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 326: 56

Answer

$K.E=4.81J$

Work Step by Step

We know that $K.E=\frac{1}{2}I\omega^2$ We plug in the known values to obtain: $K.E=\frac{1}{2}(4.30)(\frac{2\pi}{4.20})$ $K.E=4.81J$
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