Answer
The center of mass rises a distance of $~~0.23~m~~$ above the lowest position.
Work Step by Step
We can find the rotational inertia:
$I = \frac{1}{3}ML^2$
$I = \frac{1}{3}(0.42~kg)(0.75~m)^2$
$I = 0.118~kg~m^2$
We can find the rotational kinetic energy at the lowest position:
$K = \frac{1}{2}I~\omega^2$
$K = (\frac{1}{2})(0.118~kg~m^2)(4.0~rad/s)^2$
$K = 0.944~J$
Note that at the highest position, the kinetic energy is zero, and the gravitational potential energy at the highest position is equal to the kinetic energy at the lowest position.
We can use conservation of energy to find the distance that the center of mass rises above the lowest position:
$Mgh = 0.944~J$
$h = \frac{0.944~J}{Mg}$
$h = \frac{0.944~J}{(0.42~kg)(9.8~m/s^2)}$
$h = 0.23~m$
The center of mass rises a distance of $~~0.23~m~~$ above the lowest position.
