A Ferris wheel is a vertical, circular amusement ride with radius 7 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.

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

Question

Physics

Maverick

8 March, 05:15

A Ferris wheel is a vertical, circular amusement ride with radius 7 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.

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

+3

Answers (1)

Know the Answer?

Carl Butler · Answer 1

Change in momentum of the rider=803.4N

Explanation:

Fnet = mac

N - mg=mrw^2

N = mg + mrw^2

N = mg + mr (2*3.142/T) ^2

Where T = period of revolutions

r = radius of ferris wheel

m = mass of rider

N = (52*9.8) + (52*7) (2*3.142/7) ^2

N=803.4N