Question

A Ferris wheel of radius 10.0 m rotates in a vertical circle of 7.0 rev/min. A 45.0 kg girl rides in a car
alone. What (vertical) normal force would she experience when she is:

Haflway to the top
at the top
halfway to the bottom
at the bottom

Answer 1

at each stage of the Ferris wheel, the centripetal force acts towards the center of the wheel.

the angular velocity, ω, is 2*pi*f, where f is the frequency in revolutions per second. we are given 7 revolutions per minute, so f would be 7/60 revolutions per second.

plugging that into ω = 2pi * f gives us ω = 2*pi*(7/60)

centripetal force = mω^2 * r = 45.0(2*pi*(7/60))^2 * 10

Answer 2

draw the FBD at each stage of the cycle. at each stage, there's an upward normal force, and a downward gravitational force, but the relative magnitude of each force vector will be different depending on the direction of the centripetal force. The centripetal force is the net force at each stage.

I will also abbreviate centripetal force as Fc. (when you see Fc, plug in our centripetal force value from the above reply)

1) At the top, the Fc points downward, so the gravitational force must be larger than the normal force. so Fc = mg - N, solve for N.

2) At the bottom, the Fc points upward, so the normal force must be larger than the gravitational force. Fc = N - mg, solve for N

3) At either halfway point, the Fc points in the horizontal direction only (0 in the vertical direction). so Fc = 0 = N - mg, so N = mg. it's the same value for both halfway points.