Question

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

Answer 1

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=m/l dr` (as uniform rod is mentioned)

Now,tension on that part will be the Centrifugal force acting on it, i.e `dT=-dm omega^2r` (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=-m/l dr omega^2r`

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

So, `T=-(momega^2)/(2l) (x^2-l^2)=(momega^2)/(2l) (l^2-x^2)`