Magnitude of centripetal force acting on a roller coaster at the top of a loop

[XwakeriderX]

Homework Statement

A roller coaster goes through an inside vertical loop at the minimum speed required to just barely make it through the top of the vertical track. At the very top of the track, the magnitude of the centripetal force acting on the 'coaster is
a. mg.
b. 2 mg.
c. 6 mg.
d. 1/2 mg.
e. 5/2 mg

Homework Equations

Fc=(MV^2)/2

The Attempt at a Solution

My guess is there is only a normal force acting on it so its MG?

 

[tiny-tim]

Hi XwakeriderX!

A roller coaster goes through an inside vertical loop at the minimum speed required to just barely make it through the top of the vertical track. At the very top of the track, the magnitude of the centripetal force acting on the 'coaster is
a. mg.
b. 2 mg.
c. 6 mg.
d. 1/2 mg.
e. 5/2 mg
…
My guess is there is only a normal force acting on it so its MG?

I would have said that the centripetal force was zero …

there are generally only two forces on the coaster, its weight (which isn't generally centripetal) and the normal reaction force (which is always centripetal),

but "at the minimum speed required", the normal force at the top must be zero

(maybe they mean the mass times the centripetal acceleration??)

 

[XwakeriderX]

Hmm well it has to be one of those...it just doesn't make sense because at the top the centripetal should equal mg meaning its zero? or is there no centripetal so its just mass and gravity acting on it?