resolved thanks to astronuc... radius = d/2
i accidentally used the diameter as my radius
Well, let the period T = 1/f = 4.4 s.
and f = [itex]\omega[/itex] /2[itex]\pi[/itex]
and the centripetal force = Fc = m[itex]\omega^2[/itex]r = mv2/r, where r is the radius of the circular trajectory.
Also don't forget the rider is still in a gravitational field, so don't forget the rider's weight due to gravity, mg, which is always down.
See references on circular motion - http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#circ
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