Answer
(a) $v = 5.24~m/s$
(b) At the highest point, the apparent weight is 833 N.
At the lowest point, the apparent weight is 931 N.
(c) One revolution would take 14.2 seconds.
(d) At the lowest point, the apparent weight is 1760 N.
Work Step by Step
(a) $v = \frac{2\pi ~r}{t} = \frac{(2\pi)(50~m)}{60~s}$
$v = 5.24~m/s$
(b) Note that the apparent weight is equal to the normal force $F_N$ exerted on the passenger by the seat.
At the highest point:
$\sum F = \frac{mv^2}{r}$
$mg - F_N = \frac{mv^2}{r}$
$F_N = mg - \frac{mv^2}{r}$
$F_N = 882~N - \frac{(882~N)(5.24~m/s)^2}{(9.80~m/s^2)(50~m)}$
$F_N = 833~N$
At the lowest point:
$\sum F = \frac{mv^2}{r}$
$F_N - mg = \frac{mv^2}{r}$
$F_N = mg + \frac{mv^2}{r}$
$F_N = 882~N + \frac{(882~N)(5.24~m/s)^2}{(9.80~m/s^2)(50~m)}$
$F_N = 931~N$
(c) $mg = \frac{mv^2}{r}$
$v = \sqrt{gr}$
We can find the time $T$ for one revolution.
$T = \frac{2\pi ~r}{v} = 2\pi~\sqrt{\frac{r}{g}}$
$T = 2\pi~\sqrt{\frac{50~m}{9.80~m/s^2}}$
$T = 14.2~s$
(d) At the lowest point:
$\sum F = \frac{mv^2}{r}$
$F_N - mg = \frac{mv^2}{r}$
$F_N = mg + \frac{mv^2}{r}$
$F_N = 882~N + \frac{(882~N)(9.80~m/s^2)(50~m)}{(9.80~m/s^2)(50~m)}$
$F_N = 1760~N$
