He Cosmoclock Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms , each of which can function as a second hand (so that it makes one revolution every 60 seconds) A) Find the speed of the passengers when it is rotating at this rate. B) A passenger weighs 882 N at the weight - guessing both on the ground. What is his ap parent weight at the highest and lowest points on the wheel. C) What would the time be for one revolution if the passenger was to feel weightless at the top? D) What, then, would be that passenger’s apparent weight at the BOTTOM?

a. v = 5.236m/s

b. Apparent weight at the lowest point, Q = 0.111N

Apparent weight at the highest point, Q =0.099N

c. Time = 14.192s

d. Apparent weight at the bottom = Q = 0.111N

Explanation:

a. To calculate the speed of the passengers, we use the formula for angular speed

v = ωR

Where v = speed

ω = angular velocity = 2π ÷ T

where T = 60 seconds

R = Radius = Distance ÷ 2

From the above question, Distance = 100, therefore,

R = 100÷ 2 = 50

Speed(v) = (2π ÷ 60) × 100

v = 5.236m/s

b.

Apparent weight at the lowest point

Q = ω + (ω²R ÷g)

ω = 2π ÷ T  , where T = 60 seconds

R = 50m

g= acceleration due to gravity = 9.81m/s²

Q = (2π÷60) + ((2π÷60)² ×50) ÷ 9.81

Q = 0.111N

Apparent weight at the highest point

Q = ω - (ω²R ÷g)

w = 2π ÷ T  , where T = 60 seconds

R = 50m

g= acceleration due to gravity = 9.81m/s²

Q = (2π÷60) - ((2π÷60)²×50) ÷ 9.81

Q = 0.099N

c. Time for one passenger to feel weightless at the top =

√ 4π²R÷ a

Where a = acceleration due to gravity = 9.81m/s²

R = radius

T = √ 4π²×50 ÷ 9.81

T = 14.192s

d. Apparent weight at the bottom is calculated as :

Q = ω + (ω²R ÷g)

ω = 2π ÷ T  , where T = 60 seconds

R = 50m

g= acceleration due to gravity = 9.81m/s²

Q = (2π÷60) + ((2π÷60)²×50) ÷ 9.81

Q = 0.111N

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

. west