While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15 m, and you use your watch to find that each loop around takes 25 s. a. What are your speed and magnitude of your acceleration? b. What is the ratio of your apparent weight to your true weight at the top of the ride? c. What is the ratio of your apparent weight to your true weight at the bottom?
Solution Step 1 of 9 Person takes a rotation on the wheel with velocity v as shown in the figure below in a country fair. The pictorial representation of the given problem, Step 2 of 9 a. What are your speed and magnitude of your acceleration Given data, Radius of the wheel, r=15 m Time taken , dt= 25 sec To complete 1 revolution, d= 2 rad To find, Speed, v = Acceleration= Step 3 of 9 Relation between speed(v) and angular velocity() v = r Where r is the radius of the rim Substituting equation = dt v =( d ) r dt Using above data in this equation, v =( 25 sec) 15 m v = 3.768 m/s v = 3.77 m/s Hence, the speed of the person is 3.77 m/s. Step 5 of 9 To calculate the centripetal acceleration as the person on the rotating wheel is performing rotational motion. For a body in circular motion, the centripetal acceleration is the rate of change of tangential velocity as in figure below The expression for the centripetal acceleration(a ) isc v2 ac= r Where v are the speed of the person and r is the radius. Substituting r = 15 m and v = 3.77 m/s (3.77 m/s) a c 15 m 2 a c0.947 m/s a c0.95 m/s 2 Therefore, the magnitude of the acceleration of the person is 0.95 m/s .2