here is the question: A Ferris wheel is a vertical, circular amusement ride with

Question

here is the question:

A Ferris wheel is a vertical, circular amusement ride with radius 10 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 8 s. Consider a rider whose mass is 58 kg. At the bottom of the ride, what is the rate of change of the rider's momentum? kg-m/s/s At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider? N At the bottom of the ride, what is the vector force exerted by the seat on the rider? N Next consider the situation at the top of the ride. At the top of the ride, what is the rate of change of the rider's momentum? kg-m/s/s At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider? At the top of the ride, what is the vector force exerted by the seat on the rider? N A rider feels heavier if the electric, interatomic contact force of the seat on the rider is larger than the rider's weight mg (and the rider sinks more deeply into the seat cushion). A rider feels lighter if the contact force of the seat is smaller than the rider's weight (and the rider does not sink as far into the seat cushion). Does a rider feel heavier or lighter at the bottom of a Ferris wheel ride? heavier lighter Does a rider feel heavier or lighter at the top of a Ferris wheel ride? heavier lighter

Explanation / Answer

a.)

Net force on the rider at the bottom of the ferris wheel is the net change in momentum.

Net acceleration of the rider is only centripetal force which is -(w^2)r j where j is the unit vector in y-axis because there is no change in angular velocity w

The ferris wheel covers 2(pie) radian angle in 8 sec.

w=2(pie)/8=pie/4

Thus d(P)/dt=ma= -m(w^2)r j = -58(pie^2/16)(10) j = -357.773 j

d(P)/dt= -357.773 j

b.)

F(grav)=mg j =58(g) j = 568.4 j

F(grav)= 568.4 j   ..................(Equ 1)

c.)

F(by seat)+F(grav)=d(P)/dt

Thus,

F(by seat)=d(P)/dt-F(grav)= -357.773 j - 568.4 j = -926.173 j

F(by seat) = -926.173 j ..................(Equ 2)

d.)

At top the direction of the cantripetal acceleration of the rider is in negative y-axis.

Thus d(P)/dt=ma= m(w^2)r j = 58(pie^2/16)(10) j = 357.773 j

d(P)/dt= 357.773 j

e.)

Vector gravitational force remains constant

F(grav)=mg j =58(g) j = 568.4 j

F(grav)= 568.4 j   ..................(Equ 3)

f.)

F(by seat)+F(grav)=d(P)/dt

Thus,

F(by seat)=d(P)/dt-F(grav)= 357.773 j - 568.4 j = -210.627 j

F(by seat) = -210.627 j ..................(Equ 4)

g.)

Comparing Equ 1 and Equ 2-

|F(by seat)| > |F(grav)|

Thus rider would feel heavier

h.)

Comparing Equ 1 and Equ 2-

|F(by seat)| < |F(grav)|

Thus rider would feel lighter

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