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 291: 64a

Answer

$I = 0.15~kg~m^2$

Work Step by Step

We can use the parallel axis theorem to find the rotational inertia: $I = \frac{1}{2}MR^2+Md^2$ $I = \frac{1}{2}(20~kg)(0.10~m)^2+(20~kg)(0.050~m)^2$ $I = 0.15~kg~m^2$
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.