Answer
The work required to increase the rotational rate from $40.0~rad/s$ to $60.0~rad/s$ is $~~0.0560~J$
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 the rotational kinetic energy when $\omega = 40.0~rad/s$:
$K = \frac{1}{2}I~\omega^2$
$K = (\frac{1}{2})(5.6\times 10^{-5}~kg~m^2)(40.0~rad/s)^2$
$K = 0.0448~J$
We can find the rotational kinetic energy when $\omega = 60.0~rad/s$:
$K = \frac{1}{2}I~\omega^2$
$K = (\frac{1}{2})(5.6\times 10^{-5}~kg~m^2)(60.0~rad/s)^2$
$K = 0.1008~J$
We can find the increase in kinetic energy:
$\Delta K = 0.1008~J-0.0448~J = 0.0560~J$
The work required to increase the rotational rate from $40.0~rad/s$ to $60.0~rad/s$ is $~~0.0560~J$
