A thin uniform rod of length `l` and masses `m` rotates uniformly with an angularly velocity `omega` in a horizontal plane about a verticle axis passing through one of its ends determine the tension in the rot as a funtion of the distance x from the rotation axis

Explanation

1) Consider and element dx at a distance from the axis of rotation.
2) The centripetal force on this portion is
`dt= dmr omega^(2) dT=(rho A dr) r omega^(2)`
This force is provided by the tension in the rod
`int dT= overset(L) underset(r) int rho A dr omega^(2), T rho A omega^(2)overset(L) underset(r) intrdr, T= rho A omega^(2)[ (r^(2))/(2)]_(r)^(L)`
`rArr T=(rhoA omega^(2))/(2)[ L^(2)-r^(2)] T=(1)/(2) rho A omega^(2)(L^(2)-r^(2))`
(3) strain `=("stress")/(Y)` If dy is the elongation in element of length dx
`(dy)/(dx)=(T)/(AY) rArr dy=(Tdx)/(AY) rArr dy=((1)/(2) rho A omega^(2)(L^(2)-r^(2)) dr)/(AY)`
`int dy=(rho omega^(2))/(2y) overset(L) underset(0) int (L^(2)-r^(2))dr rArr DeltaL=(rho omega^(2))/(2y)[ L^(2)r-(r^(3))/(3)]_(0)^(L)`
`DeltaL=(rho omega^(2))/(2Y)[L^(2)-(L^(3))/(3)] rArr DeltaL=( rho omega^(2))/(2y)[(2L)/(3)], DeltaL-=( rho omega^(2)L^(3))/(3Y)`