Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 9 - Exercises and Problems - Page 164: 37


$$(0, (\frac{a}{12tan(36)}))$$

Work Step by Step

This question seems really difficult at first. However, we can treat the missing part of the pentagon as negative mass and use symmetry, which will make this problem easier to solve. Using symmetry, we see that the given image is symmetric about the x axis, making the x center of mass 0. Now, we treat the missing piece as negative mass. If it were part of the piece, then the center of mass would simply be 0. However, since it is not, we subtract its center of mass. We know that its center of mass is: $-2/3h$ (since it is an equilateral triangle). We now convert this to be in terms of a. Using trigonometry, the apothem is: $$ tan \theta = \frac{a/2}{h}$$ $$ h = \frac{a}{2tan(36)}$$ $$y_{cm}=-2/3 \times(\frac{a}{2tan(36)})$$ $$y_{cm}=-(\frac{a}{3tan(36)})$$ However, we must remember that there are 4 other parts of the figure. Thus, the center of mass becomes: $$y_{cm}=-1/4(\frac{a}{3tan(36)})$$ We make this positive, since we are considering a negative mass. This gives our answer: $$(0, (\frac{a}{12tan(36)}))$$
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