Find Speed in Centripetal Motion?

Question

The radius of curvature of a loop-to-loop roller coaster is 12 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.4mg. Find the speed of the roller coaster at the top of the loop. Answer: v = 12.83 m/s.

Please provide step by step instructions on how you solve this.

Answer 1

Consider a FBD of the passenger
There is gravity downward m*g
there is the centripetal force upward
m*v^2/r
The net force is 0.4*m*g upward
so
m*v^2/r-m*g=0.4*m*g
v=sqrt(12*1.4*9.8)
v=12.83 m/s

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