Essential University Physics: Volume 1 (4th Edition)

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

Chapter 10 - Exercises and Problems - Page 194: 69


a) $M=\frac{2\pi\rho_0wR^2}{3}$ b) $I=\frac{3MR^2}{5}$

Work Step by Step

a) We know that the mass is equal to the integral of the quantity of the density times the volume. Thus, we find: $M = 2\int_0^R \frac{\rho_0r}{R}\pi r^2 w dr $ $M=\frac{2\pi\rho_0wR^2}{3}$ b) Using the definition of moment of inertia, we find: $I=\frac{3MR^2}{5}$
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