#### Answer

The rotational kinetic energy of the system is 1.75 J

#### Work Step by Step

We can convert the angular velocity to units of rad/s
$\omega = (100~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega = 10.47~rad/s$
We can find the moment of inertia of the system. Note that the center of mass is 13.3 cm from the 600-gram ball and 26.7 cm from the 300-gram ball.
$I = M_1~R_1^2+M_2~R_2^2$
$I = (0.60~kg)(0.133~m)^2+(0.30~kg)(0.267~m)^2$
$I = 0.032~kg~m^2$
We can find the rotational kinetic energy of the system.
$KE = \frac{1}{2}I~\omega^2$
$KE = \frac{1}{2}(0.032~kg~m^2)(10.47~rad/s)^2$
$KE = 1.75~J$
The rotational kinetic energy of the system is 1.75 J