A Ferris wheel is a vertical, circular amusement ride with radius 8 m. Riders si

Question

A Ferris wheel is a vertical, circular amusement ride with radius 8 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 7 s. Consider a rider whose mass is 52 kg.

a) At the bottom of the ride, what is the rate of change of the rider's momentum?

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

c) At the bottom of the ride, what is the vector force exerted by the seat on the rider?

Next consider the situation at the top of the ride.

d) At the top of the ride, what is the rate of change of the rider's momentum?

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

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

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).

g) Does a rider feel heavier or lighter at the bottom of a Ferris wheel ride?

h) Does a rider feel heavier or lighter at the top of a Ferris wheel ride?

Explanation / Answer

given that

R = 8 m

m = 52 kg

T = 7 s

(a)

we know that "rate of change of momentum" is force.

For an object of mass m, going around a circle of radius R, with angular speed omega (= 2 pi / the period), the centripetal force is:
F = m *omega^2* R
F = m* (2 *pi / T)^2*R

F = 52*(2*3.14/7)^2*8

F = 334.82 N

(b)

force of gravity = -m*g = -52*9.8 = - 509.6 N

negative sign show that direction of force is always be down.

(c)

The seat pushing up must counteract both the centrifugal force (same magnitude as A) and gravity pushing down. So

Fr = 334.82 + ( - 509.6 )

Fr = -174.78 N

(d)

we know that "rate of change of momentum" is force.

For an object of mass m, going around a circle of radius R, with angular speed omega (= 2 pi / the period), the centripetal force is:
F = m *omega^2* R
F = m* (2 *pi / T)^2*R

F = 52*(2*3.14/7)^2*8

F = 334.82 N

(e)

force of gravity = -m*g = -52*9.8 = - 509.6 N

negative sign show that direction of force is always be down.

(f)

This time, the net force the seat has to counteract is the weight down negative the centrifugal force .so

Fr = 334.82 -(- 509.6)

Fr = 844.42 N

(g)

A rider feels lighter because the the electric, interatomic contact force of the seat on the rider is smaller than the rider's weight mg.

(h)

A rider feels heavier because the electric, interatomic contact force of the seat on the rider is larger than the rider's weight mg.

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