## College Physics (4th Edition)

(a) The person's apparent weight is $518.5~N$ (b) The person's apparent weight is $521.5~N$ (c) The radius of the Ferris wheel is $45.2~m$
(a) The radial acceleration is: $~a_r = \omega^2~R$ At the top, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight. $\sum F = m~a_r$ $mg-F_N = m~a_r$ $F_N = mg - m~\omega^2~R$ At the top, the person's apparent weight is: $~~mg - m~\omega^2~R$ Therefore, the person's apparent weight is $518.5~N$ because the apparent weight at the top is less than the person's actual weight. (b) At the bottom, we can find the normal force $F_N$ exerted on the person by the scale on the floor of the Ferris wheel. Note that the normal force is equal to the person's apparent weight. $\sum F = m~a_r$ $F_N - mg = m~a_r$ $F_N = mg + m~\omega^2~R$ At the bottom, the person's apparent weight is: $~~mg + m~\omega^2~R$ Therefore, the person's apparent weight is $521.5~N$ because the apparent weight at the bottom is greater than the person's actual weight. (c) We can find the radius $R$: $m~\omega^2~R = 1.5~N$ $R = \frac{1.5~N}{m~\omega^2}$ $R = \frac{1.5~N}{(520.0~N/9.80~m/s^2)~(0.025~rad/s)^2}$ $R = 45.2~m$ The radius of the Ferris wheel is $45.2~m$.