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

$I = 0.255~kg~m^2$

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. $\tau = I\alpha$ $I = \frac{\tau}{\alpha}$ $I = \frac{F~R}{(12.0~\pi)~rad/s^2}$ $I = \frac{(80.0~N)(0.120~m)}{(12.0~\pi)~rad/s^2}$ $I = 0.255~kg~m^2$