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

 

[kyle8921]

in my calculations for centripetal motion, but thanks to astronuc's correction, I have now resolved my mistake. The correct formula for centripetal motion is r = d/2, where r is the radius and d is the diameter. This is a common mistake, but it is important to remember that the radius is always half of the diameter in a circle. Thank you, astronuc, for helping me correct my error and improving the accuracy of my research.