A thin uniform metallic rod of length 0.5 m and radius 0.1 m rotates with an angular velocity 400rad/s is a horizontal plane about a vertical axis passing through one of its ends. Calculate (a) tenstion in the rod and (b) the elogation of te rod. The density of material of the rod is `10^(4)kg//m^(3)` and the 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 rotatio as shown in figure. The centripetal force acting on this element will be `dT=dmromega^(2)=(rhoAdr)romega^(2)` As this force is provided by tension in the rod (due to elasticity), so the tension i the rod at a distance r from the axis of rotation will be due to the centripetal force due to all elements between x=r to x=L i.e.,
(b). Now if dy is the elongation in the element of length dr at posittion r then strain
`(dy)/(dr)=("stress")/(Y)=(T)/(AY)=(1)/(2)(rho omega^(2))/(Y)[L^(2)-r^(2)]`
So the elongation of the whole rod
`DeltaL=(rho omega^(2))/(2y)underset(0)overset(L)int(L^(2)-r^(2))dr=(1)/(2)xx(10^(4)xx(100)^(2)(0.5)^(3))/(2xx10^(11))=(1)/(3)xx10^(-3)m`