Answered

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?

Answer :

anniwuanyanwu1

Answer:

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

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