Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 13

Answer

(a) The coordinates of the center of mass are (5.7 cm, 4.6 cm) (b) $I = 0.0066~kg~m^2$

Work Step by Step

(a) Note that mass $A$ is located at the origin. We can find the x-coordinate of the center of mass. $x_{com} = \frac{(100~g)(0)+(200~g)(0)+(200~g)(10~cm)+(200~g)(10~cm)}{100~g+200~g+200~g+200~g}$ $x_{com} = 5.7~cm$ We can find the y-coordinate of the center of mass. $y_{com} = \frac{(100~g)(0)+(200~g)(8~cm)+(200~g)(8~cm)+(200~g)(0)}{100~g+200~g+200~g+200~g}$ $x_{com} = 4.6~cm$ The coordinates of the center of mass are (5.7 cm, 4.6 cm) (b) We can find the moment of inertia about an axis located at the origin. Note that mass $C$ is located a distance of 12.8 cm from the origin. $I = \sum m_i~r_i^2$ $I = (0.10~kg)(0)^2+(0.20~kg)(0.08~m)^2 + (0.20~kg)(0.128~m)^2+ (0.20~kg)(0.10~m)^2$ $I = 0.0066~kg~m^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.