Answer
$slope = 2.8\times 10^{-5}~kg~m^2$
Work Step by Step
We can find the rotational inertia:
$I = Md^2+M(2d)^2+M(3d)^2$
$I = 14~Md^2$
$I = (14)(0.0100~kg)(0.0200~m)^2$
$I = 5.6\times 10^{-5}~kg~m^2$
We can find an expression for the rotational kinetic energy:
$K = \frac{1}{2}I~\omega^2$
We can find the slope of a plot of the kinetic energy versus the square of the rotation rate:
$slope = \frac{K}{\omega^2}$
$slope = \frac{\frac{1}{2}I~\omega^2}{\omega^2}$
$slope = \frac{1}{2}~I$
$slope = (\frac{1}{2})~(5.6\times 10^{-5}~kg~m^2)$
$slope = 2.8\times 10^{-5}~kg~m^2$
