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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.

6 years ago

Answers : (1)

vikas askiitian expert
510 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 

6 years ago
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