Answer
(a) v = 11.3 m/s
(b) $F_N = 5400~N$
Work Step by Step
(a) Let's assume that gravity provides the centripetal force and the normal force is zero.
$mg = \frac{mv^2}{r}$
$v = \sqrt{gr} = \sqrt{(9.80~m/s^2)(13.0~m)}$
$v = 11.3~m/s$
(b) We can find the normal force $F_N$.
$\sum F = \frac{mv^2}{r}$
$F_N-mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}+mg$
$F_N = \frac{(110.0~kg)(22.6~m/s)^2}{13.0~m}+(110.0~kg)(9.80~m/s^2)$
$F_N = 5400~N$
