Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 10 - Rotational Kinematics and Energy - Problems and Conceptual Exercises - Page 331: 115

Answer

$r=2.9ft$

Work Step by Step

As $a_{cp}=r\omega^2$ This can be rearranged as: $r=\frac{a_{cp}}{\omega^2}$ We plug in the known values to obtain: $r=\frac{3.0\times 9.81}{55 \frac{rev}{min}\times \frac{2\pi rad}{rev}\times \frac{1min}{60s}}$ $r=0.89m\times 3.281\frac{ft}{m}$ $r=2.9ft$
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