Answer
$I = (7.2\times 10^{-4})~kg~m^2$
Work Step by Step
The slope of the speed versus time graph is the magnitude of the acceleration.
Since the slope of the graph is $3.5~m/s^2$, the magnitude of the acceleration is $3.5~m/s^2$
We can use Equation (11-10) to find an expression for the rotational inertia:
$a = \frac{g~sin~\theta}{1+I/MR^2}$
$1+I/MR^2 = \frac{g~sin~\theta}{a}$
$I/MR^2 = \frac{g~sin~\theta}{a}-1$
$I = (\frac{g~sin~\theta}{a}-1)~MR^2$
$I = [\frac{(9.8~m/s^2)~sin~30^{\circ}}{3.5~m/s^2}-1]~MR^2$
$I = 0.4~MR^2$
We can find the rotational inertia:
$I = 0.4~MR^2$
$I = (0.4)~(0.500~kg)(0.0600~m)^2$
$I = (7.2\times 10^{-4})~kg~m^2$
