Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 15

Answer

(a) The coordinates of the center of gravity are (6 cm, 4 cm) (b) $I = 0.0020~kg~m^2$ (c) $I = 0.0013~kg~m^2$

Work Step by Step

(a) We can find the x-coordinate of the center of gravity. $x_{cog} = \frac{(100~g)(0)+(200~g)(6~cm)+(100~g)(12~cm)}{100~g+200~g+100~g}$ $x_{cog} = 6~cm$ We can find the y-coordinate of the center of gravity. Note that the y-coordinate of the 200-gram mass is 8 cm. $y_{cog} = \frac{(100~g)(0)+(200~g)(8~cm)+(100~g)(0)}{100~g+200~g+100~g}$ $y_{cog} = 4~cm$ The coordinates of the center of gravity are (6 cm, 4 cm) (b) We can find the moment of inertia about the axis that passes through mass A and is perpendicular to the page. $I = \sum m_i~r_i^2$ $I = (0.10~kg)(0.10~m)^2+(0.20~kg)(0)^2+(0.10~kg)(0.10~m)^2$ $I = 0.0020~kg~m^2$ (c) We can find the moment of inertia about the axis that passes through masses B and C. $I = \sum m_i~r_i^2$ $I = (0.10~kg)(0)^2+(0.20~kg)(0.080~m)^2+(0.10~kg)(0)^2$ $I = 0.0013~kg~m^2$
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