To solve this exercise it is necessary to apply the concepts related to the conservation of rotational, kinetic and potential energy, as well as the concepts of moments of inertia in this type of bodies.
By definition we know that,
[tex]KE_f +U_f = KE_i + U_i[/tex]
Where,
KE = Kinetic energy
U = Potential energy
Let us start by defining that the center of mass of the body is located at a distance h / 2 from the bar and that the moment of inertia of a bar is defined by
[tex]I = \frac{1}{12}Mh^2[/tex]
Where M means the mass and h the height, then,
[tex]KE_f +U_f = KE_i + U_i[/tex]
[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +mgh_f = \frac{1}{2}Mv_f^2+Mgh (\frac{h}{2})[/tex]
There is not potential energy at the beginning and also there is not Kinetic energy at the end then
[tex]\frac{1}{2}Mv^2+\frac{1}{2}I\omega +0 = 0+Mgh (\frac{h}{2})[/tex]
Replacing inertia values,
[tex]\frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{12}Mh^2)(\frac{v}{h/2})^2 = Mg \frac{h}{2}[/tex]
Re-arrange for v, we have
[tex]v = \sqrt{\frac{3gh}{4}}[/tex]
Note that the value for the angular velocity ([tex]\omega[/tex])we replace with the equivalent in tangential velocity, which is [tex]\frac{v}{R}[/tex], where v is the velocity and R the radius, at this case h/2
Therefore the center of mass hat a velocity equal to[tex]v = \sqrt{\frac{3gh}{4}}[/tex]
