University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 331: 10.36

Answer

$8.4\times10^3 kg \frac{m^2}{s}$.

Work Step by Step

First find the moment of inertia of the system, which is the sum of the individual moments. $$I=\frac{1}{2}(110kg)(4m)^2+(50kg)(4m)^2$$ $$I=1680kg\cdot m^2$$ Now find the angular momentum. $$L=I \omega=(1680kg\cdot m^2)\frac{0.80rev}{s}\frac{2 \pi rad}{rev}=8.4\times10^3 kg \frac{m^2}{s}$$
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