# 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.

509 Points
13 years ago

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