Answer
$41.4^{\circ}$
Work Step by Step
We can determine the required angle $\theta$ as follows:
We know that
$\Delta KE=\Delta PE$
$\implies \frac{1}{2}m_2^2-\frac{1}{2}mv_1^2=-mg(rcos\theta-r)$
This simplifies to:
$v_2^2=2gr(\frac{9}{8}-cos\theta)$
As $a_{u_n}=\frac{v_2^2}{r}$
$\implies a_{u_n}=\frac{2gr(\frac{9}{8}-cos\theta)}{r}=2g(\frac{9}{8}-cos\theta)$
Now $F_N=mgcos\theta-ma_{u_n}$
$\implies F_N=mgcos\theta-m(2g(\frac{9}{8}-cos\theta))$
This simplifies to:
$F_N=mg(3cos\theta-\frac{9}{4})$
$\implies 0=mg(3cos\theta-\frac{9}{4})$
This simplifies to:
$\implies \theta=41.4^{\circ}$
