Three identical thin rods each of length l and mass M one joined to form a letter H. If the H so formed is followed to fall from rest so as to allow it to rotate about one of the side, then what will be its angular velocity, when the plane of H is vertical..?

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

plz approve!

sorry couldnt get it can you plz explain it briefly

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

thank you so much noe i am clear

but how come the limit is pi/2?

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

oh thankz