Question

What is the effective gravitational acceleration at the bottom position of a Ferris wheel if the wheel has a radius of 8.0m and has a period of 10.0s? Top position? Side position? Use a coordinate system in which up is the +y direction and the line from the center of the wheel to the side position is the +x direction. Please give all answers as vectors.

Answer 1

The ferris wheel has time period of 10 s. Radius is 8 m. Thus, tangential velocity of a person sitting in it will be

v = 2*pi*r/t = 2*3.14*8/10 = 5.024 m/s

Now, centripetal acceleration = v^2/r = 5.024^2/8 = 3.16 m/s^2

Thus, at the bottom, we can write using the FBD of a person, effective acceleration = g+v^2/r = 9.81+3.16 = 12.96 m/s2

and

at the top, effective acceleration = g - v^2/r = 9.81-3.16 = 6.65 m/s2

At the side position, acceleration due to gravity effectively be the same as g = 9.81 m/s2