Answer
$48.2^{\circ}$
Work Step by Step
We can find the required angle as follows:
The balanced equation of the forces acting on the ice cube is
$-N+mgcos\theta=\frac{mv^2}{r}$............eq(1)
According to law of conservation of energy
$\Delta K.E=-\Delta U$
$\implies \frac{1}{2}mv^2=-mg(y_f-y_i)$
$\implies v^2=2g(y_i-y_f)$
$v^2=2g(r-rcos\theta)$
$\implies v^2=2gr(1-cos\theta)$
We plug in this value in eq(1) to obtain:
$N=mgcos\theta\frac{m}{r}[2gr(1-cos\theta)]$
$\implies cos\theta=2-3cos\theta$
$\implies \theta=cos^{-1}(\frac{2}{3})$
$\implies \theta=48.2^{\circ}$
