A thin uniform metallic rod of length`0.5m` and radius `0.1m` rotates with an angular velocity of `400rad//s` in a horizontal plane about a vertical axis passing through one of its ends. Calculare tension in the rod and the elongation of the rod. The density of material of the rod is `10^(4)kg//m^(3)` and the Young's modulus is `2xx10^(11)N//m^(2)`.

Explanation

(a) Consider an element of length dr at a diameter r from the axis of rotation as shown in figure. The centripetal force acting on this element will be
`dT=dmromega^(2)=(rhoAdr)omega^(2)`

As this force is provided by tension in the rod (due to elasticity) , so the tension in the rod at a distance r from the axis of rotation will be due to the centripetal force due to all elements between `x=r` to `x=L`.
i.e, `T=int_(tau)^(L)rhoAomega^(3)rdr=(1)/(2)rhoAomega^(2)[L^(2)-r^(2)]`............(i)
So, here `T=(1)/(2)xx10^(4)xxpixx10^(-2)xx(400)^(2)[((1)/(2))^(2)-r^(2)]=8pixx10^(6)[(1)/(4)-r^(2)]N`
(b) Now , if dy is the elongation in the element of length dr at position r where tension is T, by definition of Young.s models.
`(dy)/(dr)=(T)/(AY)` [as strain =`("stress")/(Y)`]
which in the light of Eqn. (i) gives
`dy=(1)/(2)(rhoomega^(2))/(2Y)[L^(2)-r^(2)]dr`
So, the elongation of the whole rod
`DeltaL=(rhoomega^(2))/(2Y)int_(0)^(L)(L^(2)-r^(2))dr=(1)/(3)(rhoomega^(2)L^(3))/(Y)`
Here `DeltaL=(1)/(3)xx(10^(4)xx(400)^(2)xx(0.5)^(3))/(2xx10^(11))=(1)/(3)xx10^(-3)m`