## University Physics with Modern Physics (14th Edition)

(a) Let's suppose the centripetal acceleration is 4.00 g. $4.00~g = \frac{v^2}{r}$ $r = \frac{v^2}{4.00~g} = \frac{(95.0~m/s)^2}{(4.00)(9.80~m/s^2)}$ $r = 230~m$ The minimum radius is 230 meters. (b) $\sum F = \frac{mv^2}{r}$ $F_N-mg = \frac{mv^2}{r}$ $F_N = \frac{mv^2}{r} + mg$ $F_N = \frac{(50.0~kg)(95.0~m/s)^2}{230~m} + (50.0~kg)(9.80~m/s^2)$ $F_N = 2450~N$ The pilot's apparent weight is 2450 N. (Note that this apparent weight is equal to 5 times the pilot's usual weight.)