Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 11 - Exercises and Problems: 52

Answer

$L=1.13Js$

Work Step by Step

As we know that $I_{system}=2\times\frac{M_{rod}L^2}{12}+4\times M_{cup}(\frac{L}{2})^2$ We plug in the known values to obtain: $I_{system}=2\times\frac{0.0757(0.326)^2}{12}+4\times(0.124)(\frac{0.326}{2})^2$ $I_{system}=14.5\times 10^{-3}Kgm^2$ We also know that $L=I\omega$ $\implies L=14.5\times 10^{-3}\times 78=1.13Js$
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