Question

A Ferris wheel 22.0 m in diameter rotates once every 12.5 s. What is the ratio of a person's apparent weight to her real weight at the top?What is the ratio of a person's apparent weight to her real weight at the bottom?

Answer 1

Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed:
a = omega^2*r

omega and r in terms of given data:
omega = 2*Pi/T
r = d/2

Thus:
a = 2*Pi^2*d/T^2

What forces cause this acceleration for the passenger, at either top or bottom?

At top (acceleration is downward):
Weight (m*g): downward
Normal force (Ntop): upward

Thus Newton's 2nd law reads:
m*g - Ntop = m*a

At top (acceleration is upward):
Weight (m*g): downward
Normal force (Nbottom): upward

Thus Newton's 2nd law reads:
Nbottom - m*g = m*a

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame:
Ntop = m*(g - a)
Nbottom = m*(g + a)


Substitute a:
Ntop = m*(g - 2*Pi^2*d/T^2)
Nbottom = m*(g + 2*Pi^2*d/T^2)

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground):
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g)
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g)

Simplify:
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2)
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2)

Data:
d:=22 m; T:=12.5 sec; g:=9.8 N/kg;

Results:
Ntop/(m*g) = 71.64%...she feels "light"
Nbottom/(m*g) = 128.4%...she feels "heavy"