Question

A Ferris wheel is a vertical, circular amusement ride with radius 8.5 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 7.3 s. Consider a rider whose mass is 73 kg.

a. At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider?

b. At the top of the ride, what is the rate of change of the rider's momentum?

c. At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider?

d. At the top of the ride, what is the vector force exerted by the seat on the rider?

Answer 1

Given data Radius, R = 8.5 m time, T = 7.3 sec Mass, m = 73 kg a) in this case parallel component of rate of change of riders momentum dp/dt = 0 kg - m/sec^2 = (0, 0, 0) kg - m/sec^2 b) at the bottom perpendicular component of rate \r\n of change of riders momentum is dp/dt (perpendicular) = F_cent = mw^2 R = m(2 pi/T)^2 R = 73 (2 pi/7.3)^2 times 8.5 = 73 times 1.0339 times 8.5 dp/dt(perpendicular) = 641.534 kg - m/sec^2 Therefore dp/dt = (0,641.534,0) kg - m/sec^2 c) at the bottom of the ride the gravitational force executed by the earth on rider is given by fg = mg = 73 times 9.81 fg = 716.13 N \r\n In vector form the force is fg = (0, - 716.13,0) N d) at the bottom of the ride met force acting on the rider is f cent = N - mg N = f cent + mg N = 641.534 + 73 times 9.81 N = 1357.664 neutrons In vector