The Cosmo Clock 21 Ferris wheel in Yokohama, 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 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger’s apparent weight at the highest point were zero? (d) What then would be the passenger’s apparent weight at the lowest point?

Solution 50E The diameter of the wheel is 100 meters. So, the radius will be 50 meters. Each hand completes a full rotation in 60 seconds. a) The person sitting at the edge of a hand will travel a distance of 2r = 2 × × 50 = 314.15 meters in 60 seconds. So, the speed will be, v = distance / time = 314.15 / 60 = 5.235 m/s. b) The weight of the person in ground is 882 N. As we know that, weight is the total normal reaction on an object, so, weight = mg = 882 N m = 82/9.8 = 90 kg . While rotating, the acceleration due to the centripetal force by rotation is, a = v /r = (5.235) / 50 = 0.548 m/s .2 At the highest point, the weight will be the normal reaction felt by the person. weight = m(g a) = 90 × (9.8 0.548) = 832.68 N At the lowest point, the normal reaction will be, weight = m(g + a) = 90 × (9.8 + 0.548) = 931.32 N Conclusion: So, at the highest point the apparent weight is832.68 N and at the lowest point the apparent weight is931.32 N . c) The apparent weight should be zero at the highest point. So, the equation...