University Physics with Modern Physics (14th Edition)

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

Chapter 9 - Rotation of Rigid Bodies - Problems - Exercises - Page 296: 9.35

Answer

$I = 8.52~kg~m/s^2$

Work Step by Step

Let $m_1$ be the mass of the disk and let $m_2$ be the mass of the outer ring. We can find $m_1$. $m_1 = (\pi~r_1^2)(3.00~g/cm^2)$ $m_1 = (\pi)(50.0~cm)^2(3.00~g/cm^2)$ $m_1 = 23.56~kg$ We can find $m_2$. $m_2 = (\pi~r_2^2-\pi~r_1^2)(2.00~g/cm^2)$ $m_2 = [(\pi)(70.0~cm)^2-(\pi)(50.0~cm)^2](2.00~g/cm^2)$ $m_2 = 15.08~kg$ We can find the moment of inertia $I$ $I = \frac{1}{2}m_1~r_1^2+\frac{1}{2}m_2~(r_1^2+r_2^2)$ $I = \frac{1}{2}(23.56~kg)(0.500~m)^2+\frac{1}{2}(15.08~kg)[(0.500~m)^2+(0.700~m)^2]$ $I = 8.52~kg~m/s^2$
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