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 9 - Exercises and Problems - Page 173: 86

Answer

$.1r$

Work Step by Step

We know that a point on a circle is given by $rsin\theta$. We know that this semi-circle ranges from $\theta = 0$ to $\theta=180$. However, we are looking for the y-center of mass, so $\theta$ ranges from $\theta = 45$ to $\theta=135$. $=\int_0^{180}rsin\theta d\theta=r\frac{2}{\sqrt{2}}$ We use this to find the center of mass: $y_{cm}=1-\frac{\sqrt2 r}{1}\times\frac{2}{\pi}=.1r$
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