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 324: 9

Answer

$\omega=1.90\frac{rad}{s}$

Work Step by Step

We know that angular speed is given as $\omega=\frac{\Delta \theta}{\Delta t}$ $\implies \omega=\frac{2\pi}{T}$ We plug in the known values to obtain: $\omega=\frac{2\pi}{0.33}$ $\omega=1.90\frac{rad}{s}$
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