Let's consider the simple pendulum as is displayed here or over there (page 10). The analysis of the second Newton's law in polar coordinates goes as follows:
$$ \vec{F} = m\frac{d^2\vec{r}}{dt^2}, \\ F_r \hat{r} + F_\theta \hat{\theta} = m\frac{d^2 (r\hat{r})}{dt^2} , \\ F_r \hat{r} + F_\theta \hat{\theta} = m(\ddot{r} - r\dot{\theta}^2) \hat{r} + m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) \hat{\theta} , \\ F_r \hat{r} + F_\theta \hat{\theta} = ma_r \hat{r} + m a_\theta \hat{\theta} . $$
Substituing the forces we get,
$$ -T + mg\cos(\theta) = ma_r = m(\ddot{r} - r\dot{\theta}^2) , \\ -mg\sin(\theta) = ma_\theta = m(r\ddot{\theta} + 2\dot{r}\dot{\theta}) $$
Considering the restrictions $r = L$ and $\dot{r} = \ddot{r} = 0$ we get
$$ -T + mg\cos(\theta) = m(- L \dot{\theta}^2) , \\ -mg\sin(\theta) = m(L\ddot{\theta}) $$ The second one is the known pendulum equation $$ \ddot{\theta} + \frac{g}{L}\sin(\theta) = 0 , $$ while the first one is a much less used equation $$ T = mL \dot{\theta}^2 + mg\cos(\theta) $$ ¿Is it the correct equation to calculate the tension? Note that this implies that $a_r \neq 0$; which in words means that the radial acceleration is different from zero which looks unphysical, ¿where is the trick? ¿Has it something to do with noninertial forces?
