Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 9 - Rotational Dynamics - Problems - Page 246: 32



Work Step by Step

We have $$\sum\tau=I\alpha$$ $$I=\frac{\sum\tau}{\alpha}$$ The net torque $\sum\tau=10N.m$, and the vase's angular acceleration $\alpha=8rad/s^2$. The total moment of inertia is $$I=1.25kg.m^2$$
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