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

Explanation

Consider an element of length dr at a distance r from the axis of rotation as shown in figure. The centripetal force acting on this element will be
`dt = d m r omega^(2) = (pAdr)r 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 he axis of rotation will be due to the centriple force due to all element between x = r to x = L.
i.e., `int_(r)^(L)pAomega^(2)rdr=(1)/(2)pAomega^(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)p[(1)/(4)-r^(2)]N`
(b) Now, if dy is the elongation in the element of lenght dr at position r where tension. T, by definition of Young.s models.
`(dy)/(dx)=(T)/(AY)["as strain"=("stress")/(Y)]`
which in the light of Eqn. (i) gives
`dy=(1)/(2)(pomega^(2))/(2Y)[L^(2)-r^(2)]dr`
So , the elongation of the whole rod
`DeltaL=(rhoomega^(2))/(2Y)int_(0)^(L)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`