Position and Displacement — Question 8

Question 8

A helical path can return horizontally without returning spatially

A particle follows $\vec r(t)=[3\cos(\pi t/(2\ \mathrm s))\hat\imath+3\sin(\pi t/(2\ \mathrm s))\hat\jmath+(t/1\ \mathrm s)\hat k]$ m for $t\geq0$. Here the trigonometric arguments mean $\pi t/(2\ \mathrm s)$, in radians.

1. Find the displacement from 0 to 2 s and its magnitude.

2. Find the displacement from 0 to 4 s. Has the particle returned to its starting point?

3. Determine all positive times when the horizontal coordinates repeat their initial values, and explain why none is a full spatial return.

Question 8 - Solution

A helical path can return horizontally without returning spatially

(a) Evaluate the half-turn endpoints.

$$\vec r(2)-\vec r(0)=(-3,0,2)-(3,0,0)=(-6,0,2)\,\mathrm m,\quad\boxed{|\Delta\vec r|=\sqrt{40}\,\mathrm m}$$ $\vec r(0)=(3,0,0)$ m and $\vec r(2\ \mathrm s)=(-3,0,2)$ m. Hence $\Delta\vec r=(-6,0,2)$ m and $|\Delta\vec r|=\sqrt{40}=2\sqrt{10}\approx6.325$ m.

(b) Keep the vertical coordinate.

$$\vec r(4)-\vec r(0)=(3,0,4)-(3,0,0)=(0,0,4)\,\mathrm m$$ At 4 s, $\vec r=(3,0,4)$ m, giving displacement $(0,0,4)$ m. The horizontal projection returns to its start after one turn, but the full position does not.

(c) Apply periodicity only where it exists.

$$\omega t=2\pi n\ \Rightarrow\ t=4n\,\mathrm s,\quad z=4n\,\mathrm m>0\quad(n=1,2,\ldots)$$ The horizontal coordinates repeat at $t=4n$ s for positive integers n. The height then equals $4n$ m, rather than zero. The trigonometric coordinates are periodic; the steadily increasing vertical coordinate prevents a positive-time return in three dimensions.