Question

Using the energy conservation law determine the following...
If the minimum velocity needed by the skater to pass the loop is 5.42m/s, what is the minimum initial height that allows the skater complete the loop?

Answer 1

So that skater does not leave the loop it is necessery that $w_n=v^2/R\geq g$ then $v_{min}=\sqrt{gR}$

and $R=v^2_{min}/g.$ As $v^2_{min}/2+mg\sdot2R=mgH$ hence $v^2_{min}/2=g(H-2v^2_{min}/g)$

hence $H=\frac{5v^2_{min}}{2g}$ then $H\approx7.5m$