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 10 - Exercises and Problems - Page 194: 78

Answer

a) The proof is below. b) $I =M(h^2+r_{cm}^2)$

Work Step by Step

a) We know that the Law of Cosines is: $C^2 = A^2 + B^2 -2AB cos \gamma$ Where $\gamma$ is the angle opposite to side C. Thus, plugging in $r_{cm}$ for A, $h$ for B, and $r$ for C, we find: $r^2 =r_{cm}^2 + h^2 -2hr_{cm} cos \gamma$ Looking at the value of gamma, this simplifies to: $r^2 =r_{cm}^2 + h^2 -2hr_{cm} $ b) As the book recommends in the problem, we take the integral of $r^2$ and use substitution to solve: $I=\int r^2 dm$ $I=\int (r_{cm}^2 + h^2 -2hr_{cm}) dm$ Taking the integral, this becomes: $I =M(h^2+r_{cm}^2)$
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