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BIO Gymnastics. We can roughly model a gymnastic tumbler

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

Solution for problem 18E 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 18E

BIO Gymnastics.? We can roughly model a gymnastic tumbler as a uniform solid cylinder of mass 75 kg and diameter 1.0 m. If this tumbler rolls forward at 0.50 rev/s, (a) how much total kinetic energy does he have, and (b) what percent of his total kinetic energy is rotational?

Step-by-Step Solution:

Solution 18E Step 1: The total kinetic energy can be given as sum of the rotational kinetic energy and translational kinetic energy. 2 Rotational kinetic energy K =r1/2Iw Where I moment of inertia along axis of rotation w angular velocity I = 1/2mR 2 Where R radius of cylinder m mass of cylinder Now since it completes 0.50 rev/sec Which means converts 0.50 rev/sec 2 radian/sec * w = 1 pi rad/sec * Step 2: 2 2 (a).Rotational kinetic energy K r 1/2(1/2mR )w K r 46.26 J 2 Translational kinetic energy K = t/2mv where v = wr 2 K t 1/2m(w r) * K t 92.52 J Total kinetic energy K.E = K + Kr t K.E = 46.26 J + 92.52 J K.E = 138.78 J

Step 3 of 3

Chapter 10, Problem 18E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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BIO Gymnastics. We can roughly model a gymnastic tumbler

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