Answer
$I = 0.20~kg~m^2$
Work Step by Step
We can find the rotational inertia:
$I = (0.20~kg)(0.50~m)^2+(0.20~kg)(0.50~m)^2+(0.20~kg)[(\sqrt{2})(0.50~m)]^2$
$I = 0.20~kg~m^2$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 10 - Rotation - Problems - Page 294: 104a
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