Answer
(a) $10m/s^2$
(b) $47^{\circ}$
(c) Please see the work below.
Work Step by Step
(a) The required centripetal acceleration can be determined as follows:
$a_c=\frac{v^2}{r}$
We plug in the known values to obtain:
$a_c=\frac{[(25mi/h)(\frac{1h}{3600s})(\frac{1m}{6.2137\times 10^{-4}mi})]^2}{12m}$
$a_c=10.4m/s^2\approx10m/s^2$
(b) We can find the required angle as follows:
The x-component of the tension is given as
$Tsin\theta=F_c$
$Tsin\theta=ma_c$.....eq(1)
and the Y-component of the tension is given as
$Tcos\theta=mg$...eq(2)
Dividing eq(1) by eq(2), we obtain:
$\frac{Tsin\theta}{Tcos\theta}=\frac{ma_c}{mg}$
$tan\theta=\frac{a_c}{g}$
$\theta=tan^{-1}\frac{a_c}{g}$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{10.4m/s^2}{9.8m/s^2})$
$\theta=47^{\circ}$
(c) We know that the weight and the centripetal force both depend on the mass but the mass drops from the equation for the angle $\theta$, and thus the angle $\theta$ is independent of the weight of the rider.
