Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems - Page 290: 44c

Answer

$I = 2.0~kg~m^2$

Work Step by Step

The two particles on the rotation axis are a distance of $~~0~~$ from the rotation axis. We can find the distance of the other two particles: $r = \sqrt{(1.0~m)^2+(1.0~m)^2} = \sqrt{2.0}~m$ We can find the rotational inertia: $I = 2\times (0.50~kg)(0)^2+2\times (0.50~kg)(\sqrt{2.0}~m)^2$ $I = 2.0~kg~m^2$
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