1st problem 2nd problem A Ferris wheel is a vertical, circular amusement ride wi

Question

1st problem

2nd problem

A Ferris wheel is a vertical, circular amusement ride with radius 9 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 8.5 s. Consider a rider whose mass is 54 kg At the bottom of the ride, what is the rate of change of the rider's momentum? =| 264.18 dp | × kg m/s/s At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider? Fgrav = | 265.56 At the bottom of the ride, what is the vector force exerted by the seat on the rider? Fby seat = |-529.74 | X 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? grav At the top of the ride, what is the vector force exerted by the seat on the rider? by seat-

Explanation / Answer

a)

here

dp / dt = F = m * omega^2 * r

F = m * ( 2 * pie / T)^2 * r

F = 54 * ( 2 * 3.14 / 8.5)^2 * 9

F = 265.28 kg m /s^2

b)

Fgav = m *g

Fgav = 54 * -9.8

Fgav = -529.2 N

c)

Fseat = dp / dt + Fgav

= 265.28 + 529.2

= 794.48 N

d)

there is

dp / dt = - 265.28 kg m/s^2

e)

Fgav = -529.2 N

f)

Fseat = Fgav - dp/dt

= -529.2 - 265.28

= -794.48 N

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