Centripetal Motion: Resolving Mistake with Diameter

[pureouchies4717]

resolved thanks to astronuc... radius = d/2

i accidentally used the diameter as my radius

 

[Astronuc]

Well, let the period T = 1/f = 4.4 s.

and f = $\omega$ /2$\pi$

and the centripetal force = F_c = m$\omega^2$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