A rigid body is made of three identical thin rods, each of length L fastened together in the form of the letter H. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of H is vertical  

 

Moment of inertia of the system about the given axis  $I=I_A+I_B+I_C$
Now as rod is thin  $I_A=\sum m\times 0^2=0$
Rod B is rotating about one end  $I_B=\frac{M{L}^2}{3}$
And for rod C all points are always at distance L from the axis of rotation, so  $I_C=\sum m{L}^2=m{L}^2$
$\therefore I=0+\left(\frac{M{L}^2}{3}\right)+M{L}^2=\frac{4M{L}^2}{3}$
So if $\omega$ is the desired angular speed, gain in kinetic energy due to rotation of H from horizontal to
vertical position. 
$K_R=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left[\frac{4}{3}M{L}^2\right]{\omega }^2$
And loss in potential energy of the system in doing so
$=0+Mg\frac{L}{2}+MgL=\frac{3}{2}MgL$
So by conservation of mechanical energy
$\left(\frac{2}{3}\right)M{L}^2{\omega }^2=\left(\frac{3}{2}\right)MgL$
$or\omega =\frac{3}{2}\sqrt{\frac{g}{L}}$