# Rolling ball problem

by lowbattery
Tags: ball, rolling
 P: 35 You're on the right way - conservation of energy: $$E_{potential}=E_{translation}+E_{rotation} \Rightarrow mg\Delta h = \frac{mv^{2}}{2}+\frac{I \omega^{2}}{2}, \Delta h = r(1-cos\alpha).$$ My result differs a bit from yours. I don't have the a parameter. Everything else is the same. Reaction force is the net force between one of the ball's weight component and the centrifugal force: $$\vec{N}=\vec{G_{y}}+\vec{F_{cf}}, G_{y}=mgcos\alpha.$$ Ball will leave the surface when reaction force becomes equal to 0: $$N=mgcos\alpha-\frac{mv^{2}}{r}=0.$$ If you plug the velocity derived from the conservation of energy here you will get that $cos\alpha=\frac{10}{17}.$