While at the county fair, you decide to ride the Ferris

Problem 52GP Chapter 6

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition

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Problem 52GP

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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

Step 6 of 9

Step 7 of 9

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