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

Explanation

(a) Consider an element of length dr at a distance .r. from the axis of rotation as shown in fig. The centripetal force acting on this element will be
`dT =dm r omega^(2) =(rho A dr) 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 from the axis of rotation due to all elements between x = r to x = L.
centripetal force due to all elements between x = r to x = L.
i.e., `T=int_(r)^(L) rhoA omega^(2) r dr=(1)/(2) rhoAomega^(2)[L^(2)-r^(2)] ......(1)`
So here `T=(1)/(2) xx 10^(4) xx pi xx 10^(-2) xx (400)^(2) [((1)/(2))^(2) -r^(2)]`
`=8pi xx 10^(6) [(1)/(4) -r^(2)]N`
`T=(1)/(2) rho A omega^(2) (L^(2)-r^(2))` elongation of small element
`de =(rho A omega^(2))/( 2Ay) (L^(2)-r^(2)) dr`, Total elongation
`e=intde=(rho omega^(2))/(2y) int_(O)^(L)(L^(2)-r^(2))dr=(rho^(2) omega^(2)L^(3))/(3y)`