Mechanics | 25122 - A thin uniform horizontal rod of mass m and -askIITians

Aniket Patra Grade: Upto college level

        A thin uniform horizontal rod of mass m and length l can rotate about a vertical axis passing through one of its end.at any instant the other end starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane.Find the angular velocity of the rod as a function of its rotation angle alpha counted relative to the initial position.

8 years ago

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vikas askiitian expert

509 Points

							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