Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 5 - Circular Motion; Gravitation - General Problems - Page 135: 69

Answer

The velocity is 28.3 m/s which is 0.409 rev/s

Work Step by Step

We equate the centripetal force to the force felt by the trainee: $\frac{mv^2}{r} = 7.45~mg$ $v^2 = 7.45~gr$ $v = \sqrt{7.45~gr} = \sqrt{(7.45)(9.80~m/s^2)(11.0~m)}$ $v = 28.3~m/s$ We can use the velocity $v$ to find the number of revolutions per second: $\frac{v}{2\pi r} = \frac{(28.3~m/s)}{(2\pi)(11.0~m)} = 0.409~rev/s$
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