Consider a roller coaster as it travels near the bottom of its track. At this point, the normal force on the roller coaster is two times its weight. If the speed of the roller coaster is 10 m/s, what is the radius of curvature of the track?

I am sure this question has something to do with the centripetal force equation. To me, there is not enough information to calculate the curvature. The only info they give is that Force is two times its weight and speed is a constant 10 m/s. Just a little "bump" to get me going would be appreciated.