A Non Fiction Trilogy

APPENDIX N: A CALCULATION OF ROTATIONAL VS LINEAR MOTION

To determine when and if a particle in rotation overtakes and hits a particle in linear motion.

Figure N-1

Let P1 be particle rotating from its end point, and P2 be particle with linear motion. If they start from the same point will P1 continue to hit P2 if their speeds are equal? At what speed for P2 > that of P1 will they eliminate contact from collision?

A complete rotation of a P1 sweeps area = pi x r^2. So P2 linearly traveling same area travels pi x r. So 3 + 14/100 boxes as in figure AN-1   give the area of the circle.   Therefore 3.14 box parts is as to 4 circle parts. Over the same time P1 must travel more than 1 box to separate from P2. But over that time of one quarter rotation of P2, at same speed, P1 would travel 3.14 / 4, which is < 1. Therefore, P1 still hits P2. The speed P2 needs to separate would be 4 / 3.14 or 1.27 times the speed of P1. If they are separated by any distance at the start of the calculation that needs to be considered also.

Now if P1 is rotating from other than the endpoint will collision situation be different. No because though it can spin twice as fast at midpoint rotation, but its reach (radius) is diminished by the same factor, so figures should be the same.

But I have considered overtaking from the quarter turn point. Is 1/8 turn or others different? Yes indeed, if one looks at some segments of P2's travel, as in orange (figure AN2).

Figure N-2

A rod, P1 still, rotating from endpoint would first contact P2 at the points the solid orange lines hit the circumference place. The area sweep by P2 is > sweep by P1 to that point, proving P2 has greater velocity at that intersection. I have shown that at a quarter turn P1 speed would be 73% of P2 at that point. Or as stated before P2 would have to go 127% of P1 speed to avoid being hit.

But for other sections P2 would be hit sooner, as the “reach” of P1 toward P2 “goes from 0 to 100%”. At the first section, the rotation of P1 gives a circumference section that is almost a straight line, or approaches so the smaller and smaller the section becomes. So, it is almost identical to the rectangle from P2’s first section. Hence, a diagonal from corner to corner represents P1’s rotation and is appoximately 50% of P2’s area swept to the common point on the circumference. For P1 to not hit P2 in the first instance, both traveling same speed, P1 must accelerate P2 by double, or halve its own speed. Therefore, when equalized (P2’s gain in speed comes from P1’s loss, principle 9), considering they both started at 1c, P1 would then be .666c and P2 would be 1.333c.