Let the length of the rod
AB be
l and mass of the rod be
m.
Moment of inertia of the uniform rod about an axis passing through one of its end
V given as,
MI=‌Rotational kinetic energy,
(KE)‌rod ‌=‌lω2Initially the rod is held, vertical, so potential energy of rod is given by
(PE)i=mg‌‌‌(∵‌=COM)Initial kinetic energy,
(KE)i=0 The rod is allowed to fall, when it just touches the ground its potential energy becomes zero, i.e.,
(PE)f=0Let, angular velocity of rod at that instant be
ω, then kinetic energy at that instant is
(KE)f=‌∕ω2From conservation of energy,
‌(KE)i+(PE)i=(KE)f+(PE)f‌⇒‌‌0+mg‌=‌lω2+0‌⇒‌‌mgl=lω2‌⇒‌‌mgl=‌ω2‌⇒‌‌ω=√‌