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

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Problem 52GP Chapter 6

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 52GP

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?

Step-by-Step Solution:

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

Chapter 6, Problem 52GP is Solved
Step 7 of 9

Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

This full solution covers the following key subjects: weight, ride, apparent, true, wheel. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. The answer to “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?” is broken down into a number of easy to follow steps, and 113 words. The full step-by-step solution to problem: 52GP from chapter: 6 was answered by , our top Physics solution expert on 03/03/17, 03:53PM. Since the solution to 52GP from 6 chapter was answered, more than 650 students have viewed the full step-by-step answer. Physics: Principles with Applications was written by and is associated to the ISBN: 9780130606204. This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6th.

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