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A 55-kg runner runs around the edge of a horizontal

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 98P Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 98P

A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner’s velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

Step-by-Step Solution:
Step 1 of 3

Solution 98P This question is based on the concept of angular momentum conservation. We shall have to find the initial and final angular momenta of the system. Given , mass of the runner m = 55 kg r Speed of the runner v = 2.8 r/s Radius of the turntable r = 3.0 m Therefore, angular momentum of the runner, L r m × v r r r L r 55 kg × 2.8 m/s × 3.0 m 2 L r 462 kg.m /s Now, moment of inertia of the turntable I = 80 kg.m 2 t Its angular velocity = 0.2s rad/s 2 Its angular momentum L = 80 kg.mt× 0.20 rad/s L = 16 kg.m /s 2 t Since the turntable is moving in the opposite direction, hence total initial momentum of 2 2 2 the system is L = L i L = r62 kg.t /s 16 kg.m /s = 446 kg.m /s L = 446 kg.m /s…..(1) i Let the final angular velocity be . f Final angular momentum of the runner L = 55 kg × (3.00) m × …..(2) rf f 2 Final angular momentum of the turntable L tf = 80 kg.m × …..(3)f Adding equations (2) and (3), we shall get the final angular momentum of the system. L f L rf + L tf L = (55 kg × (3.00) m + 80 kg.m ) 2 f f 2 2 L f (495 kg.m + 80 kg.m ) f L = 575 kg.m …..(4) f f According the principle of conservation of angular momentum and equating equations (1) and (2), we get L = L i f 2 2 446 kg.m /s = 575 kg.m . f = 0.776 rad/s f Therefore, the final angular velocity of the system will be 0.776 rad/s.

Step 2 of 3

Chapter 10, Problem 98P is Solved
Step 3 of 3

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

University Physics was written by and is associated to the ISBN: 9780321675460. The full step-by-step solution to problem: 98P from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner’s velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)” is broken down into a number of easy to follow steps, and 96 words. This full solution covers the following key subjects: runner, turntable, velocity, relative, angular. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 98P from 10 chapter was answered, more than 854 students have viewed the full step-by-step answer.

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A 55-kg runner runs around the edge of a horizontal