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
$I_B=M{L}^2/3$
$I=0+(M{L}^2/3)+M{L}^2=4M{L}^2/3$
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
$(2/3)M{L}^2{\omega }^2=(3/2)MgL$
Or     $\omega =\frac{3}{2}\sqrt{\frac{g}{L}}$