A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity omegaω about a vertical axis passing through one end. The tension in the rod at a distance x from the axis is?

1 Answer
Mar 15, 2018

Considering a small portion of drdr in the rod at a distance rr from the axis of the rod.

So,mass of this portion will be dm=m/l drdm=mldr (as uniform rod is mentioned)

Now,tension on that part will be the Centrifugal force acting on it, i.e dT=-dm omega^2rdT=dmω2r (because,tension is directed away from the centre whereas,rr 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 rr to ll)

Or, dT=-m/l dr omega^2rdT=mldrω2r

So, int_0^T dT=-m/l omega^2 int_l^xrdrT0dT=mlω2xlrdr (as,at r=l,T=0r=l,T=0)

So, T=-(momega^2)/(2l) (x^2-l^2)=(momega^2)/(2l) (l^2-x^2)T=mω22l(x2l2)=mω22l(l2x2)