Question
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?
Does a rider feel heavier or lighter at the top of a Ferris wheel ride? lighter heavier
A Ferris wheel is a vertical, circular amusement ride with radius 6 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9 s. Consider a rider whose mass is 60 kg. At the bottom of the ride, what is the rate of change of the rider's momentum? Correct: Your answer is correct. Does a rider feel heavier or lighter at the top of a Ferris wheel ride? lighter heavier Correct: Your answer is correct. , , 0 > 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 vector f capby seat = N At the top of the ride, what is the vector force exerted by the seat on the rider? Correct: Your answer is correct. , vector f capgrav = kg½m/s/s At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider? (dp^^- > )/(dt) = 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? vector f capby seat = N At the bottom of the ride, what is the vector force exerted by the seat on the rider? Correct: Your answer is correct. , vector f capgrav = kg½m/s/s At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider? (dp^^- > )/(dt) =
Explanation / Answer
The acceleration by the rotation of the ferris wheel is found by the followings:
w= 1 rev/8 sec*(2*pi)radians/1 rev=0.785 rad/sec
accel=w^2*r
accel=(0.785)^2*7= 4.32 m/s/s
The force is 50 kg * 4.32 m/s/s= 216 N
The magnitude at the top is <216, -490,0>
Because you are going backwards at the bottom the magnitude is
<-216, -490,0>
