Question

The Cosmoclock 21 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 60.0 s).

A- Find the speed of the passengers when the Ferris wheel is rotating at this rate.

B- A passenger weighs 897 N at the weight-guessing booth on the ground. What is his apparent weight at the lowest point on the Ferris wheel?

C-What is his apparent weight at the highest point on the Ferris wheel?

D-What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

E-What then would be the passenger's apparent weight at the lowest point?

Answer 1

A) Speed of passengers, v = ωR = (2π/60) * (100/2) = 5.24 m/s

B) Apparent weight is the normal reaction exerted by the wheel on the passenger.

At lowest point, N - W = mv²/R

=> Apparent weight, N = W + mv²/R = 897 + [(897 / 9.81) * 5.24² / 50] = 947 N

(C) At highest point, W - N = mv²/R

=> N = W - mv²/R = 897 - [(897 / 9.81) * 5.24² / 50] = 847 N

(D) Let T be the time period of revolution in this case.

At highest point,

N = 0 => W = mv²/R

=> v = (WR/m)^1/2 = (gR)^1/2 = (9.81 * 50)^1/2 = 22.15 m/s

v = (2π/T)R

=> T = 2πR/v = 2π * 50 / 22.15 = 14.2 s

(E) Apparent weight at lowest point,

N = W + mv²/R = W + W = 2W = 2 * 897 = 1794 N