Answer
The kinetic energy of the system is $~~98~J$
Work Step by Step
In part (a), we found that the radius of the circle is $~1.5~m$
In part (b), we found that the angular speed is $~\omega = 0.933~rad/s$
We can find the rotational inertia of the system:
$I = m_1r^2+m_2r^2$
$I = (50~kg)(1.5~m)^2+(50~kg)(1.5~m)^2$
$I = 225~kg~m^2$
We can find the kinetic energy of the system:
$K = \frac{1}{2}I\omega^2$
$K = \frac{1}{2}(225~kg~m^2)(0.933~rad/s)^2$
$K = 98~J$
The kinetic energy of the system is $~~98~J$
