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A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity w about a vertical axis passsing through one end.The tension in the rod at a distance x from the axis is
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=−mldrω2rSo, ∫T0dT=−mlω2∫xlrdr (as,at r=l,T=0)So, T=−mω22l(x2−l2)=mω22l(l2−x2)
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)
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