To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have
[tex]KE = PE[/tex]
[tex]\frac{1}{2} mv^2 = mgh[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.8)(2(55.8*10^{-2}))}[/tex]
[tex]v = 4.67m/s[/tex]
Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s