## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(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.