Question

A 30-kg child rides on a circus Ferris wheel that takes her around a vertical circular path
with a radius of 20 m every 22 s. What is the magnitude of the resultant force on the
child at the highest point on this trajectory?

Answer 1

I'm not sure from your question whether you need only the forcefrom the Ferris wheel or the total force, so I'll include bothanswers.

Neglecting gravity and only determining the force from the Ferriswheel:

F = ma

for centripetal acceleration:

a = v²/r    and     v = rω    so,

F = mω²r

We just need to determine ω (in radians per second)

ω = 2π / 22 s = 0.2856 rad/sec

F = (30kg)(0.2856rad/s)²(20m) = 48.94 N

F = 49 N     (to twosig fig)

Including gravity:

ΣF = ma
F = mg - mω²r

F = (30kg)(9.8m/s²) - 48.94 =245.06 N

F = 250 N    (to twosig figs)



Hope that helps!

Answer 2

We know that at the top of the Ferris wheel
the normal force   Fn = mv² / r -mg = m(v² / r - g )
                                
here   v = 2πr/t = 2π(20m) / 22s =5.7 m/s
                            Fn = (30kg)(5.7² / 20   - 9.8 ) =245.65N