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A circular-motion addict of mass 85 kg rides a Ferris wheel around in a vertical circle of radius 10 m at a constant speed of 5.8 m/s.
(a) What is the period of the motion?

(b) What is the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path?

(c) What is it when both go through the lowest point?

a) period = 2pi r/v = 2pi*10/5.8 = 10.8 sec

b) Draw free body diagram, and apply Newton's second law:

N+mg = mv2/r

N = m(v2/r - g) = 547 N <==Answer

c) Draw free body diagram again, and apply Newton's second law:
N-mg = mv2/r

N = m(v2/r + g) = 1119 N <==Answer

a) to get the period of the motion, find the distance involved in one complete cycle  d = 2*pi*r. since this is constant speed, the time for one complete cycle is T = d/v = 2*pi*r / v. substitute to get 10.8 s.

b) while the addict traverses the circular path, two forces act on the addict at all times. the force of gravity, Fg = mg = 833 N acts straight down to the center of the earth at all time. the centripetal force, Fc, is the other force. this arises due to centripetal acceleration, ac = v^2/r (m/s^2). Fc = m * v^2/r. The magnitude of this force is 286 N. The direction of this vector always points inward to the center of the wheel.

this means that ar the top of the wheel, both Fg and Fc point down. since Fc, the foce on the chair, points down, it takes away from the force pushing the two objects together. this means that you take the difference to get the normal force, Fg - Fc = 547 N down.

c) when at the bottom, the Fc is pointing up. this means that the normal force, the force pushing the two objects together is the sum of the two forces, Fg + Fc = 1119 N.

THANKS TO DANIEL O FOR POINTING OUT MISCUE FROM EARLIER ANSWER.