at the top most point if Rupert will not fall then normal force at the top point is almost zero for minimum speed
so here we can say
[tex]F_n + mg = m\frac{v^2}{R}[/tex]
now if
[tex]F_n = 0[/tex]
[tex]mg = \frac{mv^2}{R}[/tex]
[tex]v = \sqrt{Rg}[/tex]
[tex]v = \sqrt{25 \times 9.8}[/tex]
[tex]v = 15.65 m/s[/tex]
so above will be the minimum speed
