University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 330: 10.15

Answer

$I = 0.255~kg~m^2$

Work Step by Step

We can find the angular acceleration of the wheel. $\alpha = \frac{\omega}{t}$ $\alpha = \frac{(12.0~rev/s)(2\pi~rad/rev)}{2.00~s}$ $\alpha = (12.0~\pi)~rad/s^2$ We can find the moment of inertia of the wheel. $\tau = I\alpha$ $I = \frac{\tau}{\alpha}$ $I = \frac{F~R}{(12.0~\pi)~rad/s^2}$ $I = \frac{(80.0~N)(0.120~m)}{(12.0~\pi)~rad/s^2}$ $I = 0.255~kg~m^2$
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