physics
posted by sandy on .
Consider a roller coaster as it travels near the bottom of its track as sketched in the figure below. At this point, the normal force on the roller coaster is three times its weight. If the speed of the roller coaster is 27 m/s, what is the radius of curvature of the track?
I tried the equation
r=v^2/m*gr=27^2/3*9.81, but I keep getting it wrong. Please help!

Use 3mg=mgmv^2/r. M cancels out
so..3g=gv^2/r. then solve for r algebraically