#### Answer

(a) At the top, the ring pushes down on the rider with a force of 320 N
At the bottom, the ring pushes up on the rider with a force of 1400 N
(b) The longest rotation period that will prevent riders from falling off at the top is 5.7 seconds.

#### Work Step by Step

(a) We can find a person's speed as the ring rotates.
$v = \frac{distance}{time}$
$v = \frac{2\pi~r}{t}$
$v = \frac{(2\pi)(8.0~m)}{4.5~s}$
$v = 11.17~m/s$
We can use the centripetal force to find the normal force $F_N$ at the top.
$\sum F = \frac{mv^2}{r}$
$F_N + mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}- mg$
$F_N = \frac{(55~kg)(11.17~m/s)^2}{8.0~m}- (55~kg)(9.80~m/s^2)$
$F_N = 320~N$
At the top, the ring pushes down on the rider with a force of 320 N
We can use the centripetal force to find the normal force $F_N$ at the bottom.
$\sum F = \frac{mv^2}{r}$
$F_N - mg = \frac{mv^2}{r}$
$F_N = \frac{mv^2}{r}+ mg$
$F_N = \frac{(55~kg)(11.17~m/s)^2}{8.0~m}+ (55~kg)(9.80~m/s^2)$
$F_N = 1400~N$
At the bottom, the ring pushes up on the rider with a force of 1400 N
(b) We can find the speed when the force of gravity is equal to the centripetal force at the top.
$\frac{mv^2}{r} = mg$
$v = \sqrt{g~r}$
We can find the rotation period $T$ for this speed.
$T = \frac{distance}{speed}$
$T = \frac{2\pi~r}{v}$
$T = \frac{2\pi~r}{\sqrt{g~r}}$
$T = (2\pi)~\sqrt{\frac{r}{g}}$
$T = (2\pi)~\sqrt{\frac{8.0~m}{9.80~m/s^2}}$
$T = 5.7~s$
The longest rotation period that will prevent riders from falling off at the top is 5.7 seconds.