Answer
(a) The ratio of the apparent weight to the real weight is 0.75
(b) The ratio of the apparent weight to the real weight is 1.25
Work Step by Step
We can find the speed of a person riding on the Ferris wheel as:
$v = \frac{distance}{time}$
$v = \frac{2\pi~r}{t}$
$v = \frac{(2\pi)(11.0~m)}{12.5~s}$
$v = 5.53~m/s$
We can find the centripetal acceleration as;
$a_c = \frac{v^2}{r}$
$a_c = \frac{(5.53~m/s)^2}{12.5~m}$
$a_c = 2.45~m/s^2$
(a) Let $F_N$ be the normal force of the seat pushing up on the person. Note that $F_N$ is equal to the person's apparent weight.
$\sum F = M~a_c$
$Mg-F_N = M~a_c$
$F_N = M~(g-a_c)$
$F_N = (M)(9.80~m/s^2-2.45~m/s^2)$
$F_N = (M)(7.35~m/s^2)$
We can find the ratio of $F_N$ to $mg$:
$ratio = \frac{F_N}{mg}$
$ratio = \frac{(M)(7.35~m/s^2)}{(M)(9.80~m/s^2)}$
$ratio = 0.75$
The ratio of the apparent weight to the real weight is 0.75.
(b) We can find an expression for the normal force at the bottom as:
$\sum F = M~a_c$
$F_N-Mg = M~a_c$
$F_N = M~(g+a_c)$
$F_N = (M)(9.80~m/s^2+2.45~m/s^2)$
$F_N = (M)(12.25~m/s^2)$
We then find the ratio of $F_N$ to $mg$:
$ratio = \frac{F_N}{mg}$
$ratio = \frac{(M)(12.25~m/s^2)}{(M)(9.80~m/s^2)}$
$ratio = 1.25$
The ratio of the apparent weight to the real weight is 1.25.
