# Centripetal motion

1. Feb 12, 2006

### nick727kcin

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

i accidentally used the diameter as my radius

Last edited: Feb 12, 2006
2. Feb 12, 2006

### Staff: Mentor

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

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

and the centripetal force = Fc = 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