A uniform rod of length `l`, mass `m`, cross-sectional area `A` and Young's modulus `Y` is rotated in horizontal plane about a fixed vertical axis passing through one end, with a constant angular velocity `omega`. Find the total extension in the rod due to the tension produced in the rod.

Explanation

Tension in small element of mass `dm` is
`dT = dm r omega^(2) rArr = -rho A dr r omega^(2)`
This tension is only due to centripetal force due to all elements between `x=L` to `x=r`
`therefore T = int dT = int_(L)^(r) - rho A omega^(2)r dr = rho A omega^(2) int_(L)^(r) - rdr`
`therefore T = (1)/(2) rho A omega^(2) [L^(2) - r^(2)]`
If `'dy'` is the elogation in the element `'dr'` then
`(dy)/(dr) = (T)/(AY) [therefore "Strain" = ("stress")/(Y)]`
`int dy = int (T)/(AY)dr`
`y = int_(0)^(L) (T)/(AY) dr = int_(0)^(L) (1)/(2) (rho A omega^(2))/(AY) (L^(2) - r^(2)) dr`
Total elogation in the rod `y = (1)/(3) (rho omega^(2) L^(3))/(Y)`