Answer
$v = 5.5~m/s$
Work Step by Step
We can use the equation for centripetal force to find the speed at the bottom.
$\sum F = \frac{mv^2}{r}$
$T-mg = \frac{mv^2}{r}$
$v^2 = \frac{r~(T-mg)}{m}$
$v = \sqrt{\frac{r~(T-mg)}{m}}$
$v = \sqrt{\frac{(1.5~m)[15~N-(0.500~kg)(9.80~m/s^2)]}{0.500~kg}}$
$v = 5.5~m/s$