Answer
$I = 4.6\times 10^{-3}~kg~m^2$
Work Step by Step
We can find the rotational inertia about $O$:
$I = \sum~m_i~r_i^2$
$I = (0.023~kg)(0.12~m)^2+(0.023~kg)(0.24~m)^2+(0.023~kg)(0.36~m)^2$
$I = 4.6\times 10^{-3}~kg~m^2$
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 11 - Rolling, Torque, and Angular Momentum - Problems - Page 323: 37a
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