Circular motion/Notes

Acceleration

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

Special cases

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

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

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

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

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