Answer
(a) $19^{\circ}$
(b) $0.78N$
Work Step by Step
(a) We can find the required angle as follows:
$tan\theta=\frac{v^2}{r}$
$\implies \theta=tan^{-1}\frac{v^2}{gr}$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{(1.21)^2}{(9.8)(0.44)})$
$\theta=tan^{-1}(0.338)=19^{\circ}$
(b) We can find the required tension as
$T=\frac{mg}{cos\theta}$
We plug in the known values to obtain:
$T=\frac{(0.075)(9.81)}{cos(19^{\circ})}$
$T=0.78N$
