Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The critical speed is the speed at which the centripetal acceleration is equal to $g$. We can find an expression for the critical speed as; $a_c = \frac{v^2}{r} = g$ $v = \sqrt{g~r}$ Note that at the top of the loop, the normal force from the track is directed straight down. We can use an equation for the centripetal force to find an expression for the normal force $F_N$. We can let the speed $v = 2\sqrt{g~r}$. Therefore; $\sum F = \frac{mv^2}{r}$ $F_N+mg = \frac{(m)(2\sqrt{g~r})^2}{r}$ $F_N+mg = 4~mg$ $F_N = 3~mg$ The ratio of the normal force to the gravitational force is 3.