Answer
$I = 0.019~kg~m^2$
Work Step by Step
We can find the rotational inertia:
$I = \frac{1}{12}~M~L_1^2+\frac{1}{3}M~L_2^2$
$I = (\frac{1}{12})~(0.20~kg)~(0.40~m)^2+(\frac{1}{3})(0.20~kg)~(0.50~m)^2$
$I = 0.019~kg~m^2$
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: 100a
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