mgl/2 + mgl = 1/2 x (0 + ml^2/3 + ml^2) x w^2
 solve it and get w (angular velocity.
						

							plz approve!
						

							FIRST OF ALL, moment of inertia about one of its side will be the addition of moment of inertia of each rod about the side..   hence now moment of inertia about the first rod about its own axis is zero because its radius is zero.. now MOI of the second rod about its end is Ml²/3 and that of the third rod about the first is Ml² hence moment of inertia of the body is 4Ml²/3  .....  now  torque of gravitational force on the body is R*F=Mgl/2..... weknow that...
T=?a  hence   Mgl/2=4Ml²/3 a ......  which gives a=3g/8l..... now as  
a=vdv/ds .....,,  a=?d?/d?.....   hence... ?d?=(3g/8l).d? ....integrating both the sides..., we get ... ?²/2=3g?/8l.....putting limits.. ?²/2=3g?/16l ...... ?=v(3g?/8l)... whichis ur answer

  thanks,
 regards,
 HD.
						

							FIRST OF ALL, moment of inertia about one of its side will be the addition of moment of inertia of each rod about the side.. hence now moment of inertia about the first rod about its own axis is zero because its radius is zero.. now MOI of the second rod about its end is Ml²/3 and that of the third rod about the first is Ml² hence moment of inertia of the body is 4Ml²/3 ..... now torque of gravitational force on the body is R*F=Mgl/2..... weknow that...
T=Ia hence Mgl/2=4Ml²/3 a ...... which gives a=3g/8l..... now as 
a=vdv/ds .....,, a=wdw/d§..... hence... wdw=(3g/8l).d§ ....integrating both the sides..., we get ... w²/2=3g§/8l.....putting limits.. w²/2=3g(pi)/16l ...... w=v(3g(pi)/8l)... whichis ur answer

thanks,
regards,
HD
						

							it was initially horizontal... finnaly became vertical..!
						

							
Three identical thin roda each of length l and mass M are joined together to form a letter H. The moment of inertia of the system about one of the outer sides of H is :-