Aferris wheel 22.0m in diameter rotates once every 12.5s. what is the ratio of a persons apperenet weight to her real weight (a.) at the top, and (b) at the bottom?

Apparent weight: N = mg - F_up

So basically, you have two forces acting on the body - the force of gravity pulling it down, and the seat pushing it up.

The net force, F = m*a, and we're going to have to use centripetal acceleration in this problem, since it's a Ferris wheel.

So, centripetal acceleration is:

a = w^2 * r (w is omega in this case),

where w = 2pi/t

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%.
Nbottom/(m*g) = 128.4%

Ive seen this before! the answer is b