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 327: 66

Answer

$I=0.17Kg.m^2$

Work Step by Step

We know that $I=mR^2(\frac{2gh}{v^2}-1)$ We plug in the known values to obtain: $I=2.1(0.080)^2(\frac{2(9.81)^2(0.074)}{(0.33)^2}-1)$ $I=0.17Kg.m^2$
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