Chapter 10 - Rotation - Problems - Page 293: 87

$I = 0.054~kg~m^2$

We can find the tension in the cord: $mg~sin~\theta-T = ma$ $T = mg~sin~\theta-ma$ $T = (2.0~kg)(9.8~m/s^2)~sin~20^{\circ}-(2.0~kg)(2.0~m/s^2)$ $T = 2.70~N$ We can find the rotational inertia of the wheel: $\tau = I~\alpha$ $R~T = (I)~(\frac{a}{R})$ $I = \frac{R^2~T}{a}$ $I = \frac{(0.20~m)^2~(2.70~N)}{2.0~m/s^2}$ $I = 0.054~kg~m^2$