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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.
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
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
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