A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.

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A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.

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

Explanation:

Radius of wheel is 6m

Rider mass=96kg

He completes one revolution in 9.6s

Let get angular velocity (w)

1 Revolution =2πrad

θ=2πrad

w= θ/t

w=2π/9.6

w=0.654rad/s

Linear speed is give as

v=wr

v=0.654×6

v=3.93m/s

Centripetal acceleration a

a=rw²

a=6×0.654²

a=2.57m/s²

Acceleration due to gravity g=9.81m/s²

According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider

ΣF = ma

N-W=ma

N-mg=ma

N=ma+mg

N=m(a+g)

N=96(2.57+9.81)

N=1188.48 N

Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.

i believe your answer is d mgf2

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student