Circular motion

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This and pendulum stuff

Notes

General polar motion

For convenience we will denote:

• $$\dd{t}{\theta(t)}{t}$$ as $\theta'(t)$ and $$\dd{t}{\theta'(t)}{t}$$ as $\theta''(t)$
• $$\dd{t}{r(t)}{t}$$ as $r'(t)$ and $$\dd{t}{r'(t)}{t}$$ as $r''(t)$

Let:

• $p(t):\eq\Vecc{r(t)\ \!\cos(\theta(t))}{r(t)\ \!\sin(\theta(t))}$ be used for $p$osition - with respect to time
• $$v(t):\eq\dd{t}{p(t)}{t}$$
  • $$v(t)\eq\dd{t}{r(t)}{t}\cdot\Vecc{\cos(\theta(t))}{\sin(\theta(t))}\ +\ \dd{t}{\theta(t)}{t}\cdot r(t)\cdot \Vecc{-\sin(\theta(t))}{\cos(\theta(t))}$$

    $$\eq r'(t)\Vecc{\cos(\theta(t))}{\sin(\theta(t))}+\theta'(t)\cdot r(t)\cdot\Vecc{-\sin(\theta(t))}{\cos(\theta(t))}$$

• $a(t):\eq\dd{t}{v(t)}{t}$
  • $a(t)\eq r''(t)\Vecc{\cos(\theta(t))}{\sin(\theta(t))} +2\theta'(t)r'(t)\Vecc{-\sin(\theta(t))}{\cos(\theta(t))} + \theta''(t)r(t)\Vecc{-\sin(\theta(t))}{\cos(\theta(t))} - \big(\theta'(t)\big)^2\Vecc{r(t)\cos(\theta(t))}{r(t)\sin(\theta(t))}$

    $$\eq\frac{r''(t)}{r(t)}p(t)+\Vecc{-\sin(\theta(t))}{\cos(\theta(t))}\Big[\theta''(t)r(t)+2\theta'(t)r'(t)\Big] - \big(\theta'(t)\big)^2 p(t)$$

    $$\eq p(t)\left[\frac{r''(t)}{r(t)} - \big(\theta'(t)\big)^2\right]+\Vecc{-\sin(\theta(t))}{\cos(\theta(t))}\Big[\theta''(t)r(t)+2\theta'(t)r'(t)\Big]$$

Work to do:

1. Reduce to circular case first by setting $r(t)\eq c$ for some constant $c>0$ and handle $r(t)\eq 0$ special cases.
2. Define $\omega(t):\eq \theta'(t)$
   • Get to $p''(t)\eq -\big(\omega(t)\big)^2\ \!p(t)$