A thin uniform copper rod of length l and mass m rotates uniformly with an angular velocity `omega` in a horizontal plane about a vertical axis passing through one of its ends. Determine the tension in the rod as a function of the distance r from the rotation axis. Find the elongation of the rod.

Explanation

Let us consider an element of the rod at a distance `r` from it's rotation axis. As the element rotates in a horizontal circle of radius r, we have from Newton's second law in projection form directed toward the axis of rotation:
`T-(T+dT)=(dm)omega^2r`
or , `-dT=(m/ldr)omega^2r=m/lomega^2rdr`
At the free end tension becomes zero. Integrating the above expression we get, thus
`-underset(T)overset(0)intdT=m/lomega^2underset(r)overset(l)intrdr`
Thus `T=(momega^2)/(l)((l^2-r^2)/(2))=(momega^2l)/(2)(1-(r^2)/(l^2))`
Elongation in elemental length `dr` is given by:
`deltaxi=(sigma(r))/(E)dr=(T)/(SE)dr`
(where S is the cross sectional area of the rod and T is the tension in the rod at the considered element)
or, `deltaxi=(momega^2l)/(2SE)(1-(r^2)/(l^2))dr`
Thus the sought elongation
`xi=intdxi=(momega^2l)/(2SE)underset(0)overset(l)int(1-r^2/l^2)dr`
or, `xi=(momega^2l)/(2SE)(2l)/(3)=((Slrho))/(3SE)omega^2l^3`
`=1/2(rhoomega^2l^3)/(E)` (where `rho` is the density of the copper.)