Question

A 60kg man is on a steadily rotating Ferris wheel. The man is standing on a bathroom scale (yeah, sure, very likely, I know!) and the scale reads 45 kg at the top the wheel. Take g=10m/s

 a) What does the scale read when the man is at the bottom of the wheel? 

 b) What does the scale read when the man is level with the hub of the wheel going Assume there is sufficient friction that the man's feet do not slide off the scale. 

 c) What co-efficient of friction is needed to keep the man's feet on the scale in b)?

Answer 1

given mass of person, m = 60 kg
scale reading on the top of ferris wheel, m' = 45 kg

a. let angular speed of the ferris wheel be w
   then
   reading at the top of the wheel should be
   m'g = mg - mw^2R [ where R is radius of the ferris wheel]
   and reading, m" at the bottom of the ferris wheel must be
   m"g = mg + mw^2R
   hence
   45 = 60 - mw^2R/g
   hence
   m" = 60 + (60 - 45) = 75 kg

b. when the person is at the level of the hub of the rotation of the ferris wheel
   reading be M
   Mg = mg
   hence M = 60 kg

c. in b, normal reaction = mg
   so friction force = kmg ( where k is coefficent of static friction)
   also, centrifugal force = mw^2R = 15g
   hence
   from force balance
   kmg = 15g
   k = 15/60 = 0.25
   hence coefficenet of fricxtion between the person and the ferris wheel must be 0.25 for the person not to slip in c