Answer
(a) The time of one revolution is 6.19 seconds.
(b) The angle $\theta$ does not depend on the weight of the passenger.
Work Step by Step
(a) Note that $v = \frac{2\pi ~r}{T}$.
We can find the radius of the circle that the passenger follows.
$r = 3.00~m + 5.00~sin(30.0^{\circ})$
$r = 5.50~m$
Let $T$ be the tension force exerted on the passenger by the cable.
$T~sin(\theta) = \frac{mv^2}{r}$
$T~cos(\theta) = mg$
We can divide the first equation by the second equation.
$tan(\theta) = \frac{v^2}{gr}$
$tan(\theta) = \frac{(2\pi)^2~r}{T^2g}$
$T^2 = \frac{(2\pi)^2~r}{g~tan(\theta)}$
$T = 2\pi~\sqrt{\frac{r}{g~tan(\theta)}}$
$T = 2\pi~\sqrt{\frac{5.50~m}{(9.80~m/s^2)~tan(30.0^{\circ})}}$
$T = 6.19~s$
The time of one revolution is 6.19 seconds.
(b) $tan(\theta) = \frac{(2\pi)^2~r}{T^2g}$
$\theta = arctan(\frac{(2\pi)^2~r}{T^2g})$
For a given rotation rate $T$, we can see that the angle $\theta$ does not depend on the weight of the passenger.
