Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 11 - Exercises and Problems - Page 207: 21

Answer

Please see the work below.

Work Step by Step

We know that moment of inertia is given as $I=mr^2$ We plug in the known values to obtain: $I=(0.640)(\frac{0.90}{2})^2=0.1296Kgm^2$ Then angular speed is given as $\omega=2\pi(\frac{170}{60})=17.80\frac{rad}{s}$ Thus, the angular momentum is given as $L=I\omega$ We plug in the known values to obtain: $L=(0.1296)(17.80)=2.3kg\frac{m^2}{s}$
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