A uniform rod of mass M and lenght L is rotating uniformly with angular velocity (o about an axis as shown in figure. Assume axle as frictionless. Now a small particle of mass is gently placed at other end B of rod and it sticks to the rod. Find new angular velocity of system.
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Considering a small portion of dr in the rod at a distance r from the axis of the rod.
So,mass of this portion will be dm=mldr (as uniform rod is mentioned)
Now,tension on that part will be the Centrifugal force acting on it, i.e dT=−dmω2r(because,tension is directed away from the centre whereas,r is being counted towards the centre,if you solve it considering Centripetal force,then the force will be positive but limit will be counted from r to l)
Or, dT=−mldrω2r
So, ∫T0dT=−mlω2∫xlrdr (as,at r=l,T=0)
So, T=−mω22l(x2−l2)=mω22l(l2−x2)
So,mass of this portion will be dm=mldr (as uniform rod is mentioned)
Now,tension on that part will be the Centrifugal force acting on it, i.e dT=−dmω2r(because,tension is directed away from the centre whereas,r is being counted towards the centre,if you solve it considering Centripetal force,then the force will be positive but limit will be counted from r to l)
Or, dT=−mldrω2r
So, ∫T0dT=−mlω2∫xlrdr (as,at r=l,T=0)
So, T=−mω22l(x2−l2)=mω22l(l2−x2)
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