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?

Mar 15, 2018

Considering a small portion of $\mathrm{dr}$ in the rod at a distance $r$ from the axis of the rod.

So,mass of this portion will be $\mathrm{dm} = \frac{m}{l} \mathrm{dr}$ (as uniform rod is mentioned)

Now,tension on that part will be the Centrifugal force acting on it, i.e $\mathrm{dT} = - \mathrm{dm} {\omega}^{2} r$ (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, $\mathrm{dT} = - \frac{m}{l} \mathrm{dr} {\omega}^{2} r$

So, ${\int}_{0}^{T} \mathrm{dT} = - \frac{m}{l} {\omega}^{2} {\int}_{l}^{x} r \mathrm{dr}$ (as,at $r = l , T = 0$)

So, $T = - \frac{m {\omega}^{2}}{2 l} \left({x}^{2} - {l}^{2}\right) = \frac{m {\omega}^{2}}{2 l} \left({l}^{2} - {x}^{2}\right)$