow ##v<\omega##, we can get a solution with ##n=6##. I attach graphs that show solutions for ##n=6## and ##v/\omega## ranging from 0.6 to 0.9. The blue vertical stick is the obstacle, and we can see it is cleared by all trajectories. As we reduce the ratio ##v/\omega##, the arched curves get steeper, giving more headroom to fit the blue stick obstacle. But the loops at the bottom, which are where the lowest points move backwards, also get fatter. Once two consecutive lower loops touch, we can no longer place an obstacle between them. That happens when the ratio is 0.5.

Code:
> nrange <- 1:6
> #vrange <- c(1/(5:1), 1:5)
> vrange <- c(1:5)
> omega <- 1
> fn <- function(t) v * t - sin(omega * t) - pi * v / (omega * n)
> for (n in nrange){
+   print(paste('n = ', n, '+++++++++++++++++++++++++++++++++++++'))
+   for (v in vrange){
+     print(paste('v = ', v, '----------------------------'))
+     if (sign(fn(0) == sign(fn(2 * pi / omega ))))
+       print('No solution')  else {
+         t <- uniroot(fn, lower = 0, upper = 2 * pi / omega)\$root
+         y <- 1 - cos(omega * t )
+         if (y > 1)
+           print(paste('t = ', t, 'y = ', y)) else
+             print('No solution')
+       }
+   }
+ }
[1] "n =  1 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "t =  3.14159265358979 y =  2"
[1] "v =  2 ----------------------------"
[1] "t =  3.14159265358979 y =  2"
[1] "v =  3 ----------------------------"
[1] "t =  3.14159265358979 y =  2"
[1] "v =  4 ----------------------------"
[1] "t =  3.14159265358979 y =  2"
[1] "v =  5 ----------------------------"
[1] "t =  3.14159265358979 y =  2"
[1] "n =  2 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "t =  2.30987610774504 y =  1.67360807339406"
[1] "v =  2 ----------------------------"
[1] "t =  2.02098020353088 y =  1.43513109803117"
[1] "v =  3 ----------------------------"
[1] "t =  1.88756664213621 y =  1.31149919427964"
[1] "v =  4 ----------------------------"
[1] "t =  1.81346585986426 y =  1.24029479521041"
[1] "v =  5 ----------------------------"
[1] "t =  1.7669596061595 y =  1.19490763611046"
[1] "n =  3 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "t =  1.96896860192469 y =  1.38773424671918"
[1] "v =  2 ----------------------------"
[1] "No solution"
[1] "v =  3 ----------------------------"
[1] "No solution"
[1] "v =  4 ----------------------------"
[1] "No solution"
[1] "v =  5 ----------------------------"
[1] "No solution"
[1] "n =  4 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "t =  1.76633701251506 y =  1.19429694507934"
[1] "v =  2 ----------------------------"
[1] "No solution"
[1] "v =  3 ----------------------------"
[1] "No solution"
[1] "v =  4 ----------------------------"
[1] "No solution"
[1] "v =  5 ----------------------------"
[1] "No solution"
[1] "n =  5 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "t =  1.6267261216081 y =  1.05590063998406"
[1] "v =  2 ----------------------------"
[1] "No solution"
[1] "v =  3 ----------------------------"
[1] "No solution"
[1] "v =  4 ----------------------------"
[1] "No solution"
[1] "v =  5 ----------------------------"
[1] "No solution"
[1] "n =  6 +++++++++++++++++++++++++++++++++++++"
[1] "v =  1 ----------------------------"
[1] "No solution"
[1] "v =  2 ----------------------------"
[1] "No solution"
[1] "v =  3 ----------------------------"
[1] "No solution"
[1] "v =  4 ----------------------------"
[1] "No solution"
[1] "v =  5 ----------------------------"
[1] "No solution"

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