at any time let it makes @ angle then

force acting on the rod can be resolved into two components , one along the length & other  perpendicular to length...

Fsin@ is along the rod & F cos@ is perpendicular component of F ...

now instantaneous torque = T = (Fcos@)r = Fcos@L     ............1

(L is the length of rod & the perpendicular distance)

now we know , T  =  I(alfa)

                    alfa = T/I

                   alfa  = FLcos@/I          ................2

now we have  , alfa = dw/dt                   (w is angular velocity)

                     (alfa) = (dw/d@).(d@/dt)                    (multiplying & dividing by d@)

          d@/dt = w so

                     (alfa)d@ = wdw         .............3

now from 2 & 3

               wdw = FLcos@d@/I

now interating both sides and taking proper limits

              [w²/2] (lim w=0 to w=w) = FLsin@/I      (lim 0 to @)

             w² = 2FLsin@/I                                ( I rod about one end = mL²/3)

             w²  =6Fsin@/mL        

 or        w = [(6Fsin@)/mL]^1/2