Circular motion/Notes

Acceleration

• [math]a(t)\eq p(t)\cdot\left(\frac{r' '(t)}{r(t)}-(\theta'(t))^2\right)+\big(\theta' '(t)\cdot r(t)+2\theta'(t)\cdot r'(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/math], or:
  • substituting in [ilmath]p(t)[/ilmath] by it's definition:
    • [math]a(t)\eq \left(\frac{r' '(t)}{r(t)}-(\theta'(t))^2\right)\cdot\left[\begin{array}{r}r(t)\cdot\cos(\theta(t)) \\ r(t)\cdot\sin(\theta(t))\end{array}\right]+\big(\theta' '(t)\cdot r(t)+2\theta'(t)\cdot r'(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/math]
    • However in many special cases it is useful to consider the first form with [ilmath]p(t)[/ilmath] in it.

Special cases

1. unchanging radius, [ilmath]r(t):\eq r_0\in\mathbb{R}_{>0} [/ilmath]
   • obviously, now [ilmath]r'(t)\eq 0[/ilmath] and [ilmath]r' '(t)\eq 0[/ilmath], thus:
     • [ilmath]a(t)\eq -(\theta' '(t))^2\cdot p(t)+\big(\theta' '(t)\cdot r(t)\big)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right][/ilmath]

       [ilmath]\eq \theta' '(t)\left(r(t)\cdot\left[\begin{array}{r}-\sin(\theta(t)) \\ \cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

     • But notice:
       • [ilmath]a(t)\eq \theta' '(t)\left(\left[\begin{array}{r}-r(t)\cdot\sin(\theta(t)) \\ r(t)\cdot\cos(\theta(t))\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

         [ilmath]\eq \theta' '(t)\left(\left[\begin{array}{r} -p_y(t)\\p_x(t)\end{array}\right]-\theta' '(t)\cdot p(t)\right)[/ilmath]

There must be a geometric interpretation for this! As the vector here is [ilmath]p(t)[/ilmath] reflected in the line [ilmath]x\eq 0[/ilmath]!

• TODO: This is wrong actually
  as the [ilmath]x[/ilmath] component comes from cos in [ilmath]p(t)[/ilmath], not sin. Which is not a reflection in [ilmath]x\eq 0[/ilmath], a matrix might be able to represent this linear combination better
  • I've corrected the formula Alec (talk) 18:56, 13 September 2018 (UTC)